POWER ELECTRONICS
for elektroda
VOL.1
CONTENTS
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INTRODUCTION
SIMPLE DIODE CIRCUITS
SIMPLE SCR CIRCUITS
FULLY-CONTROLLED 1-PH SCR BRIDGE RECTIFIER
FULLY-CONTROLLED 3-PH SCR BRIDGE RECTIFIER
SEMI-CONTROLLED RECTIFIER CIRCUITS
SWITCH-MODE POWER SUPPLY
INTRODUCTION TO POWER
ELECTRONICS
DEFINITION
MAIN TASK OF POWER ELECTRONICS
RECTIFICATION
DC-TO-AC CONVERSION
DC-TO-DC CONVERSION
AC-TO-AC CONVERSION
ADDITIONAL INSIGHTS INTO POWER
ELECTRONICS
STRUCTURE OF THE ONLINE TEXT ON
POWER ELECTRONICS
DEFINITION
Power electronics refers to control and conversion of electrical
power by power semiconductor devices wherein these devices
operate as switches. Advent of silicon-controlled rectifiers,
abbreviated as SCRs, led to the development of a new area of
application called the power electronics. Prior to the
introduction of SCRs, mercury-arc rectifiers were used for
controlling electrical power, but such rectifier circuits were part
of industrial electronics and the scope for applications of
mercury-arc rectifiers was limited. Once the SCRs were
available, the application area spread to many fields such as
drives, power supplies, aviation electronics, high frequency
inverters and power electronics originated.
MAIN TASK OF POWER ELECTRONICS
Power electronics has applications that span the whole field of
electrical power systems, with the power range of these
applications extending from a few VA/Watts to several MVA /
MW.
The main task of power electronics is to control and convert
electrical power from one form to another. The four main forms
of conversion are:
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Rectification referring to conversion of ac voltage to dc
voltage,
DC-to-AC conversion,
DC-to DC conversion and
AC-to-AC conversion.
"Electronic power converter" is the term that is used to refer to a
power electronic circuit that converts voltage and current from
one form to another. These converters can be classified as:
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Rectifier converting an ac voltage to a dc voltage,
Inverter converting a dc voltage to an ac voltage,
Chopper or a switch-mode power supply that converts a dc
voltage to another dc voltage, and
Cycloconverter and cycloinverter converting an ac voltage
to another ac voltage.
In addition, SCRs and other power semiconductor devices are
used as static switches.
RECTIFICATION
Rectifiers can be classified as uncontrolled and controlled
rectifiers, and the controlled rectifiers can be further divided into
semi-controlled and fully-controlled rectifiers. Uncontrolled
rectifier circuits are built with diodes, and fully-controlled
rectifier circuits are built with SCRs. Both diodes and SCRs are
used in semi-controlled rectifier circuits.
There are several rectifier circuits rectifier configurations. The
popular rectifier configurations are listed below.
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Single-phase semi-controlled bridge rectifier,
Single-phase fully-controlled bridge rectifier,
Three-phase three-pulse, star-connected rectifier,
Double three-phase, three-pulse star-connected rectifiers
with inter-phase transformer (IPT),
Three-phase semi-controlled bridge rectifier,
Three-phase fully-controlled bridge rectifier and
Double three-phase fully-controlled bridge rectifiers with
IPT.
Apart from the configurations listed above, there are seriesconnected and 12-pulse rectifiers for delivering high power
output.
Power rating of a single-phase rectifier tends to be lower than 10
kW. Three-phase bridge rectifiers are used for delivering higher
power output, up to 500 kW at 500 V dc or even more. For low
voltage, high current applications, a pair of three-phase, threepulse rectifiers interconnected by an inter-phase
transformer(IPT) is used. For a high current output, rectifiers
with IPT are preferred to connecting devices directly in parallel.
There are many applications for rectifiers. Some of them are:
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Variable speed dc drives,
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Battery chargers,
DC power supplies and Power supply for a specific
application like electroplating
DC-TO-AC CONVERSION
The converter that changes a dc voltage to an alternating voltage
is called an inverter. Earlier inverters were built with SCRs.
Since the circuitry required to turn the SCR off tends to be
complex, other power semiconductor devices such as bipolar
junction transistors, power MOSFETs, insulated gate bipolar
transistors (IGBT) and MOS-controlled thyristors (MCTs) are
used nowadays. Currently only the inverters with a high power
rating, such as 500 kW or higher, are likely to be built with
either SCRs or gate turn-off thyristors(GTOs). There are many
inverter circuits and the techniques for controlling an inverter
vary in complexity.
Some of the applications of an inverter are listed below:
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Emergency lighting systems,
AC variable speed drives,
Uninterrupted power supplies, and
Frequency converters.
DC-TO-DC CONVERSION
When the SCR came into use, a dc-to-dc converter circuit was
called a chopper. Nowadays, an SCR is rarely used in a dc-to-dc
converter. Either a power BJT or a power MOSFET is normally
used in such a converter and this converter is called a switchmode power supply. A switch-mode power supply can be of one
of the types listed below:
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Step-down switch-mode power supply,
Step-up chopper,
Fly-back converter and
Resonant converter.
The typical applications for a switch-mode power supply or a
chopper are:
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DC drive
Battery charger and
DC power supply.
AC-TO-AC CONVERSION
A cycloconverter or a cycloinverter converts an ac voltage, such
as the mains supply, to another ac voltage. The amplitude and
the frequency of input voltage to a cycloconverter tend to be
fixed values, whereas both the amplitude and the frequency of
output voltage of a cycloconverter tend to be variable. On the
other hand, the circuit that converts an ac voltage to another ac
voltage at the same frequency is known as an ac-chopper.
A typical application of a cycloconverter is to use it for
controlling the speed of an ac traction motor and most of these
cycloconverters have a high power output, of the order a few
megawatts and SCRs are used in these circuits. In contrast, low
cost, low power cycloconverters for low power ac motors are
also in use and many of these circuit tend to use triacs in place
of SCRs. Unlike an SCR which conducts in only one direction, a
triac is capable of conducting in either direction and like an
SCR, it is also a three terminal device. It may be noted that the
use of a cycloconverter is not as common as that of an inverter
and a cycloinverter is rarely used.
ADDITIONAL INSIGHTS INTO POWER
ELECTRONICS
There are several striking features of power electronics, the
foremost among them being the extensive use of inductors and
capacitors. In many applications of power electronics, an
inductor may carry a high current at a high frequency. The
implications of operating an inductor in this manner are quite a
few, such as necessitating the use of litz wire in place of singlestranded or multi-stranded copper wire at frequencies above 50
kHz, using a proper core to limit the losses in the core, and
shielding the inductor properly so that the fringing that occurs at
the air-gaps in the magnetic path does not lead to
electromagnetic interference. Usually the capacitors used in a
power electronic application are also stressed. It is typical for a
capacitor to be operated at a high frequency with current surges
passing through it periodically. This means that the current
rating of the capacitor at the operating frequency should be
checked before its use. In addition, it may be preferable if the
capacitor has self-healing property. Hence an inductor or a
capacitor has to be selected or designed with care, taking into
account the operating conditions, before its use in a power
electronic circuit.
In many power electronic circuits, diodes play a crucial role. A
normal power diode is usually designed to be operated at 400 Hz
or less. Many of the inverter and switch-mode power supply
circuits operate at a much higher frequency and these circuits
need diodes that turn ON and OFF fast. In addition, it is also
desired that the turning-off process of a diode should not create
undesirable electrical transients in the circuit. Since there are
several types of diodes available, selection of a proper diode is
very important for reliable operation of a circuit.
Analysis of power electronic circuits tends to be quite
complicated, because these circuits rarely operate in steadystate. Traditionally steady-state response refers to the state of a
circuit characterized by either a dc response or a sinusoidal
response. Most of the power electronic circuits have a periodic
response, but this response is not usually sinusoidal. Typically,
the repetitive or the periodic response contains both a steadystate part due to the forcing function and a transient part due to
the poles of the network. Since the responses are nonsinusoidal,
harmonic analysis is often necessary. In order to obtain the time
response, it may be necessary to resort to the use of a computer
program.
Power electronics is a subject of interdisciplinary nature. To
design and build control circuitry of a power electronic
application, one needs knowledge of several areas, which are
listed below.
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Design of analogue and digital electronic circuits, to build
the control circuitry.
Microcontrollers and digital signal processors for use in
sophisticated applications.
Many power electronic circuits have an electrical machine
as their load. In ac variable speed drive, it may be a
reluctance motor, an induction motor or a synchronous
motor. In a dc variable speed drive, it is usually a dc shunt
motor.
In a circuit such as an inverter, a transformer may be
connected at its output and the transformer may have to
operate with a nonsinusoidal waveform at its input.
A pulse transformer with a ferrite core is used commonly
to transfer the gate signal to the power semiconductor
device. A ferrite-cored transformer with a relatively higher
power output is also used in an application such as a high
frequency inverter.
Many power electronic systems are operated with negative
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feedback. A linear controller such as a PI controller is used
in relatively simple applications, whereas a controller
based on digital or state-variable feedback techniques is
used in more sophisticated applications.
Computer simulation is often necessary to optimize the
design of a power electronic system. In order to simulate,
knowledge of software package such as MATLAB and the
know-how to model nonlinear systems may be necessary.
The study of power electronics is an exciting and a challenging
experience. The scope for applying power electronics is growing
at a fast pace. New devices keep coming into the market,
sustaining development work in power electronics.
SIMPLE DIODE CIRCUITS
This chapter describes two simple diode
circuits:
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A SINGLE DIODE CIRCUIT
A DIODE CIRCUIT WITH A FREE-WHEELING
DIODE
A SINGLE DIODE CIRCUIT
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
INTERACTIVE SIMULATION
PSPICE SIMULATION
MATLAB SIMULATION
MATHCAD SIMULATION
SUMMARY
This page describes a single diode circuit. Most of the power electronic applications operate at a
relative high voltage and in such cases, the voltage drop across the power diode tends to be small.
It is quite often justifiable to use the ideal diode model. An ideal diode has zero conduction drop
when it is forward-biased and has zero current when it is reverse-biased. The explanation and the
analysis presented below is based on the ideal diode model.
CIRCUIT OPERATION
A circuit with a single diode and an RL load is shown above. The source vs is an alternating
sinusoidal source. If vs = E * sin (wt), vs is positive when 0 < wt < π, and vs is negative when π <
wt <2π. When vs starts becoming positive, the diode starts conducting and the source keeps the
diode in conduction till wt reaches π radians. At that instant defined by wt = π radians, the current
through the circuit is not zero and there is some energy stored in the inductor. The voltage across an
inductor is positive when the current through it is increasing and it becomes negative when the
current through it tends to fall. When the voltage across the inductor is negative, it is in such a
direction as to forward-bias the diode. The polarity of voltage across the inductor is as shown in the
sketches shown below.
When vs changes from a positive to a negative value, there is current through the load at the
instant wt = π radians and the diode continues to conduct till the energy stored in the inductor
becomes zero. After that the current tends to flow in the reverse direction and the diode blocks
conduction. The entire applied voltage now appears across the diode.
MATHEMATICAL ANALYSIS
An expression for the current through the diode can be obtained as shown below. It is assumed
that the current flows for 0 < wt < β, where β > π . When the diode conducts, the driving
function for the differential equation is the sinusoidal function defining the source voltage. During
the period defined by β < wt < 2π, the diode blocks current and acts as an open switch. For this
period, there is no equation defining the behaviour of the circuit. For 0 < wt < β , the equation (1)
defined below applies.
Given a linear differential equation, the solution is found out in two parts. The homogeneous
equation is defined by equation (2). It is preferable to express the equation in terms of the angle θ
instead of 't'. Since θ = wt, we get that dθ = w.dt. Then equation (2) then gets converted to
equation (3). Equation (4) shown above is the solution to this homogeneous equation and is called
the complementary integral.
The value of constant A in the complimentary solution is to be evaluated later.
The particular solution is the steady-state response and equation (5) expresses the particular
solution. The steady-state response is the current that would flow in steady-state in a circuit that
contains only the source, the resistor and the inductor shown in the circuit above, the only element
missing being the diode. This response can be obtained using the differential equation or the
Laplace transform or the ac sinusoidal circuit analysis. The total solution is the sum of both the
complimentary and the particular solution and it is shown as equation (6). The value of A is
obtained using the initial condition. Since the diode starts conducting at wt = 0 and the current
starts building up from zero, i(0) = 0. The value of A is expressed by equation (7).
Once the value of A is known, the expression for current is known. After evaluating A, current can
be evaluated at different values of wt, starting from wt = π. As wt increases, the current would keep
decreasing. For some value of wt, say β , the current would be zero. If wt > β , the current would
evaluate to a negative value. Since the diode blocks current in the reverse direction, the diode stops
conducting when wt reaches β. Then an expression for the average output voltage can be obtained.
Since the average voltage across the inductor has to be zero, the average voltage across the resistor
and the average voltage at the cathode of the diode are the same. This average value can be
obtained as shown in equation (8).
INTERACTIVE SIMULATION
The operation of the circuit can be simulated as shown below. In order to simulate, the solution for
current is presented in the following form, where τ = (wL)/R. Then
Again it is preferable to normalize. Here E is set to unity and E/R is also set to unity. Then
vs = sin (wt).
vo = i for 0 < wt < β,
vL = vs - i for 0 < wt < β .
To solve the expression, all we need to know is then the ratio τ. The applet shown below simulates
this circuit. You have to key-in the ratio τ and then click on the button next to it. Do not key-in a
NaN.
The next page presents the same circuit with an additional diode.
PSPICE SIMULATION
For simulation using Pspice, the circuit used is shown below. Here the nodes are numbered. The ac
source is connected between nodes 1 and 0. The diode is connected between nodes 1 and 2 and the
inductor links nodes 2 and 3. The resistor is connected from 3 to the reference node, that is, node 0.
The Pspice program is presented below.
* First Chapter: Half-wave Rectifier with RL Load
* A problem to find the diode current
VIN 1 0 SIN(0 340V 50Hz)
D1 1 2 DNAME
L1 2 3 31.8MH
R1 3 0 10
.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)
.TRAN 10US 60.0MS 20.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The diode is described using the MODEL statement. The TRAN statement simulates the transient
operation for a period of 60 ms at an interval of 10 µs. The OPTIONS statement sets limits for
tolerances. The output can be
viewed on the screen because of the PROBE statement. A snapshot of output is presented below.
MATLAB SIMULATION
The Matlab program used is re-produced below.
% Program to simulate the half-wave rectifier circuit
% Enter the peak voltage, frequency, inductance L in mH and
resistor R
disp('Typical value for peak voltage is 340 V')
peakV=input('Enter Peak voltage in Volts>');
disp('Typical value for line frequency is 50 Hz')
freq=input('Enter line frequency in Hz>');
disp('Typical value for Load inductance is 31.8 mH')
L=input('Enter Load inductance in mH>');
disp('Typical value for Load Resistance is 10.0 Ohms')
R=input('Enter Load Resistance in Ohms>');
w=2.0*pi*freq;
X=w*L/1000.0;
if (X<0.001) X=0.001; end;
Z=sqrt(R*R+X*X);
loadAng=atan(X/R);
A=peakV/Z*sin(loadAng);
tauInv=R/X;
for n=1:360;
theta=n/180.0*pi;
X(n)=n;
cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*theta);
if (cur>0.0)
Vind(n)=peakV*sin(theta)-R*cur;
iLoad(n)=cur;
Vout(n)=peakV*sin(theta);
else
Vind(n)=0;
iLoad(n)=0;
Vout(n)=0;
end;
end;
plot(X,iLoad)
title('The diode current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,Vout)
title('Voltage at cathode')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,Vind)
title('Inductor Voltage')
xlabel('degrees')
ylabel('Volts')
grid
The plots obtained for the typical values mentioned are shown below.
It can be seen from the waveform of voltage across the inductor is that the area above the x-axis at
0 V is equal to its area below the x-axis. It can be seen that the matlab program is relatively
simple.
MATHCAD SIMULATION
The simulation of this circuit is in a file called halfrec1.mcd. You can download it by clicking on
the image below. You need MathCad program to open this file. If you open this file with MathCad
program, you can change the parameters and see how the waveforms of output voltage, diode
current and voltage across the inductor.
Alternatively, you can view the mathcad file by clicking on the image below. Then you see the file
in HTML format, but you cannot change the parameters.
This page has described the circuit of a half-wave rectifier. It has been simulated using different
programs. The next page is on a half-wave rectifier with a free-wheeling diode.
A DIODE CIRCUIT WITH A FREE-WHEELING DIODE
CIRCUIT DIAGRAM
CIRCUIT OPERATION
MATHEMATICAL_ANALYSIS
SIMULATION
PSPICE SIMULATION
MATLAB SIMULATION
MATHCAD SIMULATION
SUMMARY
CIRCUIT DIAGRAM
The circuit shown above differs from the circuit described in the previous page, which had only diode D1. This circuit
has another diode, marked D2 in the circuit shown above. This diode is called the free-wheeling diode. The circuit
operation is described next. The explanation is based on the assumption that the reader knows how the circuit without a
free-wheeling diode operates.
CIRCUIT OPERATION
Let the source voltage vs be defined to be E*sin (wt). The source voltage is positive when 0 < wt < π radians and it is
negative when π < wt < 2π radians. When vs is positive, diode D1 conducts and the voltage vc is positive. This in turn
leads to diode D2 being reverse-biased during this period. During π < wt < 2π, the voltage vc would be negative if
diode D1 tends to conduct. This means that D2 would be forward-biased and would conduct. When diode D2 conducts,
the voltage vc would be zero volts, assuming that the diode drop is negligible. Additionally when diode D2 conducts,
diode D1 remains reverse-biased, because the voltage vs is negative.
When the current through the inductor tends to fall, it starts acting as a source. When the inductor acts as a source, its
voltage tends to forward bias diode D2 if the source voltage vs is negative and forward bias diode D1 if the source
voltage vs is positive. Even when the source voltage vs is positive, the inductor current would tend to fall if the source
voltage is less than the voltage drop across the load resistor.
During the negative half-cycle of source voltage, diode D1 blocks conduction and diode D2 is forced to conduct. Since
diode D2 allows the inductor current circulate through L, R and D2, diode D2 is called the free-wheeling diode. We can
say that the current free-wheels through D2.
MATHEMATICAL ANALYSIS
An expression for the current through the load can be obtained as shown below. It can be assumed that the load current
flows all the time. In other words, the load current is continuous. When diode D1 conducts, the driving function for the
differential equation is the sinusoidal function defining the source voltage. During the period defined by π < wt < 2π,
diode D1 blocks current and acts as an open switch. On the other hand, diode D2 conducts during this period, the driving
function can be set to be zero volts. For 0 < wt < π , the equation (1) shown below applies.
For the negative half-cycle of the source, equation (2) applies. As in the previous case, the solution is obtained in two
parts. The expressions for the complementary integral and the particular integral are the same. The expression for the
complementary integral is presented by equation (3). The particular solution to the equation (1) is the steady-state
response and is presented as equation (4). The total solution is the sum of both the complimentary and the particular
solution. For 0 < θ < π, where wt = θ, the total solution is presented as equation (5).
The difference in solution is in how the constant A in complementary integral is evaluated. In the case of the circuit
without free-wheeling diode, i(0) = 0, since the current starts building up from zero at the start of every positive halfcycle. On the other hand, the current-flow is continuous when the circuit contains a free-wheeling diode also. Since the
input to the RL circuit is a periodic half-sinusoid function, we expect that the response of the circuit should also be
periodic. That means, the current through the load is periodic. It means that i(0) = i(2π).
Since the current through the load free-wheels during π < θ < 2π , we get equation (6). We use ( θ - π ) for the elapsed
period in radians instead of θ itself, since the free-wheeling action starts at θ = π . From the total solution, we can get
i(π) from equation (7) by substituing θ = π. To obtain A, the following steps are necessary. From the total solution,
obtain an expression for i(0) by substituting 0 for θ. Also obtain an expression for i(π) by substituting π for θ in
equation (7). Using this expression for i(π) in equation (6), obtain i(2π) by letting θ = 2π . Since i(0) = i(2π), we can
obtain A from equation (8). In equation (8), the terms containing constant A are grouped on the left-hand side of
equation and the other terms on the right-hand side.
SIMULATION
The operation of the circuit can be simulated as shown below. During 0 < θ < π , the expression for current is presented
as equation (9). During π < θ < 2π , the expression for current is shown as equation (10).
The voltage across the inductor is obtained to be
vL(θ ) = vs(θ) - R*i(θ) , for 0 < θ < π and
= - R*i(θ) , for π < θ < 2π .
vs(θ) = E*sin (θ).
It is preferable to normalize vs with respect to E and the current with respect to E/R and then the only value required to
be known for solving for the current is τ.
PSPICE SIMULATION
The circuit that is used for Pspice simulation is shown below. The nodes are numbered and the components have been
labeled.
A Pspice program to simulate the circuit shown above is presented now.
* Half-wave Rectifier with free-wheeling diode and with RL Load
* A problem to find the diode current
VIN 1 0 SIN(0 340V 50Hz)
D1 1 2 DNAME
L1 2 3 31.8MH
R1 3 0 10
D2 0 2 DNAME
.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)
.TRAN 10US 60.0MS 20.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The waveforms obtained are presented now. Since the program specifies that the waveforms be displayed from the
second cycle, there is no output for the first 20 ms. The waveform of voltage at the cathode of both diodes is shown
below.
The waveform of current through diode D1 is presented next.
The waveform of current through diode D2 is shown below.
The waveform of current through load resistor is shown below. It is the sum of both diode currents.
The waveform of voltage across the inductor is shown below.
The advantage with Pspice is the simplicity of the program. In addition, the devices used are also simulated using the
spice models.
MATLAB SIMULATION
A matlab program for simulating the half-wave rectifier with a free-wheeling diode is presented below.
% Program to simulate the half-wave rectifier circuit
% The circuit has a free-wheeling diode
% Enter the peak voltage, frequency, inductance L in mH and
resistor R
disp('Typical value for peak voltage is 340 V')
peakV=input('Enter Peak voltage in Volts>');
disp('Typical value for line frequency is 50 Hz')
freq=input('Enter line frequency in Hz>');
disp('Typical value for Load inductance is 31.8 mH')
L=input('Enter Load inductance in mH>');
disp('Typical value for Load Resistance is 10.0 Ohms')
R=input('Enter Load Resistance in Ohms>');
w=2.0*pi*freq;
X=w*L/1000.0;
if (X<0.001) X=0.001; end;
Z=sqrt(R*R+X*X);
tauInv=R/X;
loadAng=atan(X/R);
A1=peakV/Z*sin(loadAng);
A2=peakV/Z*sin(pi-loadAng)*exp(-pi*tauInv);
A3=(A1+A2)/(1-exp(-2.0*pi*tauInv));
Ampavg=0;
AmpRMS=0;
for n=1:360;
theta=n/180.0*pi;
X(n)=n;
if (n<180)
cur=peakV/Z*sin(theta-loadAng)+A3*exp(-tauInv*theta);
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
else
A4=peakV/Z*sin(pi-loadAng)*exp(-(theta-pi)*tauInv);
cur=A4+A3*exp(-tauInv*theta);
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
end;
if (n<180)
Vind(n)=peakV*sin(theta)-R*cur;
Vout(n)=peakV*sin(theta);
diode2cur(n)=0;
diode1cur(n)=cur;
else
Vind(n)=-R*cur;
Vout(n)=0;
diode2cur(n)=cur;
diode1cur(n)=0;
end
iLoad(n)=cur;
end;
plot(X,iLoad)
title('The Load current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,Vout)
title('Voltage at cathode')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,Vind)
title('Inductor Voltage')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,diode1cur)
title('Diode 1 current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,diode2cur)
title('Diode 2 current')
xlabel('degrees')
ylabel('Amps')
grid
AmpRMS=sqrt(AmpRMS);
[A,message]=fopen('outhfr2.dat','w');
fprintf(A,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(A)
The responses obtained for the typical specified values are shown below.
The average and rms values of load current are presented below.
Avg Load Cur=
1.082254e+001
RMS Load Cur=
13.954542
MATHCAD SIMULATION
This circuit can be simulated using MathCad as shown next. Click on DOWNLOAD to download the file containing the
program and then open it using MathCad. Click on VIEW ONLY to view this file in HTML format.
SUMMARY
It has been shown how the half-wave rectifier with a free-wheeling diode can be simulated using different
software packages. The next page shows how a half-wave controlled rectifier circuit operates.
SIMPLE SCR CIRCUITS
This chapter describes two SCR circuits which
are:
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A SINGLE SCR CIRCUIT
A SINGLE-SCR CIRCUIT WITH A FREE-WHEELING
DIODE
A SINGLE SCR CIRCUIT
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
MATLAB SIMULATION
MATHCAD SIMULATION
SUMMARY
This page describes a circuit with a single SCR. It is similar to the single diode circuit, the difference being that an SCR
is used in place of the diode. Most of the power electronic applications operate at a relative high voltage and in such
cases, the voltage drop across the SCR tends to be small. It is quite often justifiable to assume that the conduction drop
across the SCR is zero when the circuit is analysed. It is also justifiable to assume that the current through the SCR is
zero when it is not conducting. It is known that the SCR can block conduction in either direction. The explanation and
the analysis presented below is based on the ideal SCR model. It is also assumed that the reader knows how an SCR
operates.
CIRCUIT OPERATION
A circuit with a single SCR and an RL load is shown above. The source vs is an alternating sinusoidal source. If vs = E
* sin (wt), vs is positive when 0 < wt < π, and vs is negative when π < wt <2π. When vs starts becoming positive, the
SCR is forward-biased but remains in the blocking state till it is triggered. If the SCR is triggered at when wt = α, then α
is called the firing angle. When the SCR is triggered in the forward-bias state, it starts conducting and the positive
source keeps the SCR in conduction till wt reaches π radians. At that instant, the current through the circuit is not zero
and there is some energy stored in the inductor at wt = π radians. The voltage across an inductor is positive when the
current through it is increasing and it becomes negative when the current through the inductor tends to fall. When the
voltage across the inductor is negative, it is in such a direction as to forward-bias the SCR.
There is current through the load at the instant wt = π radians and the SCR continues to conduct till the energy stored in
the inductor becomes zero. After that the current tends to flow in the reverse direction and the SCR blocks conduction.
The entire applied voltage now appears across the diode.
MATHEMATICAL ANALYSIS
An expression for the current through the SCR can be obtained as shown below. It is assumed that the current flows for
α < wt < δ, where δ > π . When the SCR conducts, the driving function for the differential equation is the sinusoidal
function defining the source voltage. Outside this period, the SCR blocks current and acts as an open switch. For this
period, there is no equation defining the behaviour of the circuit. For α < wt < δ , equation (1) applies. Given a linear
differential equation, the solution is found out in two parts. The homogeneous equation is given by equation (2), where
α is the firing angle. The value of constant A in the complimentary solution is to be evaluated later. The particular
solution is the steady-state response and is diplayed as equation (3). The total solution is the sum of both the
complimentary and the particular solution and is presented as equation (4). The value of A is obtained using the initial
condition. Since the SCR starts conducting at wt = α and the current starts building up from zero, i(α) = 0. In the
expression above τ = wL/R. Then A can be expressed as in equation (5).
Once the value of A is known, the expression for current is known. When the firing angle α and the extinction angle δ
are known, the average output voltage at the cathode of the SCR can be evaluated as shown in equation (6).
The average load current can be obtained by dividing the average load voltage by the load resistance, since the average
voltage across the inductor is zero.
SIMULATION
The operation of the circuit can be simulated as shown below. In order to simulate, the solution for current is presented
in the following form, where τ = (wL)/R. Then
Again it is preferable to normalize. Here E is set to unity and E/R is also set to unity. Then
vs = sin (wt).
vo = vs - i for α < wt < δ, and
vL = vs - i for α < wt < δ .
To solve the expression, all we need to know is then the ratio τ. The applet shown below simulates this circuit. You
have to key-in the ratio τ and then click on the button next to it. Do not key-in a NaN. Enter the ratio τ in the left textfield to the left of the click button and the firing angle in degrees in the textfield to its right.
The next page presents the same circuit with a free-wheeling diode.
PSPICE SIMULATION
The program below presents a PSPICE program. The circuit used is shown below.
The PSPICE program shown below presents the SCR as a subcircuit. This model of the SCR has been described in the
book SPICE FOR POWER ELECTRONICS AND ELECTRIC POWER (Muhammed H.Rashid, Prentice-Hall , 1993,
pages 148-160).
* Half-wave Rectifier with RL Load
* A problem to find the SCR current
VIN 1 0 SIN(0 340V 50Hz)
XT1 1 2 5 2 SCR
VP 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)
L1 2 3 31.8MH
R1 3 0 10
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 20.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The waveforms obtained are presented below.
The voltage waveform at the cathode of the SCR
The load current waveform
The inductor voltage waveform
The voltage waveform acros the SCR
The voltage waveform of the pulse source used for triggering the SCR
MATLAB SIMULATION
The Matlab program used for simulation is presented below.
% Program to simulate the half-wave controlled rectifier circuit
% Enter the peak voltage, frequency, inductance L in mH and
resistor R
disp('Typical value for peak voltage is 340 V')
peakV=input('Enter Peak voltage in Volts>');
disp('Typical value for line frequency is 50 Hz')
freq=input('Enter line frequency in Hz>');
disp('Typical value for Load inductance is 31.8 mH')
L=input('Enter Load inductance in mH>');
disp('Typical value for Load Resistance is 10.0 Ohms')
R=input('Enter Load Resistance in Ohms>');
disp('Typical value for Firing angle is 30.0 degree')
fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');
fangRad=fangDeg/180.0*pi;
w=2.0*pi*freq;
X=w*L/1000.0;
if (X<0.001) X=0.001; end;
Z=sqrt(R*R+X*X);
tauInv=R/X;
loadAng=atan(X/R);
A=peakV/Z*sin(loadAng-fangRad);
Ampavg=0;
AmpRMS=0;
for n=1:360;
theta=n/180.0*pi;
X(n)=n;
if (n<fangDeg)
cur=0.0;
Vind(n)=0;
iLoad(n)=0;
Vout(n)=0;
else
cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad));
if (cur>0)
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
Vind(n)=peakV*sin(theta)-R*cur;
iLoad(n)=cur;
Vout(n)=peakV*sin(theta);
else
Vind(n)=0;
iLoad(n)=0;
Vout(n)=0;
end;
end;
end;
plot(X,iLoad)
title('The Load current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,Vout)
title('Voltage at cathode')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,Vind)
title('Inductor Voltage')
xlabel('degrees')
ylabel('Volts')
grid
AmpRMS=sqrt(AmpRMS);
[A,message]=fopen('hwavec1.dat','w');
fprintf(A,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(A)
The waveforms obtained for the typical specified values are displayed now.
The output file containing the values of average load current and the RMS load current is presented below.
Avg Load Cur= 8.481852e+000 RMS Load Cur= 12.878237
MATHCAD SIMULATION
The MathCad program can be downloaded by clicking on the image below.
This program can also be viewed in the HTML format. Click on the image below to view this file.
SUMMARY
This page has described how the half-wave controlled rectifier circuit operates. The next page shows how the behaviour
of this circuit can be changed by adding a free-wheeling diode.
A SINLGE-SCR CIRCUIT WITH A FREE-WHEELING DIODE
CIRCUIT DIAGRAM
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
MATLAB SIMULATION
MATHCAD SIMULATION
SUMMARY
CIRCUIT DIAGRAM
The circuit shown above differs from the circuit described in the previous page, which had only a single SCR. This
circuit has a free-wheeling diode in addition, marked D in the circuit shown above. The circuit operation is described
next. The explanation is based on the assumption that both the diode and the SCR are ideal. It means that the voltage
drop across the device while in conduction is zero and the leakage current in the blocking state is zero.
CIRCUIT OPERATION
The source vs is an alternating sinusoidal source. If vs = E * sin (wt), vs is positive when 0 < wt < π, and it is negative
when π < wt <2π. When vs starts becoming positive, the SCR is forward-biased but remains in the blocking state till it
is triggered. If the SCR is triggered at when wt = α, then α is called the firing angle. When the SCR is triggered in the
forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till wt reaches π radians. At
that instant, the current through the circuit is not zero and there is some energy stored in the inductor at wt = π radians.
In the absence of the free wheeling diode, the inductor would keep the SCR in conduction for part of the negative cycle
till the energy stored in it is discharged. But when a free wheeling diode is present as shown in the circuit, the current has
a path that offers almost zero resistance. Hence the inductor discharges its energy during π < wt < (2π + α) through the
free wheeling diode. When there is a free wheeling diode, the current through the load tends to be continuous, at least
under ideal conditions. When the diode conducts, the SCR remains reverse-biased, because the voltage vs is negative.
MATHEMATICAL ANALYSIS
An expression for the current through the load can be obtained as shown below. It can be assumed that the load current
flows all the time. In other words, the load current is continuous. When the SCR conducts, the driving function for the
differential equation is the sinusoidal function defining the source voltage. During the period defined by π < wt < (2π +
α), the SCR blocks current and acts as an open switch. On the other hand, the free wheeling diode conducts during this
period, and the driving function can be set to be zero volts. For α < wt < π , equation (1) applies whereas equation (2)
applies for the rest of the cycle.
As in the previous cases, the solution is obtained in two parts. The expressions for the complementary integral and the
particular integral are the same. The expression for the complementary integral is presented as equation (3). The
particular solution is the steady-state response and is presented as equation (4).
The total solution is the sum of both the complimentary and the particular solution for α < wt < π .
The difference in solution is in how the constant A in complementary integral is evaluated. In the case of the circuit
without free-wheeling diode, i(α) = 0, since the current starts building up from zero when the SCR is triggered during the
positive half-cycle. On the other hand, the current-flow is continuous, when there is a free wheeling diode. Since the
input to the RL circuit is a periodic function, we expect that the response of the circuit should also be periodic. That
means, the current through the load is periodic. It means that
i(α) = i(2π + α).
Since the current through the load free-wheels during π < θ < (2π + α) , we get that,
i(θ) = i(π) *exp[-(θ − π)/τ] , where τ = wL/R.
We use ( θ - π ) for the elapsed period in radians instead of θ itself, since the free-wheeling action starts at θ = π . From
the total solution, we get i(π).
i(π) = A*exp(- π/τ) + (E/Z)*sin (π - α).
To obtain A, the following steps are necessary. From the total solution, obtain an expression for i(α) by substituting α
for θ. From the expression for the free-wheeling period, obtain i(2π + α) by letting θ = (2π + α) . Since i(α) = i(2π +
α), we can obtain A to be:
SIMULATION
The operation of the circuit can be simulated as shown below. During 0 < θ < π , the expression for current is:
i(θ) = A * exp [ -(θ − α)/τ) + (E/Z)*sin (θ − β ),
where τ = wL/R, β = atan (τ), Z2 = R2 + (wL)2 = R2*(1 + τ2) and α is the firing angle.
During π < θ < (2π + α) , the expression for current is:
i(θ) = A*exp {- (θ − α) /τ} + (E/Z)*sin (π - β)*exp {- (θ − π) /τ} .
The voltage across the inductor is obtained to be
vL(θ ) = vs(θ) - R*i(θ) , for α < θ < π and
= - R*i(θ) , for π < θ < (2π + α) .
vs(θ) = E*sin (θ).
It is preferable to normalize vs with respect to E and the current with respect to E/R and then the only value that is
required to be known for solving for the current is τ. The applet shown below simulates this circuit. You have to key-in
the ratio τ in the text-field to the left of the button and the firing angle in degrees in the text-field to the right of the button
and then click on the button next to it. Do not key-in a NaN.
PSPICE SIMULATION
The circuit used for Pspice simulation is shown below.
The Pspice program is presented next.
* Half-wave Rectifier with a free-wheeling diode
* A problem to find the SCR current
VIN 1 0 SIN(0 340V 50Hz)
XT1 1 2 5 2 SCR
VP 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)
L1 2 3 31.8MH
R1 3 0 10
D2 0 2 DNAME
.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 20.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The waveform of voltage at the cathode of SCR
The waveform of current through the load
The waveform of voltage across the SCR
The waveform of current through the SCR
The waveform of current through the free-wheeling diode
The waveform of voltage across the inductor
MATLAB SIMULATION
The Matlab program is presented below.
% Program to simulate the half-wave controlled rectifier circuit
% The circuit has a free-wheeling diode
% Enter the peak voltage, frequency, inductance L in mH and
resistor R
disp('Typical value for peak voltage is 340 V')
peakV=input('Enter Peak voltage in Volts>');
disp('Typical value for line frequency is 50 Hz')
freq=input('Enter line frequency in Hz>');
disp('Typical value for Load inductance is 31.8 mH')
L=input('Enter Load inductance in mH>');
disp('Typical value for Load Resistance is 10.0 Ohms')
R=input('Enter Load Resistance in Ohms>');
disp('Typical value for Firing angle is 30.0 degree')
fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');
fangRad=fangDeg/180.0*pi;
w=2.0*pi*freq;
X=w*L/1000.0;
if (X<0.001) X=0.001; end;
Z=sqrt(R*R+X*X);
tauInv=R/X;
loadAng=atan(X/R);
k1=exp(R*(pi+fangRad)/X);
k2=exp(R*(fangRad-pi)/X);
A=peakV/Z*(sin(pi-loadAng)+sin(loadAng-fangRad)*k1)/(k1-k2);
Ampavg=0;
AmpRMS=0;
Cur180=peakV/Z*sin(pi-loadAng)+A*k2;
for n=1:360;
theta=n/180.0*pi;
X(n)=n;
if (n<fangDeg)
cur=Cur180*exp(-(pi+theta)*tauInv);
Vind(n)=-R*cur;
iLoad(n)=cur;
Vout(n)=0;
VSCR(n)=peakV*sin(theta);
curSCR(n)=0;
diodecur(n)=cur;
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
elseif ((n>=fangDeg) & (n<180))
cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad));
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
Vind(n)=peakV*sin(theta)-R*cur;
iLoad(n)=cur;
Vout(n)=peakV*sin(theta);
VSCR(n)=0;
curSCR(n)=cur;
diodecur(n)=0;
else
cur=Cur180*exp((pi-theta)*tauInv);
Vind(n)=-R*cur;
iLoad(n)=cur;
Vout(n)=0;
VSCR(n)=peakV*sin(theta);
curSCR(n)=0;
diodecur(n)=cur;
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
end;
end;
plot(X,iLoad)
title('The Load current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,Vout)
title('Voltage at cathode')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,Vind)
title('Inductor Voltage')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,VSCR)
title('SCR Voltage')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,curSCR)
title('SCR Current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,diodecur)
title('diode Current')
xlabel('degrees')
ylabel('Amps')
grid
AmpRMS=sqrt(AmpRMS);
[C,message]=fopen('hwavec2.dat','w');
fprintf(C,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(C)
The output file is re-produced below.
Avg Load Cur=
1.009749e+001
RMS Load Cur=
Next the plots obtained for the typical specified values are displayed.
13.337142
This page has described the operation of a half-wave rectifier wirh a free-wheeling diode. Next we take up the study of
single-phase, full-wave, fully-controlled rectifier.
FULLY-CONTROLLED 1-PH SCR
BRIDGE RECTIFIER
This chapter describes the operation of a singlephase fully-controlled SCR bridge rectifier
circuit and two applications. The pages to
follow contain:
●
●
●
●
●
●
OPERATION WITH A PURELY RESISIVE LOAD
OPERATION WITH AN RL LOAD
OPERATION WITH SOURCE INDUCTANCE
OPERATION WITH AN RLC LOAD AND SOURCE
INDUCTANCE
AN APPLICATION: A BATTERY CHARGER
AN APPLICATION: A TWO-QUADRANT DC DRIVE
FULLY-CONTROLLED SINGLE-PHASE SCR BRIDGE RECTIFIER:
RESISTIVE LOAD
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
PERFORMANCE PARAMETERS FOR CONVERTERS
SIMULATION
PSPICE SIMULATION
MATHCAD SIMULATION
SUMMARY
This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a resistive load. The
operation of this circuit can be understood more easily when the load is purely resistive. The analysis in this page is
based on the assumption that the SCRs are ideal controlled switches. It means that when the SCRs are ON, the ON-state
drops are zero. In the OFF- state, the leakage current is assumed to be zero. The main purpose of the fully-controlled
bridge rectifier circuit is to provide a variable dc voltage from an ac source.
CIRCUIT OPERATION
The circuit of a single-phase fully-controlled bridge rectifier circuit is shown in the figure above. The circuit has four
SCRs. It is preferable to state that the circuit has two pairs of SCRs, with S1 and S4 forming one pair and, S2 and S3
the other pair. For this circuit, vs is a sinusoidal voltage source. When it is positive, SCRs S1 and S4 can be triggered
and then current flows from vs through SCR S1, load resistor R, SCR S4 and back into the source. In the next halfcycle, the other pair of SCRs conducts. Even though the direction of current through the source alternates from one
half-cycle to the other half-cycle, the current through the load remains unidirectional.
The main purpose of this circuit is to provide a variable dc output voltage, which is brought about by varying the firing
angle. Let vs = E sin wt, with 0 < wt < 360o. If wt = 30o when S1 and S4 are triggered, then the firing angle is said to be
30o. In this instance, the other pair is triggered when wt = 210o.
When vs changes from a positive to a negative value, the current through the load becomes zero at the instant wt = π
radians, since the load is purely resistive and the SCRs cease to conduct. After that there is no current flow till the other
pair is triggered. The conduction through the load is discontinuous.
The operation of the circuit is illustrated by animating the functioning of this circuit. Key in the firing angle in degrees
and click the button. The source voltage and the output voltage waveforms are also displayed.
MATHEMATICAL ANALYSIS
The analysis is relatively simple when the load is purely resistive. The aims of the analysis are:
i.
ii.
iii.
iv.
v.
vi.
vii.
To obtain the average output voltage as a function of firing angle,
To obtain the rms output voltage as a function of firing angle,
To obtain the ripple factor of output voltage,
To obtain the rms line current,
To obtain the fundamental component of line current,
To obtain the Displacement power factor and power factor of line current and
To obtain the total harmonic distortion(THD) in line current.
The average value of the output voltage is obtained as follows. Let the supply voltage be vs = E*Sin (θ ), where θ varies
from 0 to 2π radians. Since the output waveform repeats itself for every half-cycle, the average output voltage is
expressed as a function of α, the firing angle, as shown in equation (1). The r.m.s. value of output voltage is obtained as
shown in equation (2). The ripple factor in output voltage can defined in two ways. The definition followed in this text as
follows. The maximum average output voltage occurs at a firing angle of 0o. Let it be Vom. Then the ripple factor RF(α)
is defined as shown in equation (3).
The alternate definition uses Vo,avg(α) as the denominator instead of Vom. If Vo,avg(α) is used as the denominator, then
RF(α) can tend to infinity. It is more logical to express the ripple content as a fraction of the maximum average voltage.
The variation of average output voltage, rms output voltage and ripple factor with the firing angle have been shown
below. The plots shown below have been normalized with respect to Vom. For example, when the firing angle is 90o, the
average output is shown to be 0.5. It means that the actual average output voltage is 0.5Vom. It can also be seen that
when the firing angle is 0o, the r.m.s. output voltage is about 1.1Vom and the ripple factor is about 0.48. The ripple factor
increases as the firing angle increases. It increases to 0.658 at a firing angle of 65o and then it falls as the firing angle
increases further. At the firing angle of 65o, the r.m.s. ripple voltage in the output is 0.658Vom. For a sinusoidal source of
240 V r.m.s., the maximum r.m.s. ripple voltage works out to be 142 V.
PERFORMANCE PARAMETERS FOR CONVERTERS
The role of a rectifier circuit is to produce a dc output voltage from an ac sinusoidal source. The rectifier's output is not
pure d.c. and it contains ripple superimposed on its d.c. content and the current drawn from the ac source is not
sinusoidal either. It contains some fundamental component and harmonics. For the output voltage, its ripple content is
the performance criterion. It would be desirable to obtain the amplitude frequency spectrum of the output of the rectifier
in order to design a suitable filter circuit. However there is no mention of how to design a filter in this page.
As far as the ac source is concerned, the distortion in the source current is the performance criterion. The total harmonic
distortion(THD) or the harmonic factor, the amplitude frequency spectrum of the source current, the apparent power
factor and the displacement power factor have also to be computed. In addition, the crest factor is also to be computed, to
facilitate the selection of the proper SCR.
Given a periodic function f(t) with a period of T, f(t) can be described by a trigonometric Fourier series, as shown in
equation (4). The coefficients are defined as shown in equations (5), (6) and (7).
In the equations above, * is used to in place of the product sign and it should not be confused with the same symbol used
for indicating convolution integrals. The source frequency, f, is taken as the fundamental and hence wo = 2πf. It is
preferable to express the above equation in terms of angle θ, where θ = wot. If T is the cycle period, woT = 2π fT = 2π,
since f = 1/T. Then the equations for the Fourier coefficients can be expressed as shown in equations (8), (9) and (10).
In the case of the full-wave bridge rectifier circuit, the period of the output is only half that of the input sinusoidal source
and hence the output contains a dc component and even harmonics only. The source current has half-wave symmetry. A
waveform defined as f(t) over a cycle is said to have half-wave symmetry if it satisfies equation (11). A waveform with
half-wave symmetry contains a fundamental component and odd harmonics only.
Let the Fourier coefficients for the output voltage be av0(α), av2n(α) and bv2n(α) , where α is the firing angle and these
coefficients are evaluated as shown in equations (12), (13) and (14). Since the output repeats itself twice for every cycle
of source voltage, it contains only even harmonics. Then we obtain the amplitude of the harmonic as shown in equation
(16).
Let the Fourier coefficients for the source current be acur0(α), acur2n(α) and bcur2n(α) . The line current waveform has
half-wave symmetry and contains only odd harmonics. When the SCRs are assumed to be ideal, load current iLine(θ) is
defined by equation (17).
For the case where the load is purely resistive, the r.m.s. source current can be computed from the value obtained for the
output voltage, as shown in equation (22). Then the total harmonic distortion is defined by equation (23). Let the r.m.s
current when the firing angle is 0o be Irms,max. Since the waveform of the source current is purely sinusoidal when the
firing angle is 0o, the crest factor can be taken to be square root of 2. The program that simulates the operation of this
circuit computes the various values for a given firing angle and displays them in a suitable manner.
The displacement power factor, DPF, is the cosine of the angle by which the fundamental component of the line current
lags the source voltage. Then apparent power factor can be estimated as shown in equation (24).
SIMULATION
The firing angle has to be keyed-in. Then click on Compute button. The plots for voltage have been normalized with
respect to Vom and the currents with respect to Irms,max. The statistical details related to the output voltage have been
normalized with respect to Vom and those related to the source current with respect to Irms,max. The amplitude of each
harmonic in the output voltage has been normalized with respect to E which is the amplitude of the source voltage ,
whereas the amplitude of each harmonics in the source current has been normalized with respect to E/R.
PSPICE SIMULATION
The circuit used for Pspice simulation is shown below. All the nodes other than the datum node are connected to the
datum node through a 1 MΩ resistor. A floating node or a node that tends to float can be a problem for Pspice
simulation.
The program is presented below.
* Full-wave Bridge
VIN 1 0 SIN(0 340V
XT1 1 2 5 2 SCR
XT2 0 2 6 2 SCR
XT3 3 0 7 0 SCR
XT4 3 1 8 1 SCR
VP1 5 2 PULSE(0 10
VP2 6 2 PULSE(0 10
VP3 7 0 PULSE(0 10
VP4 8 1 PULSE(0 10
R1 2 3 10
R2 1 0 1MEG
R3 2 0 1MEG
R4 3 0 1MEG
Rectifier with a resistive load
50Hz)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 20.0MS 10US
.FOUR 50 V(2,3) I(VIN)
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The waveforms obtained are presented below.
The voltage waveform across the load resistor
The voltage waveform across SCR1
The average load current
The Frequency spectrum of line current
The Frequency spectrum of output voltage
SUMMARY
This page has described the operation of a fully-controlled bridge rectifier circuit with a purely resistive load. The next
page describes how the circuit behaves when the load has a reactive and a resistive component.
FULLY-CONTROLLED SINGLE-PHASE SCR BRIDGE RECTIFIER: RL LOAD
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
PERFORMANCE PARAMETERS
SIMULATION
PSPICE SIMULATION
MATHCAD SIMULATION
MATLAB SIMULATION
SUMMARY
This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a load consisting of
both a resistor and an inductor in series. The analysis in this page is based on the assumption that the SCRs are ideal
controlled switches. The main purpose of the fully-controlled bridge rectifier circuit is to provide a variable dc voltage
from an ac source.
CIRCUIT OPERATION
The circuit of a single-phase fully-controlled bridge rectifier circuit is shown in the figure above. The circuit has four
SCRs. It is preferable to state that the circuit has two pairs of SCRs, with S1 and S3 forming one pair and, S2 and S4
the other pair. For this circuit, vs is a sinusoidal voltage source. When it is positive, the SCRs S1 and S3 can be
triggered and then current flows from vs through SCR S1, load inductor L, load resistor R, SCR S3 and back into the
source. In the next half-cycle, the other pair of SCRs conducts. Even though the direction of current through the
source alternates from one half-cycle to the other half-cycle, the current through the load remains unidirectional.
The main purpose of this circuit is to provide a variable dc output voltage, which is brought about by varying the firing
angle. Let vs = E sin wt, with 0 < wt < 360o. If wt = 30o when S1 and S3 are triggered, then the firing angle is said to be
30o. In this instance the other pair is triggered when wt= 210o.
When vs changes from a positive to a negative value, the current through the load does not fall to zero value at the
instant wt = π radians, since the load contains an inductor and the SCRs continue to conduct, with the inductor acting as
a source. When the current through an inductor is falling, the voltage across it changes sign compared with the sign that
occurs when its current is rising. When the current through the inductor is falling, its voltage is such that the inductor
delivers power to the load resistor, feeds back some power to the ac source under certain conditions and keeps the SCRs
in conduction forward-biased. If the firing angle is less than the load angle, the energy stored in the inductor is sufficient
to maintain conduction till the next pair of SCRs is triggered. When the firing angle is greater than the load angle, the
current through the load becomes zero and the conduction through the load becomes discontinuous. Usually the
description of this circuit is based on the assumption that the load inductance is sufficiently large to keep the load current
continuous and ripple-free.
The operation of the circuit is illustrated by animating the functioning of this circuit. Key in the firing angle in degrees
and click the button. The source voltage, and the bridge output voltage are also displayed. It is assumed here that the load
inductance is quite large. The animation is correct only if the firing angle is less than 90o. The programs under
simulation section will run correctly for any firing angle.
MATHEMATICAL ANALYSIS
Analysis by hand is based on the assumption that the load inductance is sufficiently large to keep the load current ripplefree. Programs written for computer simulation are not based on this assumption and they simulate the operation based
on the parameter keyed-in.
When the load current is continuous, the average value of the output voltage is obtained as follows. Let the supply
voltage be vs = E*Sin (θ ), where θ varies from 0 to 2π radians. Since the output waveform repeats itself for every halfcycle, the average output voltage is expressed in equation (1) as a function of α, the firing angle.
If the load current is continuous, the r.m.s. value of output voltage is obtained as shown in equation (2).
The maximum average output voltage occurs at a firing angle of 0o. Let it be Vom. Then the ripple factor RF(a) is
defined as shown in equation (3). Equations (4) and (5) apply when the conduction is discontinuous.
The variation of average output voltage, r.m.s. output voltage and the ripple factor with the firing angle have been shown
below, based on the assumption that the load inductance is large. The plots shown below have been normalized with
respect to Vom. For example, when the firing angle is 60o, the average output is shown to be 0.5. It means that the actual
average output voltage is 0.5Vom. It can also be seen that when the firing angle is 0o, the r.m.s. output voltage is about
1.1Vom and the ripple factor is about 0.48. The ripple factor increases as the firing angle increases.
If the firing angle is highly retarded or if the load inductance is not sufficiently large, conduction of load current is not
continuous. Let us consider the positive half-cycle. When θ equals the firing angle, the SCRs S1 and S3 are triggered. Let
the load current start from zero at θ = α and let the load current fall to zero when θ = π + β , before the next pair of SCRs
is triggered. It means that β < α . Then the average output voltage is computed as shown in equation (4), whereas the
r.m.s. output voltage is computed as shown in equation (5).
The value of β can be found out by iteration, as shown in earlier pages. For the case when the conduction is
discontinuous, the plots of the average output voltage, the r.m.s. output voltage and the ripple factor are illustrated below.
Key-in the ratio of wL/R, where L is the load inductance, R is the load resistance and w is the angular frequency of the
source. In practice this ratio can vary from a low value to about 5. The load angle can be defined to be tan-1(wL/R).
When the firing angle is less than the load angle, the conduction is continuous. As explained below, the load current is
discontinuous if the firing angle is greater than the load angle.
PERFORMANCE PARAMETERS
In this section, the expressions for the load current are first derived. Then an expression for the line current can be
obtained. The r.m.s line current, the r.m.s. value of the fundamental of line current, the THD in line current, the
frequency spectrum of line current, the frequency spectrum and the ripple content of the bridge output voltage and the
frequency spectrum and the ripple content of the voltage across the load resistor are also determined.
For the case of discontinuous conduction, let the load current in each half cycle flow from θ = α till θ = π + β , where β <
α. During this period, the differential equation that applies to the load current is calculated according to equation (6),
where α is the firing angle, (π +β) is the extinction angle and the conduction angle within a cycle is (π + β − α ). The
solution to the above equation is obtained as given shown in equation (7). In equation (7), τ represents wL/R,the ratio of
load reactance to load resistance. φ is the load angle, defined to be tan-1(wL/R). When the conduction is discontinuous,
the load current starts from zero value at θ = α . Hence equation (8) is obtained. When φ < α , A has a negative value and
we have discontinuous conduction when φ < α . It means that the firing of the SCRs is retarded and α > φ .
When the conduction is continuous, equation (9) applies. The solution to equation (9) is presented as equation (10).
Since the signal applied to the load circuit is a periodic signal, the load current is also periodic, after a transient period. It
means that equation (11) is valid for continuous conduction.
For continuous conduction, φ > α and A is positive. When φ = α, the conduction is about to become continuous and the
value of A should satisfy both expressions for A. It is seen that φ = α, A = 0 according to both expressions for A. When φ
= α , the current drawn from the source is sinusoidal and equals (E/Z)*Sin (wt - φ), where Z is defined as shown in
equation (12).
Once the load current is defined for a half-cycle, the expression for line current can be obtained. It equals the load current
when SCRs S1 and S3 are conducting and it is the negative of the load current when the other pair of SCRs is conducting.
Further analysis can be carried out using Fourier series and the DPF and PF of the line current can be determined.
When the circuit is switched on initially, the load current may settle down to a periodic response after a few output
cycles. The time constant of the load is L/R and a time period corresponding to five times the constant should elapse
before the load current becomes periodic. For example if the load time constant is 20 ms, a time period of 100 ms should
pass before the load current becomes periodic. This time period corresponds to five input voltage cycles at 50 Hz and six
input cycles at 60 Hz.
SIMULATION
The plots for voltage have been normalized with respect to Vom, the maximum average output voltage and the plots for
line current with respect to Irms,max, which is 0.707E/R. To see the periodic response, key-in firing angle in degrees in the
box on the left-hand side and the ratio τ in the box on the right-hand side and then click start button. The boxes have
default values filled-in.
The statistical details related to the output voltage have been normalized with respect to Vom and those related to the
source current with respect to Irms,max. The amplitude of each harmonic in the output voltage has been normalized with
respect to E which is the amplitude of the source voltage , whereas the amplitude of each harmonics in the source current
has been normalized with respect to E/R. The applet to follow displays the statistical details.
The applet below displays the transient response. As the ratio gets larger and larger, the load current requires larger
period to elapse before it becomes periodic.
PSPICE SIMULATION
The circuit used for simulation is shown below.
The program is presented below.
* Full-wave Bridge
VIN 1 0 SIN(0 340V
XT1 1 2 5 2 SCR
XT2 0 2 6 2 SCR
XT3 4 0 7 0 SCR
XT4 4 1 8 1 SCR
VP1 5 2 PULSE(0 10
VP2 6 2 PULSE(0 10
VP3 7 0 PULSE(0 10
VP4 8 1 PULSE(0 10
L1 2 3 31.8M
R1 3 4 10
R2 1 0 1MEG
R3 2 0 1MEG
R4 4 0 1MEG
Rectifier with a resistive load
50Hz)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 0.0MS 10US
.FOUR 50 V(2,4) I(VIN)
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The responses obtained are shown below.
The waveform of voltage at the output of bridge
The waveform of load current
The waveform of voltage across the SCR
The waveform of voltage across the inductor
The waveform of line current
The average load current
The rms line current
Fourier frequency spectrum of voltage at the output of bridge
Fourier frequency spectrum of line current
MATLAB SIMULATION
The Matlab program is presented below.
% Program to simulate the full-wave fully-controlled bridge
rectifier
% Simulation at a specified firing angle
% Enter the peak voltage, frequency, inductance L in mH and
resistor R
disp('Typical value for peak voltage is 340 V')
peakV=input('Enter Peak voltage in Volts>');
disp('Typical value for line frequency is 50 Hz')
freq=input('Enter line frequency in Hz>');
disp('Typical value for Load inductance is 31.8 mH')
L=input('Enter Load inductance in mH>');
disp('Typical value for Load Resistance is 10.0 Ohms')
R=input('Enter Load Resistance in Ohms>');
disp('Typical value for Firing angle is 30.0 degree')
fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');
fangRad=fangDeg/180.0*pi;
w=2.0*pi*freq;
X=w*L/1000.0;
if (X<0.001) X=0.001; end;
Z=sqrt(R*R+X*X);
tauInv=R/X;
loadAng=atan(X/R);
k1=peakV/Z;
k2=2.0*k1*sin(loadAng-fangRad)/(1.0-exp(-pi*tauInv));
k3=k1*sin(loadAng-fangRad);
if (fangRad<loadAng)
A=k2;
sw=1;
else
A=k3;
sw=2;
end;
Ampavg=0;
AmpRMS=0;
for n=1:360;
theta=n/180.0*pi;
X(n)=n;
if (sw==1);
if (n<fangDeg);
cur=k1*sin(pi+theta-loadAng)+A*exp(-tauInv*(pi+thetafangRad));
vbr(n)=peakV*sin(theta+pi);
vind(n)=vbr(n)-R*cur;
iLoad(n)=cur;
vSCR(n)=peakV*sin(theta);
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
elseif ((n>=fangDeg) & (n<(180+fangDeg)));
cur=k1*sin(theta-loadAng)+A*exp(-tauInv*(thetafangRad));
vbr(n)=peakV*sin(theta);
vind(n)=vbr(n)-R*cur;
iLoad(n)=cur;
vSCR(n)=0;
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
else (n>=(180+fangDeg));
cur=k1*sin(theta-pi-loadAng)+A*exp(-tauInv*(theta-pifangRad));
vbr(n)=peakV*sin(theta-pi);
vind(n)=vbr(n)-R*cur;
iLoad(n)=cur;
vSCR(n)=peakV*sin(theta);
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
end;
else
if (n<fangDeg);
cur=k1*sin(pi+theta-loadAng)+A*exp(-tauInv*(pi+thetafangRad));
if (cur>0);
vbr(n)=peakV*sin(theta+pi);
vind(n)=vbr(n)-R*cur;
iLoad(n)=cur;
vSCR(n)=peakV*sin(theta);
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
else;
vbr(n)=0.0;
vind(n)=0.0;
iLoad(n)=0.0;
vSCR(n)=peakV*sin(theta)/2.0;
end;
elseif ((n>=fangDeg) & (n<(180+fangDeg)));
cur=k1*sin(theta-loadAng)+A*exp(-tauInv*(thetafangRad));
if (cur>0);
vbr(n)=peakV*sin(theta);
vind(n)=vbr(n)-R*cur;
iLoad(n)=cur;
vSCR(n)=0;
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
else;
vbr(n)=0.0;
vind(n)=0.0;
iLoad(n)=0.0;
vSCR(n)=peakV*sin(theta)/2.0;
end;
else (n>=(180+fangDeg));
cur=k1*sin(theta-pi-loadAng)+A*exp(-tauInv*(theta-pifangRad));
vbr(n)=peakV*sin(theta-pi);
vind(n)=vbr(n)-R*cur;
iLoad(n)=cur;
vSCR(n)=peakV*sin(theta);
Ampavg=Ampavg+cur*1/360;
AmpRMS=AmpRMS+cur*cur*1/360;
end;
end;
end;
plot(X,iLoad)
title('The Load current')
xlabel('degrees')
ylabel('Amps')
grid
pause
plot(X,vbr)
title('Bridge Output volt')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,vind)
title('Inductor Voltage')
xlabel('degrees')
ylabel('Volts')
grid
pause
plot(X,vSCR)
title('SCR Voltage')
xlabel('degrees')
ylabel('Volts')
grid
AmpRMS=sqrt(AmpRMS);
[C,message]=fopen('fwrl1ph1.dat','w');
fprintf(C,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);
fclose(C);
The responses obtained for the typical specified values have been displayed below.
The results obtained with a firing angle of 60o have been displayed below. The other values are the same as for the
previous set. When the firing angle is retarded this much, the load current is discontinuous.
SUMMARY
This page has described the operation of a fully-controlled bridge rectifier circuit with an RL load. Even though the
circuit has been simulated using Pspice, Mathcad and Matlab, it is difficult to reach the level of interactive programming
that can be achieved with Java or C by using these packages. Next page describes the operation of this circuit when the
source inductance is also taken into account.
OPERATION WITH SOURCE INDUCTANCE:
1-PH. CONTROLLED RECTIFIER
CIRCUIT DIAGRAM
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
MATHCAD SIMULATION
SIMULATION USING C
SUMMARY
CIRCUIT DIAGRAM
This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a load consisting of
both resistance and inductance. In addition, the inductance of ac source supplying power to the circuit is taken into
account.
CIRCUIT OPERATION
The presence of source inductance changes the way the circuit operates during commutation. The word commutation
refers to switching conduction from one pair of SCRs to the other. Let vs = E sin wt, with 0 < wt < 360o. Let the load
inductance be large enough to maintain a steady current through the load. Let firing angle α be 30o. Let SCRs S2 and S4
be in conduction before wt < 30o. When S1 and S3 are triggered at wt = 30o, there is current through the source
inductance, flowing in the direction opposite to that marked in the circuit diagram and hence commutation of current
from S2 and S4 to S1 and S3 would not occur instantaneously. The source current changes from - iL to iL due to the whole
of the source voltage being applied across the source inductance. When S1 is triggered with S4 in conduction, the current
through S1 would rise from zero to iL and the current through S4 would fall from iL to zero. Similar process occurs with
the SCRs S2 and S3. During this period, the current through S3 would rise from zero to iL and , the current through S2
would fall from iL to zero. The duration of the process of commutation is usually referred to as commutation overlap
period.
The operation of the circuit is illustrated with the help of an animation. There is no scope for setting any variable for
animation, since the purpose is only to show how the circuit operates. It is assumed that the load inductance is large to
keep the load current at a steady value. You can observe how the currents through the devices and the line current change
during commutation overlap. The running of the program can be halted by clicking on the Stop button. It can be resumed
by clicking on the Run button. By clicking on the One Step button, you can step through the process. To start or run the
process for one more cycle, click on the Start or Repeat button.
MATHEMATICAL ANALYSIS
BASED ON THE ASSUMPTION THAT LOAD INDUCTANCE IS INFINITE
When the load inductance is infinite, we can assume that the load current is continuous and steady without ripple. Let
the firing angle be α. Let the commutation overlap period last from wt = α till wt = β.
During the period, α < wt < β, the output voltage is zero because all the SCRs are in conduction. If the SCRs are ideal,
the drop across an SCR in conduction is zero and hence the output voltage is zero. During β < wt < ( π + α), the output
voltage equals E*Sin (wt) and then the average output voltage can be obtained as shown in equation (1).
This expression is not very useful because the value of β is required to be known. The value of α is known, since it is the
firing angle, whereas the value of β is not likely to be known. Here β is the angle at which process of commutation
overlap ends and the duration of commutation depends on the firing angle, the value of source reactance and the load
impedance and value of β is variable and unknown. Hence it is preferable to derive an alternate expression. It is possible
to derive an alternate expression for the case when the load reactance is large enough to ensure that the load current
remains steady without ripple at a given firing angle. Let us assume that the load current be I and the firing angle be α.
Let the line current change from - I to + I during commutation overlap when wt changes from α to β. From the
waveforms shown above, the area of volt-seconds lost to output due to commutation overlap is computed as shown in
equation (2).
Since this area is lost over π radians, the average value of output voltage lost due to commutation is calculated as shown
in equation (3). It is known that the average output voltage with no commutation overlap is (2E/π)*Cos (α). By
subtracting the voltage lost due to commutation, we can get the average output voltage taking into account the effect of
commutation overlap, as shown in equation (4).
The waveforms appear as shown below. Key-in a firing angle less than 90o and then press the Start button.
ANALYSIS WITH A FINITE LOAD INDUCTANCE
With a finite load inductance, the conduction through the load can be either continuous or discontinuous. Let angle ϕ be
defined as shown in equation (5). If the firing angle α is greater than ϕ, then the current through the load is discontinuous
and the analysis is similar to that used for the circuit without a source inductance. When α < ϕ , then the conduction is
continuous. The analysis is carried out as follows. In these equations, α is the firing angle, β is the angle at which
commutation overlap ends and ϕ has been defined above. Let iL be the load current and is is the source current. Let the
source voltage vs = E *Sin (θ), where θ = wt and 0 < θ < 2π . Then the equation that is applicable during β < θ < (π + α)
can be expressed as shown in equation (6). During this period, the line current is equal to load current as defined by
equation (7).
During β < θ < (π + α), the voltage that appears as output is almost the negative of the source voltage and hence equation
(8) is used and the line current has the same magnitude as the load current, but its polarity is opposite to that of load
current as indicated by equation (9).
There are two instances of switching in one input cycle and commutation overlap occurs immediately after triggering
either pair of SCRs. The pair consisting of S1 and S3 is triggered at wt = α and commuation overlap lasts from wt = α
till wt =β and the output voltage is zero during this period as indicated by equation (10). Similarly after S2 and S4 are
triggered at wt = (π + α), the output voltage is zero from wt = (π + α) till wt = (π + β), as indicated by equation (11).
During the commuation overlap, the entire input voltage is applied across the source inductance, as indicated by
equations (12) and (13).
The average and rms output voltage can be obtained as shown by equations (14) and (15). The maximum output voltage
that can occur is indicated by equation (16) neglecting the loss in output that may occur due to commutation overlap.
Then the ripple factor of output voltage can be expressed as shown by equation (17). If the ripple factor is multiplied by
Vom, the rms value of ripple content in output voltage is obtained.
The fundamental component of the source current can be determined and then the THD, DPF and apparent power factor
can be determined. The programs for simulation have been based on the equations displayed above.
SIMULATION
The program in the applet to follow displays the bridge output waveform, the load current, the line current, the voltage
across the load inductance, the voltage across the source inductance, the voltage across SCR S1, and the currents through
S1 and S4. Key-in the ratio wLL/RL, the ratio wLs/RL and the firing angle in degrees and then click on the Start button.
The program to follow produces the statistics for this circuit and needs the same inputs as the applet above.
The next applet displays the transient response of the load current. If the load inductance is large, the transient response
may last for a few cycles before it settles down to a periodic response. The load current starts building up from zero from
the instant one of the pairs of SCRs is triggered first. The response shown starts from this instant.
The next applet shows how the average bridge output voltage, the rms output voltage, and the ripple factor vary with
firing angle for a given set of wLL/RL and wLs/RL. This program may take some time to run.
PSPICE SIMULATION
The circuit used for simulation is presented below.
The Pspice program is presented first.
* Full-wave Bridge
VIN 9 0 SIN(0 340V
XT1 1 2 5 2 SCR
XT2 0 2 6 2 SCR
XT3 4 0 7 0 SCR
XT4 4 1 8 1 SCR
VP1 5 2 PULSE(0 10
VP2 6 2 PULSE(0 10
VP3 7 0 PULSE(0 10
VP4 8 1 PULSE(0 10
L1 2 3 31.8M
Rectifier with RL Load and source Inductance
50Hz)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
L2
R1
R2
R3
R4
1
3
1
2
4
9
4
0
0
0
1.6M
10
1MEG
1MEG
1MEG
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 0.0MS 10US
.FOUR 50 V(2,4) I(VIN)
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The waveforms obtained are now presented.
The Voltage Output of the bridge
The Expanded view of Voltage Output of the bridge to show the commutation overlap
The waveform of Load Current
The waveform of line current
The expanded waveform of line current
The waveform of voltage across the line/source inductance
It can be seen that there is close agreement between the results obtained from the Applets and the Mathcad program.
SIMULATION USING C
Instead of a Matlab program, a C program is presented below. You can download this file by clicking on the image
displayed below. This file, called fcb_1pha.cpp, can be complied just as a C program and no feature of C++ has been
used in this program. This program produces an output file called tran_v1.csv and stores in it in the same directory in
which the source file is located. This file can be opened by a spreadsheet program like EXCEL and the plots displayed
can be obtained.
C Program
// Simulation of a single-phase fully-controlled bridge
// rectifier circuit.
//
//
//
//
//
//
//
//
//
//
//
This program produces the transient response, the periodic
response. Commuation overlap angle is also computed.
Calculations normalized. The average voltage that would
occur for diode-bridge rectifier with a resistive load is
taken to be unity. The average current through the load
resistor with a pure resistive load on the diode bridge
rectifier is taken as the unity current. The parameters to
be fed are just three: the ratio of load inductive reactance
to the load resistance, wL1/R , the ratio of source reactance
to the load resistance to load resistance wL2/R and the firing
angle in degrees.
#include <math.h>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<iostream.h>
void
void
void
void
void
kreateRespFile(void);
tranResponse(void);
OneCycle(void);
computeOneStep(void);
produceEntries(void);
const double pi=3.1415926;
const double deg_rad=pi/180.0;
const double step=pi/720.0;
double L1,L2,alpha,fang,elapseAng,cycleAng,tote_knt;
double cur_load,cur_line,outVolt,vSource,vInput;
double OverLapangle;
int commute,toggle,yes_Entry,modeSw;
FILE *fnew;
char *tstr,*p;
int main(void)
{
float ka1,kb2,kc3;
printf(" \n");
printf("
Load Time Constant in radians = ");
scanf("%f",&ka1);
L1=(double)ka1;
if (L1<0.05) L1=0.05;
printf(" \n");
printf("
Line Reactance Time Constant in radians = ");
scanf("%f",&kb2);
L2=(double)kb2;
printf(" \n");
printf("
Firing angle in degrees = ");
scanf("%f",&kc3);
alpha=(double)kc3;
fang=alpha*deg_rad;
alpha=0.0;
printf(" \n");
// Initialise though not necessary for global variables
elapseAng=0.0 ; cycleAng=0.0;
cur_load=0.0; cur_line=0.0; outVolt=0.0; tote_knt=0.0;
OverLapangle=0.0; commute=0; yes_Entry=0;
// Create a file for entering transient response
fnew=fopen("tran_v1.csv","w+r");
tranResponse();
fclose(fnew);
return 0;
}
void kreateRespFile(void)
{
int n1,n2;
tstr="Angle,InputV,VoutBr,LoadCur,LineCur,IndVolt";
p=strtok(tstr,",");
fprintf(fnew,p); fprintf(fnew,",");
n2=6;
for (n1=0;n1<(n2-1);n1++)
{
p=strtok(NULL,",");
if (n1!=(n2-2))
{
fprintf(fnew,p); fprintf(fnew,",");
}
else
{
fprintf(fnew,p); fprintf(fnew,"\n");
}
}
}
void tranResponse(void)
{
double d1;
int NumCycles,count;
d1=(6.0*(L1+L2))/(2.0*pi)+0.5;
NumCycles=(int)d1 + 1;
kreateRespFile();
modeSw=1;
for (count=0;count<=NumCycles;count++)
OneCycle();
}
void OneCycle(void)
{
int knt;
double dknt;
dknt=0.0;
for (knt=0;knt<1440;knt++)
{
cycleAng=dknt*step;
vInput=pi/2.0*sin(cycleAng);
if ((cycleAng<fang) || (cycleAng>(fang+pi)))
{
if (cycleAng<fang)
vSource=pi/2.0*sin(cycleAng+pi);
else vSource=pi/2.0*sin(cycleAng-pi);
toggle=0;
}
else
{
vSource=pi/2.0*sin(cycleAng);
toggle=1;
}
if (modeSw!=toggle)
{
commute=1;
if (L2<0.0005) commute=0;
modeSw=toggle;
}
computeOneStep();
dknt+=1.0;
}
}
void computeOneStep(void)
{
double dLoadI,dLineI;
switch (commute)
{
case 0:
dLoadI=(vSource-cur_load)/(L1+L2)*step;
cur_load=cur_load+dLoadI;
if (cur_load<0.00001) cur_load=0.0;
if (toggle==0) cur_line=-cur_load;
else cur_line=cur_load;
outVolt=cur_load+(vSource-cur_load)*L1/(L1+L2);
break;
case 1:
dLoadI=(0.0-cur_load)/L1*step;
cur_load=cur_load+dLoadI;
if (cur_load<0.00001) cur_load=0.0;
dLineI=vInput/L2*step;
cur_line=cur_line+dLineI;
if (toggle==1)
{
if (cur_line>cur_load) commute=0;
}
if (toggle==0)
{
if ((cur_line+cur_load)<0.0) commute=0;
}
outVolt=0.0;
break;
}
tote_knt+=1.0;
elapseAng=tote_knt*step;
alpha=tote_knt/4.0;
yes_Entry+=1;
if (yes_Entry==4) produceEntries();
if (yes_Entry==4) yes_Entry=0;
}
void produceEntries(void)
{
double negate;
gcvt(alpha,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(vInput,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(outVolt,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(cur_load,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(cur_line,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
if (toggle==0) negate = 1.0; else negate=-1.0;
gcvt(vInput-negate*outVolt,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
}
The responses obtained for L1=1.5, L2=0.2 and α = 30o are produced below.
Another C program is presented below. This program carries out the harmonic analysis. This file, called fcb_1phb.cpp,
can be downloaded. It can also be compiled as a C program. It produces an output file, called harm_v1.csv.
C Program
// Simulation of a single-phase fully-controlled bridge
// rectifier circuit.
// This program obtainss the periodic response first and
// then computes the ripple components in output current
// and voltage and the harmonic components in the line current.
// The parameters to be fed are just three: the ratio of load
// inductive reactance to the load resistance, wL1/R , the
// ratio of source reactance to the load resistance to load
// resistance wL2/R and the firing angle in degrees.
#include <math.h>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
void periodicResponse(void);
void harmonicAnalysis(void);
void OneCycle(void);
void computeOneStep(void);
void kreateRespFile(void);
void produceEntries(void);
const double pi=3.1415926;
const double deg_rad=pi/180.0;
const double step=pi/720.0;
double L1,L2,fang,cycleAng;
double cur_load,cur_line,outVolt,vSource,vInput;
double OverLapangle,VoAvg,VoRMS,ILoadAvg,ILoadRMS;
double ILineAvg,ILineRMS,THD,RFVolt,RFCur;
int commute,toggle,modeSw;
double anLoadCur[26],bnLoadCur[26],cnLoadCur[26];
double anLineCur[26],bnLineCur[26],cnLineCur[26];
double anOutVolt[26],bnOutVolt[26],cnOutVolt[26];
FILE *fnew;
char *tstr,*p;
void main(void)
{
float ka1,kb2,kc3;
printf(" \n");
printf(" Load Time Constant in radians = ");
scanf("%f",&ka1);
L1=(double)ka1;
if (L1<0.05) L1=0.1;
printf(" \n");
printf(" Line Reactance Time Constant in radians = ");
scanf("%f",&kb2);
L2=(double)kb2;
printf(" \n");
printf(" Firing angle in degrees = ");
scanf("%f",&kc3);
fang=deg_rad*(double)kc3;
printf(" \n");
// Initialise though not necessary for global variables
cycleAng=0.0;
cur_load=0.0; cur_line=0.0; outVolt=0.0;
OverLapangle=0.0; commute=0; modeSw=0; toggle=0;
// obtain the transient response.
periodicResponse();
harmonicAnalysis();
printf("Close window by clicking on x button at top right corner \n");
}
void periodicResponse(void)
{
double startVal;
// Calculate one cycle of response and repeat till the
// response becomes periodic.
do
{
startVal=cur_load;
OneCycle();
} while (fabs(startVal-cur_load)>0.0005);
}
void OneCycle(void)
{
int knt;
double dknt;
dknt=0.0;
for (knt=0;knt<1440;knt++)
{
cycleAng=dknt*step;
vInput=pi/2.0*sin(cycleAng);
if ((cycleAng<fang) || (cycleAng>(fang+pi)))
{
if (cycleAng<fang) vSource=pi/2.0*sin(cycleAng+pi);
else vSource=pi/2.0*sin(cycleAng-pi);
toggle=0;
}
else
{
vSource=pi/2.0*sin(cycleAng);
toggle=1;
}
// To record commutation overlap, the routine described below is used.
// When commute is 1, overlap ensues. It is set to to 1 whenever
// a pair of SCRs is triggered at the set firing agnle. It is resset
// by the compute_one_step routine when overlap ends. This can be
// checked from the values of load current and line current.
if (modeSw!=toggle)
{
commute=1;
if (L2<0.0005) commute=0;
modeSw=toggle;
}
computeOneStep();
dknt+=1.0;
}
}
void computeOneStep(void)
{
double dLoadI,dLineI;
switch (commute)
{
case 0:
dLoadI=(vSource-cur_load)/(L1+L2)*step;
cur_load=cur_load+dLoadI;
if (cur_load<0.00001) cur_load=0.0;
if (toggle==0) cur_line=-cur_load;
else cur_line=cur_load;
if (cur_load>0.0) outVolt=cur_load+(vSource-cur_load)*L1/(L1+L2);
else outVolt=0.0;
break;
case 1:
dLoadI=(0.0-cur_load)/L1*step;
cur_load=cur_load+dLoadI;
if (cur_load<0.00001) cur_load=0.0;
dLineI=vInput/L2*step;
cur_line=cur_line+dLineI;
if (toggle==1)
{
if (cur_line>cur_load) commute=0;
}
if (toggle==0)
{
if ((cur_line+cur_load)<0.0) commute=0;
}
outVolt=0.0;
break;
}
}
void harmonicAnalysis(void)
{
int m;
double theta,dblm;
int knt;
double dknt;
dknt=0.0;
for (m=0;m<26;m++)
{
anLoadCur[m]=0.0; bnLoadCur[m]=0.0; cnLoadCur[m]=0.0;
anLineCur[m]=0.0; bnLineCur[m]=0.0; cnLineCur[m]=0.0;
anOutVolt[m]=0.0; bnOutVolt[m]=0.0; cnOutVolt[m]=0.0;
}
OverLapangle=0.0;
ILoadAvg=0.0; ILoadRMS=0.0; RFCur=0.0;
ILineAvg=0.0; ILineRMS=0.0; THD=0.0;
VoAvg=0.0; VoRMS=0.0; RFVolt=0.0;
for (knt=0;knt<720;knt++)
{
cycleAng=dknt*step;
vInput=pi/2.0*sin(cycleAng);
if ((cycleAng<fang) || (cycleAng>(fang+pi)))
{
vSource=pi/2.0*sin(cycleAng-pi);
toggle=0;
}
else
{
vSource=pi/2.0*sin(cycleAng);
toggle=1;
}
if (modeSw!=toggle)
{
commute=1;
if (L2<0.0005) commute=0;
modeSw=toggle;
}
computeOneStep();
if ((toggle==1) && (commute==1)) OverLapangle=OverLapangle+step;
dblm=0.0;
for (m=0;m<13;m++)
{
theta=2.0*dblm*cycleAng;
while (theta>(2.0*pi)) theta=theta-2.0*pi;
anLoadCur[2*m]=anLoadCur[2*m]+cur_load*cos(theta)*step;
bnLoadCur[2*m]=bnLoadCur[2*m]+cur_load*sin(theta)*step;
anOutVolt[2*m]=anOutVolt[2*m]+outVolt*cos(theta)*step;
bnOutVolt[2*m]=bnOutVolt[2*m]+outVolt*sin(theta)*step;
theta=theta+cycleAng;
if (theta>(2.0*pi)) theta=theta-2.0*pi;
anLineCur[2*m+1]=anLineCur[2*m+1]+cur_line*cos(theta)*step;
bnLineCur[2*m+1]=bnLineCur[2*m+1]+cur_line*sin(theta)*step;
dblm=dblm+1.0;
}
ILoadRMS=ILoadRMS+cur_load*cur_load*step;
VoRMS=VoRMS+outVolt*outVolt*step;
ILineAvg=ILineAvg+fabs(cur_line)*step;
ILineRMS=ILineRMS+cur_line*cur_line*step;
dknt+=1.0;
}
for (m=0;m<13;m++)
{
anLoadCur[2*m]=2.0/pi*anLoadCur[2*m];
bnLoadCur[2*m]=2.0/pi*bnLoadCur[2*m];
anOutVolt[2*m]=2.0/pi*anOutVolt[2*m];
bnOutVolt[2*m]=2.0/pi*bnOutVolt[2*m];
anLineCur[2*m+1]=2.0/pi*anLineCur[2*m+1];
bnLineCur[2*m+1]=2.0/pi*bnLineCur[2*m+1];
cnLoadCur[2*m]=anLoadCur[2*m]*anLoadCur[2*m]
+ bnLoadCur[2*m]*bnLoadCur[2*m];
cnLoadCur[2*m]=sqrt(cnLoadCur[2*m]);
cnLineCur[2*m+1]=anLineCur[2*m+1]*anLineCur[2*m+1]
+ bnLineCur[2*m+1]*bnLineCur[2*m+1];
cnLineCur[2*m+1]=sqrt(cnLineCur[2*m+1]);
cnOutVolt[2*m]=anOutVolt[2*m]*anOutVolt[2*m]
+ bnOutVolt[2*m]*bnOutVolt[2*m];
cnOutVolt[2*m]=sqrt(cnOutVolt[2*m]);
}
VoRMS=sqrt(VoRMS/pi);
ILoadRMS=sqrt(ILoadRMS/pi);
ILineAvg= ILineAvg/pi;
ILineRMS= sqrt(ILineRMS/pi);
ILoadAvg=anLoadCur[0]/2.0;
VoAvg=anOutVolt[0]/2.0;
THD=sqrt(ILineRMS*ILineRMS-cnLineCur[0]*cnLineCur[0]/2.0);
RFVolt=sqrt(VoRMS*VoRMS-VoAvg*VoAvg);
RFCur=sqrt(ILoadRMS*ILoadRMS-ILoadAvg*ILoadAvg);
fnew=fopen("harm_v1.csv","w+r");
kreateRespFile();
produceEntries();
fclose(fnew);
}
void kreateRespFile(void)
{
int n1,n2;
tstr="harmNo,LoadCur,OutVolt,harmNo,LineCur";
p=strtok(tstr,",");
fprintf(fnew,p); fprintf(fnew,",");
n2=5;
for (n1=0;n1<(n2-1);n1++)
{
p=strtok(NULL,",");
if (n1!=(n2-2))
{
fprintf(fnew,p); fprintf(fnew,",");
}
else
{
fprintf(fnew,p); fprintf(fnew,"\n");
}
}
}
void produceEntries(void)
{
int m;
for(m=0;m<13;m++)
{
itoa(2*m, p, 10);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(cnLoadCur[2*m],5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(cnOutVolt[2*m],5,p);
fprintf(fnew,p);
fprintf(fnew,",");
itoa(2*m+1, p, 10);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(cnLineCur[2*m+1],5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
}
fprintf(fnew,"\n");
p="LoadIAvg";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(ILoadAvg,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
p="LoadIRMS";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(ILoadRMS,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
p="RFCur";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(RFCur,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
fprintf(fnew,"\n");
p="VoAvg";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(VoAvg,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
p="VoRMS";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(VoRMS,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
p="RFVolt";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(RFVolt,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
fprintf(fnew,"\n");
p="ILineAvg"; fprintf(fnew,p);
fprintf(fnew,",");
gcvt(ILineAvg,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
p="ILineRMS";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(ILineRMS,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
p="THD";
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(THD,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
fprintf(fnew,"\n");
p="OverLapAngle";
fprintf(fnew,p);
fprintf(fnew,",");
if (OverLapangle>0.0001) OverLapangle=OverLapangle+step/2.0;
gcvt(OverLapangle,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
}
The results obtained are presented below.
SUMMARY
This page has described how the fully-controlled bridge rectifier circuit behaves when the source has inductance. In the
next page, it is decribed how the operation of this circuit changes when it has a filter capacitor.
OPERATION WITH RLC LOAD: 1-PH. CONTROLLED RECTIFIER
CIRCUIT DIAGRAM
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
SIMULATION USING C
SUMMARY
CIRCUIT DIAGRAM
The aim of this page is to illustrate the effect of a smoothing capacitor across the load resistor, in addition to a dc link
inductor. When both an inductor and a capacitor are used as filter elements, the ripple in output voltage is reduced by a
fair amount. The inductor reduces the ripple in the current output of the bridge rectifier, whereas the capacitor reduces
further the ripple in the voltage across the load resistor.
CIRCUIT OPERATION
The operation of the circuit has been described in the earlier pages. One pair of SCRs conducts at a time, producing a
bridge output voltage containing significant ripple content. Since the current through an inductor cannot change
suddenly, the inductor in the dc link tries to maintain the current passing through it as a steady value, which in turn
means that the ripple content of the current through it reduces. Nevertheless it does contain some ripple and this ripple in
the inductor current causes some ripple in the capacitor voltage, whereas the current through the load can be more or less
a steady value. The effectiveness of the filter circuit depends on the values of the inductor and the capacitor. The larger
they are, the higher the reduction in the filter content is. Of particular interest is the comparison of the bridge output
voltage and the capacitor voltage. The ripple in the capacitor voltage is almost out-of-phase with the ripple in the bridge
output voltage. This aspect is important when the output voltage across the load resistor is to be controlled in closed
loop.
MATHEMATICAL ANALYSIS
The analysis of this circuit is slightly more complex. The differential equations that describe the operation of the circuit
are presented below. Because of the filter capacitor, the current in the inductor can become discontinuous for a light load,
even if the firing angle is not high. Hence the differential equations are described for the two cases separately.
DISCONTINUOUS CONDUCTION
When the current through the load is discontinuous, the load current starts building up from zero value when one of the
pair of SCRs is triggered and it falls to zero before the next pair of SCRs is triggered. Let the firing angle be α and let
the current through the inductor become 0 when wt = π + β, where β < α. That is, there is current flow during α < wt <
(π + β) SCRs S1 and S3 in conduction and during (π + α) < wt < (2π + β) SCRs S2 and S4 in conduction. This means
that ther will no current flow during (π + β) < wt < (π + α) and β < wt < α. Let the supply voltage vs be E*Sin (θ),
where θ = wt. Let the voltage across the capacitor be vC(θ) and the current through the inductor be iL. Then equations (1)
and (3) are for the periods when there would no current flow. When SCRs S1 and S3 are in conduction, both the line
current and the load current have the same magntitude and polarity and equation (2) applies. When SCRs S2 and S4 are
in conduction, the line current is the negative of the load current and equation (4) is to be used. The current through the
capacitor is the difference of the dc link inductor current and the current through the load resistor, as shown in equation
(5).
CONTINUOUS CONDUCTION
For continuous conduction, it is sufficient if the equations are described for half-a-cycle only. In the other half-cycle, the
only difference is in the source current waveform. Since the source current has half-wave symmetry, it is sufficient if it is
described over half-a-cycle. Let the firing angle be α. Let the commutation overlap angle be δ. Then it means the SCRs
are triggered when θ = α or when θ = π + α, the process of commutation ends δ radians later. Let the source current be is.
Then during α < θ < ( α + δ ), the entire source voltage is applied across the source inductor, as shown in equation
(6). During this period, the output voltage of the bridge is zero and the voltage across the inductor is then the voltage
across the output capacitor. Since the voltage across the output capacitor tends to reverse the current through the
inductor, equation (7) describes the current-voltage relationship in the inductor for this period. During ( α + δ ) < θ <
(π + α) , the voltage across the source inductor and the dc link inductor is defined by equation (8). During this period,
the line current and the load current have both the same magnitude and polarity, as shown by equation (9). Equation (10)
defines the current-voltage relationship of the output capacitor. As shown in equation (10), capacitor current is the
difference between the inductor current and the current through the load resistor.
SOLUTION
The solution of the equations for both the cases is carried out using numerical technique.
SIMULATION
In this section, the circuit is simulated. You have to key-in the firing angle in degrees, the ratio wLL/RL, the ratio wLs/RL
and the average load current fraction. The ratios are specified for the nominal load conditions, assuming that the average
output voltage is at its maximum and that the average load current is at its rated value. The setting for load current
fraction allows for simulation under different load conditions. A setting of 1.0 means that the simulation is for rated load
condition. When the setting is 0.25, it means that the average load current set is 25% of the rated load current. The applet
below generates the waveforms for the parameters keyed-in.
The parameters for the next applet are the same as those for the applet above. This applet prints out the statistics.
PSPICE SIMULATION
The pspice program used for simulation is as follows:
* Full-wave Bridge
VIN 9 0 SIN(0 340V
XT1 1 2 5 2 SCR
XT2 0 2 6 2 SCR
XT3 4 0 7 0 SCR
XT4 4 1 8 1 SCR
VP1 5 2 PULSE(0 10
VP2 6 2 PULSE(0 10
VP3 7 0 PULSE(0 10
VP4 8 1 PULSE(0 10
L1 2 3 31.8M
L2 1 9 1.6M
R1 3 4 10
C1 3 4 318.5u
R2 1 0 1MEG
R3 2 0 1MEG
Rectifier with RL Load and source Inductance
50Hz)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
1667U 1N 1N 100U 20M)
11667U 1N 1N 100U 20M)
R4 4 0 1MEG
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 0.0MS 10US
.FOUR 50 V(2,4) I(VIN)
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The results obtained are presented below.
The waveform of bridge output voltage
The waveform of bridge output current
The waveform of voltage across the load
The waveform of filter capacitor current
The waveform of line current
The waveform of dc link inductor voltage
SIMULATION USING C
//
//
//
//
//
//
//
//
//
//
//
//
//
Simulation of a single-phase fully-controlled bridge
rectifier circuit.
This program produces the transient response, the periodic
response. Commuation overlap angle is also computed.
Calculations normalized. The average voltage that would
occur for diode-bridge rectifier with a resistive load is
taken to be unity. The average current through the load
resistor with a pure resistive load on the diode bridge
rectifier is taken as the unity current. The parameters to
be fed are just three: the ratio of load inductive reactance
to the load resistance, wL1/R , the ratio of source reactance
to the load resistance to load resistance wL2/R and the firing
angle in degrees.
#include <math.h>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<iostream.h>
void
void
void
void
void
kreateRespFile(void);
tranResponse(void);
OneCycle(void);
computeOneStep(void);
produceEntries(void);
const double pi=3.1415926;
const double deg_rad=pi/180.0;
const double step=pi/720.0;
double L1,L2,alpha,fang,elapseAng,cycleAng,tote_knt;
double cur_load,cur_line,outVolt,vSource,vInput,capVolt;
double OverLapangle,Cap,capCur,dcLinkCur,Res;
int commute,toggle,yes_Entry,modeSw;
FILE *fnew;
char *tstr,*p;
int main(void)
{
float ka1,kb2,kc3,kd4,ke5;
printf(" \n");
printf("
Load Time Constant in radians = ");
scanf("%f",&ka1);
L1=(double)ka1;
if (L1<0.05) L1=0.05;
printf(" \n");
printf("
Line Reactance Time Constant in radians = ");
scanf("%f",&kb2);
L2=(double)kb2;
printf(" \n");
printf("
Cap. Filter Time Constant in radians
= ");
scanf("%f",&kc3);
Cap=(double)kc3;
if (Cap<0.1) Cap=0.1;
if (Cap>10.0) Cap=10.0;
printf(" \n");
printf("
Load Resistance in p.u.
= ");
scanf("%f",&ke5);
Res=(double)ke5;
if (Res<0.1) Res=0.1;
if (Res>10.0) Res=10.0;
printf(" \n");
printf("
Firing angle in degrees = ");
scanf("%f",&kd4);
alpha=(double)kd4;
fang=alpha*deg_rad;
alpha=0.0;
printf("
\n");
// Initialise though not necessary for global variables
elapseAng=0.0 ; cycleAng=0.0; capCur=0.0; capVolt=0.0;
cur_load=0.0; cur_line=0.0; outVolt=0.0; tote_knt=0.0;
OverLapangle=0.0; commute=0; yes_Entry=0; dcLinkCur=0.0;
cur_load=0.0;
// Create a file for entering transient response
fnew=fopen("tran_v3.csv","w+r");
tranResponse();
fclose(fnew);
return 0;
}
void kreateRespFile(void)
{
int n1,n2;
tstr="Angle,InputV,VoutBr,dcLinkCur,LineCur,IndVolt,LoadVolt,capCur";
p=strtok(tstr,",");
fprintf(fnew,p); fprintf(fnew,",");
n2=8;
for (n1=0;n1<(n2-1);n1++)
{
p=strtok(NULL,",");
if (n1!=(n2-2))
{
fprintf(fnew,p); fprintf(fnew,",");
}
else
{
fprintf(fnew,p); fprintf(fnew,"\n");
}
}
}
void tranResponse(void)
{
double d1;
int NumCycles,count;
d1=(6.0*(L1+L2+Cap))/(2.0*pi)+0.5;
NumCycles=(int)d1 + 1;
kreateRespFile();
modeSw=1;
for (count=0;count<=NumCycles;count++)
OneCycle();
}
void OneCycle(void)
{
int knt;
double dknt;
dknt=0.0;
for (knt=0;knt<1440;knt++)
{
cycleAng=dknt*step;
vInput=pi/2.0*sin(cycleAng);
if ((cycleAng<fang) || (cycleAng>(fang+pi)))
{
if (cycleAng<fang)
vSource=pi/2.0*sin(cycleAng+pi);
else vSource=pi/2.0*sin(cycleAng-pi);
toggle=0;
}
else
{
vSource=pi/2.0*sin(cycleAng);
toggle=1;
}
if (modeSw!=toggle)
{
commute=1;
if (L2<0.0005) commute=0;
modeSw=toggle;
}
computeOneStep();
dknt+=1.0;
}
}
void computeOneStep(void)
{
double dLoadI,dLineI,doutV;
switch (commute)
{
case 0:
dLoadI=(vSource-dcLinkCur)/(L1+L2)*step;
doutV=(dcLinkCur-capVolt/Res)*step/Cap;
dcLinkCur=dcLinkCur+dLoadI;
capVolt=capVolt+doutV;
if (capVolt<0.0) capVolt=0.0;
if (dcLinkCur<0.00001) dcLinkCur=0.0;
if (toggle==0) cur_line=-dcLinkCur;
else cur_line=dcLinkCur;
outVolt=cur_load+(vSource-cur_load)*L1/(L1+L2);
cur_load=capVolt/Res;
capCur=dcLinkCur-cur_load;
break;
case 1:
dLoadI=(0.0-cur_load)/L1*step;
doutV=(dcLinkCur-capVolt/Res)*step/Cap;
dcLinkCur=dcLinkCur+dLoadI;
capVolt=capVolt+doutV;
if (capVolt<0.0) capVolt=0.0;
if (dcLinkCur<0.00001) dcLinkCur=0.0;
dLineI=vInput/L2*step;
cur_line=cur_line+dLineI;
if (toggle==1)
{
if (cur_line>dcLinkCur) commute=0;
}
if (toggle==0)
{
if ((cur_line+dcLinkCur)<0.0) commute=0;
}
if (dcLinkCur<0.00001) outVolt=capVolt;
else outVolt=0.0;
cur_load=capVolt/Res;
capCur=dcLinkCur-cur_load;
break;
}
tote_knt+=1.0;
elapseAng=tote_knt*step;
alpha=tote_knt/4.0;
yes_Entry+=1;
if (yes_Entry==4) produceEntries();
if (yes_Entry==4) yes_Entry=0;
}
void produceEntries(void)
{
double negate;
gcvt(alpha,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(vInput,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(outVolt,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(dcLinkCur,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(cur_line,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
if (toggle==0) negate = 1.0; else negate=-1.0;
gcvt(vInput-negate*outVolt,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(capVolt,5,p);
fprintf(fnew,p);
fprintf(fnew,",");
gcvt(capCur,5,p);
fprintf(fnew,p);
fprintf(fnew,"\n");
}
The plots obtained for L1=1.0, L2=0.1, R=1.0, C=1.0 and α =30o have been displayed below.
SUMMARY
This page has described the effect of adding a filter capacitor. It can also be seen that no attempt is made to obtain the
periodic response by determining the coefficients, since it is easier to use the numerical technique to obtain the solution.
The next page describes an application of this circuit.
A BATTERY CHARGER
CIRCUIT DIAGRAM
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
CIRCUIT DIAGRAM
The circuit diagram of a battery charger is shown above. The ac source and the
inductance in series with it can represent a transformer and its leakage inductance as
viewed from its secondary. In a battery charger circuit, the current supplied by the
source is usually discontinuous and the load across the battery is usually very light.
CIRCUIT OPERATION
The behaviour of a fully-controlled bridge circuit has been described in the previous
pages. This page describes how the battery-charger can be controlled in closed-loop.
Traditionally systems use controllers such as a PI controller or a PDF controller for
feedback control. But a battery has a very large time constant and control of a system
with a large time constant is not easy. Hence the controller described in this page is a
rule-based controller. It can be implemented either with traditional logic gates, counters
and ADCs or by using a micro-controller and some more ICs to complement its
operation.
The battery is modeled mainly as a very large capacitor and a series resistor RB. The
capacitance used in the model is not as large as the value that can represent the stored
AH capacity of the battery, since that would lead to a very large period for simulation.
The capacitance used is large enough to make the behaviour of the system similar to that
of a real battery-charger, but small enough to get the simulation performed in a
reasonable time.
Firstly, a method by which the phase-angle can be controlled is explained and then the
rule-based controller is described. A synchronizing voltage, usually the output from a
secondary winding of a transformer with its primary connected to the mains supply, is
fed to a zero-crossing detector. In the scheme described here, two logic signals are
developed; logic signal A is set to high level when the synchronizing voltage crosses
zero, becomes positive and stays positive, and logic signal B is set to high level when the
synchronizing voltage crosses zero, becomes negative and stays negative. If the
synchronizing signal is described to be E*Sin (θ), it would be preferable if A is at 1 from
θ = 5o to θ = 175o and B is at 1 from θ = 185o to θ = 355o . Then a logic NOR signal,
A NOR B , can be obtained. The NOR signal can be used to reset a ramp generator
when the synchronizing signal crosses zero in either direction. The ramp output can then
be compared with a control voltage VC and the output C of comparator can be ANDed
with either A or B. The logic signal, A AND C, can be used to generate a firing signal
for SCRs S1 and S3. It can be seen that these SCRs are triggered when the source
voltage Vs is positive, provided that the synchronizing signal is in phase with the mains
supply voltage. The logic signal, B AND C, can be used to generate a firing signal for
SCRs S2 and S4. It can be seen that these SCRs are triggered when the source voltage Vs
is negative.
The scheme outlined can be implemented in different ways. One of the methods that can
be followed is described next.
From the mains supply, a synchronizing signal can be obtained. If a single-polarity
supply is to be used, CMOS opamps, suitable for operation with a single power supply,
can be used. The resistors have to be chosen to suit the requirements. For example, let
the peak of synchronizing voltage be 10 V. If the ramp is to start at 5o, then the
magnitude of ramp signal would be 10 * Sin (5o), which is 0.871 V. If VCC is 12 V,
then R1 and R2 can be selected such that the voltage across R2 is 0.871 V. The values of
R3 and R4 are selected that the ratio (R3/R4) is very small, of the order of about 0.01.
The circuit to generate the ramp signal is shown above. Whenever the signal, A NOR B,
is at 1, the transistor is switched ON and it discharges the capacitor. When the NOR
signal becomes zero, the boot-strap integrator generates a ramp signal at its output. If
the ramp signal is be called y(t), then
where t starts from zero from the instant the transistor stops conducting. The values of R
and C should be chosen such that when the synchronizing voltage varies from 10 * Sin
(5o) to 10 * Sin (175o), the ramp signal varies from 0 V to about 10 V. The time
corresponding to 170o depends on the frequency. At 50 Hz, the corresponding time
period is 9.444 ms and it is 7.87 ms at 60 Hz. When the ramp signal varies in this
manner, the control voltage can be varied from 0 V to 10 V to vary the firing angle. For
the boot-strap integrator also, a CMOS opamp can be used . It would be preferable to
use a CMOS opamp as the comparator too. From the comparator output and the logic
signals A and B, the signals that can be used to generate firing pulse for the SCRs is
described next.
A firing signal for an SCR can be generated in several ways. For high power SCRs, the
best technique is to use a pulse transformer with a ferrite core. But in an application
such as the battery-charger with a single-phase fully-controlled bridge rectifier circuit,
the rating of the SCRs is not likely to be in the range of hundreds of amperes and it may
be preferable to use an opto-coupler with light-activated SCR detector.
One opto-coupled light-activated triggering IC is required for each SCR. It would be
preferable to connect a resistor between the gate and the cathode of light-activated SCR.
Its value can be in the region of several kilo-ohms and it depends on the IC used. The
light-activated SCR in turn triggers the main SCR as shown below. For the SCRs in the
bridge rectifier circuit also, it would be preferable to connect a resistor between the gate
and the cathode, and its value can be about 500 Ω. In addition, it would be necessary to
connect a snubber circuit from the anode to cathode of each SCR. The values of
components to be used for the snubber circuit can be normally obtained from the
datasheet of the SCR. It should also be remembered that the blocking voltage of the
light-activated SCR should be the same as that of the main SCR.
The source that drives the bridge circuit is usually the secondary of a transformer. The
leakage inductance of the transformer as viewed from the secondary terminals acts as the
source inductance for the circuit. With a light load, the bridge output tends to be
discontinuous. The next section describes how the circuit operates.
MATHEMATICAL ANALYSIS
The equations are normalized. The nominal voltage of the battery is taken to be 1 p.u
and the nominal rms current of the secondary of the transformer is taken to be 1 p.u. For
example, let the nominal voltage of the battery bank be 72 V. Let the rms secondary
voltage be 100 V and its rated rms current be 10 A. Then the peak secondary voltage
equals (1.414*100/72), that is 1.964 p.u. If the drop in terminal voltage of the battery
bank when it delivers 10 A is 1 V, then its source impedance is set to be 1.414/72, that is
0.0196 p.u.
The load current drawn by the load resistor should be relatively low, less than 0.25 p.u.
That is, 0.3 p.u corresponds to 4.24 A the corresponding load resistance is 72/4.24, that
is 17 Ω.
The simulation routine is explained below. All setting are in per unit notation.
Initializing routine:
Set the value of peak secondary voltage
Set the value of fractional load
Set the source inductance.
Set the peak battery current
Set the maximum average battery current
Set the initial battery voltage(internally set)
Set the time constant of the battery(set to 300 radians internally)
Set the source resistance of the battery
Set the firing angle to 175o (internal setting)
Reset to zero the battery current. peak charging current.
Reset to zero the rms secondary current. average battery charging current.
Go to pre-amble routine
Pre-amble routine:
Print the battery voltage
Print the average battery voltage
Print the peak charging current
Print the rms secondary current
Set the rule-based controller as follows:
if (average Battery Current =(1.1*maximum average battery Current))
increase the firing angle by 1o
else
{
if ((batteryVolt<=1.1) and (average Battery Current <maximum average
battery Current))
decrease the firing angle by 1o
if ((batteryVolt<=1.1) and (average Battery Current =maximum average
battery Current))
do no change the firing angle
if ((batteryVolt1.1) and (batteryVolt<=1.15)) do no change the firing angle
if (batteryVolt1.15) increase the firing angle by 1o
}
if (peakBtyCurpeakCurSet) increase the firing angle by 2o
Reset to zero peak charging current, average battery current and rms secondary
Current.
Go to One-Output-Cycle routine.
One-Output-cycle routine:
angle=0;
while ( angle =0 and angle <180 )
{
compute increments in battery voltage and source current
compute increments to rms secondary current, average battery charging current
set the peak battery charging current appropriately
increment angle by step size
}
SIMULATION
The applet below demonstrates how a rule-based battery charger would function.
Next page describes how a dc motor can be controlled in closed-loop.
A TWO-QUADRANT DC DRIVE
PERFORMANCE CHARACTERISITCS
BLOCK DIAGRAM
CLOSED-LOOP CONTROL
SIMULATION
PERFORMANCE CHARACTERISTICS
This page describes how a separately-excited DC motor can be controlled in closed-loop with a
single-phase fully-controlled rectifier supplying dc source to its armature. The operation of a
DC motor is described briefly at first.
A symbolic representation of a separately-excited DC motor is shown above. The resistance of
the field winding is Rf and its inductance is Lf, whereas the resistance of the armature is Ra and
its inductance is La. In the description of the motor, the armature reaction effects are ignored.
It is justifiable since the motor used has either interpoles or compensating winding to minimize
the effects of armature reaction. The field current is described by equation (1). If a steady
voltage Vf is applied to the field, the field current settles down to a constant value, as shown in
equation (2). When the field current is constant, the flux induced by the field winding remains
constant, and usually it is held at its rated value φ. If the voltage applied to the armature is va,
then the differential equation that is to be applied to the armature circuit is shown in equation
(3). In steady-state, equation (4) applies. The voltage, ea, is the back e.m.f. in volts. In a
separately-excited DC motor, the back e.m.f is proportional to the product of speed of motor w
rad/s and the field φ Webers, as shown by equation (5).
In equation (5), Km is a coefficient and its value depends on the armature winding. If the
armature current in steady-state be Ia, then the power P that is supplied to the armature is EaIa.
This electric power is converted to mechanical power by the armature of the DC motor. Let the
torque developed by the armature be Te, the unit for torque being Nm (Newton-metre). Then
power and torque can be related as shown in equation (6). On canceling the common term on
both sides, the torque Te developed by the armature is obtained as presented in equation (7).
If the instantaneous armature current is ia, then equation (8) applies. Torque has been denoted
by Te in both equations.
The speed of the motor can be controlled by varying Va and holding Vf constant at its rated
value. Then as the voltage applied to the armature is raised, the armature current increases
first. As the armature current increases, the torque developed by motor increases and hence the
speed of motor increases. The drop across the armature resistance tends to be small and hence
the motor speed rises almost proportionately with the voltage applied to the armature. But there
is a limit to the voltage that can be applied to the armature and that limit is the rated voltage of
the armature voltage. The speed of the motor corresponding to the rated armature voltage and
the rated field voltage is its rated speed. Thus the speed of a motor can be varied below its rated
speed by controlling the armature voltage. It would be desirable that the motor should be able
to develop as high as a torque as possible and hence the voltage rated applied to the field is held
at its rated value. Applying higher than the rated voltage to either the field or the armature is
not recommended. When the rated voltage is applied to the field, the flux would be near the
saturation level in the poles. If a voltage higher than its rated voltage is applied to the field, the
flux would saturate and there would not be any significant increase in the torque that the motor
can deliver. On the other hand, this would only result in increased losses in the winding. Since
the total heat which the DC motor can dissipate is fixed due to its surface area and cooling
system, increased losses from the excitation system would mean that the other losses would
have to reduce, implying that the armature current cannot be at its rated level and the maximum
torque that the motor can deliver may reduce. Increasing the armature voltage above its rated
value is not recommended because the insulation of the armature is designed for operation of
the motor with the rated voltage applied to its armature. Moreover, the torque that the motor
can deliver depends on the armature current and the field current. If the motor is operated
continuously, the maximum armature current should not be higher than its rated value. When
the armature current and the field voltage are at their rated level, the motor generates the rated
torque. Hence the maximum torque the motor can deliver continuously over a long period of
time is its rated torque when its speed is varied from a low value to its rated speed. Over this
period, 0 < w < wr, where wr is its rated speed, the power output is given by:
The maximum torque which the motor can deliver continuously is called Te,max cont. What is
being referred to here is the maximum torque the motor can deliver, and not the actual torque
the motor delivers. The actual torque the motor delivers depends on the mechanical load
connected to its shaft. If the speed of the motor is to be increased beyond its rated value, the
voltage applied to the armature can be held at its rated value and the field can be weakened by
reducing the voltage applied to it. When the speed of the motor is in this manner, the maximum
power that can be supplied to the armature is fixed, since both the voltage applied to the
armature and the armature current cannot exceed the rated level over a long period. That means
the maximum torque the motor can develop above the rated speed is:
The plots of Te,max cont and the maximum power Pa,max can be plotted as a function of rotor
speed asshown below. The rated values of speed, torque and power to the armature have been
set equal to unity.
A separately-excited dc motor can be controlled, either by varying the voltage applied to the
field winding or by varying the voltage applied to the armature. This page describes how the
motor can be controlled by varying the armature voltage and it is assumed that the field is
excited by a constant voltage, equaling the rated voltage of the field winding. It means that the
discussion to follow assumes that the field current remains steady at its rated value.
BLOCK DIAGRAM
The block diagram of a dc drive is shown above. It does not show all details. The DC motor
has not been represented in the form of a block diagram and the details of the load the motor
drives have also not been shown. The block diagram functions as follows.
For the system described here, the output of the system is the speed of the motor. Hence when
this system is to be controlled in closed-loop, the parameter that is to be set is what that speed
should be. It is denoted to be Ωref. In order to control the speed in closed-loop, we need a
feedback signal too. It can be obtained in several ways. A digital tacho or an analogue
tachogenerator can be used. It is assumed that an analogue tachogenerator is used here. It is
coupled to the motor shaft and its output voltage varies linearly with its speed. Let the speed
feedback signal be Ωf. This signal can be compared with the speed reference signal and the
error can be processed by a controller. The controller can be of one of several types. It can be
an integral controller, or a PI controller and PDF (pseudo-derivative feedback) controller or a
PID controller or a rule-based fuzzy logic controller. Here both the controllers used are PI
(proportional plus integral) controllers. A PI controller can lead to fast response and zero-error
for a step input.
The PI controller for speed has as its input the error between the two signals, Ωref and Ωf. If the
speed feedback signal Ωf is lower than the reference signal Ωref , it means that the DC motor
speed is running below the set speed and it needs to be accelerated. In order to accelerate the
motor, it should develop greater torque. To develop greater torque, its armature current has to
increase. Hence the output of speed controller is set to function as the reference signal for
armature current. It will be a voltage corresponding to armature current with an appropriate
coefficient linking the two quantities. When Ωf < Ωref, the difference causes the output of speed
controller to increase. Since the output of speed-controller is set to function as the armature
current reference signal, an increase in the value of speed-controller output would in turn lead to
an increase in the armature current.
The rectifier circuit is made up of SCRs and the SCRs have a current rating. Hence it is
necessary to ensure that the current through the SCRs remains within a safe level. Hence the
output of speed controller is limited at both ends. Its maximum value corresponds to the safe
level for SCRs. It is not normally the rated current of the motor and it is usually set at a value
ranging from 1.5 times to 2 times the rated armature current. The reason is that the motor may
have to develop more than the rated torque under transient conditions to achieve fast response.
In order to ensure that the motor armature current remains within its rated value, another
supervisory loop may be used. Another option is to use a circuit-breaker. The instantaneous
trip action in the circuit breaker can be due to magnetic effect and the overload trip can be due
to thermal action. A bi-metallic strip within the circuit-breaker expands due to temperature and
would trip circuit-breaker. The lower limit on the output of speed-controller would correspond
to zero current in the armature, since the motor current in this scheme cannot be in the reverse
direction.
The current controller has two inputs, the reference current signal which is the output of the
speed controller and a feedback signal proportional to the armature current. The feedback signal
can be obtained in several ways. A current transformer can be introduced in the path of ac
current from the ac supply. Another option would be to use a DC current transducer that makes
use of a Hall-effect sensor or isolated opamp. The transducer used produces a voltage
proportional to the current in the armature. The difference between these two signals is
processed by another PI controller and its output is also limited to correspond to 0o and 180o
firing angle. The output of current controller may vary between 0 V and 10 V, with 0 V
corresponding to 180o firing angle and 10 V corresponding 0o firing angle. If the firing angle
be α and the output of current controller VC, then
As the output voltage of current controller increases due to the difference between the reference
signal and the feedback signal corresponding to armature current, the firing angle is advanced
towards 0o and the average output voltage of the bridge rectifier increases. This in turn leads to
increased torque generation and the motor accelerates.
If the speed reference is brought down suddenly, the current in the motor cannot be reversed
and hence the motor slows down due to friction and the load. This process can be slow.
The question that can be raised is whether we need the current loop. The answer is that it
improves the performance. If there is a change in the supply voltage even by a small amount,
the output of the bridge circuit tends to a fall a bit for the same firing angle. The reduction in
output voltage causes a large change in the armature current, with the speed remaining more or
less constant. The current loop comes into action, correcting the firing angle to the required
value. The time constant of the armature, due to its inductance and resistance, tends to be of the
order of a few tens of ms and the mechanical time constant, due to the moment of inertia of
motor and load and the friction, is of the order of a few tenths of a second. If a current
controller is not used, the speed would have to change before the speed controller can come into
action. Since the mechanical time constant is about at least 10 times greater, there would be a
significant change in speed if there be no current controller.
Normally a filter may be necessary in the feedback circuit for speed. The tacho signal usually
contains a small ripple superimposed on its dc content. The frequency of the ripple is usually
dependent on the speed and the lower the speed is the lower is the frequency of this ripple.
Hence the time constant of the filter may have to be set to correspond to the lowest speed at
which the motor would be required to run. Usually the motor speed does not have to vary over
a range larger than 0.1 p.u to 1 p.u. Since the power output varies proportionately with the
speed, there is usually no justification to run the motor at an extremely low speed. The next
section describes how the simulation is carried out. The routines are explained with the help of
pseudo-code that can be understood by a reader with some knowledge of one of the
programming languages such as C, PASCAL, BASIC, Fortran or Matlab.
CLOSED-LOOP CONTROL
This section explains how the simulation can be carried out.
Initialize_Routine:
Set and get Parameters
Initialize controller Outputs
Go to Calculation_Routine
Calculation_Routine
Execute One_Cycle_Routine
Plot the results
One_Cycle_Routine
Set angle to zero.
For (angle = 0; angle <360o; angle +stepsize)
{
Set the SCR pair based on firing angle and angle.
Go to Next_Values
}
Next_Values
Calculate the increments in
armature current
motor speed
tacho filter output
speed controller output
current controller output
Add the increments
Compute the firing angle
The pseudo-routine presents only the main steps.
SIMULATION
Before selecting the type of response, set the value of the selected parameter. When you select
a parameter, the textfield shows the default value set inside the program. Change the parameter
value if you want to and then you must click on the SET VALUE button for the change to take
effect. You can go from one type of response to another after the present calculations are
carried out. When you have selected a new type of response, you must click on Click to Start.
If you click on Reset button, initializing routine is carried out and the motor speed is set to zero,
and the other values are also reset. The program has been written assuming that the frequency
of operation is 50 Hz. If the frequency is different, the parameter values should be scaled
suitably.
Next page describes how a three-phase fully-controlled bridge rectifier circuit operates.
FULLY-CONTROLLED 3-PH SCR
BRIDGE RECTIFIER
This chapter describes the operation of a threephase fully-controlled SCR bridge rectifier
circuit and two applications. The pages to
follow contain:
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●
●
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●
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OPERATION OF A 3-PHASE FULLY-CONTROLLED
RECTIFIER
OPERATION WITH A RESISTIVE LOAD
OPERATION WITH AN RL LOAD
OPERATION WITH AN RL LOAD AND SOURCE
INDUCTANCE
AN APPLICATION: A DC POWER SUPPLY
AN APPLICATION: A FOUR-QUADRANT DC DRIVE
OPERATION OF A 3-PHASE FULLY-CONTROLLED RECTIFIER
CIRCUIT OPERATION
SYNCHRONIZING SIGNALS
MATHEMATICAL ANALYSIS
SIMULATION
MATHCAD SIMULATION
CIRCUIT OPERATION
The operation of a 3-phase fully-controlled bridge rectifier circuit is described in this page. A three-phase
fully-controlled bridge rectifier can be constructed using six SCRs as shown below.
The three-phase bridge rectifier circuit has three-legs, each phase connected to one of the three phase
voltages. Alternatively, it can be seen that the bridge circuit has two halves, the positive half consisting of
the SCRs S1, S3 and S5 and the negative half consisting of the SCRs S2, S4 and S6. At any time when there
is current flow, one SCR from each half conducts. If the phase sequence of the source be RYB, the SCRs
are triggered in the sequence S1, S2 , S3 , S4, S5 , S6 and S1 and so on.
The operation of the circuit is first explained with the assumption that diodes are used in place of the SCRs.
The three-phase voltages vary as shown below.
Let the three-phase voltages be defined as shown below.
It can be seen that the R-phase voltage is the highest of the three-phase voltages when θ is in the range
from 30o to 150o. It can also be seen that Y-phase voltage is the highest of the three-phase voltages when θ
is in the range from 150o to 270o and that B-phase voltage is the highest of the three-phase voltages when θ
is in the range from 270o to 390o or 30o in the next cycle. We also find that R-phase voltage is the lowest
of the three-phase voltages when θ is in the range from 210o to 330o. It can also be seen that Y-phase
voltage is the lowest of the three-phase voltages when θ is in the range from 330o to 450o or 90o in the next
cycle, and that B-phase voltage is the lowest when θ is in the range from 90o to 210o. If diodes are used,
diode D1 in place of S1 would conduct from 30o to 150o, diode D3 would conduct from 150o to 270o and
diode D5 from 270o to 390o or 30o in the next cycle. In the same way, diode D4 would conduct from 210o
to 330o, diode D6 from 330o to 450o or 90o in the next cycle, and diode D2 would conduct from 90o to
210o. The positive rail of output voltage of the bridge is connected to the topmost segments of the envelope
of three-phase voltages and the negative rail of the output voltage to the lowest segments of the envelope.
At any instant barring the change-over periods when current flow gets transferred from diode to another,
only one of the following pairs conducts at any time.
Period, range of θ
Diode Pair in conduction
30o to 90o
D1 and D6
90o to 150o
D1 and D2
150o to 210o
D2 and D3
210o to 270o
D3 and D4
270o to 330o
D4 and D5
330o to 360o and 0o to 30o
D5 and D6
If SCRs are used, their conduction can be delayed by choosing the desired firing angle. When the SCRs are
fired at 0o firing angle, the output of the bridge rectifier would be the same as that of the circuit with
diodes. For instance, it is seen that D1 starts conducting only after θ = 30o. In fact, it can start conducting
only after θ = 30o , since it is reverse-biased before θ = 30o. The bias across D1 becomes zero when θ = 30o
and diode D1 starts getting forward-biased only after θ =30o. When vR(θ) = E*Sin (θ), diode D1 is reversebiased before θ = 30o and it is forward-biased when θ > 30o. When firing angle to SCRs is zero degree, S1
is triggered when θ = 30o. This means that if a synchronizing signal is needed for triggering S1, that signal
voltage would lag vR(θ) by 30o and if the firing angle is α, SCR S1 is triggered when θ = α + 30o. Given
that the conduction is continuous, the following table presents the SCR pair in conduction at any instant.
Period, range of θ
SCR Pair in conduction
α + 30o to α + 90o
S1 and S6
α + 90o to α + 150o
S1 and S2
α + 150o to α + 210o
S2 and S3
α + 210o to α + 270o
S3 and S4
α + 270o to α + 330o
S4 and S5
α + 330o to α + 360o and α + 0o to α + 30o
S5 and S6
The operation of the bridge-rectifier is illustrated with the help of an applet that follows this line. You can
set the firing angle in the range 0o < firing angle < 180o and the instantaneous angle. The applet displays
the SCR pair in conduction at the chosen instant. The current flow path is shown in red colour in the
circuit diagram. The instantaneous angle can be either set in its text-field or varied by dragging the scrollbar button. The rotating phasor diagram is quite useful to illustrate how the circuit operates. Once the
firing angle is set, the phasor position for firing angle is fixed. Then as the instantaneous angle changes, the
pair that conducts is connected to the thick orange arcs. One way to visualize is to imagine two brushes
which are 120o wide and the device in the phase connected to the brush conducts. The brush that has
"Firing angle " written beside it acts as the brush connected to the positive rail and the other acts as if it is
connected to the negative rail. This diagram illustrates how the rectifier circuit acts as a commutator and
converts ac to dc. The output voltage is specified with the amplitude of phase voltage being assigned unity
value.
SYNCHRONIZING SIGNALS
To vary the output voltage, it is necessary to vary the firing angle. In order to vary the firing angle, one
commonly used technique is to establish a synchronizing signal for each SCR. It has been seen that zero
degree firing angle occurs 30o degrees after the zero-crossing of the respective phase voltage. If the
synchronizing signal is to be a sinusoidal signal, it should lag the respective phase by 30o and then the
circuitry needed to generate a firing signal can be similar to that described for single-phase. Instead of a
single such circuit for a single phase rectifier, we would need three such circuits.
When the 3-phase source supply connected to the rectifier is star-connected, the line voltages and the phase
voltages have a 30o phase angle difference between them, as shown below.
The line voltage can also be obtained as:
This line voltage lags the R-phase voltage by30o and has an amplitude which is 1.732 times the amplitude
of the phase voltage. The synchronizing signal for SCR S1 can be obtained based on vRB line voltage. The
synchronizing signals for the other SCRs can be obtained in a similar manner.
To get the synchronizing signals, three control transformers can be used, with the primaries connected in
delta and the secondaries in star, as shown below.
For S1, voltage vS1 is used as the synchronizing signal. Voltage vS2 is used as the synchronizing signal for
SCR S2 and so on. The waveforms presented by the synchronizing signals are as shown below. The
waveforms do not show the effect of turns ratio, since any instantaneous value has been normalized with
respect to its peak value. For example, let the primary phase voltage be 240 V and then its peak value is
339.4 V. The primary voltage is normalized with respect to 339. V. If the peak voltage of each half of
secondary is 10 V, the secondary voltage are normalized with respect to 10 V.
MATHEMATICAL ANALYSIS
Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of single-phase
bridge rectifier circuit. We are interested in output voltage and the source current. The average output
voltage, the rms output voltage, the ripple content in output voltage, the total rms line current, the
fundamental rms current, THD in line current, the displacement power factor and the apparent power factor
are to be determined. In this section, the analysis is carried out assuming that the load current is a steady
dc value.
AVERAGE OUTPUT VOLTAGE
Before getting an expression for the output voltage, it is preferable to find out how the output voltage
waveform varies as the firing angle is varied. In one cycle of source voltage, six pairs conduct, each pair
for 60o. This means that the period for output waveform is one-sixth of the period of line voltage. The
output waveform repeats itself six times in one cycle of input voltage. The waveform of output voltage can
be determined by considering one pair. It is seen that when vR(θ) = E* Sin (θ), SCR S1 and S6 conduct
when θ varies from 30o + α to 90o + α , where α is the firing angle. Then
The waveform of output can be plotted for different firing angles. The applet below takes in the firing
angle as an input and plots the output. The peak line-to-line voltage is marked as 'U' and the applet starts
with the instant an SCR is fired and displays the output waveform for one input cycle period.
The average output voltage of the bridge circuit is calculated as follows, with a change in variable, where θ
= α + 60o.
In the expression above, U is the peak line-to-line voltage, whereas E is the amplitude of phase voltage of 3phase supply.
RMS OUTPUT VOLTAGE
The rms output voltage is calculated as follows:
The ripple factor of the output voltage is then:
The applet below displays the average output voltage, the rms output voltage and the ripple factor for the
case of continuous conduction through the load.
It is seen that the average output voltage is negative when firing angle exceeds 90o. It means that power
flow is from the dc side to the ac source. When the firing angle is kept in the region 0o < α < 90o, this
circuit is said to be operating in the rectifier region. When the firing angle is kept in the region 90o < α <
180o, this circuit is said to be operating in the inverter region. When the circuit operates in the rectifier
region, the net power flow is from the ac source to the dc link. In the inverter region, the net power flow is
in the reverse direction. To operate in the inverter region, it is necessary to have a dc source present in the
dc link which can provide the power that is fed back to the ac source.
RMS LINE CURRENT
The rms line current is relatively easy to find out if the dc current is ripple-free and steady. The load
current is ripple-free if the inductance in the dc link is relatively large. To maintain load current at any
firing angle, it is necessary that the dc link should contain a voltage source. Given that the resistance of the
load circuit is zero, the voltage source should equal the average output voltage of the bridge circuit. The
waveforms shown below are based on the assumption that these conditions are met. It has been shown that
if vR(θ) = E*Sin (θ), SCR S1 conducts when θ varies from α + 30o to α + 90o and that SCR S4 conducts
when θ varies from α + 210o to α + 270o. If the amplitude of dc load current is assigned to be unity, the
line current waveform is then a rectangular pulse, remaining at + 1 from α + 30o to α + 150o, at - 1 from α
+ 210o to α + 330o, and zero elsewhere. The amplitude of the fundamental in line current is then 3.464/π (
which evaluates to nearly 0.78) and the amplitude of other odd harmonics is 3.464/nπ, where n is the odd
harmonic number. When the dc load current is steady and has a magnitude of unity, the rms line current is
obtained as shown in equation (5). The rms value of the fundamental is obtained as shown in equation (6).
Equation (6) is based on how trigonometric Fourier coefficients are defined for waveforms with quarterwave symmetry. When the line current is a rectangular and symmetric, the phase current is the same as the
line current and the fundamental component of the phase current lags the phase voltage by an angle equal to
the firing angle. Hence the displacement power factor is expressed as shown by equation (7). Since the line
current is not sinusoidal, the apparent power factor, usually referred to just as the power factor in most of
the texts, is less than DPF and is represented by equation (8). Since the line current is not sinusoidal, the
distortion component in the line current has to be computed. This component, called the THD( Total
Harmonic Distortion ), is calculated as shown in equation (9).
OPERATION WITH A RESISTIVE LOAD
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
SUMMARY
CIRCUIT OPERATION
The three-phase fully-controlled bridge rectifier with a purely resistive load is shown above. With any single-controlled
rectifier circuit, the load current is unidirectional. Hence when the output voltage of the bridge tends to become
negative, the load current is zero and the conduction is discontinuous, with the output voltage clamped to zero volts.
The operation of this circuit has been described in detail in the previous page.
MATHEMATICAL ANALYSIS
Since the load current tends to be discontinuous, two expressions for the average output voltage can be derived, one for
the continuous mode and the other for the discontinuous mode. In the continuous mode, the average output voltage is as
shown in equation (1). The conduction becomes discontinuous when the firing angle exceeds 60o and remains less than
120o. Then the average voltage is obtained as shown in equation (2).
In equations (1) and (2), the amplitude of phase voltage is designated as E and the amplitude of line voltage is designaed
as U.
The rms voltage is computed as follows. When α < 60o, the conduction is continuous and the expression for the rms
voltage is presented in equation (3), whereas equation (4) expresses the rms voltage obtained when firing angle α > 60o.
The ripple factor can be found out as defined in the previous page.
The line/phase current can be defined as follows. Let R-phase voltage be defined to be
vR(θ) = E*Sin (θ).
Then the R-phase current iR(θ) is defined as follows, when the conduction is continuous.
When the conduction is discontinuous,
From this definition of phase current, the rms line current, the rms of the fundamental in line current and its THD can be
found out.
The applet below displays the average output voltage, the rms output voltage, its ripple factor, the rms line current, the
fundamental rms content in line current and its THD as a function of firing angle. The peak average output voltage,
(3U/π), is taken to be unity and (3U/πR) is set equal to unity, where R is the load resistor.
SIMULATION
The applet shown below displays the source voltage, the output voltage, the current in R-phase and the voltage across
SCR S1. In addition, the relevant statistical details are also displayed. To run the applet, key-in the firing angle and then
click on Start Button.
PSPICE SIMULATION
The Pspice program for simulation of this three-phase rectifier circuit is presented below. The model used for the SCRs
is the same as defined for the single-phase fully-controlled bridge rectifier. The three-phase brdige rectifier contains six
SCRs and it is necessary to define six pulse sources, one for each SCR. The pulse sources have been defined for a firing
angle of 30o and the frequency of the three-phase source is 50 Hz. At any time two SCRs need to conduct, one from the
top half and another bottom half and hence only if one SCR is triggered at a time, conduction may never get established.
To overcome this problem, two SCRs are triggered at the same time. For example, when SCR S2 is to be triggered, SCR
S1 is also triggered. In the same way, when SCR S3 is to be triggered, SCR S2 is also triggered and so on. In order to
effect this in program, one voltage-controlled voltage source is defined for each SCR. The dependent source defined for
SCR S1 is dependent on two sources, the pulse source that defines when SCR S1 is to be triggered and the pulse source
that defines when SCR S2 is to be triggered.
* Three-phase Full-wave Fully-Controlled Bridge Rectifier
VA 1 0 SIN(0 340V 50Hz)
VB 2 0 SIN(0 340V 50Hz 0 0 -120)
VC 3 0 SIN(0 340V 50Hz 0 0 -240)
XT1 1 4 11 4 SCR
XT3 2 4 13 4 SCR
XT5 3 4 15 4 SCR
XT4 5 1 14 1 SCR
XT6 5 2 16 2 SCR
XT2 5 3 12 3 SCR
RP 4 0 100K
RN 5 0 100K
R1 4 5 10
VP1
VP2
VP3
VP4
VP5
VP6
RP1
RP2
RP3
RP4
RP5
RP6
EP1
EP2
EP3
EP4
EP5
EP6
21
22
23
24
25
26
21
22
23
24
25
26
11
12
13
14
15
16
0
0
0
0
0
0
0
0
0
0
0
0
4
3
4
1
4
2
PULSE(0
PULSE(0
PULSE(0
PULSE(0
PULSE(0
PULSE(0
100K
100K
100K
100K
100K
100K
poly(2)
poly(2)
poly(2)
poly(2)
poly(2)
poly(2)
10
10
10
10
10
10
3333.3U 1N 1N 100U 20M)
6666.7U 1N 1N 100U 20M)
10M 1N 1N 100U 20M)
13333.3U 1N 1N 100U 20M)
16666.7U 1N 1N 100U 20M)
0M 1N 1N 100U 20M)
(21,0)
(22,0)
(23,0)
(24,0)
(25,0)
(26,0)
(22,0)
(23,0)
(24,0)
(25,0)
(26,0)
(21,0)
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 0.0MS 10US
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The responses obtained for a load resistance of 10Ω are presented below.
The Output Voltage Waveform
The Line Current(phase A) Waveform
The Voltage(SCR S1) Waveform
SUMMARY
This page has described the operation of the three-phase fully-controlled bridge rectifier with a resistive load. The next
page describes how this rectifier functions when the load contains an inductor also.
OPERATION WITH RL LOAD
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
MATHCAD SIMULATION
SUMMARY
CIRCUIT OPERATION
The circuit of the three-phase fully-controlled bridge rectifier circuit with an RL load and a voltage source has been
shown above. The purpose of providing a dc source in the dc link is to illustrate how two-quadrant operation can take
place. When the firing angle is held between 0o and 90o, the circuit operates in the rectifier region. When the firing
angle is held between 90o and 180o, the circuit can operate in the inverter region if the source E is sufficiently negative.
In the inverter region, the power flow is from the dc link to the ac 3-phase source. The inductor reduces the ripple in dc
link current.
MATHEMATICAL ANALYSIS
When the current flow in the dc link is continuous, the average output voltage can be calculated as outlined in the
previous page. It is difficult to estimate what the average output would be if there is a source present in the dc link and
the conduction is discontinuous. If there be no source in the dc link, the average output voltage for discontinuous
conduction is expressed by equation (1). The analysis of the circuit is along the lines described for the single-phase
controlled rectifier circuit. In order to get an expression for the line/phase current, it is necessary to get an expression
for the load current. The expression for load current is obtained from the expression for output voltage. The output
voltage is described by equation (2).
The differential equation that describes the load current is expressed by equation (3). The solution is of the form
expressed by equation (4). The impedance of load is Z and the load angle is φ. They are defined as shown in equation
(5).
In the equation above, w is the angular frequency in radians/second corresponding to the source frequency. When the
load current is continuous, equation (6) is valid. Using equation (6), equation (7) is obtained. Solving for A, we get
equation (8). When the conduction is discontinuous, iL(0) = 0 and then A is evaluated as shown in equation (9).
Once the value of A is known, the load current can be found out. From the load current, an expression for the line
current can be obtained. When SCR S1 is ON, the line current equals the load current. The line current is the negative
of load current when SCR S4 is ON, and it is zero when neither S1 nor S4 is ON. From the expression of the load
current, the rms value of line current, the rms value of the fundamental component of line current, the THD in line
current, the harmonic spectrum of line current, the DPF and the apparent power factor can be determined as outlined in
the earlier pages.
SIMULATION
The parameters to be keyed in are the firing angle, the value of source in the dc link (between 1.0 and -1.0 preferably)
and the ratio of load reactance to load resistance. The load reactance is wL, where w is the angular frequency in rad/s
corresponding to the ac source frequency. Click on Start Button for the program to respond.
PSPICE SIMULATION
* Three-phase Full-wave Fully-Controlled Bridge Rectifier
VA 1 0 SIN(0 340V 50Hz)
VB 2 0 SIN(0 340V 50Hz 0 0 -120)
VC 3 0 SIN(0 340V 50Hz 0 0 -240)
XT1 1 4 11 4 SCR
XT3 2 4 13 4 SCR
XT5 3 4 15 4 SCR
XT4 5 1 14 1 SCR
XT6 5 2 16 2 SCR
XT2 5 3 12 3 SCR
RP 4 0 100K
RN 5 0 100K
R1 4 6 10
L1 6 5 31.8M
VP1
VP2
VP3
VP4
VP5
VP6
RP1
RP2
RP3
RP4
RP5
RP6
EP1
EP2
EP3
EP4
EP5
EP6
21
22
23
24
25
26
21
22
23
24
25
26
11
12
13
14
15
16
0
0
0
0
0
0
0
0
0
0
0
0
4
3
4
1
4
2
PULSE(0
PULSE(0
PULSE(0
PULSE(0
PULSE(0
PULSE(0
100K
100K
100K
100K
100K
100K
poly(2)
poly(2)
poly(2)
poly(2)
poly(2)
poly(2)
10
10
10
10
10
10
3333.3U 1N 1N 100U 20M)
6666.7U 1N 1N 100U 20M)
10M 1N 1N 100U 20M)
13333.3U 1N 1N 100U 20M)
16666.7U 1N 1N 100U 20M)
0M 1N 1N 100U 20M)
(21,0)
(22,0)
(23,0)
(24,0)
(25,0)
(26,0)
(22,0)
(23,0)
(24,0)
(25,0)
(26,0)
(21,0)
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 0.0MS 10US
.PROBE
.FOUR 50 V(4,5) I(VA) I(L1)
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The results of the Fourier series analysis are presented below.
The Fourier Series Spectrum of the output voltage
The Fourier Series Spectrum of the Line Current (Phase A)
The Fourier Series Spectrum of the Load Current
It can be seen that the lowest harmonic frequency in the output voltage is at 300 Hz, the sixth harmonic, whereas there is
hardly any harmonic present in the load current because of the relatively large inductance in the load circuit. On the
other hand, the 5-th and the 7-th harmonics are visibly high in the line current.
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
MATHCAD SIMULATION
SIMULATION USING C
SUMMARY
CIRCUIT OPERATION
The circuit of a three-phase fully-controlled bridge rectifier with source inductance is presented
above. The presence of source inductance introduces an additional mode of operation when the
firing angle is less than a certain value. Let us assume that SCRS S1 and S2 are in conduction when
SCRS S3 is triggered. Then the current from the source does not transfer from S1 to S3
instantaneously, and the transfer of current, called commutation, takes a while. During this
commutation overlap, both S1 and S3 conduct in addition to S2. SCR S1 continues to conduct till
the current through S3 rises to equal the dc link current.
The effects of commutation overlap are:
i. A slight reduction in output voltage,
ii. A notch in the supply voltage to the circuit during commutation overlap.
When the source has inductance, other loads connected to this source along with the controlled
rectifier are supplied voltages with notches in them and some of these loads can be sensitive to
these notches and they may operate improperly. Hence in order to reduce the magnitude of
notches, it is mandatory in some countries for the rectifier to be provided with an inductance in
series with each of its three-phase input lines. If these inductors are much larger than the source
inductance, the notch voltages are absorbed by these inductances and the other loads connected to
the same 3-phase source are not supplied with distorted voltages. The internal inductances
connected in series with the source are sometimes referred to as 4% inductances. If the inductor is
such that the voltage drop across it is about 4% of the phase voltage at rated current, it is normally
sufficient to reduce the notches at the source terminals to an acceptable level.
MATHEMATICAL ANALYSIS
When there is an inductor in series with each input line, it is necessary to find out its effect. We
need to find out:
a. The reduction in output voltage.
b. The duration of commutation overlap.
c. The relationship between the firing angle and the commutation overlap.
REDUCTION IN OUTPUT VOLTAGE
Calculations by hand are carried out assuming that the dc link current remains steady without any
ripple. The source voltages at its terminals and the output voltage appear as shown below,
assuming that the inductances belong to the source.
It is seen that there are six notches in one input cycle. The reduction in average output voltage can
be found out as follows. Let SCR S1 be in conduction and let S3 be triggered. Let the current
through the dc link be IDC. Then current throughY-phase has to rise from zero to IDC, whereas
current through R-phase has to fall from - IDC to zero. On the other hand, loop current iLOOP
marked in the sketch below has to rise from zero to IDC. This means that during commutation
current through Y-phase would rise from zero to IDC and the volt-second area the output misses out
is L2IDC, that absorbed by the inductor in the Y-phase.
From the volt-seconds lost per commutation, we can find the total volt-seconds lost in one input
cycle. Since there are six commutations per cycle, the total volt-seconds lost per cycle is expressed
as shown by equation (1). Dividing this area by the time corresponding to one cycle, we get the
average voltage reduction in output. The time corresponding to one input cycle is 1/f, where f is the
line frequency. Then the average reduction in output voltage is obtained as shown in equation (2).
In equation (2), we make use of the relation that the angular frequency, w = 2πf. It is to be noted
that commutation overlap occurs only when there is continuous conduction through the load and
the average output voltage is expressed by equation (3). In equation (3), U is the peak line-to-line
voltage and α is the firing angle.
COMMUTATION OVERLAP ANGLE
The commutation of commutation overlap depends on:
a. the firing angle,
b. the dc link current and
c. the source inductance or the inductance in series with each phase.
To find out the commutation overlap, it is sufficient to analyse one commutation. Let SCR S5 be in
conduction and let SCR S1 be triggered at a firing angle of α. As seen earlier, the loop current
iLOOP builds up from zero to IDC. This means that the current in the inductance in R-phase builds
up to IDC, whereas it decreases to zero in the inductance in the B-phase. Let the commutation last
for an angle µ. Then during commutation, the voltage across the source inductance is expressed as
shown in equation (4). In equation (4), the loop current is denoted as ‘i’. Since θ = wt, the equation
for commutation overlap can be represented as shown in equation (5). During this interval, the loop
current changes from zero to IDC. Hence equation (6) defines how current in R phase changes.
The solution of equation (6) is presented in equation (7). Equation (7) can be re-arranged and
presented as shown in equation (8). Solving for µ, we obtain equation (9).
It is seen that overlap angle µ increases
a. as the firing angle moves closer to either 0o or 180o ,
b. as the dc link current becomes larger, and
c. as the source inductance gets larger.
The above equation has been obtained based on the assumption that the dc link current remains
steady, which happens only when the dc link inductance is relatively large. In practice, it is not true
and hence the above equation yields only an approximate result.
The rest of mathematical analysis follows the familiar route. The items of interest are:
a. RMS output voltage of bridge circuit,
b. Average output voltage of bridge circuit,
c. Ripple Factor of output voltage of the bridge,
d. RMS output voltage/voltage across load resistor,
e. Average output voltage (across load resistor),
f. Ripple factor of output voltage (across load resistor),
g. RMS line current,
h. RMS value of fundamental component in line current,
i. THD in line current,
j. Displacement power factor,
k. Apparent power factor and
l. Harmonic analysis.
In order to simulate it is necessary to have an expression for line current and load current. Let us
consider one output cycle, starting from the instant SCR S1 is triggered till SCR S2 is triggered.
During this period, the R-phase voltage is defined as shown in equation (10). When SCR S1 is
triggered, we have commutation overlap till the load current is transferred from SCR S5 to SCR S1
and let the commutation overlap angle be µ. During commutation overlap, the current through SCR
S1 rises from zero to load current. At the end of commutation overlap, line current is obtained from
the expression in equation (11). During the period of commuation overlap, the voltage source that
is seen by the dc link circuit is described as shown in equation (12), where vOL represents the
source during overlap period.
On simplifying equation (12), we get equation (13). During this period, the source appears to have
a source inductance equalling 1.5L2 . We get this value because the path through SCR S6 contains
L2 whereas the path through both S1 and S5 appears to have an equivalent inductance of value
equal to 0.5L2. The current through the load at the instant when SCR S1 is triggered can be
expressed as shown in equation (14).. The values of Z and φ in equation (14) are expressed in
equation (15).
The constant A in equation (14) is to be evaluated and τ2 = tan (φ). At the end of commutation, this
current would be equal to the line current. Hence we obtain equation (16). The terms used in
equation (17) are defined in equation (18).
Equation (16) is the first of the three equations we need in order to obtain a solution and this
equation equates the line current to the load current at the end of commutation. Equation (17)
expresses the load current at the end of commutation, using another constant B and B is also to be
evaluated. Another equation can be formed as shown in equation (19), which equates the line
current at the end of commuation to the load current. Unlike equation (16), equation (19) makes
use of constant B. Equations (16) and (19) are two of the three equations required to solve for A, B
and δ. The third equation is obtained as follows. The load current at the instant when θ = π /3 can
be computed in two ways, one from equation (16) and the other from equation (19) and these two
expressions can be equated to yield equation (20).
From these three equations, the three unknowns, A, B and µ, can be obtained. These equations
have been used in the program written for simulation.
SIMULATION
Three applets are presented in this section. The first applet animates the circuit. The only purpose
is to illustrate the sequence of operation of this circuit. It does not take in any parameter. To run it,
click on the RUN button. The SINGLE STEP button allows the user to step through, the PAUSE
button allows the user to stop the program and the RESET button allows the user to view the
simulation once more. To view the simulation once more, click on the RESET button and then
click on the RUN button. During the simulation, the user can ask the program to pause for a while,
then step through for a while and allow it to run through to the end of its cycle.
The second applet takes in these parameters:
a. the firing angle in degrees,
b. the ratio of load reactance to load resistance (load reactance evaluated at line frequency),
c. the ratio of source reactance to load resistance(preferably below 0.1 p.u), and
d. the value of dc link source.
The program allows you to view either the waveforms or the statistics. The values of load and line
reactance are to be entered in per unit. For example, if the line reactance is entered as 0.05 p.u.,
rated load current would cause a drop of 4% of phase voltage across the line reactance.
An example is presented now to explain how the per unit values can be set. Let a 3 phase, 415 V,
50 Hz source supply power to the converter. Then the maximum average voltage that can be
obtained is obtained as shown below. Let the nominal rated dc link current be 100 A. Then the
nominal load resistance or the base impedance for the system is computed as shown below. It is
also shown how the line reactance can be obtained, if its value in p.u. is known. Given that the
current through the load is free of ripple, the rated RMS line current is obtained as illustrated
below. From the total rms line current, inclusive of both the fundamental component and the
harmonic components, the fundamental rms component is obtained as illustrated below.
Given that the line inductance is 1 mH, its p.u. value is obtained as shown below.
Given that the dc link inductance is 10 mH, its p.u. value is computed as illustrated below.
Usually the line inductance is called as the 4% reactor, implying that when the line current is at its
rated value, the rms value of the fundamental component of voltage across the line reactor is 4% of
the phase voltage. For example, if the rms phase voltage is 240 V, the drop across 4% reactor at
rated current would be 9.6 V. When the line voltage is 415 V, the phase voltage is calculated as
shown below.The drop across the line inductor can now be stated as a fraction of the phase voltage
as given below.
This means that if the drop across the line inductor is to be 4% of phase voltage, the inductance
should be 0.4 mH and not 1 mH.
The third applet takes in these parameters:
a. the ratio of load reactance to load resistance (load reactance evaluated at line frequency),
b. the ratio of source reactance to load resistance(preferably below 0.1 p.u), and
c. the value of dc link source.
Within the program the firing angle is varied and the plots of rms bridge output voltage, the
average bridge output voltage, its RF, the rms line current, the fundamental rms component of line
current, the THD in line current, the average output current, its RF and the overlap angle as a
function of firing angle are displayed. The value of dc source in the dc link is assigned to be zero
in this applet.
SIMULATION USING C
A C-program called, ss_resp.cpp, can be downloaded by clicking on the image below. It can be
compiled as a C program and executed.
The program was run with the following values:
Time constant of the Link inductor in radians: 1.00
Time constant of the Source Inductance in radians : 0.03
Firing Angle = 30o.
The results obtained are presented below.
Another C-program called, harm3ph.cpp, can be downloaded by clicking on the image below. It
can be compiled as a C program and executed.
The results obtained for the same parameters are presented below.
harmNo
LoadCur
0
1.6874
OutVolt harmNo LineCur
1.6872
1 0.93105
2 2.69E-08 1.16E-08
3 2.31E-08
4 2.75E-08 2.94E-08
5 0.19975
6 0.029517
7
0.1743
0.1142
8 2.85E-08 2.82E-08
9 1.85E-08
10 2.85E-08 3.00E-08
11 0.080921
12
0.00639 0.074642
13 0.061509
14 2.87E-08 2.90E-08
15 1.82E-08
16 2.86E-08 2.97E-08
17 0.048484
18 0.0023099 0.040382
19 0.040233
20 2.87E-08 2.91E-08
21 1.84E-08
22 2.87E-08 2.94E-08
23 0.032741
24 0.0009234 0.021504
25 0.02832
LoadIAvg
0.84371
LoadIRMS
0.84398
RFCur
0.021433
VoAvg
0.84362
VoRMS
0.85546
RFVolt
0.14184
ILineAvg
0.56248
ILineRMS
0.68481
THD
0.684811
OverLapAngle 0.085085
SUMMARY
This page has explained how source inductance leads to commutation overlap and how it affects
the output voltage. Next page how a DC power supply can be built using the 3-phase fullycontrolled bridge rectifier.
AN APPLICATION: A DC POWER SUPPLY
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
CIRCUIT OPERATION
A three-phase fully-controlled bridge circuit is a much more suitable circuit to be used for generating a
variable dc output voltage than the single-phase fully-controlled bridge circuit, on account of two
reasons, which are:
a. reduced ripple content in its output and
b. much higher ripple frequency.
Both these factors lead to an LC filter which is relatively small and economical. This page describes
how such a power supply can be built and controlled.
An inductor in the dc link reduces ripple in the output current of the bridge circuit, whereas the
capacitor absorbs the ripple in output voltage. The inductor has to be designed such that it does not
saturate even when it carries the maximum current. This means that it should have an airgap in the path
of flux. The ripple current through the capacitor can also be significant. Hence it needs to be checked
from the datasheet that the capacitor chosen has the required ripple current rating. For such an
application, an electrolytic capacitor is normally chosen and its voltage rating should also be adequate.
We can have a block diagram to describe the operation of this dc power supply obtained using a threephase fully-controlled bridge rectifier. The output voltage Vo is varied by varying the firing angle α.
The firing angle in turn is controlled by voltage VC, which is the output of a PI controller. The inputs
to the PI controller are a voltage named Vref representing the desired output voltage and the output
voltage Vo of the bridge circuit. If Vo is less than the desired output voltage, the resultant error causes
the output, VC, to increase, which in turns should advance firing angle. As the firing angle is
advanced, the output voltage of the bridge circuit increases. The next section describes how the block
diagram can be analysed, leading to simulation of the system.
MATHEMATICAL ANALYSIS
The simulation program is based on the pseudo-code displayed below.
Start block:
Set the values of load reactance, line reactance, capacitor, load fraction.
Set the desired output voltage, PI controller parameters
Set the current firing angle to be 120o.
Set base reference angle to 60o.
Set Commute = 0. ( Indicates no commutation overlap exists at start. )
Go to Loop Routine.
Set theta to zero.
Loop Routine
Call Compute routine.
Increment theta.
If {(theta + base reference angle) = (next firing angle) }
[
current firing angle = next firing angle.
base reference angle = next firing angle - 60o.
Set Commute = 1 . ( Indicates next SCR is triggered)
]
Execute Loop Routine
Compute Routine:
If (Commute == 0)
[
compute next value of link inductor current.
]
else
[
compute next value of triggered SCR current.
compute next value of link inductor current.
if (SCR current link inductor current) Reset Commute to 0.
]
Compute next value of capacitor voltage.
Compute next value of PI controller's output.
Compute next firing angle.
The equations used in the Compute Routine are obtained as follows.
When there is commutation overlap, the output voltage behind the source inductance is expressed by
equation (1). In equation (1), µ is the overlap angle and U is the amplitude of line voltage. If the
output voltage be vo(θ) during this period, the differential equation for the current through the dc link
inductance is presented as equation (2). During commutation overlap, the current in the SCR just
triggered on is described by equation (3). The commutation overlap ends when the current through this
SCR equals the dc link inductor current. When there is no overlap, the bridge output voltage is
described by equation (4).
The differential equation that describes the dc link inductor current is then described by equation (5).
The differential equation for the capacitor voltage is easily obtained and it is expressed as equation (6).
Next the equations relating to closed-loop control are described. Let the output of PI controller be
vC(θ) and it is expressed by equation (7). In equation (7), A is a constant to be evaluated, K is the
proportional gain of the controller and T is its time constant. The above equation is represented as
equation (8) , which is more convenient for use in simulation. In equation (8), Vref is the desired output
voltage. The output of the controller is normally checked to ensure that it is within the set limits.
From the output of the controller, the firing angle, α can be obtained. The maximum output voltage of
the controller should correspond to zero degree firing angle and the minimum to 120o firing angle.
Hence we get the following equation for firing angle. This means that the range for vC(θ) is from 0 to
VCmax.
The simulation program uses the above equations and displays the results in a graphical format.
SIMULATION
The applet below can be run with the default parameters. To set any parameter, click on the arrow
pointing downwards beside the Peak Source Voltage, a menu would appear. The default value of the
parameter highlighted appears in the textfield for Set Value. To change the parameter, select the
parameter and then click within the editable textfield for Set Value. In order to change this parameter,
you must click on Set Value button. You can set the desired response to be one of three responses.
An example is presented now to explain how the per unit values can be set. Let a 3 phase, 415 V, 50
Hz source supply power to the converter. Then the maximum average voltage that can be obtained is
presented in equation (10). Let the nominal rated dc link current be 100 A. Then the nominal load
resistance or the base impedance for the system is assigned as shown by equation (11). Given that the
current through the load is free of ripple, the rms line current is obtained according to equation (12) and
this value includes both the fundamental component and the harmonic components. The fundamental
rms component is obtained as illustrated by equation (13). Given that the dc link inductance is 10 mH,
its p.u. value is obtained as shown by equation (14).
Given that the line inductance is 1 mH, its p.u. value is obtained as shown by equation (16). Usually the
line inductance is called as the 4% reactor, implying that when the line current is at its rated value, the
rms value of the fundamental component of voltage across the line reactor is 4% of the phase voltage.
For example, if the rms phase voltage is 240 V, the drop across 4% reactor at rated current would be 9.6
V. When the line voltage is 415 V, the phase voltage is obtained as shown by equation (16). The drop
across the line inductor can now be stated as a fraction of the phase voltage as given by equation
(17).This means that if the drop across the line inductor is to be 4% of phase voltage, the inductance
should be 0.4 mH and not 1 mH.
It is possible to set the load resistance to a value other than its nominal value. The nominal value of
load resistance is 5.6 Ω. If the load resistance is to be 10 Ω, then set Load Fraction as shown by
equation (18).
Some explanation is offered here to indicate how the values for different parameters can be set in the
applet shown below. The rated dc output voltage corresponds to 1 per unit. If the rated dc output
voltage is 400 V, the peak voltage of 415 V, 3 phase, 50 Hz supply is 1.467 p.u.. Let the nominal load
current through the load resistor be 100 A. Then Rnom is 4 Ω. If the dc link inductance is 10 mH, its
time constant in radians is (2πfLLink/Rnom) . Given that frequency is 50 Hz, it works out to be 0.7854
radians or p.u., since it expresses the ratio of the reactance over the base impedance. The per cunit
value to be set for the line reactance is also obtained similarly. Given the value of the filter capacitance,
its per unit value is obtained as (1/2πfCRnom). The applet allows the dc output voltage to be set to a
value other than unity and the load fraction can also be varied.
INTRODUCTION
CIRCUIT OPERATION
CURRENT LOOP
THE PER UNIT NOTATION
SPEED LOOP
PARAMETERS IN PER UNIT NOTATION
CURRENT CONTROLLER DESIGN
SPEED CONTROLLER DESIGN
FIELD CONTROLLER DESIGN
LOGIC FOR FOUR-QUADRANT OPERATION
SWITCH-OVER OF CONTROL FROM ARMATURE TO FIELD
SIMULATION OF THE FOUR-QUADRANT DRIVE
SUMMARY
INTRODUCTION
This page describes how a separately-excited dc shunt motor can be operated in either direction in either
of the two modes, the two modes being the motoring mode and the regenerating mode. It can be seen
that the motor can operate in any of the four quadrants and the armature of the dc motor in a fast fourquadrant drive is usually supplied power through a dual converter. The dual converter can be operated
with either circulating current or without circulating current. If both the converters conduct at the same
time, there would be circulating current and the level of circulating current is restricted by an inductor. It
is possible to operate only one converter at any instant, but switching from one converter to the other
would be carried out after a small delay. This page describes the operation of a dual converter operating
without circulating current.
As shown in Fig. 1, the motor is operated such that it can deliver maximum torque below its base speed
and maximum power above its base speed. To control the speed below its base speed, the voltage
applied to the armature of motor is varied with the field voltage held at its nominal value. To control the
speed above its base speed, the armature is supplied with its rated voltage and the field is weakened. It
means that an additional single-phase controlled rectifier circuit is needed for field control. Closed-loop
control in the field-weakening mode tends to be difficult because of the relatively large time constant of
the field.
The power circuit of the dual-converter dc drive is shown in Fig. 2.
Each converter has six SCRs. The converter that conducts for forward motoring is called the positive
converter and the other converter is called the negative converter. Instead of naming the converters as
positive converter and negative converter, the names could have been forward and reverse converters.
The field is also connected to a controlled-bridge in order to bring about field weakening.
The circuit shown above can be re-drawn as shown in Fig. 3. Usually an inductor is inserted in each line
as shown in Fig. 3 and this inductor reduces the impact of notches on line voltages that occur during
commutation overlap.
CIRCUIT OPERATION
The operation of the circuit in the circulating-current free mode is not very much different from that
described in the previous pages. In order to drive the motor in the forward direction, the positive
converter is controlled. To control the motor in the reverse direction, the negative converter is
controlled. When the speed of motor is to be changed fast from a high value to a low value in the
forward direction, the conduction has to switch from the positive converter to the negative converter.
Then the direction of current flow changes in the motor and it regenerates, feeding power back to the
source. When the speed is to be reduced in the reverse direction, the conduction has to switch from the
negative converter to the positive converter. It is seen that conduction has to switch from one converter
to the other when the direction of motor rotation is to change, so that regeneration can occur. During
regeneration, the direction torque developed by motor is opposite to that of the motoring torque. Thus
the regenerating torque acts as the breaking torque and the motor decelerates fast.
At the instant when the switch from one converter to the other is to occur, it would be preferable to
ensure that the average output voltage of either converter is the same. Let the firing angle of the positive
converter be αP, and the firing angle of the negative converter be αN . If the peak line voltage be U, then
equation (1) should apply. Equation (1) leads to equation (2). Then the sum of firing angles of the two
converters is π, as shown in equation (3).
In a dual-converter, the firing angles for the converter are changed according to equation (3). But it
needs to be emphasized that only one converter operates at any instant.
When the speed of the motor is to be increased above its base speed, the voltage applied to the armature
is kept at its nominal value and the phase-angle of the single phase bridge is varied such that the field
current is set to a value below its nominal value. If the nominal speed of the motor is 1500 rpm, then the
maximum speed at which it can run cannot exceed a certain value, say 2000 rpm. Above this speed, the
rotational stresses can affect the commutator and the motor can get damaged.
Next it is shown how the operation of motor can be represented by means of a block diagram. This
approach can be helpful in designing the closed-loop system.
CURRENT LOOP
Let the field excitation be assumed to remain constant at its nominal level. Let the voltage applied to
armature be va volts, the back e.m.f. eb volts and rotor speed wr rad/s. The back emf is expressed by
equation (4), where Km is the coefficient relating speed of motor to its back emf. If Ra be the resistance
of armature and La its inductance, then the applied armature voltage equals the sum of the motor back
e.m.f, the drop across its armature resistance and the drop across the armature inductance, as shown in
equation (5). In equation (5), va is the voltage applied to the armature and ia is the current though the
armature. The above equation can be represented in terms of Laplace transform, leading to equation (6).
The block diagram shown in Fig. 4 represents equation (6).
Given that the excitation of the motor is constant and that the effects of armature reaction are negligible
either due to interpoles or series compensation winding, the torque output Me can be expressed as shown
in equation (7). If the load torque be ML N-m, the combined polar moment of inertia of motor and load
be J kg.m2 and its friction coefficient be B N-m/rad/sec, then the torque output of motor equals the
expression on the right-hand side of equation (8). Equation (8) can be represented in terms of Laplace
transform, as shown in equation (9), where the Laplace transform of w, the motor speed, is assigned to
be Ω(s). A block diagram, as shown in Fig. 5, can now be drawn based on equations (4), (6), (7) and (9).
It can be seen that unit for Km is N-m/A.
With the load torque set to zero, a transfer function linking current Ia(s) and the input voltage Va(s) can
be obtained. It is expressed in equation (10).
The reason for obtaining this transfer function is to facilitate the design of a controller for controlling the
armature current. The two output parameters of interest are the torque and the speed. The armature
current is selected as one of the state-variables to be controlled in closed-loop, since the torque output
varies linearly with it. It is preferable that the variable to be controlled by negative feedback is a variable
that reflects some energy stored in a system. Here the armature current reflects the energy stored in the
inductance in the armature circuit. If the motor has a compensating winding and/or a compound
winding, the inductance of this winding should be added to La. In some drives, an additional inductor is
used in series with the armature and this value should also be added to La. Let G1(s) reflect the transfer
function in equation (10) and equation (11) shows the expression for G1(s). The part of the closed-loop
system that is usually used for controlling the armature current is shown in Fig. 6.
The block diagram in Fig. 6 is now described. If the armature current is to be controlled in closed-loop,
it is necessary to have a current reference signal, marked as IR(s) in Fig. 6. This signal is internally
generated, most often as the output of the controller for speed and it is shown later how that is achieved.
It is possible to use a controller other than a proportional plus gain(PI) controller. A PDF controller(a
pseudo-derivative controller) or a PID-controller can be used. But a PI controller is often sufficient,
since the integrating part of the PI controller leads to zero steady-state error for a step input and the
proportional gain can be adjusted to yield fast response and stability. The output of the current controller
is often a voltage which sets the firing angle for the fully-controlled bridge circuit. Since the gain KB(α)
is negative, the sign of both the proportional gain KI and the integrating time-constant TI should be
negative in order to keep the loop-gain of the system represented by block diagram in Fig. 6 as negative.
A variation in the output of the current controller does not change the firing angle instantaneously since
the SCRs in the bridge are triggered in a sequence at an interval of 60o on the average and there is a
delay before the change in the output of the current controller has an effect on the firing angle. This
delay can be classified as a transportation lag and it can be approximated by a first-order transfer
function, as shown in equation (12). In equation (12), y has been used in place of sTD.
For a system with 50 Hz input source, one-sixth of a cycle is about 3.3 ms and then the delay TD can be
set to be half of that value, that is 1.67 ms. What is carried out is an approximation to facilitate the
design of current controller.
As the firing angle α increases, the instant of triggering of each SCR is delayed more and more from its
reference point corresponding to 0o firing angle. When the current flow in the dc link is continuous, the
average output voltage of the bridge changes from its positive maximum average value to its negative
maximum average value, as α is allowed to very from 0o to 180o. In order to ensure that the loop gain is
negative, it is necessary that the gain due to controlled rectifier circuit is inverted. It is explained later
how it can be brought about for practical realization.
Another point to be noted is that the gain of the controlled rectifier is not constant and it varies with
firing angle. Let the maximum average output voltage be Vom. Then equation (13) shows how the
average output voltage at any firing angle, α, is obtained. The actual gain of the controlled rectifier is
defined by equation (14). If the rated armature voltage, VRA, is assigned to be the base voltage, then the
gain of the controlled rectifier in per unit notation can be defined as in equation (15).
In equation (15), KA defines the ratio of maximum average output voltage of the bridge to the rated
armature voltage. It is seen that the gain varies and hence the controller has to be designed such that it
operates properly over this range of variation.
THE PER UNIT NOTATION
It is better to design the current loop first before the outer loop design is attempted. But it is necessary to
describe the per unit notation that is adopted here. Let the rated armature voltage VRA be the base
voltage. Then equation (16) is valid. As shown in equation (17), the rated armature current IRA is chosen
to be the base current.
Per Unit Value of Rated Armature Voltage = 1 (16)
Per Unit Value of Rated Armature Current = 1 (17)
Then the base impedance for the armature circuit is obtained as shown in equations(18) whereas
equation (19) shows how the per unit value of armature resistance can be obtained if it is ra Ω. Given
that the inductance present in the armature circuit is La H, the voltage across it is obtained as shown in
equation (20). Equation (21) is obtained by dividing both sides of equation (20) by VRA. Equation (21)
uses symbol τa, representing the time constant of the armature circuit and it is defined by equation (22).
For a 3-phase controlled-bridge rectifier circuit, the maximum average output voltage that can be
obtained at 0o firing angle is shown in equation (23). Then the amplitude of line voltage of 3-ph supply
is described by equation (24). The per unit value of the peak line voltage is obtained from equation (25).
We have seen so far how voltages, currents and impedances related to armature circuit can be expressed
in per unit values. Next, it is shown how the torque developed, moment of inertia J and friction
coefficient B can be expressed in per unit values. Let the torque developed by motor be Me N-m. Then
when the motor is operating with the nominal or the rated flux, the torque developed by motor is defined
by equation (26), where ia is the armature current. Also, let wr be the armature shaft speed in rad/s. Then
the per unit value of the torque developed is expressed as shown in equation (27), where IRA is the rated
armature current.
The per unit value of moment of inertia is obtained as follows. Let ΩR be the rated shaft speed in rad/s,
and the moment of inertia of motor and the coupled load be J kg-m2. Let the torque required to
accelerate this moment of inertia be MJ. Then equation (28) can be used to relate J and MJ. Dividing
both sides of equation (28) by the rated torque, we get equation (29). From equation (29), it is seen how
the per unit value of the moment of inertia can be obtained. Similarly, we can get an expression for
friction coefficient, as shown by equation (30).
It is necessary to state how the parameters for the current controller should be specified. The gain, KI , is
just a ratio whereas the time constant, TI , should be specified in seconds.
SPEED LOOP
Before the design of speed loop is to be attempted, the current loop should be approximated by a suitable
transfer function. The block diagram in Fig. 6 can be expressed by a transfer function, say G2(s). Then
G2(s) = Ia(s)/IR(s). Using G2(s), the speed loop can be represented as shown in Fig. 7. Again a PI
controller is used and it may be necessary to provide a filter in the path of speed feedback signal. The
time constants are to be specified in seconds. In per unit notation, the value of KM marked in Fig. 7
would be 1.
PARAMETERS IN PER UNIT NOTATION
At first, typical per unit values are obtained from the datasheet of a dc motor. Then it is shown how the
current loop can be designed. Next, the design is verified by simulation. Finally, the speed loop design is
illustrated.
In the per-unit calculations, a per unit value of 1 is assigned to the rated armature voltage, the rated
motor speed and the rated armature current. From the values of the rated armature voltage and the rated
motor speed, obtain Km, the coefficient for the motor. Then other per unit values can be obtained as
outlined earlier. The applet displayed below computes the per unit values given the actual values. The
textfields contain default values and the applet computes and displays the per unit values when the
Compute Button is clicked.
CURRENT CONTROLLER DESIGN
The transfer function G1(s) expressed as equation (11) can expressed in terms of per unit values and then
Va(s) and Ia(s) marked in Fig. 6 would be values in per unit notation. Conversion of equation (11) such
that it conforms to the per unit notation is explained below.
Both the numerator and the denominator of expression in equation (11) can be divided by BRa. The ratio
of J/B can be represented as the mechanical time constant, τm. The resulting expression for G1(s)
presented as equation (31). Then equations (32) and (33) explain how equation (31) can be converted
such that it is in per unit notation.
The block diagram shown in Fig. 6 conforms to per unit notation. Here G1(s) is expressed by equation
(34) and Va(s) and Ia(s) marked in Fig. 6 are values in per unit notation. Using (34), the transfer
function, G2(s) representing the block diagram in Fig. 6 can be represented as shown below. In this case,
the controlled-rectifier is assigned to have its highest gain, KA. It is logical to do so, because the system
designed for stability at gain KA would be stable at lower gains too.
The second applet in this page finds the location of the poles and zeros of the closed-loop system in Fig.
6, given the necessary data. It is seen for a wide range of controller parameters, the zeros are located
such that they cancel almost two poles. The other two poles are located away from the origin. It is seen
that the design of current controller is fairly easy.
SPEED CONTROLLER DESIGN
It is necessary to design the speed controller next. To design the speed controller, it is necessary to
represent the transfer function G2(s) suitably. It is found that for a wide range of values, two of the zeros
of G2(s) are located near two of the poles and the other two poles are away from the origin. Hence while
designing the speed loop, G2(s) is set equal to unity. The simulation of the drive presented later would
show whether this simplification is justified.
The design of the speed controller is carried out based on the assumption that the motor is on no load. A
variable drive system tends to exhibit oscillatory behaviour under no load conditions and hence the
design based on no load condition is assumed to be justified. Here the output of the speed controller is
not clamped, whereas there would be limits on the output of speed controller. The output of speed
controller corresponds to armature current and it is necessary to limit the peak value of speed controller
in order to protect the SCRs used in the bridge. The applet presented later for simulating the drive has
limits imposed on the output of the speed controller.
FIELD CONTROLLER DESIGN
The block diagram for closed-loop operation with the field controller in action turns out to be somewhat
complex. The interaction that occurs within a separately-excited DC motor is first presented in Fig. 8.
The block diagram in Fig.8 is now described. The field current, marked as IF, produces magnetic flux in
the motor and the back e.m.f of the motor is then proportional to the product of the field current and the
speed of the motor. This statement is based on the assumption that the field flux in the motor is not
saturated and that the field flux varies linearly with the field current. If the field current is in per unit
notation, where the field current corresponding to the rated current equals unity, then the back emf can
be shown to be equal to Km × iF × wR, where both Km and wR are also in per unit representing the motor
coefficient and the speed of the motor. Once the back e.m.f and the applied voltage are known, the
armature current can be obtained as shown in Fig. 8. From the values of armature current and field
current, the torque output of motor is obtained and the speed of the motor changes as shown.
For design of field controller, the block diagram in Fig. 8 is too complex. The design is carried out using
a simplified or a simplistic block diagram and the performance of the controller is evaluated using the
final simulation program, which uses a model that is reasonably close to real system.
The design of field controller is based on the block diagram shown in Fig. 9.
It is easy to represent the block diagram in Fig. 9 in per unit notation. The gain of the controlled bridge
for the field circuit is KF. Its value equals the ratio of the maximum rate of bridge output voltage to the
rated voltage of the field circuit and normally the value of KF is likely to be near 1.2. The delay due to
firing circuit is again approximated by TD2, and it is set equal to (1/4f), where f is the frequency of the ac
source. Then the field current is obtained in per unit value and it can be made equal to the torque,
assuming that the armature circuit has comparatively a small time constant and that the armature current
stays at the rated value. The friction coefficient, the mechanical time constant and the time constant of
the filter in the speed feedback signal are the same signals used for design of the speed controller. The
applet below can be used to design the field controller. This applet runs somewhat slowly. The poles and
zeros are calculated for the block diagram shown in Fig.9, whereas the step response is obtained using
the block diagram in Fig. 10.
The design of field controller is somewhat difficult because both the field circuit time constant and the
mechanical time constant are relatively large.
The applet displayed below shows step response of the drive with the field controller.
LOGIC FOR FOUR-QUADRANT OPERATION
For four-quadrant operation, the scheme outlined in this page makes use of two converters, called the
positive converter and the negative converter. It has also been shown that the sum of the firing angles of
the two converters should be π radians. Hence the synchronizing signals for the SCRs in both converters
can be obtained as shown in Fig. 11.
It is also necessary to find out when the switch from the positive converter to the negative converter or
vice-versa can be made. One possible method is outlined in Fig. 12. Based on the polarity of current
reference signal, a logic signal, called W can be developed. A comparator can be designed to yield an
output of 1 (W =1 ) when the current reference signal is positive and an output of 0 (W =0 ) when the
current reference signal is negative. Along with W, another signal can be derived based on the armature
current. A signal, called Enable, can be produced such that Enable is 1 when the armature current is
zero. When Enable is 1, the output of a flip-flop can be set. Output Q takes on the polarity of W signal.
When both W and Q are at logic 1, the positive converter is allowed to be triggered. When both W and
Q are at logic 0, the negative converter is allowed to be triggered.
The rest of the control arrangement for generating the firing signals is shown in Fig. 13. The block
diagram shows that there are active limits on the firing angle. The voltage output of the bridge can be
sensed and when it is at about 1.05 times the rated armature voltage, the firing angle may not be allowed
to become any smaller.
The variation of firing angle towards either 0o or 180o can be blocked, avoiding further rise in the output
voltage. It is better to have a provision that would allow for varying the limiting values to accommodate
changes in source voltages. It would not be difficult to implement such a scheme for a micro-controller
based control system.
SWITCH-OVER OF CONTROL FROM ARMATURE TO FIELD
Field control is necessary above the base speed of the motor and the field current has to vary only over a
limited range, say from 0.6 pu to 1 pu. The signal that sets the active limits on the firing angle for
armature control can be used also for field control. The block diagram related to field control is shown in
Fig. 14.
When the active limit is not set, the firing angle of the field controller is set such that the field current
remains at the nominal value. When the active limits are applied to the armature control circuit, the
firing angle is varied such that the field current gets adjusted to the value required for the speed
reference set.
SIMULATION OF THE FOUR-QUADRANT DRIVE
Before selecting the type of response, set the value of the selected parameter. When you select a
parameter, the textfield shows the default value set inside the program. Change the parameter value if
you want to and then you must click on the SET VALUE button for the change to take effect. You can
go from one type of response to another after the present calculations are carried out. When you have
selected a new type of response, you must click on Click to Start. If you click on Reset button,
initializing routine is carried out and the motor speed is set to zero, and the other values are also reset. It
is possible to see the effect of step changes in speed or load or unbalance in source or unbalance in firing
circuit.
SEMI-CONTROLLED
RECTIFIERS
This chapter describes two semi-controlled
rectifier circuits:
●
●
HALF-CONTROLLED SINGLE-PHASE BRIDGE
RECTIFIER
HALF-CONTROLLED THREE-PHASE BRIDGE
RECTIFIER
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION
PSPICE SIMULATION
MATHCAD SIMULATION
SUMMARY
CIRCUIT OPERATION
A fully-controlled rectifier circuit contains only controlled-rectifiers, whereas a semi-controlled rectifier circuit is made
up of both controlled and uncontrolled rectifiers. Due to presence of diodes, free-wheeling operation takes place without
allowing the bridge output voltage to become negative. In a semi-controlled rectifier, control is effected only for positve
output voltage, and no control is possible when its output voltage tends to become negative since it is clamped at zero
volt. This page describes the operation of a single-phase half-controlled rectifier.
A semi-controlled full-wave bridge rectifier can be configured in a few ways. They are shown below.
The circuit in Configuration 1 contains two SCRs and two diodes. When source Vin is positive, SCR S1 can be triggered
at a firing angle called α and then current flows out of the source through SCR S1 first, then through the load and returns
via diode D3. If
then SCR S1 and diode D3 conduct during α < wt < π. When π < wt < 2π, Vin is negative and SCR S2 is normally
triggered when wt = π + α. During π < wt < (π + α) , the output of the bridge circuit would have been negative if we had
used a fully-controlled bridge rectifer and if the current flow was continuous. But here we have two diodes D3 and D4
instead of two SCRs. When the output of the bridge tends to becomes negative just after wt exceeds π, diode D4 tends to
get forward-biased and it starts conducting. Then diode D3 is reverse-biased and it stops conducting. During π < wt < (π
+ α) , the devices in conduction are SCR S1 and diode D4 and the output of the bridge is clamped at zero, assuming that
the on-state drops across devices in conduction is zero. During ( π + α) < wt < 2π , the devices in conduction are SCR S2
and diode D4. SCR S2 and diode D3 would conduct during 0 < wt < α .
The circuit in configuration 1 has SCRs as the devices in the top-half and diodes as the devices in the bottom-half.
Instead, it it is possible to use SCRs as the devices in the bottom-half and diodes as the devices in the top-half.
It is also possible to build a semi-controlled full-wave bridge rectifier as shown by the circuit in configuration 2.
The behaviour of the circuit is the same as described earlier. In this circuit, SCR S1 and diode D3 conduct during α < wt
< π . During π < wt < (π + α) , the devices in conduction are diodes D3 and D4 and the output of the bridge is clamped at
zero. During (π + α) < wt < 2π , the devices in conduction are SCR S2 and diode D4. Diodes D3 and D4 would conduct
during 0 < wt < α .
Yet another configuration is available for semi-controlled bridge rectifier, as shown by the circuit in configuration 3.
In this circuit, SCRs S1 and S3 conduct during α < wt < π. During π < wt < (π + α) , the device in conduction is diode D
and the output of the bridge is clamped at zero. During (π + α) < wt < 2π , the devices in conduction are SCRs S2 and S4.
Diode D would conduct during 0 < wt < α .
MATHEMATICAL ANALYSIS
The aim of analysis is to obtain the following values:
1. The average output voltage of the bridge as a function of firing angle.
2. The rms output voltage of the bridge as a function of firing angle.
3. The ripple factor of output voltage of the bridge as a function of firing angle.
4. The rms line current as a function of firing angle and the ratio wL/R.
5. The fundamental rms line current as a function of firing angle and the ratio wL/R.
6. The THD in line current as a function of firing angle and the ratio wL/R.
The Average Output Voltage
The ripple factor is defined then as
Next it is shown how the line current is to be analysed. An expression for load current over half-a-cycle can be obtained
first. The load current during α < wt < π can be defined as follows.
where
From the expression for load current,
The load current during π < wt < (π + α) can be defined as follows.
When the load current is repetitive, we have that
That is,
and
Hence we obtain that
Once A is known, the total rms value of line current and the rms value of its fundamental component can be estimated.
Let
and
Then the rms line current given t and a is obtained as follows.
To obtain the rms value of the fundamental component of the line current, we obtain the trigonometric Fourier series
coefficients of the fundamental component. The line current has half-wave symmetry and hence these coefficients are
obtained as follows.
Then
We obtain the rms value of fundamental component as:
Total harmonic distortion in line current is then
SIMULATION
The applet shown below simulates this circuit. The parameters to be keyed-in are the ratio of load reactance to load
resistance and the firing angle in degrees.
PSPICE SIMULATION
The semi-controlled bridge rectifier that has been simulated has four SCRs and a single free-wheeling diode. The
program is presented below.
* Full-wave Bridge Rectifier with a resistive load
VIN 1 0 SIN(0 340V 50Hz)
XT1 1 2 5 2 SCR
XT2 0 2 6 2 SCR
XT3 4 0 7 0 SCR
XT4 4 1 8 1 SCR
VP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)
VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)
VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)
VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)
D1 4 2 DNAME
L1 2 3 31.8M
R1 3 4 10
R2 1 0 1MEG
R3 2 0 1MEG
R4 4 0 1MEG
.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)
* Subcircuit for SCR
.SUBCKT SCR 101 102 103 102
S1 101 105 106 102 SMOD
RG 103 104 50
VX 104 102 DC 0
VY 105 107 DC 0
DT 107 102 DMOD
RT 106 102 1
CT 106 102 10U
F1 102 106 POLY(2) VX VY 0 50 11
.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0)
.MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0)
.ENDS SCR
.TRAN 10US 60.0MS 0.0MS 10US
.FOUR 50 V(2,4) I(VIN)
.PROBE
.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)
.END
The waveforms obtained are displayed.
The waveform of bridge output voltage
The waveform of load current
The waveform of mains current
The waveform of current through the free-wheeling diode
INTRODUCTION
CIRCUIT OPERATION
MATHEMATICAL ANALYSIS
SIMULATION WITH AN APPLET
MATHCAD SIMULATION
SUMMARY
INTRODUCTION
This page describes the operation of the three-phase semi-controlled rectifier. When a semi-controlled
rectifier is used, the output of the bridge is controlled when it remains positive. If it tends to become
negative, the output is clamped to zero volt through the free-wheeling action of the diodes.
Fig. 1
The circuit of the semi-controlled bridge rectifier is shown in Fig. 1, and this circuit contains three-
SCRs and three diodes. It is possible to configure the circuit in two more ways. For example, the tophalf can contain the diodes and the bottom half the SCRs. Alternatively, six SCRs can be used as in
the case of the fully-controlled bridge rectifier and an additional diode can be connected from the
negative rail to the positive rail of the bridge, with the anode connected to the negative rail and the
cathode to the positive rail. The operation of this circuit is different from that of the circuit displayed
above.
CIRCUIT OPERATION
The operation of the semi-controlled rectifier circuit displayed in Fig. 1 is now explained. The
waveform of the bridge output voltage for a firing angle of 30o is shown in Fig. 2. .It can be seen that
the trace of the positive rail of output voltage is that of a controlled rectifier, since only the top-half of
the bridge is controlled. The trace of the bottom rail is that of an uncontrolled bridge rectifier, since
the bottom half contains only diodes.
Fig. 2
Let the 3-phase supply be defined as shown in equation (1).
Given that the firing angle is 30o, SCR S1 is triggered when wt = 60o . The conduction range of the
SCRs in the top half can be is now expressed in equation (2).
As long as the firing angle α remains less than 60o, the expression for output voltage over one output
cycle can be expressed as follows.
Substituting for vR, vY, and vB from equation (1), we get that
When the firing angle α is higher than 60o, the expression for output voltage over one output cycle
can be expressed as follows, if conduction through the load is continuous.
Substituting for vR, and vB from equation (1), we get that
Depending on the firing angle, the bridge output repeats itself every 120o. The applet shown below
plots the output voltage, given the firing angle. It is assumed that conduction through the load
impedance is continuous.
Fig. 3
If the firing angle α is less than 60o, when SCR S1 is triggered, SCR S1 and diode D6 conduct during
the period (α + 30o) ≤ wt < 90o. Figure 3 shows the path of conduction that would exist during this
period. The pairs that conduct vary depending on the firing angle. Table 1 show the pairs that would
conduct when the firing angle α is less than 60o, whereas Table 2 shows the pairs that would conduct
when the firing angle α is greater than 60o.
MATHEMATICAL ANALYSIS
The aims of anlysis are:
1.
2.
3.
4.
5.
6.
7.
8.
9.
To obtain an expression for the average bridge output voltage as a function of firing angle,
To obtain an expression for the rms bridge output voltage as a function of firing angle,
To obtain an expression for the ripple factor of bridge output voltage as a function of firing
angle,
To obtain an expression for the instantaneous load current as a function of firing angle,
To obtain an expression for the instantaneous line current as a function of firing angle,
To obtain an expression for the rms value of the fundamental of line current as a function of
firing angle,
To obtain an expression for the rms line current as a function of firing angle,
To obtain an expression for the THD of line current as a function of firing angle,
To carry out harmonic analysis of line current, bridge output voltage and load current at a given
firing angle.
TABLE 1: PAIRS IN CONDUCTION WHEN α < 60o
S1 – D6
(α + 30o) ≤ wt < 90o
S1 – D2
90o ≤ wt < (α + 150o)
S 3 – D2
(α + 150o) ≤ wt < 210o
S3 – D4
210o ≤ wt < (α + 270o)
S 5 – D4
(α + 270o) ≤ wt < 330o
S5 – D6
330o ≤ wt < (α + 390o)
TABLE 2: PAIRS IN CONDUCTION WHEN α > 60o
S1 – D2
(α + 30o) ≤ wt < 210o
S1 – D4
210o ≤ wt < (α + 150o)
S 3 – D4
(α + 150o) ≤ wt < 330o
S3 – D6
330o ≤ wt < (α + 390o)
S 5 – D6
(α + 270o) ≤ wt < 450o
S5 – D2
90o ≤ wt < (α + 30o)
Average Output Voltage
When the conduction through the load is continuous, the average bridge output voltage is obtained as
shown below.
Given that the firing angle is less than 60o,
If the firing angle is greater than 60o and the conduction is continuous,
The maximum output voltage occurs when α = 0o and let it be Vdm:
From equation (7) and (9), we obtain that
RMS Output Voltage
The expression for the rms output voltage is found separately for the two cases. The assumption here
is that the conduction is continuous. When the firing angle is greater than 60o,
When the firing angle is less than 60o,
Ripple Factor of the Bridge Output Voltage
The ripple factor, RF(α), of the bridge output voltage can be computed as follows:
Since both VRMS(α) and VDC(α) are known, the ripple factor can be computed.
Instantaneous Load Current
An expression for the instantaneous load current as a function of firing angle can be obtained through
a somewhat tedious process.
When the firing angle is less than 60o, the instantaneous bridge output voltage expressed by equation
(4) is reproduced below.
The above expression can be written with the origin shifted to the instant of triggering an SCR. Then
Let the load angle be:
The expression for load current during 0o ≤ θ ≤ (60o – α) can be expressed to be:
In the expression, A1 is a constant to be evaluated. Then
The expression for load current during (60o – α) ≤ θ ≤ 120o can be expressed to be:
In equation (20), A2 is a constant to be evaluated. Then
Another expression for the load current can be obtained when the output cycle ends.
When the load current is periodic, then
From equations (18), (19), (21) , (22) and (23), we can determine A1 and A2. From equations (19) and
(21), we get that
On simplifying the above expression, we get that
From equations (18), (22) and (23), we get that
On simplifying the above expression, we get that
Substituting for A2 from equation (26),
Then A2 can be determined from equation (25).
Since both A1 and A2 are known, the expression for load current has been obtained for the case when
the firing angle is less than 60o.
When the firing angle is greater than 60o,
Then the load current can be expressed to be:
In the above equation, A3 and A4 have to be determined. The conditions we have are:
On solving for A3, we obtain that
Now an expression for the load current is known for any firing angle.
Instantaneous Load Current
An expression for the instantaneous line current as a function of firing angle can be obtained from the
expression for the load current. Here an expression for current through R-phase over an input cycle is
obtained with the origin coinciding with the triggering of SCR S1. When SCR S1 is in conduction, the
line current is equal to the load current Then
When the firing angle is less than 60o,
When the firing angle is greater than 60o,
RMS Line Current
Since the line current iL(θ) is known over one input cycle, the rms line current can be obtained.
RMS Value of the Fundamental of Line Current
By performing Fouries series analysis of the line current, the rms value of the fundamental can be
obtained. From the rms line current and the rms value of the fundamental of line current, THD can be
computed.
SWITCH-MODE POWER SUPPLY
Power electronics deals with four forms of power conversion.
1.
2.
3.
4.
ac-dc conversion called rectification
ac-ac conversion ,
dc-ac conversion and
dc-dc conversion.
DC-DC converters were referred to as choppers earlier, when
SCRs were used. Nowadays, IGBTs and MOSFETs are the
devices used for dc-dc conversion and these circuits can be
classified as switch mode power supply circuits. The
abbreviation or acronym for switch mode power supply is
SMPS.
A switch mode power supply circuit is versatile. It can be used
to:
1.
2.
3.
4.
5.
step down an unregulated dc input voltage to produce a
regulated dc output voltage using a circuit known as Buck
Converter or Step-Down SMPS,
step up an unregulated dc input voltage to produce a
regulated dc output voltage using a circuit known as Boost
Converter or Step-Up SMPS,
step up or step down an unregulated dc input voltage to
produce a regulated dc output voltage ,
invert the input dc voltage using usually a circuit such as
the Cuk converter, and
produce multiple dc outputs using a circuit such as the fly-
back converter.
A switch mode power supply is a widely used circuit nowadays
and it is used in a system such as a computer, television receiver,
battery charger etc. The switching frequency is usually above 20
kHz, so that the noise produced by it is above the audio range. It
is also used to provide a variable dc voltage to armature of a dc
motor in a variable speed drive. It is used in a high-frequency
unity-power factor circuit.
This chapter describes the basics, operation and design of
switched-mode power supplies. The pages to follow contain:
●
●
●
●
STEP-DOWN /BUCK CONVERTER: IDEAL CIRCUIT
STEP-DOWN /BUCK CONVERTER: PRACTICAL
CIRCUIT
STEP-UP SWITCH MODE POWER SUPPLY /IDEAL
BOOST CONVERTER
PRACTICAL BOOST CONVERTER AND UNITYPOWER FACTOR RECTIFIER
INTRODUCTION
BASIC CIRCUIT OPERATION
DISCONTINUOUS OPERATION
CONTROL BY PULSE-WIDTH MODULATION
CLOSED-LOOP CONTROL
MICRO-CONTROLLER IMPLEMENTATION
IMPLEMENTATION USING A MULTIPLIER
SUMMARY
INTRODUCTION
A buck converter or step-down switch mode power supply can also be called a switch mode regulator.
Popularity of a switch mode regulator is due to its fairly high efficiency and compact size and a switch mode
regulator is used in place of a linear voltage regulator at relatively high output, because linear voltage regulators
are inefficient . Since the power devices used in linear regulators have to dissipate a fairly large amount of
power, they have to be adequately cooled, by mounting them on heatsinks and the heat is transferred from the
heatsinks to the surrounding air either by natural convection or by forced-air cooling. Heatsinks and provision
for cooling makes the regulator bulky and large. In applications where size and efficiency are critical, linear
voltage regulators cannot be used.
A switch mode regulator overcomes the drawbacks of linear regulators. Switched power supplies are more
efficient and they tend to have an efficiency of 80% or more. They can be packaged in a fraction of the size of
linear regulators. Unlike linear regulators, switched power supplies can step up or step down the input voltage.
The buck converter is introduced in this page using the evolutionary approach. Let us consider the circuit in Fig.
1, containing a single pole double-throw switch.
For the circuit in Fig. 1, the output voltage equals the input voltage when the switch is in position A and it is
zero when the switch is in position B. By varying the duration for which the switch is in position A and B, it can
be seen that the average output voltage can be varied, but the output voltage is not pure dc. The output voltage
contains an average voltage with a square-voltage superimposed on it, as shown in Fig. 2.. Usually the desired
outcome is a dc voltage without any noticeable ripple content and the circuit in Fig. 1 is to be modified.
The circuit in Fig. 1 can be modified as shown in Fig. 3 by adding an inductor in series with the load resistor.
An inductor reduces ripple in current passing through it and the output voltage would contain less ripple content
since the current through the load resistor is the same as that of the inductor. When the switch is in position A,
the current through the inductor increases and the energy stored in the inductor increases. When the switch is in
position B, the inductor acts as a source and maintains the current through the load resistor. During this period,
the energy stored in the inductor decreases and its current falls. It is important to note that there is continuous
conduction through the load for this circuit. If the time constant due to the inductor and load resistor is relatively
large compared with the period for which the switch is in position A or B, then the rise and fall of current
through inductor is more or less linear, as shown in Fig. 3.
The next step in evolutionary development of the buck converter is to add a capacitor across the load resistor
and this circuit is shown in Fig. 4. A capacitor reduces the ripple content in voltage across it, whereas an
inductor smoothes the current passing through it. The combined action of LC filter reduces the ripple in output
to a very low level.
The circuit in Fig. 4 contains a single-pole double-throw switch. It is a difficult configuration to realize using
power semiconductor devices. On the other hand, an understanding of the circuit in Fig. 4 leads to a realizable
and simple configuration. When the switch is in position A, the current through the inductor and it decreases
when the switch is in position B. It is possible to have a power semiconductor switch to correspond to the switch
in position A. When switch is in position B, the inductor current free-wheels through it and hence a diode can be
used for free-wheeling operation. Then only the power semiconductor switch needs to be controlled, and in
practice, a pulse-width modulating IC is used. The circuit that results is shown in Fig. 5.
Generally any basic switched power supply consists of five standard components:
a.
b.
c.
d.
e.
a pulse-width modulating controller,
a transistor switch,
an inductor ,
a capacitor and
a diode.
Control by pulse-width modulation, usually effected by an IC, is necessary for regulating the output. The
transistor switch is the heart of the switched supply and it controls the power supplied to the load. Power
MOSFETs are more suited than BJTs at power outputs of the order of 50 W. Transistors chosen for use in
switching power supplies must have fast switching times and should be able to withstand the voltage spikes
produced by the inductor.
An inductor is used in a filter to reduce the ripple in current. This reduction occurs because current through the
inductor cannot change suddenly. When the current through an inductor tends to fall, the inductor tends to
maintain the current by acting as a source. Inductors used in switched supplies are usually wound on toroidal
cores, often made of ferrite or powdered iron core with distributed air-gap to minimize core losses at high
frequencies.
A capacitor is used in a filter to reduce ripple in voltage. Since switched power regulators are usually used in
high current, high-performance power supplies, the capacitor should be chosen for minimum loss. Loss in a
capacitor occurs because of its internal series resistance and inductance. Capacitors for switched regulators are
chosen on the basis of effective series resistance (ESR). Solid tantalum capacitors are the best in this respect.
For very high performance power supplies, sometimes it is necessary to parallel capacitors to get a low enough
effective series resistance.
The diode used in a switched regulator is usually referred to as free-wheeling diode or sometimes as a catch
diode. The purpose of this diode is not to rectify, but to direct current flow in the circuit and to ensure that there
is always a path for the current to flow into the inductor. It is also necessary that this diode should be able to
turn off relatively fast. Diodes known as the fast recovery diodes are used in these applications.
Most of the switched supplies needs a minimum load, in order to ensure that the inductor carries current always.
If the current flow through the inductor is not continuous, regulation may become poorer.
The buck converter or SMPS can be controlled in two ways, known as :
1.
2.
Constant-frequency operation or pulse-width modulation control
Variable-frequency operation or control by frequency modulation
With pulse-width modulation control, the regulation of output voltage is achieved by varying the duty cycle of
the switch, keeping the frequency of operation constant. Duty cycle refers to the ratio of the period for which the
power semiconductor is kept ON to the cycle period. Usually control by pulse width modulation is the preferred
method since constant frequency operation leads to optimization of LC filter and the ripple content in output
voltage can be controlled within the set limits. On the other hand, if the load on the converter is below a certain
level, voltage regulation of output becomes a problem and in such a case, control by frequency modulation is to
be preferred.
When control by frequency modulation is to be achieved, the ON period of the power semiconductor switch is
kept constant and the frequency of operation is varied to effect voltage regulation. Design of LC filter is not
easy in such a case.
If a micro-controller is used instead of a specific PWM IC, it is possible to switch from one mode of control to
the other depending on the load conditions.
BASIC CIRCUIT OPERATION
The operation of the buck converter is explained first. This circuit can operate in any of the three states as
explained below. The first state corresponds to the case when the switch is ON. In this state, the current through
the inductor rises, as the source voltage would be greater than the output voltage, whereas the capacitor current
may be in either direction, depending on the inductor current and the load current. When the inductor current
rises, the energy stored in it increases. During this state, the inductor acquires energy.
When the switch is closed, the elements carrying current are shown in red colour in Fig. 6, whereas the diode is
in gray, indicting that it is in the off state. In Fig. 6(a), the capacitor is getting charged, whereas it is discharging
in Fig. 6(b).
FIG. 6(a):
FIG. 6(b):
Fig. 6.: Buck Converter : First State
The equations that govern the operation of the circuit in the first state are shown below.
The second state relates to the condition when the switch is off and the diode is ON. In this state, the inductor
current free-wheels through the diode and the inductor supplies energy to the RC network at the output. The
energy stored in the inductor falls in this state. In this state, the inductor discharges its energy and the capacitor
current may be in either direction, depending on the inductor current and the load current Figure 7 illustrates the
second state.
FIG. 7(a):
FIG. 7(b):
Fig. 7.: Buck Converter : Second State
The equations that govern the operation of the circuit in the second state are shown below.
When the switch is open, the inductor discharges its energy. When it has discharged all its energy, its current
falls to zero and tends to reverse, but the diode blocks conduction in the reverse direction. In the third state, both
the diode and the switch are OFF and Fig.8 illustrates the third state. During this state, the capacitor discharges
its energy and the inductor is at rest, with no energy stored in it. The inductor does not acquire energy or
discharge energy in this state.
FIG. 8:
Fig. 8.: Buck Converter : Third State
The equation that governs the operation of the circuit in the third state is shown below.
When the circuit receives a periodic signal, the response of the circuit also becomes periodic. Here it is assumed
that the source voltage remains constant with no ripple, and the frequency of operation is kept fixed with a fixed
duty cycle. If the RC time constant due to the load resistor and the filter capacitor is very large compared to the
cycle period of the switching frequency, the output voltage is more or less constant, with no noticeable ripple.
When both the input voltage and the output voltage are constant, the current through the inductor rises linearly
when the switch is ON and it falls linearly when the switch is OFF. Under this condition, the current through the
capacitor also varies linearly when it is getting charged or discharged.
The responses obtained for a particular set of parameters are displayed in Fig. 9. The values of parameters used
are:
Source voltage = 100 V dc,
Switching frequency = 20 kHz,
L = 500 µH,
C = 500 µF,
R = 10 Ω, and
duty cycle = 0.5.
The value of 1 in a voltage plot in Fig. 9 corresponds to 100 V and the value of 1 in a current plot corresponds to
10 A. Figure 9 displays the responses over one cycle.
Fig.9: Periodic Response with a dc voltage Input
An expression for the average output voltage can be obtained as follows. It is assumed that there is continuous
conduction in the inductor. Given that the cycle period is T, the ON-period is DT, and the source voltage is E,
In eqn. (6), D stands for the duty cycle. The same expression for output voltage can be obtained in another way.
When the responses in the circuit are periodic, the inductor current is the same at the beginning and end of a
cycle. That is,
Equation (7) can be expressed as follows:
When the switch is ON, vL(t) = E - Vo,avg and when the diode is conducting,
vL(t) = - Vo,avg. Therefore
On evaluation,
The change in inductor current when the switch is On can be determined as follows. Let the change in inductor
current be ∆I, as shown in Fig. 10. In this figure, the change in output voltage has been exaggerated for sake of
clarity. When the switch is ON, the voltage across the inductor can be expressed as:
When the output voltage remains steady at Vo,avg, the inductor current linearly during the ON period of the
switch. Then
During the ON-period, the inductor current rises from (Vo,avg/R - ∆I/2) to (Vo,avg/R + ∆I/2). That is,
The capacitor current iC is expressed as follows:
Now an assumption is made to find out the change in output voltage. It is assumed that the capacitor gets
charged for half of the cycle period and gets discharged during the other half , as shown in Fig. 11. Since the
current through the capacitor varies linearly, the average charging current is half of its peak value of the
triangular waveform. The peak value of its triangular waveform is shown to be ∆I/2. Hence
If a periodic signal has zero dc value over its cycle period, its average is defined based only on its positive part
and hence the average capacitor current is obtained as shown above. For a capacitor
Based on the average charging current and half of the cycle period as the charging period, we get the change in
output voltage ∆V as:
Using equation (12), the above equation can be expressed as:
Assuming that the ripple in output voltage is sinusoidal, the rms value of ripple content in output voltage is:
Note that the variation in output voltage is not shown to be sinusoidal in Fig. 10.. Even though the variation
appears to be triangular, equation (20) gives a better approximation of the rms value of ripple content in output
voltage.
Given that source voltage = 100 V dc, switching frequency = 20 kHz, L = 500 µH, C = 500 µF, R = 10 Ω, and
duty cycle = 0.5, the results obtained are:
Vo,avg = 50 V,
∆I = 2.5 A,
∆V = 31.25 mV, and
Vrms,ripple = 11.05 mV.
In order that the capacitor current and the inductor current vary linearly, it is necessary that the RC time constant
should be relatively large, equal to about four or five times the cycle period. When the RC time constant is much
smaller than the cycle period, the responses obtained are not linear. To illustrate, Fig. 12 displays another set of
responses. The only change is that a 1 µF capacitor is used in place of the 500 µF capacitor.
When the RC time constant is small, the output voltage contains noticeable ripple. In addition, the ripple in
output voltage appears to be sinusoidal, justifying the equation used for finding out the rms value of the ripple
content in output voltage. Another aspect can be noticed in Fig. 12. It is that the ripple in output voltage goes
through its negative half-cycle when the switch is ON. When this circuit is to be controlled by negative
feedback, the feedback at the ripple frequency would become positive and closed-loop control effected without
taking this aspect into account would produce a larger ripple in the output.
Fig.12: Periodic Response with a dc voltage Input
Transient Response
Fig. 13 Transient Response
Let us assume that the output voltage is zero and that there is no current through the inductor at start. If the
source voltage is connected suddenly and the switch is turned ON and OFF at a fixed frequency with a preset
duty cycle, the transient response of the circuit lasts for several cycles before it settles down to periodic
response. The inrush current through the inductor is quite high, several times the maximum current that can flow
under settled conditions. The transient response obtained over the first 10 ms with source voltage = 100 V dc,
switching frequency = 20 kHz, L = 500 µH, C = 500 µF, R = 10 Ω, and duty cycle = 0.5, is shown in Fig. 13 .
Even after 200 cycles, the response is still in the transient state.
Effect of ripple in input voltage
Usually the input to an SMPS happens to be an unregulated dc voltage provided by a rectifier-filter circuit. Such
a filter contains significant ripple content at double the line frequency. The first applet in this page illustrates the
response obtained when the input voltage contains ripple. In this program, the ripple frequency, the peak-topeak ripple and the source voltage can be set. If the source voltage = 100 V, peak-to-peak ripple voltage = 20 V,
and the ripple frequency = 100 Hz, then the input voltage falls from 110 V to 90 V in the first 7.5 ms and rises
from 90 V to 110 V in the remaining 25% of the input cycle period. The input voltage falls linearly in the
initially for 75% of the cycle period from (E + Vrip,pk-to-pk/2) to (E - Vrip,pk-to-pk/2) and then rises linearly during
the remainder of the input cycle period.
When the input voltage varies cyclically, the response of the circuit is periodic over its input cycle period and it
is not periodic not over the period corresponding the switching frequency. In addition, there is significant
overshoot in the inductor current when the input voltage is rising linearly. It can also be seen that the output
voltage has nearly the same waveform as the input voltage, which is only to be expected. Since the duty cycle is
kept fixed, the output voltage would tend to rise as the input voltage rises. The response obtained with a peak-topeak ripple voltage of 20 V at 100 Hz is shown in Fig. 14. It becomes clear from the response that closed-loop
control is necessary to maintain the output voltage when the input voltage has some ripple content. The closedloop control circuit has to be designed with care, since the duty cycle has to be continually varied to maintain
the output voltage at its set value.
Fig. 14: Periodic Response over One Input Cycle Period
Effect of Step Change in Load
When there is a step change in load, the circuit goes through transient response before it settles back to periodic
response. Here the circuit is allowed to be a settled state with a load resistance of 10 Ω. Then the load resistance
is changed to 5 Ω and the transient response that is obtained is presented in Fig. 15. In Fig. 15, '1' on the axis for
corresponds to 10 A.
When the load resistor was 10 Ω, the load current would have been 5 A given that the duty cycle is 0.5 and the
input voltage is 100 V. It can be seen that after the step change in load resistor, the output voltage dips first and
then recovers. If the change is in the other direction from 5 Ω to 10 Ω, the output voltage rises first before
falling back, as shown in Fig. 16. For this figure, '1' on the axis for corresponds to 20 A.
Effective closed-loop control would reduce the transients. The under-damped nature of inductor current
response can be improved.
Fig. 15: Effect of Step Load Change: Increase in Load
DISCONTINUOUS OPERATION
When the load resistor becomes high, the buck converter circuit operates in the discontinuous mode. The
inductor current falls to zero when the switch is open and remains at zero till the switch becomes ON in the next
cycle. Such a mode is the desirable mode of operation when variable frequency of operation is employed for
voltage control. Additionally, discontinuous operation puts less stress on the diode and the circuit operates well.
This aspect will be explained later.
The effect of load resistance on the inductor current is shown in Fig. 17. Given that the conduction is
continuous, the average load current is Vo,avg/R. As the load resistance increases, the average load current
decreases, but on the other hand, the change in inductor current as defined by equation (12) remains constant.
As indicated by equation (13), the inductor current remains within the range specified below.
Fig. 16: Effect of Step Load Change: Reduction in Load
When the inductor current is continuous, the value of Vo,avg is as defined by equation (10) and it is not so if the
conduction is discontinuous. After substituting for ∆I from equation (12), we get that
Hence a critical resistance, say RC can be defined that makes the circuit operate at boundary of continuous
conduction, a situation illustrated in Fig. 17. When the load resistance is higher than the conduction in inductor
is discontinuous and it is continuous if the load resistance is less than RC .
It can be seen from equation that at the boundary of continuous and discontinuous conduction,
Discontinuous operation is illustrated in Fig. 18. In Fig. 18, the duty cycle for the switch is defined to D1, and
the duty cycle for the diode is defined to D2. It means that in each cycle, the switch conducts for a time interval
equal to D1T and the diode conducts for a time interval equal to D2T, where T is the cycle period. It can be seen
that for discontinuous conduction
At the boundary of continuous and discontinuous conduction, (D1 + D2) = 1. Given that the discontinuous, there
is no current in the inductor for a time interval equal to (1 - D1 -D2)T.
The average load current is the average of the inductor current in Fig. 18. From Fig. 18,
Substituting for ∆I from equation (24), we get that
Solving for D2,
When R > RC, D2 can be obtained from equation (27), assuming that D1, f and L are known. Since the average
of the inductor voltage over a cycle is zero, we obtain from Fig. 18 that,
Then
Substituting for D2 from equation (27),
The second applet shows how D2 and Vo,avg vary when R > RC.
The first applet is shown below. It has a pull-down list of items, of which the only first item is displayed. On
clicking on the downward arrow, the list is displayed. If any item of the list is high-lighted, its default value is
displayed in the window adjacent to the label with the caption Set Value.. If one scrolls down or up the list, the
default value of the item high-lighted is displayed in the window adjacent to the label with the caption Set
Value. To change the default value of an item, highlight it first and then click inside the text-field window
adjacent to the label with the caption Set Value. A line cursor would appear and the value can be changed. After
changing the value, click on the Set Value label and then the program recognizes the change. It can be verified
that the program has made the change out by inspecting the pull-down list again.
The default values of items in pull-down list are:
Input Voltage,dc avg = 100
Input Ripple Volt., pk-pk = 0
Input Ripple Frequency = 100
Switching Frequency,kHz = 20
Inductance, microHenry = 500
Capacitor, microFarad = 500
Load Resistance, Ohms = 10
Duty Cycle= 0.5
Slow Response,0-200 = 0.
It is possible to see one of the four responses:
Periodic response over one output cycle,
Transient response over one output cycle,
Periodic response over one input cycle, and
Transient response over one input cycle.
Initially, the program sets '1' on the voltage axis of the plots equal to 100 V and sets '1' on the current axis to 10
A. If periodic response over one output cycle is chosen, the program assumes that the peak-to-peak ripple in
input voltage is zero. The response can be slowed by varying the parameter called Slow Response. This
parameter is effective only for periodic or transient response over one output cycle.
If the load resistance is changed from 10 Ω to 5 Ω, still '1' on the current axis would correspond to 10 A. If the
Reset button is clicked, the program would make changes so that '1' on the voltage axis of the plots equal to
input voltage and sets '1' on the current axis to (input voltage/load resistance).
When Start/Continue button is clicked, the program responds to the visible response type and starts with the
values of inductor current and the capacitor voltage it had from the previous calculation. Initially they are set to
zero values. If the Reset button is clicked, the program would make them zero again.
FIRST APPLET
The second applet takes in three parameters, namely the switching frequency, the inductor value and the duty
cycle and plots the duty cycle of the diode and the output voltage as a function of the load resistance varying
from RC to (11 × RC).
SECOND APPLET
CONTROL BY PULSE-WIDTH MODULATION
The pulse-width modulator controls the semiconductor switch and is a complex part of a switched regulator.
Nowadays, switched regulator uses a pulse-width modulator integrator circuit. The principle of control by pulse-
width modulation is illustrated in Fig. 19. The simplified functional diagram of a typical pulse-width modulator
is shown in Fig.19a, whereas the waveforms in Fig. 19b explain the operation.
The pulse-width modulator circuit consists of a saw-tooth generator, an error amplifier, and a comparator. The
frequency of saw-tooth generator can usually be set by choosing proper values of an RC network. The error
amplifier compares the reference voltage and the feedback signal. The feedback signal is obtained using a
voltage divider network across the output of the SMPS circuit. For example, let the feedback signal be Vf and
the reference voltage be Vref. Then
The output of the error amplifier is compared with the saw-tooth waveform and when this voltage is greater than
the output of sawtooth generator, the output of the comparator would be at logic '1'. When the output of
comparator is at logic '1', the switch in the SMPS circuit can be kept in the ON state. When the comparator is at
logic '0', the switch in the SMPS circuit can be kept in the OFF state.
If the output voltage tends to be greater than that indicated by equation (32), the output voltage of the error
amplifier would fall and the duration for which the output of comparator remains at logic '1' would decrease.
Thus the duty cycle of the switch reduces and the output of the SMPS would fall, according to equation (10).
Thus it can be seen that the negative feedback control maintains the output at the desired value. For negative
feedback control, the feedback signal should be applied to the inverting input of the error amplifier.
When control by frequency modulation is desired, the ON-period is kept constant, but the frequency is varied in
order to bring about regulation. Such a technique is necessary if the load on the regulator tends to become very
low. It is difficult to make the ON-period below a certain time duration and when this limit is reached, control
by pulse width modulation becomes impossible. Then the duty cycle is reduced by keeping the ON period fixed
and increasing the cycle period. The value of minimum ON period depends on the transistor switch.
CLOSED-LOOP CONTROL
In order to regulate the output voltage, a controller is needed to be designed. For this purpose, the power circuit
is first represented by a transfer function.
The passive part of the power circuit is shown in Fig. 20. Since the DC output of the SMPS is defined by
equation (10), the effective input to the passive circuit in Fig. 20 can be stated to be E.D(s), whereas D(s) is the
Laplace transform of the output of the error amplifier shown in Fig. 19. In the design procedure outlined below,
D(s) is taken to be the output of the controller, varying between 0 and 1.
For the circuit in Fig. 20,
where
The block diagram of the SMPS with a PI controller is shown in Fig. 21. The transfer functions of the different
blocks in Fig. 21 are as follows:
The PI controller has two inputs, the reference signal Vref corresponding to the desired output voltage and the
feedback signal Vfdb. The error signal, the difference, e, between Vref and Vfdb is fed to the PI controller and the
output of the PI controller is D(s), the signal that sets the duty cycle. Then
In equation (35), K is the proportional gain and T is the integrating time constant. The output of the PI controller
varies between 0 and 1 and has two limits, one corresponding to the lowest duty cycle and the other
corresponding to the highest duty cycle.
There is a filter in the feedback path. The filter time constant is about 4 times the cycle period corresponding to
the switching frequency of the SMPS. Then the ripple at the switching frequency gets filtered out and the
feedback signal is essentially a dc signal. Otherwise, the ripple in the output at the switching frequency has
nearly 180o phase shift with respect to the fundamental of the input square pulse to the SMPS and then feedback
at switching frequency becomes positive, which in turn leads to amplification of the ripple content. The filter
avoids this problem, without seriously affecting its dynamic performance.
The transfer function of the power circuit is obtained as follows. Since the output varies linearly with the duty
cycle, equation (33) can be presented as follows:
For designing the closed loop system, the poles due to the source can be ignored. Since the power circuit is very
much under-damped, derivative feedback is required for stable operation. The differentiating circuit provides
the feedback signal. For simulation, the derivative signal is obtained based on the capacitor current. The
derivative feedback voltage is set to be kd.iC, where iC is expected to be its per unit value.
The third applet presented below allows the user to design a suitable PI controller. The parameters to be set are:
switching frequency in kHz, gain of the PI controller, its integrating time-constant , the inductance, the
capacitance, the load resistance, the time constant of the filter in the feedback path and the derivative
coefficient.
THIRD APPLET
The feedback arrangement used for actual simulation contains additional circuitry. It is necessary to limit the
inductor current since it is seen from the open loop response that the inductor current can be several times the
rated current, where the rated load current is the nominal current rating of the SMPS. It is a parameter that can
be set in the program.. In this program, the current limit is set at 1.5 times the rated load current, and when the
inductor current exceeds the current limit, the current in excess of the current limit is amplified and added to the
feedback signal. In addition, a derivative feedback signal, obtained using the capacitor current, is added to the
feedback signal. Derivative feedback improves damping of the system and without it, the system is oscillatory in
spite of a well-designed PI controller. While designing the PI controller using the third applet, the switching
circuit is replaced by a linear amplifier and this approximation is necessary for the design of the PI controller.
But in reality, the square-wave voltage input to the power circuit makes the system to be oscillatory since the
power circuit is heavily under-damped. The feedback arrangement used is shown in Fig. 22.
When the derivative feedback coefficient is set at unity, the derivative feedback amounts to 10% of the rated
voltage when the capacitor current equals the rated current.
The fourth applet that is presented below simulates the circuit. It has a pull-down list of items, of which the only
first item is displayed. On clicking on the downward arrow, the list is displayed. If any item of the list is highlighted, its default value is displayed in the window adjacent to the label with the caption Set Value.. If one
scrolls down or up the list, the default value of the item highlighted is displayed in the window adjacent to the
label with the caption Set Value. To change the default value of an item, highlight it first and then click inside
the text-field window adjacent to the label with the caption Set Value. A line cursor would appear and the value
can be changed. After changing the value, click on the Set Value label and then the program recognizes the
change. It can be verified that the program has made the change out by inspecting the pull-down list again.
The default values of items in pull-down list are:
Input Voltage,dc avg = 100
Input Ripple Volt., pk-pk = 0
Input Ripple Frequency = 100
Switching Frequency,kHz = 20
Inductance, microHenry = 500
Capacitor, microFarad = 500
Load Resistance, Ohms = 10
Output Voltage set at = 50
Gain of the PI controller = 5.
Time-constant of the PI controller in µs =500
Time-constant of the Filter in µs =40
Derivative Feedback Coefficient = 1.0
Rated Current in Amps = 10
Slow Response,0-200 = 0.
It is possible to see one of the five responses:
Periodic response over one output cycle,
Transient response over one output cycle,
Periodic response over one input cycle,
Transient response over one input cycle and
Statistics
Initially, the program sets '1' on the voltage axis of the plots equal to 100 V and sets '1' on the current axis to 10
A. If periodic response over one output cycle is chosen, the program assumes that the peak-to-peak ripple in
input voltage is zero. The response can be slowed by varying the parameter called Slow Response. This
parameter is effective only for periodic or transient response over one output cycle.
If the load resistance is changed from 10 Ω to 5 Ω, still '1' on the current axis would correspond to rated current.
If the Reset button is clicked, the program would make changes so that '1' on the voltage axis of the plots equal
to input voltage and sets '1' on the current axis to (input voltage/load resistance).
When Start/Continue button is clicked, the program responds to the visible response type and starts with the
values of inductor current and the capacitor voltage it had from the previous calculation. Initially they are set to
zero values. If the Reset button is clicked, the program would make them zero again.
FOURTH APPLET
MICRO-CONTROLLER IMPLEMENTATION
Nowadays the use of a micro-controller is popular. But control of a circuit switching at 20 kHz tends to be
difficult, essentially due to delays involved in A/D conversion and some mathematical operations. Hence the
approach to closed-loop control is a combination of rule-based logic and action of a PID controller. The delay
that occurs from the instant the A/D converter samples an input signal to the instant when the corresponding
digital output is available for use tends to be of the same order as the switching cycle period.
At 20 kHz, the switching cycle period is 50 µS and the delay involved due to the use of an A/D converter is
about 40 µS. This means that no more than one sample can be obtained during one switching cycle period. If
this sampling is done in an asynchronous manner, closed-loop operation can be difficult. For example, the
inductorcurrent can vary considerably over a cycle period and asynchronous sampling of inductor current may
not lead to stable closed-loop control. It is preferable to sample the inductor current at a predetermined instant
of the cycle period. For example, at the start of a switching cycle, a sample and hold circuit can be used to
store the instantaneous value of inductor current and the A/D converter can convert this value. In the fifth applet
displayed below, the A/D converter delay is set to be equal to one switching cycle period. When that is so, the
sampling occurs at the beginning of each cycle. The inductor current and the capacitor voltage are sampled
once in two cycles, since conversion period for each sample equals one switching cycle period.
If the sampling is to be done at a faster rate, the sampling has to be asynchronous and then an analogue filter has
to be used to reduce the variations in the feedback signal corresponding to inductor current before it is used as
the input to the A/D converter. Alternatively, fast A/D converters such as
flash A/D converter with the sampling period of the order of 1 µs can be used.
The scheme suggested for controlling the SMPS with a micro-controller is outlined with the help of a pseudocode presented below.
Initialize:
Output Voltage Count = 0 // Output of A/D converter
Inductor Current Count =0 // Output of A/D converter
Old Current Count = 0
// Previous current count required for derivative feedback
swCycleCount = 100
// Set the period of switching frequency as a number of micro-controller clock frequency.
atod_delay =100
// Set the A/D converter's conversion period and computing delay as a number of // micro-controller clock frequency.
lowCount= 5;
highCount=95;
dutyCycle=lowCount;
// Set the limits for duty cycle and set the duty cycle at its lowest value
on_off=true;
// boolean value indicating switch is ON when it is true
rampCount=0;
// Used with dutyCycle and swCycleCount for setting on_off to either true or false value
atod_Count=0;
feedback Voltage=0
integrator Output=0
//Output of an integrator used in feedback control
Go to Main Loop
Main Loop:
Call Parameters Subroutine
Increment ramp Count
If (rampCount<dutyCycle) on_off = true;
Else on_off = false;
If (rampCount == swCycleCount)
{
on_off = true
rampCount=0;
}
Increment atod_Count
Increment rampCount
If (atod_Count>atod_delay) call NextValues subroutine
Return to Main Loop
Parameters Subroutine:
Begin
Desired Output Voltage=50.0
// numerical setting using thumb-wheel switches/KeyPad
Current Limit = 125%
// Set in software, using KeyPad or thumb-wheel switches
kd=20
// Derivative feedback coefficient in Ohms.
// Feedback in Volt equals (kd × Inductor Current).
dkI=0.1
// Integrating Coefficient effective when output voltage is within
// a close band of ± 5% of source voltage from the desired output voltage.
End subroutine
NextValues Subroutine
Begin
atod_Count=0;
Inductor Current Count = output of A/D output
Ouput Voltage Count = output of A/D output
if (Inductor Current Count <=Current Limit)
{
feedback Voltage= Ouput VoltageCoun +kd*( Inductor Current Count - Old Current Count);
if (Desired Output Voltage>(feedback Voltage+5%)) dutyCycle++;
if (Desired Output Voltage<(feedback Voltage-5%)) dutyCycle--;
if (Desired Output Voltage>(feedback Voltage-5%)) AND (Desired Output Voltage<(feedback Voltage+5
{
integrator output+=
dkI*(Desired output Voltage-Ouput Voltage Count)
dutyCycle+= integrator output;
if (dutyCycle<=lowCount) dutyCycle=lowCount
if (dutyCycle>=lowCount) dutyCycle=highCount
}
else
{
integrator output =0;
}
else
{
dutyCycle = dutyCycle-2;
if (dutyCycle<=lowCount) dutyCycle=lowCount;
}
End Subroutine
The fifth applet presented below simulates the operation of the switch mode step-down power supply controlled
by a micro-controller. The applet has default values for the parameters listed below.
Input Voltage,dc avg = 100
Input Ripple Volt., pk-pk = 0;
Input Ripple Frequency = 100
Inductance, microHenry = 500;
Capacitor, microFarad = 500 ;
Load Resistance, Ohms = 10;
Desired Output Voltage = 50;
Rated Current, Amp = 10;
Micro's Clock Freq, MHz = 2 ;
Switching Period:Clock Cycles = 100;
A/D Delay :Clock Cycles = 100;
Derivative FeedBack Coef. =20;
Integrating Coefficient = 0.1;
This applet has the same structure as the fourth applet and is hence not described any further.
FIFTH APPLET
IMPLEMENTATION USING A MULTIPLIER
Implementation using a multiplier is an attractive option since this scheme can deliver nearly desired output
voltage in spite of fluctuations in the source voltage. The scheme to be used is shown in Fig. 23.
The inputs to the multiplier are the source voltage or a part thereof and the output of the opamp in the feedback
loop. Since the output of the multiplier should equal Vref representing the desired output voltage, the output of
the opamp reflects the duty cycle. This signal can be compared with the output of a sawtooth generator and a
pulse to turn on the switch can be obtained. If Vref can be modified to include the effects of drop in the switch,
the output of the SMPS can equal the set value. The sixth applet simulates the SMPS controlled in this manner.
SUMMARY
This page has described the operation of a switch mode step-down power supply circuit. For description in this
page, it has been assumed that the components used in the power circuit are ideal, whereas they are not ideal in
reality. The next page goes into details of the practical limitations of the power circuit components.
INTRODUCTION
DIODE REVERSE RECOVERY CURRENT
TURN-ON TRANSIENT IN A DIODE
TRANSIENT PROCESSES IN A MOSFET
LOSSES IN ENERGY STORAGE ELEMENTS
CONTROL BY PWM
CURRENT MODE CONTROL: IDEAL CIRCUIT
SUMMARY
INTRODUCTION
The previous page on the step-down switch-mode power supply circuit was described and
analyzed assuming that the components used are ideal. The real diode and the MOSFET have
some limitations and they have to be taken into account for designing a practical power supply.
First, the characteristics of the diode are outlined first.
DIODE REVERSE RECOVERY CURRENT
Even if a fast-recovery or an ultra-fast recovery diode is used for free-wheeling operation, the
reverse-recovery characteristic of the diode imposes some constraints. The circuit that can be
used to test the reverse recovery characteristic of a diode is shown in Fig. 1. An applet, named
the first applet, simulates the reverse recovery property of the diode and is displayed below.
The operation of the test circuit is explained at first. The time constant due to load inductance
and load resistance should be several times the time corresponding to the switching frequency.
Then the current through the remains more or less steady and its value is given by:
When the MOSFET is ON, the load current flows through the transistor. When it is turned off,
the current flows through the snubber circuit initially and then the free-wheeling diode, marked
as DUT (device under test). When the MOSFET is switched ON again, the source tends to
reverse the current through the diode. The current through the diode falls rapidly and reverses
before it becomes zero. The reverse recovery has been simulated using the model of the diode
shown in Fig. 1.
The reverse-recovery transient process depends on the diode itself and it also depends on the
junction temperature of the diode, the forward current prior to being reverse biased, the rate of
fall forward current and the source voltage that applies to the reverse bias to the diode. As the
value of any one of these parameters rises with the other parameters remaining unchanged, the
reverse recovery transient process becomes worse, reflected by an increase in the peak reverse
current. The reverse recovery turn-off period starts once the diode current becomes negative
and lasts till the reverse diode current increases in ampltidue first and then decays to about
10% of its reverse peak current. It represents the time that has to elapse before the diode
recovers ability to block reverse voltage.
The manufacture specifies the reverse recovery transient period and the maximum reverse
current for a set defined by its forward current, the rate of fall of forward current and the
junction temperature. For many diodes, the maximum rate of fall forward current is specified
to be 100 A/µs. This means the test inductance L should set be such that
If reverse recovery transient is to be minimized, it would be preferable to select L such that
where the rated current of diode is Irated,diode and diF/dt is the permissible ,axi,u, rate of fall of
diode current. It is difficult to select an inductor based on trr, its reverse recovery transient
period, because the inductor chosen based on the diF/dt rating of the diode is more
conservative. For example, the data sheet for a diode with 30 A rating may have its diF/dt
rating to be 100 A/µs and its turn-off time at 100o C of junction temperature may be specified
to be 150 ns. If the diode is carrying 30 A when it is suddenly reverse-biased, a time interval
of 300 ns is required for its current to fall to zero and then a further 150 ns
would need to elapse before the diode is off. It is seen why the inductor is to be selected based
on the diF/dt rating of the diode. It is better that the inductor restricts the rate of fall of diode
current to a value lower than its diF/dt rating in order to restrict the reverse-recovery current.
The applet simulates the reverse recovery transient. It is an approximation of the behaviour of
the diode. No accuracy for this model is claimed, but it does show the dependence of recovery
transient on the parameters mentioned. The model developed is a mathematical model, based
purely on the turn-off time, load current, rated current, junction temperature, and the maximum
rate of fall of current of the diode. No physical model has been developed, as is the common
practice. The process that occurs within a diode during turn-off and turn-on is a complex one
and it is difficult to develop a physical model that is stable as well as realistic. On the other
hand, the mathematical model is easy to develop. When the MOSFET is turned on, the current
through the diode falls, its rate of fall restricted by the inductor. In the process, diode current
reduces and becomes negative. The peak reverse current that can occur is calculated based on
the load current, the rate of fall of diode current,and the junction temperature. Once this peak
reverse current is reached, the diode voltage builds up, depending on the reverse voltage rating,
the junction temperature and the turn-off time and the diode current can be calculated once the
diode voltage is determined.
The pull-down choice menu contains rated current of the diode, its turn-off period, its turn-on
period, load current, source voltage, test inductance in nH, test resistance in Ohms and the
junction temperature as its items. To change the value of a menu-item, pull down the menu,
highlight the item, change its value in the adjacent text-field and then click on Set Value
button. When the Run/More button is clicked, the program displays the diode current and the
voltage across the diode for a period corresponding to 12 times the turn-off period. If the diode
had not turned off by then, click again on the Run/More button, and the process that occurs for
the next 12 times the turn-off period would be displayed. Whenever the rate of fall of forward
current increases due to either lower test inductance or higher source voltage, the total period
required for the diode to turn-off decreases, but there would a significant increase in the peak
reverse current. In the plots shown by the applet, unit on the y-axis represents the rated current
for the top plot and it represents the rated reverse voltage for the bottom plot.
APPLET FOR REVERSE RECOVERY TRANSIENT IN A DIODE
TURN-ON TRANSIENT IN A DIODE
The turn-on transient in a diode is not as predominant as its turn-off process is, but nonetheless
it affects the performance of the circuit in which it is placed. The circuit for simulating the turnon transient process is shown in Fig.2 and the diode model used is the same as shown in Fig. 1.
In a practical circuit, it is often difficult to notice the turn-on transient process, for two reasons.
Firstly, the turn-on process lasts only for a brief period. Secondly, the transient process of some
other device can often mask the turn-on process of a diode. For example, the turn-off delay of
the MOSFET can mask the turn-on process of the diode. The simulation, shown in the second
applet, makes the assumption that the transistor is ideal and simulates the turn-on process in an
approximate manner. It is necessary to have a snubber capacitor across the FET as shown in
Fig. 2. Again a mathematical model is used. When the FET is turned off, the snubber
capacitor voltage builds up to the source voltage due to load current charging it. After that, the
diode current rises at a predetermined rate, depending on the rated current, the turn-on period
and the junction temperature. After the diode current becomes equal to the load current, the
diode voltage exponentially decays, based on its turn-on period. It is assumed that the diode
current remains constant.
In some circuits, such as the flyback converter to be described in one of the pages to follow,
the turn-on delay of the diode is crucial. The only practical remedy to address this problem is
to connect an RC snubber circuit across the diode.
APPLET FOR TURN-ON TRANSIENT IN A DIODE
TRANSIENT PROCESSES IN A MOSFET
For a MOSFET, three transient processes can be identified. The first is the turn-on transient
process associated with the MOSFET and the second is its turn-off process and the third is the
turn-off transient process associated with the body-diode of the MOSFET. The turn-on
transient of the body diode is relatively fast and can be ignored.
For a fast MOSFET, the turn-on transient process is characterized by two time periods, one is
the turn-on time delay and the second is the rise time. During the turn-on time delay, the gateto-source voltage builds up to its threshold value and during the rise time, the device current
rises to about 90% of its final value. There is a further delay before the drain-to-source voltage
becomes equal to its conduction drop. The turn-on delay time can be reduced to some extent by
a stiff gate drive signal. When the MOSFET is turned off, there is a delay period corresponding
to the period in which the gate voltage reduces to its threshold level. It is followed by the crossover period that includes a delay period and a fall period. During the delay period, the drain-tosource voltage rises from its conduction value to its blocking value. After this delay period has
elapsed, the fall period follows during which the current through the MOSFET decreases. For a
fast MOSFET, the total turn-on transient lasts for about 50 ns, whereas the turn-off process
lasts for about 100 ns.
The turn-on characteristic of a MOSFET is shown below. The voltage and the current
associated with the device can change in this manner and the datasheets display a similar turnon characteristic. When the MOSFET is placed in a circuit, the rise in current and the fall in
voltage get modified due to the influence of the external components.
The turn-off characteristic of a MOSFET is shown below. The voltage and the current
associated with the device can change in this manner and the datasheets display a similar turnoff characteristic. When the MOSFET is placed in a circuit, the fall in current and the rise in
voltage get modified due to the influence of the external components.
The body diode is comparatively slow. It turns on quite fast, but its turn-off process is quite
slow, of the order of 500 ns. Hence when the MOSFET is to be operated at high frequency, it is
preferable to use an external diode. In such a case, an additional diode may be required, that
has to be connected in series with the MOSFET.
A circuit that can be used for simulating the transient processes in a MOSFET is shown in Fig.
3. Here Q1 is turned on and off, whereas Q2 remains off. When Q1 is turned off, the load
current is diverted through diode D2, the body diode of Q2 . When Q1 is turned on, both the
turn-on transient of Q1 and the reverse recovery transient of D2 occur. The simulation that is
displayed as applet 3 is again quite approximate. The purpose is to illustrate how MOSFET
functions as a switch. An air-cored inductor, labeled L3 in Fig.3, is necessary to be used to
reduce the reverse recovery current of the diode. The time constant due L3 and R3 should be
much less compared with the cycle period corresponding to switching frequency. The turn-on
process of the MOSFET is quite slow, due to the slow turn-off of the body diode D2. In
applications where the MOSFET has to be switched on and off at higher frequency of the order
of 20 kHz and above, it is the practice to bypass the body diode by an external diode. In this
application, such a technique is unnecessary because the body diode does not have to conduct
at all.
When the applet for simulating the dynamic characteristics of a MOSFET is run, the
waveforms produced may appear to be bizarre. When the applet is run with the default values
to simulate the turn-on process of the MOSFET, the waveforms produced are presented below.
The logic behind the waveforms is as follows. When the MOSFET is turned on by applying a
gate pulse,
theere is a turn-on delay. During this period, the voltage across the FET remains equal to the
source voltage and there is no current through the FET. Then the rise period follows, during
which the FET current rises and the FET voltage is determined by the external circuit. Here
when the current the FET rises, inductor L3 absorbs the voltage and the voltage across the FET
is zero. The current through the FET equals the load current well before the rise period
elapses and hence the voltage across the FET becomes equal to the source voltage once the rate
of rise FET current is zero. At the end of the rise period, the body diode of Q2 contains still a
large number of charge carriers and it is predisposed to conduct in the reverse direction. As the
reverse recovery current increases, the FET voltage falls in some piecewise linear fashion.
There is some behind the simulation, even though its accuracy is questionable.
Now the waveforms obtained for the turn-off process are explained.
When the MOSFET is turned off, there is a storage delay period during which both the current
and the voltage of the FET do not change. After that, the FET voltage rises, with its current
remaining unchanged. Then the FET current falls. As the current falls, the voltage across L3
changes polarity and the FET voltage is the sum of the source voltage and the voltage across
the inductor L3.
The losses in a FET are due to three factors:
1.
2.
3.
Conduction losses due to on-state voltage,
Turn-on losses and
Turn-off losses.
The fourth applet simulates the step-down SMPS with PWM control in the open loop. The aim
of this simulation is to estimate the losses in the switch and the diode and to show how the
modifications to the power circuit lead to reliable operation, whether the conduction is
continuous or discontinuous.
LOSSES IN ENERGY STORAGE ELEMENTS
The non-ideal properties of filter capacitor at the output and the inductor affect the
performance of the circuit. The inductor is not lossless because of the winding resistance and
the core loss. The model of the inductor can be changed to a series network containing a
resistor reflecting its losses and an inductor. Due to the losses in the inductor, there is a slight
reduction in the output voltage and also a drop in the efficiency of the circuit. Given a duty
cycle and a steady input voltage, the output voltage is has been obtained in the previous page
as:
The average inductor current is the same as the average load current. Then the average inductor
current IL is:
If the internal resistance of the inductor is Rind, the drop in output voltage is approximately:
The effect of the ESR of capacitor is to increase the ripple content in output. From the
previous page, the change in capactor current from its higheest value to its lowest value is:
The worst-case increase in peak-to-peak ripple in output voltage can be obtained by just adding
the ripple due to ESR with the previously obtained value. If ESR of capavitor be Rcap, then
The actual peak-to-peak ripple would be less than the value stated above.
CONTROL BY PWM
It is necessary to modify the step-down SMPS if it is to operate at relatively high voltage in the
continuous conduction mode. The problem with continuous conduction occurs when the
MOSFET is turned on with the diode still in conduction. To explain this aspect, Fig. 4 is
presented. When the MOSFET is turned on, the diode can act as a short circuit till it recovers.
To overcome this problem, the circuit in Fig. 4 is to be modified.
The modifed circuit is presented in Fig. 5. This circuit, when designed properly, would well
whether the conduction is continuous or discontinuous. This circuit contains a few additional
components. The components added are an additional diode marked as D1 and an RC
dissipating circuit for D1. Diode D3, inductor L2 and capacitor C2 are the same components,
marked as D, L and C respectively in Fig. 4. The operation of the circuit in Fig. 5 is explained
now. The circuit in Fig. 5 does not show the snubber circuit required for the diodes D1 and D2
and possibly MOSFET. Some MOSFETs do not need a snubber circuit, whereas a diode
needs a snubber circuit. There is a separate page on design of snubber circuit.
The conduction path that exists when the MOSFET is ON is shown in red colour in Fig. 6. The
current flow is through the MOSFET and the inductors.
When the MOSFET is turned off, the set of components in conduction varies. In mode 1
following immediately after the turn-off of the MOSFET, diode D1, inductor L1, capacitor C1,
resistor R1 and components L2 C2 and RL are in conduction. The RC circuit in series with D1
conducts, the time constant associated with C1 and R1 being very small compared with the onoff periods of the MOSFET. During this mode, inductor L1 discharges its energy to R1 and
C1. The components in conduction in mode 1 are shown in Fig. 7.
When mode 1 is over after inductor L1 has discharged its energy, mode 2 follows. Here only
D2 continues to conduct, but L1 and D1 are not in conduction. This mode is illustrated in Fig.
8. During this phase, C1 would discharge any energy it may have acquired into R1. After the
end of the period corresponding to the switching frequency, the MOSFET is turned ON again
and the circuit reverts to the state shown in Fig. 9.
The circuit lasts in the state shown in Fig. 9 till the current through L1 becomes equal to that
through L2. After that, the circuit reverts to the state shown in Fig. 6.
When the MOSFET is switched on, current through L1 rises gradually and current through D2
falls gradually. If inductor L1 is sufficiently large, there would be hardly any reverse recovery
transient due to diode D2. Typically L1 should be such that the rate of rise current is less than
50% of the maximum rate of fall specified for diode D2. Inductor L1 can even be an air-core
inductor, whereas inductor L2 has a ferrite-core with an air gap.
The complete power circuit along with the required snubber circuits is shwon in Fig. 10. The
design of power circuit is described in a separate page. The snubber circuit for the MOSFET
may not be required if a MOSFET that does not need a snubber is chosen.
The fourth applet presented below simulates the circuit in Fig. 5. It is assumed that the turn-on
delay of diodes is negligible.The pull-down menu contains two items that are not shown
explicitly in Fig.5. They are the equivalent-series resistance of capacitor C2 and the internal
resistance of inductor L2. In practice, even with a fixed duty cycle, a fixed switching
frequency and a steady-input voltage, the ripple content in output tends to be higher than the
calculated value, mainly due to the ESR of C2. The default value of ESR of C2 is set to be
zero. To see its effect, the type of response should be set as Statistics. Then when the program
is run, the peak-to-peak ripple in output voltage is displayed. The program has to be run a few
times in this mode before the peak-to-peak ripple in output settles down to its periodic value.
For example, when ESR of C2 is increased to 0.1 Ω , it can be seen that the peak-to-peak
ripple that results is much higher.
The internal resistance of L2 reflects the winding resistance and the core loss and the losses in
inductor L2 tend to reduce the efficiency of this converter. The default value of internal
resistance has been set to be zero. Its realistic value can be calculated by assuming a quality
factor of about 100 at the switching frequency. For the value of inductor L2, an appropriate
value of internal resistance of L2 can be set to be about 0.5 Ω. To see its effect, the type of
response should be set as Statistics in the program.
When the type of response is Statistics, the program should be run a few times before the
results become repetitive. Even then, the sum of output power and all the losses may not add
upto the input power, due to errors in modelling and the presence of energy storage elements.
The loss calculation of diode D1 in particular is quite incorrect, because the turn-on losses of a
diode have been ignored. Even though the average current of D1 turns out to be small, its turn-
on losses are quite significant and it is advisable to use the same diode selected to be used as
D2. These diodes should be fast-switching diodes with a low reverse period of about 50 ns. It
is likely that there will be some time delay before the selected response is displayed. In the
meantime, the message that computing is going on will be flashed on the screen.
CURRENT MODE CONTROL: IDEAL CIRCUIT
Current mode control of a SMPS is a popular technique built into several PWM integratedcircuits nowadays. This topic should have been covered in the previous page, but has been held
over and presented now.
The block diagram of the current-mode control is similar to that described the previous page on
the ideal step-down SMPS circuit. The PI controller illustrated in Fig. 22 of the previous page
can used be as it is. In the previous page, closed-loop control was effected by comparing the
output of the PI controller with a ramp signal (Fig. 19). For current-mode control, the output of
the PI controller should be compared with the signal reflecting the current though the inductor
L2, with this signal suitably scaled. This signal is compared with the PI controller output and
when the signal corresponding to inductor current tends to exceed the PI controller output, the
MOSFET is turned off for the rest of the output cycle. At the start of each cycle, the MOSFET
should be turned on.
With just this type of control, the pulse-width tends to fluctuate from one cycle to the next,
leading to oscillations in output. Ideally duty cycle should be equal to the ratio of desired
output voltage to the source voltage, but the intersection of the PI controller output and the
signal reflecting inductor current may not always occur after a time lapse from the start of a
cycle that corresponds to the desired duty cycle. To overcome this problem, a component
proportional to the time elapsed from the instant the cycle starts can be added to the signal
corresponding to the inductor current and this sum can be compared with the output of the PI
controller. Let us say that the maximum of the PI controller output be 10 V and let the signal
corresponding to inductor current be 10 V when the inductor current is at nominal rated value.
Let also a ramp voltage be generated such that it rises from 0 V to 10 V from the start to the
end of a cycle. Then a fraction of the ramp voltage can be added to the signal reflecting
current, the fraction being equal to the duty cycle which in turn is the ratio of desired output
voltage to the source voltage. When the source voltage tends to vary over an input cycle, this
fraction computed as the ratio of desired output voltage to the source voltage automatically
gets adjusted and the correction to the signal reflecting the inductor current becomes the right
adjustment.
INTRODUCTION
ANALYSIS OF THE IDEAL CIRCUIT
CONTINUOUS CONDUCTION
DISCONTINUOUS CONDUCTION
CLOSED-LOOP CONTROL USING PWM
IMPLEMENTATION USING A MULTIPLIER
CURRENT-MODE CONTROL
MICRO-CONTROLLER IMPLEMENTATION
SUMMARY
INTRODUCTION
The boost converter, also known as the step-up converter, is another switching
converter that has the same components as the buck converter, but this converter
produces an output voltage greater than the source. The ideal boost converter has the
five basic components, namely a power semiconductor switch, a diode, an inductor,
a capacitor and a PWM controller. The placement of the inductor, the switch and the
diode in the boost converter is different from that of the buck converter. The basic
circuit of the boost converter is shown in Fig. 1.
The operation of the circuit is explained now. The essential control mechanism of
the circuit in Fig. 1 is turning the power semiconductor switch on and off. When the
switch is ON, the current through the inductor increases and the energy stored in the
inductor builds up. When the switch is off, current through the inductor continues to
flow via the diode D, the RC network and back to the source. The inductor is
discharging its energy and the polarity of inductor voltage is such that its terminal
connected to the diode is positive with respect to its other terminal connected to the
source. It can be seen then the capacitor voltage has to be higher than the source
voltage and hence this converter is known as the boost converter. It can be seen that
the inductor acts like a pump, receiving energy when the switch is closed and
transferring it to the RC network when the switch is open.
When the switch is closed, the diode does not conduct and the capacitor sustains the
output voltage. The circuit can be split into two parts, as shown in Fig. 2. As long as
the RC time constant is very much larger than the on-period of the switch, the output
voltage would remain more or less constant.
When the switch is open, the equivalent circuit that is applicable is shown in Fig. 3.
There is a single connected circuit in this case.
ANALYSIS OF THE IDEAL CIRCUIT
Analysis of the circuit is carried out based on the following assumptions. The circuit
is ideal. It means when the switch is ON, the drop across it is zero and the current
through it is zero when it is open. The diode has zero voltages drop in the
conducting state and zero current in the reverse-bias mode. The time delays in
switching on and off the switch and the diode are assumed to be negligible. The
inductor and the capacitor are assumed to be lossless.
1.
2.
3.
4.
5.
The responses in the circuit are periodic. It means especially that the inductor
current is periodic. Its value at the start and end of a switching cycle is the
same. The net increase in inductor current over a cycle is zero. If it is non-zero,
it would mean that the average inductor current should either be gradually
increasing or decreasing and then the inductor current is in a transient state and
has not become periodic.
It is assumed that the switch is made ON and OFF at a fixed frequency and let
the period corresponding to the switching frequency be T. Given that the duty
cycle is D, the switch is on for a period equal to DT, and the switch is off for a
time interval equal to (1 - D)T.
The inductor current is continuous and is greater than zero.
The capacitor is relatively large. The RC time constant is so large, that the
changes in capacitor voltage when the switch is ON or OFF can be neglected
for calculating the change in inductor current and the average output voltage.
The average output voltage is assumed to remain steady, excepting when the
change in output voltage is calculated.
The source voltage VS remains constant.
Inductor Current with Switch Closed
When the switch is closed, the equivalent circuit that is applicable is shown in Fig. 2.
The source voltage is applied across the inductor and the rate of rise of inductor
current is dependent on the source voltage VS and inductance L. The differential
equation describing this condition is:
If the source voltage remains constant, the rate of rise of inductor current is positive
and remains fixed, so long as the inductor is not saturated. Then equation (1) can be
expressed as :
The switch remains ON for a time interval of DT in one switching cycle and hence
DT can be used for ∆t. The net increase in inductor current when the switch is ON
can be obtained from equation (2) to be:
Inductor Current with Switch Open
When the switch is open, the circuit that is applicable is shown in Fig. 3. Now the
voltage across the inductor is:
Given that the output voltage is larger than the source voltage, the voltage across the
inductor is negative and the rate of rise of inductor current, described by equation
(5), is negative. Hence if the switch is held OFF for a time interval equal to (1 - D)T,
the change in inductor current can be computed as shown in equation (6)
.
The change in inductor current reflected by equation (6) is a negative value, since Vo
> VS. Since the net change in inductor current over a cycle period is zero when the
response iL(t) is periodic, the sum of changes in inductor current expressed by (4)
and (6) should be zero. That is,
On simplifying equation (7), we get that
It has been stated that when iL(t) is periodic, the net change in inductor current over
a cycle is zero. Since change in inductor current is related to its volt-seconds, the net
volt-seconds of the inductor has to be zero. The expression for the net volt-seconds
can be obtained from equation (7) and it can be seen that the numerator of equation
(7) should be zero. That is,
The value of D varies such that 0 < D < 1 and it can be seen from equation (8) that
output voltage is greater than the source voltage, and hence this circuit is called the
boost converter. The output voltage has its lowest value when D = 0 and then the
output voltage equals the source voltage. When D approaches unity, output voltage
tends to infinity. Usually D is varied such that 0.1 < D < 0.9 .
The waveforms of inductor voltage and inductor current are shown in Fig. 4. These
waveforms are drawn assuming that both the output and the source voltage remain
steady. These waveforms illustrate how the inductor voltage is related to its current.
Output Voltage Ripple with Switch Closed
In this sub-section, the change in output voltage is calculated. It needs to be
emphasized that the peak-to-peak ripple in output voltage is quite small for a welldesigned circuit. For the inductor, the net change in inductor current over a cycle is
zero when iL(t) is periodic. For the capacitor, the net change in capacitor voltage
over a cycle is zero when it is periodic. When the switch is closed, the equivalent
circuit in Fig. 2 shows that the boost converter is split into two sub-circuits, with the
loop currents decoupled from each other. When the switch is closed, the output
voltage is sustained by the capacitor. During this period, the capacitor discharges
part of its stored energy and it re-acquires this energy when the switch is open.
When the switch is open, part of the inductor current charges the capacitor since the
inductor current usually remains larger than the current through the load resistor.
From Fig. 2,
When current through a capacitor charges it up, its rate of rise of capacitor voltage is
positive since the capacitor voltage is increasing. When the switch is open, the
capacitor is discharging its energy with its voltage falling and the current through the
capacitor is then a negative value. The output voltage remains positive and hence the
output current is positive and it is the negative of the capacitor current, as can be
seen from Fig. 2. Since the change in output voltage is quite small, it can be assumed
that the load current remains constant at its average value and equation (10) can be
now expressed as:
When the capacitor current is constant, its voltage changes linearly with time. Here
the period for which the switch is closed is DT and the DT can be used in place of
∆t. The peak-to-peak ripple in output voltage expressed as ∆vo and it is then
expressed as:
Equation (12) yields the value of the peak-to-peak ripple in output voltage. In
equation (12), 1/f replaces T since T is the reciprocal of switching frequency.
Figure 5 shows how the capacitor current and voltage vary over a cycle. The ripple
in output voltage is exaggerated in Fig. 5, whereas in practice it would be much
smaller. If the output voltage is drawn to scale, the ripple in output voltage would
not be noticeable.
Expression for Average Inductor Current
The average inductor current can be found out by equating the power drawn from
the source to the power delivered to the load resistor. Again the ripple in output
voltage is ignored and it is assumed justifiably that the output voltage remains steady
at its average value. Power Po absorbed by load resistor is then:
It can be seen from the circuit in Fig. 1 that the current drawn from the source flows
through the inductor. Hence the average value of inductor current is also the average
value of source current. Let the average inductor current be IL. Then power PS
supplied by the source is then:
After equating equations (13) and (14), we get the average inductor current as:
Since load current Io is:
Using equations (8) and (16), equation (15) can be re-presented as:
Since 0 < D < 1, it can be seen from equation (17) that IL > Io.
CONTINUOUS CONDUCTION
The analysis thus far is based on the assumption that the current through the inductor
is continuous. The inductor current varies over a cycle, varying between a minimum
value and a maximum value. The minimum and maximum values can be expressed
in terms of its mean value and its change as expressed in equation (3). That is,
and
It is shown in Fig. 6 how the maximum and the minimum inductor current can be
obtained. It is also shown that as the load resistor becomes greater, the average
inductor current reduces, but the peak-to-peak ripple in inductor current does not
change. It has to be so and expression for ∆IL in equation (3) does not indicate any
term reflecting the load resistor.
For continuous conduction,
At the boundary of continuous and discontinuous conduction,
Another expression for IL is now obtained. Substituting for Vo in equation (15) the
expression in equation (8), we obtain that
Substituting for IL from the equation above and for ∆iL from equation (3), equation
(18) becomes:
and
From equations (23) and (24), the condition for continuous conduction is:
Equation (25) can be interpreted as follows, assuming that only one of the four
parameters is varied at a given time with the other three parameters remaining
unchanged.
The circuit tends to become discontinuous,
i.
ii.
iii.
iv.
if the switching frequency f is decreased, or
if the duty cycle D is reduced, or
if the load resistance increases, or
if the inductance used has lower value.
DISCONTINUOUS CONDUCTION
When the conduction is discontinuous, the voltage across the inductor is zero for
part of the cycle since there is no current through the inductor. Let D1T be the time
for which the switch is ON in one cycle and let D2T be the period for which the
diode conducts. Since the conduction is discontinuous,
An expression for the output voltage can be obtained in terms of source voltage,
duty cycle D1 of the switch and duty cycle D2 of the diode. Since the net change in
inductor current is over a cycle, the net volt-seconds area associated with the
inductor is zero. The waveforms relevant to the inductor when the conduction is
discontinuous are shown in Fig. 7. From Fig. 7,
On simplifying, an expression for Vo can be obtained. Then
The value of D1, the duty cycle of the switch, is usually known, but the period for
which the diode conducts is an unknown quantity depending on the other circuit
parameters. The value of D2 can be determined in several ways. Here it is
determined using the power balance between the input and output. When the circuit
is ideal, the input power equals output power. Let the average source current be IS
and the average output current be Io. Then
Using equation (28), we get that
The average source current be IS can be obtained from Fig. 7. The average source
current is the same as the average inductor current. Let the peak inductor current be
∆IL and the period for which this current flows is (D1T + D2T). This period is the
base of the triangle that defines the inductor current. The average inductor current is
obtained as the area of this triangle divided by the cycle period. We have that
Equating equations(30) and (31),
From equation (3),
Substituting for ∆IL from equation (33) in equation (32), we get that
Equation (34) can be re-written as:
Solving for D2,
Equation (36) states how D2 varies as a function of R, D1 , f and L. Once D2 is
known, Vo can be obtained from equation (28).
It is possible to get an expression for Vo as a function of R, D1 , f and L. For this, we
equate the average load current with the average diode current. The average output
current can be obtained from the average output voltage and the load resistor. The
average diode current is:
Using the expression for ∆IL from equation (33), and replacing the L.H.S. by the
average load current,
Hence we obtain that
By substituting for D2 from equation (36) in the above equation, we can get an
expression for Vo/VS. Alternatively, equation (28) can be re-written as:
Using the expression for D2 from equation (39) in equation (40),
That is,
Solving for the ratio of output to source voltage and taking the positive root of the
expression on the R.H.S. of equation (42),
Equation (43) states how (Vo/VS) varies as a function of R, D1 , f and L
Two applets are presented below, the first applet simulates the behaviour of the ideal
circuit in open loop, whereas the second applet is about the discontinuous mode of
operation.
The first applet presents four types of responses. When the circuit is switched on at a
fixed duty cycle with no energy stored in either the inductor or capacitor initially,
the transient inductor current happens to be large. If the input voltage has any ripple
content, its effect can be seen by selecting either the periodic response or the
transient response over one input cycle.
The second applet displays two sets of curves. The first set illustrates how the ratio
of output voltage to input voltage varies as a function of the ratio of load resistance
to critical resistance for different duty cycles of the switch, where the critical
resistance is calculated from equation (25). The critical resistance at a given duty
cycle can be stated to be:
The second set of curves illustrates how the duty cycle of the diode varies as a
function of the ratio of load resistance to critical resistance for different duty cycles
of the switch. The values of critical resistance at various values of duty cycle are
also displayed. The parameters to be set initially are the frequency of operation and
the value of inductor.
CLOSED-LOOP CONTROL USING PWM
Under this section, the closed-loop control of the ideal circuit is considered. For both
the buck converter and the boost converter, the output voltage increases as the duty
cycle of the switch increases and hence the same PWM circuit can be used. The
block diagram is also similar, but the transfer function is only slightly different. The
block diagram is shown in Fig. 8.
A more detailed block diagram is shown in Fig. 9.
The transfer function of the circuit in open loop is obtained as outlined below by
using an approximate method. Even though it is difficult to justify the
approximation, it is presented. The output voltage is expressed by equation (8). Let
the quiescent duty cycle be assumed to be D and let the small increment to duty
cycle be d. Then equation (8) can be presented as:
If d<< (1-D), then the above equation can be approximated using the binomial
expansion and retaining only the first two terms. That is,
The change in output, denoted as ∆Vo, can be expressed to be
The transfer function can then be obtained as:
The design of the controller is based on the block diagram in Fig. 10. The third
applet has been developed based on this block diagram. The difference between the
inductor current and the load current is the capacitor current and the current through
the capacitor is proportional to the derivative of the capacitor voltage and the
derivative feedback is obtained as shown. . The third applet presented below
displays the location of poles and zeros for the selected parameters. If any poles
happen to be in the right-half side of s-plane, the program may not work
For simulation of the power circuit using PWM, closed loop scheme used is shown
in Fig. 9. As long as the inductor current is less than the maximum current, the
feedback signal is the sum of the output voltage of the filter circuit and a derivative
signal obtained from the inductor current and the load current. When the inductor
voltage exceeds the set maximum value, the feedback signal contains an additional
element proportional to the amount by which the inductor current exceeds the
maximum current. The output of the PI controller determines the duty cycle.
The simulation of the power circuit with closed-loop control and pulse-width
modulation mode is presented in the fourth applet. One of the five responses can be
selected and the program interface is similar to many of the preceding applets. For
all the applets to follow, transient responses are displayed without any delay,
whereas there can be quite a delay before the periodic responses or the statistical
details are displayed.
IMPLEMENTATION USING A MULTIPLIER
It is possible to use a multiplier for boost converter also. Using equation (8), an
expression for the duty cycle can be obtained to be:
The above equation is valid if the conduction is continuous. The above equation can
be realized using a multiplier and a few opamps as shown in Fig. 11. It needs to be
mentioned that components should be chosen suitably, reflecting the required
scaling. For example, if the multiplier yields an output of 10 V when both its inputs
are at 10 V, an input of 10 V should correspond to the maximum input voltage. If the
maximum duty cycle is set to be 0.9 and it corresponds to 9 V, then an input of 9 V
to the opamp in the feedback path should correspond to the maximum desired output
voltage. In this case, the ramp signal with which the output of this circuit is
Power Electronic Circuits
Section 1 - Introduction to Power Electronics
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Objectives
History and Applications of Power Electronics
Power Semiconductor Devices
Ratings and Characteristics of Power Devices
Control Characteristics of Power Devices
Classification of Power Semiconductor Switching Devices
Types of Power Electronic Circuits
Design of Power Electronic Equipment
Peripheral Effects
Section 2 - Power Semiconductor Diodes
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Objectives
Introduction
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The Junction Diode
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V-I Diode Characteristics
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Reverse Recovery Characteristics
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Power Diodes Types
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Effects of Forward and Reverse Recovery Time
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Series Connected Diodes
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Parallel Connected Diodes
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SPICE Diode Model
Section 3 - Diode Circuits and Rectifiers
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Objectives
Introduction
Freewheeling Diodes
Laplace Transform used to solve equations - Revision
Performance Parameters of Rectifiers
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Single Phase Half-Wave Rectifier (R Load)
❍
Example 3.1
Single Phase Half-Wave Rectifier (RL Load)
❍
Example 3.2
Single Phase Full Wave Rectifiers - Center Tapped Transformer
❍
Example 3.3
Bridge Rectifier
Solving Exact and Non-Exact First Order Linear Differential Equations - Revision
Three Phase Bridge Rectifier
❍
Example 3.4
Rectifier Filter Circuits
❍
Example 3.5
❍
Example 3.6
❍
Example 3.7
❍
Example 3.8
Effects of Source Inductance
Section 4 - Thyristors
●
Objectives
Introduction
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Thyristor Characteristics
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Two - Transistor Model of Thyristor
●
Thyristor Turn-on
●
di/dt Protection
dv/dt Protection
❍
Example 4.1
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Thyristor Turn-Off
●
Thyristor Types
●
Series Operation of Thyristors
❍
Example 4.2
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Parallel Operation of Thyristors
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Thyristor firing circuits
Section 5 - Controlled Rectifiers
●
●
Objectives
Introduction
Single Phase Half-Wave Thyristor Converter With R Load
❍
Example 5.1
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Single Phase Semiconverters
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Single-Phase Semiconverter With RL Load
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Single-Phase Full Converter
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Single-Phase Full Converter with RL load
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Three-Phase Half-Wave Converter
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Three-Phase Semiconverter
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Three-Phase Full Converter
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Fourier Transform - Revision
Section 6 - AC Voltage Controllers
●
Objectives
Introduction
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On-off Control
●
Single-Phase Unidirectional Controller
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Single-Phase Bidirectional Controller With R Load
●
Single-Phase Bidirectional Controller With RL Load
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Single-Phase Transformer Tap Changers
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Three-Phase Full-Wave Controller
Power Electronic Circuits
Section 1 - Introduction to Power Electronics
Objectives
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●
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Learn the scope and application of power electronics
Become familiar with different types of power semiconductor devices, their
characteristics and use
Understand the fundamental techniques of power conversion
History & Application of Power Electronics
History of Power Electronic Devices
Power electronics began with the introduction of the mercury arc rectifier in 1900.
This was followed by the first electronic revolution which began in 1948 with the
invention of the silicon transistor.
The second electronic revolution began in 1958 with the development of the
thyristor. This caused the beginning of a new era for power electronics, since many
power semiconductor devices and power conversion techniques were introduced
using thyristors.
Next, was the microelectronics revolution which gave the ability to process a huge
amount of data in a very short time. The power electronics revolution which
merges power electronics and microelectronics provides the ability to control large
amounts of power in a very efficient manner.
Definition of Power Electronics
Power Electronics may be defined as the application of solid-state electronics for
the control and conversion of electric power. Power Electronics is based on the
switching of power semiconductor devices whose power handling capabilities and
switching speeds have improved tremendously over the years.
Power Electronics is presently playing an important role in modern technology and
is used in a variety of high power products e.g. Motor controls, heat controls, light
controls and power supplies.
Power Electronic Circuits
Section 1 - Introduction to Power Electronics
EE33D Page
Power Semiconductor Devices
Power semiconductor devices can be broken up into five different groups:
1. Power Diodes
2. Thyristors
3. Power Bipolar Junction Transistors (BJT)
4. Power MOSFETs
5. Insulated Gate Bipolar Transistors (IGBT) and Static Induction Transistors
(SITs)
Power Diodes
Diode Operation
A diode is a two terminal device consisting of an anode and a cathode. The diode
conducts when its anode voltage is more positive than that of the cathode. If the
cathode voltage is more positive than its anode voltage, the diode is said to be in
the blocking mode.
Diode Types
There are three types of power diodes:
1. General Purpose
2. High-Speed (or fast recovery) - used for high frequency switching of power
converters
3. Schottky - have low on state voltage and very small recovery time, typically
nanoseconds
Thyristors
Thyristor Operation
A thyristor is a three terminal device consisting of an anode, a cathode and a gate.
It is physically made up of four layers of alternate p-type and n-type silicon
semiconductor. The terminals connected to the ending p-type and the n-type layers
are the anode and cathode respictively. This configuration will give three p-n
junctions. When the anode is held more positive than the cathode, two of the p-n
junctions are foward biased, offering very little resistance, and one is reverse
biased, offering high resistance.
When a small current is passed through the gate to cathode circuit, and the anode
is at a higher potential than the cathode, the thyristor conducts current from anode
to cathode. In other words when triggered the thyristor has aproximately the same
characteristics as a single diode. Once the thyristor has been turned on, the gate
circuit looses control of the thyristor and the forward voltage drop across the device
is very small in the region of 0.5 to 2V.
Once on, the device loses control over the anode current, and the only way to turn
it off is to reduce the anode current below some value referred to as the holding
value. This can be acheived in one of two ways; by making the anode potential
equal or less than the cathode potential, due to the sinusoidal nature of an ac
voltage which is called line commutation or by the use of an auxilixry as in the case
of forced-commutation.
Thyristors can be subdivided into eight groups:
1. Forced-commutated thyristors
2. Line-commutated thyristors
3. Gate turn off thyristors (GTO) - self turned-off device.
4. Reverse conducting thyristors (RCT) - can be considered as a thyristor with an
inverse parallel diode, used for high speed switching.
5. Static induction thyristors (SITH) - self turned-off device.
6. Gate assisted turn-off thyristors (GATT) - used for high speed switching.
7. Light activated silicon-controlled rectifier (LASCR)
8. MOS-controlled thyristors (MCTs)
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●
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Triacs can be considered as two thyristors connected in inverse
parallel and having only one gate terminal. The current flow through
the triac can be controlled in either direction. They are used in simple
heat, light and motor controls
GTOs and SITHs are self turned-off devices. They are turned on by
the application of a short positive gate pulse and turned-off by a
short negative gate pulse and do not require a commutation circuit
Because of the nature of the construction of a thyristor, there exists
some capacitance between the anode and the gate. If a sharply
rising voltage is applied to the thyristor, the associated inrush of
charge can switch on te thyristor. These surges can be the result of
switching in circuits, and can be accomadated for by providing RC
circuits for diverting this surge.
All p-n junctions have a leakage current which increases with
increasing temperature. If the temperatture of operation of a
thyristor were allowed to rise too much, the leakage current could
rise enough to turn on thr thyristor. A possible use of this feature is
in the manufacture of a switching system, which needs to be turned
on exceeding certain temperatures.
Bipolar Power Transistors
These are three terminal devices consisting of emitter, base and collector which is
operated as a switch in the common emitter configuration. These devices are
turned-on when the base-emitter junction is forward biased with the base current
sufficiently large to drive the device into saturation. Under these conditions, the
collector-emitter voltage drops in a range of 0.5 to 1.5V. If the base-emitter
junction is reversed biased the device switches to the off or non-conducting state.
Power MOSFETs
These are three terminal devices consisting of gate, source and drain. It is a fast
switching device with a high input impedance and has applications in high switching
frequency circuits. MOSFETs are available in both N-channel and P-channel types.
To turn-on an N-channel device, the drain is made positive with respect to the
source and a small positive voltage is applied to the gate with respect to the
source. The device is turned-off when the gate voltage is removed
Power MOSFETs possesses faster switching speeds than power BJTs.
IGBTs
The insulated gate bipolar transistor (IGBT) is a three terminal device consisting of
gate, emitter and collector. It combines the low on-state voltage drop
characteristics of the BJT with the excellent switching characteristics and high input
impedance of the MOSFET. They are available in current and voltage ratings much
higher than those found in MOSFETs.
To turn-on the N-channel IGBT the collector must be at a positive potential with
respect to the emitter and a positive gate potential will turn-on the device. The
removal of this positive gate voltage would turn-off the device
Section 1 - Introduction to Power Electronics
Ratings & Characteristics of Power Devices
The power ratings of some semiconductors are shown in table 1.1, while table 1.2
displays the characteristics and symbols of some power devices
Table 1.1 - Ratings of Power Semiconductor Devices
Table 1.2 - Characteristics & Symbols of Some Power Devices
Control Characteristics of Power Devices
Power semiconductor devices can be operated as switches by applying control
signals to the gate of thyristors, base of power transistors to the gate of
power MOSFETs and to the gate of IGBTs. The required output voltage is
obtained by varying the conduction time of these devices. Figure 1.1 shows the
output voltage and control characteristics of a few power switching devices.
Figure 1.1 Output voltage and control characteristics of a few power switching
devices
Classification of Power Semiconductor Switching Devices
Power semiconductor devices can be classified as follows:
1. Uncontrolled turn on and off (Diode)
2. Controlled turn on and uncontrolled turn off (SCR)
3. Controlled turn on and off characteristics (BJT, MOSFET, GTO, SITH, IGBT, SIT
MCT)
4. Continuous gate signal requirement (BJT, MOSFET, IGBT, SIT)
5. Pulse gate requirements (SCR, GTO, MCT)
6. Bipolar voltage-withstanding capability (SCR, GTO)
7. Unipolar voltage-withstanding capability (BJT, MOSFET, GTO, IGBT, MCT)
8. Bidirectional current capability (TRIAC, RCT)
9. Unidirectional current capability (SCR, GTO, BJT, MOSFET, MCT, IGBT, SITH,
SIT, Diode)
Section 1 - Introduction to Power Electronics
Ronald De Four
Main Page
EE33D Page
Types of Power Electronic Circuits
The switching characteristics of power devices permit the control and conversion of
electric power from one form to another. These converters are called static power
converters and consist of a matrix of switches. Using a combination of these devices
allows us to create circuit configurations that allow us to convert between a.c. and d.c.
signals.
The resulting power electronic circuits are classified into six types:
1. diode rectifiers
2. ac-dc converters (controlled rectifiers)
3. ac-ac converters (ac voltage controllers)
4. dc-dc converters (dc choppers)
5. dc-ac converters (inverters)
6. static switches
Diode Rectifiers
A diode rectifier circuit converts ac voltage into a fixed dc voltage. Figure 1.2 shows a
diode rectifier circuit.
Figure 1.2 Single Phase Diode Rectifier Circuit
Since diodes have a much greater conductivity in one direction than the other, when
connected in series with an alternating supply and load, they will produce a direct
component of the current during one half cycle of the supply. This property is exploited in
the main application of diodes - as rectifiers.
In figure 1.2, the load is connected to the center tap of a transformer. The center tap is
such that during the positive phase of the input, the ouput of the secondary is divided
equally between the two halves. During this phase, the diode D1 is conducting while D2 is
non-conducting. So across the load we will get a positive half cycle.
During the negative half cycle, the diode D2 is conducting while the diode D1 is not, and
again we will get a positive half cycle across the load. In this way each diode conducts on
alternate cycles, passing current through the load in the same direction.
Ac-dc Converters
An ac-dc converter with two naturally commutated thyristors are shown in figure 1.3.
Figure 1.3 Single Phase ac-dc Thyristor Converter
The circuit shown is similar to the diode rectifier but with thyristors used in the place of
diodes. Recall that thyristors need to be fired to turn on. If each thyristor is fired
alternately the same effect would be obtained as with the diode rectifier. The average
output voltage is controlled by varying the conduction time or firing delay angle α of the
thyristors. These converters are also called controlled rectifiers.
●
Thyristors need to be fired to be turned on, and each one is fired at the
same point in the appropriate half cycle. This pont, or delay into the cycle,
is termed the firing angle.
Ac-ac Converters
These converters ore used to produce a variable ac output voltage from a fixed ac
source. Figure 1.4 shows a single phase converter using a TRIAC. The output voltage is
controlled by varying the conduction time of firing/delay angle α of the TRIAC. These
converters are also known as ac voltage controllers.
Figure 1.4 Single Phase ac-ac TRIAC Converter
Dc-dc Converters
A dc-dc converter is also known as a chopper or switching regulator. A transistor chopper
is shown in figure 1.5. The average output voltage is obtained by controlling the
conduction time t of transistor Q1. If T is the chopping period, then t1 = δT. δ is called the
duty cycle of the chopper.
Figure 1.5 Dc-dc Converter
Dc-ac Converters
A dc-ac converter is also known as an inverter. A single phase transistor inverter is
shown in figure 1.6. We need to switch the transistors on and off in pairs, which is why
we did not choose thyristors for this circuit. Transistors M1 and M2 conduct for one-half
cycle while M3 and M4 conduct for the other half cycle.
When transistors M1 and M2 are on the voltage VO appears across the load. When the
other pair of transistors is on the voltage appears across the load, but in the opposite
direction thereby producing an alternating output voltage waveform. The frequency of the
output voltage is controlled by varying the conduction time of the transistors.
Figure 1.6 Single Phase dc-ac Converter
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Between the time when one pair of transistors is on and the other pair is
on, there is a period during which both are off, and the output across the
load is zero.
Static Switches
Power devices can be used to operate as static switches or contactors, When the supply
to these switches is an ac supply, they are called ac static switches and when the supply
is dc, they are called dc static switches.
Section 1 - Introduction to Power Electronics
Design of Power Electronic Equipment
Power electronic equipment design can be divided into four main areas:
1. Design of power circuits
2. Protection of power devices
3. Determination of control strategy
4. Design of logic and gating circuits
Peripheral Effects
●
Power converters operate on the basis of switching power semiconductor
devices on and off. This switching action of converters introduce current and
voltage harmonics into:
1. The supply system
2. The output of converters
●
The problems caused by these harmonics are:
1. Distortion of the output voltage
2. Distortion of the supply voltage
3. Interference with communication and signaling circuits
4. Reduction of input power factor
●
The following methods can be used to solve or reduce harmonic problems
caused by power converters
1. Use of input and output filters on power converters
2. Choice of control strategy used
3. Grounded shielding
Section 2 - Power Semiconductor Diodes
Ronald De Four
Main Page
EE33D Page
Objectives
●
Identify different types of power semiconductor diodes
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Understand the importance of forward and reverse recovery times
●
Describe the methods of connecting diodes in series and in parallel for
applications that require higher voltages and currents
Introduction
Function of power semiconductor diodes in power electronic circuits:
1. Switches in rectifiers
2. Freewheeling in switching regulators
3. Charge reversal of capacitor & energy transfer between components
4. Voltage isolation
5. Energy feedback from load to power source
6. Trapped energy recovery
Section 2 - Power Semiconductor Diodes
Junction Diode
Recall that a n-type semiconductor is one in which there are free electrons and the
same number of fixed positive ions. Also recall that a p-type semiconductor is one
in which there are fixed negative ions and the same number of free moving holes.
Consider a crystal, one half of which has p-type impurity added, and the other half
has n-type. Initially the p-type semiconductor has free moving holes, and the ntype has free moving electrons, hence each region is initially neutral. Because of
random movement in nature and due to the difference in concentration of carriers,
some holes will diffuse across to the n-type material. Likewise some electrons will
diffuse across to the p-type material.
The movement of electrons and holes is only restricted to the area immediately
around the junction. Becasue of this some point will be reached when the area of
the p-type semiconductor closest to the junction will have a build-up of electrons
repelling the movement of any more electrons. Similarly, the area of the n-type
semiconductor closest to the junction will have a build up of holes repelling the
movement of more holes. These positive and negative charges are concentrated
near the junction and therefore form a potential barrier between the two regions.
Figure 2.1 pn-junction and diode symbol
Forward Biased Diode
Consider a potential difference applied across the diode, the positive of the supply
connected to the p-type material, and the negative connected to the n-type
material. Once this potential difference is greater than that created by the
concentration of holes and electrons at the junction, the orientation of the magnetic
field will produce a drift of holes towards the n-type conductor and electrons
towards the p-type conductor. At the junction, free electrons and holes will
combine. For each combination, at the p-type terminal, an electron is freed and
flows to the supply creating a free hole which moves towards the junction. Similarly
at the n-type terminal an electron enters the region from the supply and moves
towards the center junction.
With this in mind a diode is said to be forward biased or conducting when the
anode potential is positive with respect to the cathode. In this state, the diode has
a small forward voltage drop across it. The magnitude of this voltage drop depends
on:
-> the manufacturing process and
-> the junction temperature.
Reverse Biased Diode
When the cathode potential is positive with respect to the anode, the diode is said
to be reverse biased. In such a configuration, the holes will be attracted to the
negative electrode, and the electrons will be attracted towards the positive
electrode. This will create an area at the junction void of free holes or electrons.
This area is called the depletion layer in which there are no charge carriers to
facilitate current flow.
In practice, due to thermal agitation, some carriers build up sufficient velocity to
jump the gap. This causes a small reverse or leakage current. Since this current is
due to the effects of heat, the higher the temperature, the greater this leagake
current will be.
In addition, the magnitude of the reverse current increases in magnitude with
reverse voltage until the avalanche or zener voltage is reached.
Section 2 - Power Semiconductor Diodes
V-I Diode Characteristics
The steady-state V-I characteristics of a diode is shown in figure 2.2(a). In most
cases a diode can be considered to behave as an ideal diode with characteristics
shown in figure 2.2(b).
Figure 2.2 V-I characteristics of a diode
The practical characteristics shown in figure 2.2(a) can be expressed by the
Shockley diode equation, which is given by:
..............(2.1)
whereID = current through the diode in amperes
VD = diode voltage with anode positive w.r.t cathode in volts
IS =
15A
leakage (or reverse saturation) current, usually in the range 10-6 to 10-
n = empirical constant known as emission coefficient or ideality factor which
varies between 1 and 2 and depends on material used and physical
construction
VT = thermal voltage constant and is given by:
..............(2.2)
whereq = electron charge given as 1.6022 * 10-19 coulomb (C)
T = absolute temperature in kelvin K = 273 + ° C
k = Boltzmann's constant given as 1.3806 * 10-3 J/K
At a specified temperature, the leakagae current, Is is a constant for a specified
diode. In addition at a specified temperature, VT can be calculated and is also a
constant. Therefore for a given diode at a given temperature:
..............(2.1.a)
Examination of the diode characteristics in figure 2.2(a) reveals three distinct
regions:
1. Forward-biased region, where VD > 0
2. Reversed-biased region, where VD < 0
3. Breakdown region, where VD < -VZK
Forward-biased Region
In this region, VD > 0. The diode current ID is small if VD is less than the threshold
voltage or cut-in voltage or turn-on voltage written as VTD. This voltage is small
and is usually in the range 0.5V to 0.7V. The diode conducts fully if VD is higher
than VTD.
The diode equation can be simplified if VD > 0.1 volts
For VD = 0.1 V, n = 1 and VT = 25.8 mV, equation 2.1 can be used to
obtain the corresponding value of diode current ID.
with 2.1% error.
For VD > 0.1 volts which is usually the case,
ID >> IS
Hence the diode equation can be approximated to within 2.1% error to
..............(2.3)
Reverse-biased Region
In this region, VD < 0 volts and if |VD| >> VT, or in other words if the magnitude
of the diode voltage is much greater than the thershold voltage (which is generally
the case), the exponential term in equation 2.1 becomes very small compared to
unity and the diode current ID can be written as:
..............(2.4)
Equation 2.4 indicates that the diode current in the reverse direction is constant
and equal to IS.
Breakdown Region
When the reverse voltage exceeds the breakdown voltage VBR, the diode is said to
be in the breakdown region. In this region, the reverse current increases rapidly for
small increases in reverse voltage beyond VBR. Diode operation in the breakdown
region is not destructive, provided that the power dissipation is within a safe level
as specified by the manufacturer.
It is generally advisable to implement circuitry which will limit the
reverse current in the breakdown region. This is because in this region,
although it may be a non-destructive region, it should be noted that the
slightest change in VD would cause a large change in ID which could
damage the device.
Section 2 - Power Semiconductor Diodes
Reverse Recovery Characteristics
The current in a forward-biased junction diode is made up of Majority carriers and
Minority carriers. Once there is a forward current, there will be free minority
carriers. A forward conducting diode whose forward current has been reduced to
zero, will continue to conduct for some small time after due to minority carriers
stored in the pn-junction and carriers stored in the bulk semiconductor material.
The forward current in a diode goes to zero if the diode goes from foward biased to
reverse biased, or in other words V goes from +ve to -ve. According to the
characteristics of a diode, ignoring the leakage current, when reverse biased there
should be no reverse current once the reverse voltage does not exceed in
magnitude to the breakdown voltage. However, in practice, the diode does exhibit
a reverse characteristic for a short space of time due to the free carriers. These
minority carriers require some finite time, the reverse recovery time, to recombine
with opposite charges in order to be neutralized. This time is called the reverse
recovery time.
●
Two reverse recovery characteristics exists. They are:
1. Soft recovery
2. Abrupt recovery
as shown in Figure 2.3
Figure 2.3 Reverse recovery characteristics
trr = reverse recovery time, measured as the time between the initial zero crossing
of the diode current to the time when this current reaches 25% of the peak reverse
current.
IRR = maximum reverse current
ta = time between zero crossing and the maximum reverse current and it is due to
the charge stored in the depletion region of the junction
tb = time between maximum reverse current IRR and 25% of the of the maximum
reverse current IRR and is due to charge stored in the bulk semiconductor material
●
The reverse recovery time is measured from the initial zero crossing from
forward conduction to reverse blocking condition of the diode current to 25%
of the maximum reverse current IRR. Its magnitude depends on:
1. junction temperature
2. rate of fall of forward current
3. forward current prior to commutation
From the graph it can be seen that,
..............(2.5)
..............(2.6)
= softness factor (SF)
Reverse Recovery Charge
This is the amount of charge carriers that flow across the diode in the reverse
direction due to changeover from forward conduction to reverse blocking condition.
Its value is determined from the area enclosed by the path of the reverse recovery
current (Recall ∆ Q = ∆ I ∆ t). That is
..............(2.7)
..............(2.8)
From equations 2.6 and 2.8 we get
..............(2.9)
If tb is negligible in comparison to ta which is usually the case, then
Hence equation 2.9 becomes
..............(2.10)
and
..............(2.11)
Ideally diodes should not have a reverse recovery time, and it is possible to
construct such a diode. However, the manufacturing cost of such a diode would be
quite high for such a feature which in most cases has minor consequences.
Electrical Home
Section 2 - Power Semiconductor Diodes
Power Diodes Types
Power diodes can be classified into three categories depending on the recovery
characteristics. These types are:
1. Standard or general-purpose diodes
2. Fast-recovery diodes
3. Schottky diodes
General Purpose Diodes
These diodes have a generally high reverse recovery time, typically around 25 µs.
Used in low speed applications e.g. rectifiers and converters with frequencies up to
1 kHz, where the recovery time is not critical.
Fast-recovery Diodes
These diodes have a low recovery time, typically around 5 µs. They are used in
applications where the recovery time is of critical importance such as dc-dc and dcac converters.
Schottky Diodes
●
●
The barrier potential is accomplished with a contact between a metal and a
semiconductor. This barrier simulates the behavior of a pn-junction
Recovery charge of this diode is much less than the equivalent pn-junction
diode. It is due only to junction capacitance and is independent to reverse
di/dt
●
It has a relatively low forward voltage drop
●
The leakage current is higher than that of a pn-junction diode
●
They are mainly used in high current low voltage power supplies
Section 2 - Power Semiconductor Diodes
Effects of Forward and Reverse Recovery Time
Consider the chopper circuit shown below without a di/dt limiting inductor. This
circuit would be used to display the effects of forward and reverse recovery time.
Figure 2.4 Chopper circuit without di/dt limiting inductor
The switch SW is turned on at time t = 0 and remains on long enough for a steady
load current to flow. This load current is given by
..............(2.12)
Under these conditions, the freewheeling diode Dm is reverse-biased. Say switch
SW is opened at time
t = t1. Inductor L will now discharge its stored energy through diode Dm.
Switch SW is turned on again at time t = t2. Now at time t2, diode D1 would
conduct almost instantaneously, and diode Dm goes instantaneously from foward
baised to reverse biased. As discussed earlier, there would be free carriers which
would result in a reverse current. This reverse current, IRR, is directly proportional
to di/dt. And since the fall of foward current of Dm would be very high, this reverse
current would also be high, as seen from equation 2.13.
..............(2.13)
Since diode Dm would have free carriers, it would conduct in the reverse direction.
This would cause it to behave as a short in the circuit. This short would cause the
rate of rise in the foward current od D1 to be very high and consequently increate
the reverse current of Dm. This run-away effect could damage the diodes.
This problem can be overcome by connecting a current limiting inductor in series
with diode D1.
Section 2 - Power Semiconductor Diodes
Series-Connected Diodes
In high-voltage applications one commercially available diode is unable to meet the
required voltage ratings of a circuit. Connecting diodes in series increases the
reverse blocking capabilities of the diodes. Each diode must carry the same leakage
current, and have the same blocking voltage. However, in reality even two diodes
of the same part number will not have the same characteristics due to tolerances in
the production process. This is shown for two diodes of the same part number
connected in series in Figure 2.5, along with their v-i charactieristics. This gives rise
to problems when diodes are connected in series, since the blocking voltages will
differ slightly.
Figure 2.5 Two Series Connected Diodes in Reverse Bias
From the graphs it can be seen that in the forward-biased condition, both diodes
conduct the same amount of current and the forward voltage drop for each diode
would be almost equal. In the reversed-biased condition, however, where each
diode has to carry the same leakage current, the blocking voltage would differ
significantly as shown in figure 2.5b.
This problem is solved by forcing equal voltage sharing by connecting a resistor
across each diode as shown in figure 2.6.
Figure 2.6 Series-Connected Diodes with Steady-State Voltage Sharing
Due to the equal voltage sharing the leakage current of each diode would be
different as shown in figure 2.6b.
The relationship between the resistors for equal voltage sharing is developed below
In this arrangement, the total leakage current must be shared by a diode and a
resistor. Hence
..............(2.14)
but we need to get,
..............(2.15)
Now we know that,
and
Hence using equation 2.14 under conditions of equal voltage sharing yields
..............(2.16)
Hence the relationship between the two resistors for equal voltage sharing is given
by equation 2.16.
Section 2 - Power Semiconductor Diodes
Parallel-Connected Diodes
In high power applications, diodes are connected in parallel to increase the current
carrying capability in order to meet circuit requirements. In parallel operation of
diodes, current sharing depends on the magnitude of their forward voltage drops.
Uniform current sharing can be achieved either by the use of equal inductances or
by connecting current sharing resistors, the later option may not be practical due to
power losses incurred by the resistive components. Selecting diodes with equal
forward voltage drops would minimise the unequal sharing of current.
Figure 2.7 Parallel-Connected Diodes
For steady-state current sharing, the circuit of figure 2.7a with series resistors are
used.
Dynamic current sharing is achieved with the use of coupled inductors as indicated
by figure 2.7b. If the current through diode D1 rises, then the voltage across
inductor L1 (Recall VL = L di/dt) increases, causing a voltage of opposite polarity to
be induced across inductor L2. This causes a low impedance path for current flow
through diode D2, and more current is shifted through this diode.
The disadvantage of using current sharing devices under dynamic conditions is that
the inductors would generate voltage spikes and would be expensive and bulky.
Section 2 - Power Semiconductor Diodes
SPICE Diode Model
The SPICE model for a diode is shown in figure 2.8 below.
Figure 2.8 SPICE Model for Diode
●
●
●
The diode current ID depends on the diode voltage VD and is represented by a
current source.
Resistance RS is the bulk resistance of the semiconductor material and
depends on the amount of doping.
Capacitance Cd is a nonlinear function of the diode voltage vd and is
represented by
where qd is the depletion layer charge.
Section 3 - Diode Circuits And Rectifiers
Objectives
●
●
Understand the importance of freewheeling diodes in power electronic circuits.
Examine the harmonic distortion of voltage and current on the load and supply
caused by diode rectifiers.
●
Examine the performance parameters of rectifiers.
●
Select the ratings of diodes for rectifier circuits.
●
Select the parameters of simple input and output filters for diode circuits.
Introduction
Diodes are widely used in power electronics circuits for the conversion of electric
power and power processing. Ac-dc converters, usually called rectifiers, utilize
diodes and provide a fixed output dc voltage. The diodes considered in the
application circuits to follow would be considered to be ideal, i.e. having negligible
reverse recovery time and forward voltage drop.
Section 3 - Diode Circuits And Rectifiers
Freewheeling Diodes
Say we have a circuit with an inductive load in it. From the instant that electrical
power is supplied to the circuit (by the closing of a switch say), the inductive load
will accumulate stored energy. If an attempt is made to open the switch this energy
will arc across the contacts of the switch, and could cause damage to the circuit
components. Freewheeling diodes are placed across inductive loads to provide a
path for the release of energy stored in the load when the load voltage drops to
zero.
Figure 3.1 shows a diode circuit with a freewheeling diode. Diode Dm is the
freewheeling diode. The circuit operation is divided into two modes. Mode 1, which
begins when the switch is closed, and Mode 2, which begins when the switch is
opened.
Mode 1
Mode 1 begins when the switch is closed at time t = 0. The equivalent circuit for
this mode is shown in figure 3.1b.
Examination of the equivalent circuit for Mode 1, using voltage simple analysis,
reveals the following:
Now we know that for an inductor
Therefore
..............(3.1)
Figure 3.1 Circuit With Freewheeling Diode
We need to obtain an equation for the current in the circuit. This can be obtained
by taking the Laplace transform of the equation, making I(s) the subject of the
formulae, and then taking the inverse Laplace.
Recall
The Laplace Transform of a function F(t) is denoted by L{F(t)} and
is defined by
......(L.1)
The Integral above is a function of the parameter s; therefore we
call that function F(s) such that F(s) = L{F(t)}
An important property of the Laplace function is its linearity, similar
to the Differential function. If F1(t) and F2(t) have Laplace
transforms and if c1 and c2 are arbitrary constants,
......(L.2)
An important Theorem (that will not be proven here) in working in
the Laplace plane is as follows:
If F(t) is continuous for t≥ 0 and is also of exponential order as t →
∞ , and if F´(t) is of "Class A"*, then
L{F´(t)} = sL{F(t)} - F(0)
......(L.3)
Where a function of Class A is one that is
a. Sectionally Continuous over every finite interval in the range t ≥
0
b. Of exponential order as t → ∞
We can also say, from L.3, that
L{F′′(t)} = sL{F′(t)} - F′(0)
L{F′′(t)} = s[sL{F(t)} - F(0)] - F′(0)
......(L.4)
F′′(s) = s²F(s) - sF(0) - F′(0)
and the process can be repeated as many times as we wish for more
derivatives.
Let us assume that the initial conditions are zero, i.e i(t=0)=0. The Laplace
transform can be applied to equation 3.1, using L.4 yielding
..............(3.2)
Solving for I1(s) yields
..............(3.3)
where
..............(3.4)
The inverse transform of equation (3.3) in the time domain yields,
..............(3.5)
When the switch is opened at time t = t1 at the end of this mode, the current
becomes
..............(3.6)
If t1 is sufficiently long such that the exponential term in equation 3.6 becomes
negligible, then the steady-state current is given by
..............(3.7)
Mode 2
This mode begins when the switch is opened and the load current starts to flow
through the freewheeling diode Dm. The voltage equation for this mode is given by
..............(3.8)
with initial conditions
Applying the Laplace transform to equation 3.8 yields
and solving for I2 yields
..............(3.9)
where
Applying the inverse Laplace transform to equation 3.9 yields
..............(3.10)
Equation 3.10 is the freewheeling current which decays exponentially to zero at
time t = t2.
Section 3 - Diode Circuits And Rectifiers
Performance Parameters of Rectifiers
The performance of rectifiers are evaluated using the following parameters:
1. The average value of output (load) voltage given as Vdc
2. The average value of output (load) current given by Idc
3. The output dc power given by Pdc = VdcIdc
4. The rms value of output voltage given as Vrms
5. The rms value of output current given as Irms
6. The output ac power given by Pac = VrmsIrms
7. The efficiency or rectification ratio of a rectifier given by η =
8. The output voltage consists of two components an ac component and a dc
component. The effective or (rms) value of the ac component of output
voltage is given by
9. The form factor which is a measure of the shape of the output voltage is given
by
10. The ripple factor which is a measure of the ripple content is given by
By substituting the equation for the effective value of the qac component of
the output voltgae into the ripple factor equation, we can express the ripple
factor as
11. The transformer utilization factor is defined as
where Vsand Isare the rms voltage and rms current of the transformer
secondary respectively.
Figure 3.2 - Waveforms of Input Voltage & Current
Consider the waveforms in figure 3.2
vsis the sinusoidal input voltage
isis the instantaneous input current
is1is the fundamental component of is
12. The displacement angle φ is the angle between fundamental components of
input current and voltage
13. The displacement factor (DF) or Displacement Power Factor (DPF) is defined as
14. The harmonic factor (HF) also known as total harmonic distortion (THD) is a
measure of the distortion of a waveform. The harmonic factor of the input
current is given as
Where both currents are recorded as rms values
15. The crest factor is a comparison of the peak input current to its rms value. It is
given as
For a pure sinusoidal input current and voltage, power factor is defined as
the cosine of the load angle φ, i.e.
where the voltages and currents are stated in rms values and
where
For a rectifier circuit, the input power factor is given by
Power factor in rectifier circuits is related to harmonic voltages and
currents given in the diagrams
where
Hence,
An ideal rectifier should have η = 100%, Vac = 0, RF = 0, TUF = 1, HF =
THD = 0, and PF = PDF = 1.
Section 3 - Diode Circuits And Rectifiers
Single Phase Half-Wave Rectifier ( R Load)
A rectifier is a circuit that converts an ac voltage to a dc voltage. A single-phase
half-wave rectifier, shown in figure 3.3, is the simplest type, and not commonly
used in many applications. It is however, useful in understanding its operation as
they can be applied even to the complex rectifier circuits.
Figure 3.3 - Single Phase Half-Wave Rectifier
During the positive half cycle of input voltage, diode D1 conducts and the input
voltage appears across the load. During the negative half cycle of input voltage, the
diode is in the reverse biased condition, so it does not conduct and the output
voltage is zero. A point to note is that when in foward baising, there is a voltage
drop, VTD, across the diode equal to the turn on voltage. This voltage is small,
usually 0.5 to 0.7 Volts, and is generally assumed to be zero volts.
Disadvantages of this rectifier:
1. Dc output voltage is discontinuous and contains harmonics.
2. Input current is not sinusoidal
The performance of this half wave rectifier with resistive load is examined in
Example 3.1
Electrical Home
Section 3 - Diode Circuits And Rectifiers
Example 3.1
Consider the circuit below with a purely resistive load R. Determine
a. the efficiency
b. the ripple factor
c. the transformer utilization factor
d. the peak inverse voltage (PIV) of diode D1
e. form factor
f. the crest factor of the input current
Solution
Now for time interval ½T ≤ t ≤ T, vL(t) = 0 hence
But
and
∴
For the output voltage from the diode rectifier,
∴ The rms value of the output voltage is given by
a. Efficiency,
Pdc = VdcIdc
and
Pac = VrmsIrms
∴
b. Form factor
c. Ripple factor
d. Transformer utilization factor
RMS value of transformer secondary voltage Vs is given by
The rms value of the transformer secondary current is the same as the rms
value of the load current
Note:VsIs= Volt-ampere rating (VA) of the transformer
e. The peak inverse voltage is the peak reverse voltage seen by the diode and is
equal to Vm.
f. Crest factor
These results reveal
a. the rectifier has a high ripple factor of 121%
b. a low efficiency of 40.4%
c. a poor TUF of 0.286
d. the transformer must carry a dc current which may result in dc saturation
problems of the transformer core.
Section 3 - Diode Circuits And Rectifiers
Single Phase Half-Wave Rectifier ( RL Load)
The half wave rectifier with an inductive load (RL) is shown in figure 3.4.
Figure 3.4 - Half-Wave Rectifier With RL Load
The operation of the circuit is as follows:
As in the case of a resistive load, the diode turns on when its anode is positive w.r.t
its cathode, and the foward voltage is greater than the threshold voltage. Assuming
a turn-on voltage of zero volts, the voltage across the load is the same as the
positive half cycle of the ac source.
During the interval 0 to π/2
The source voltage vs increases from zero to its positive maximum, while the
voltage across the inductor vL opposes the change of current through the load. It
must be noted that the current through an inductor cannot change instantaneously,
hence the current gradually increases until it reaches its maximum value. The
current does not reach its peak when the voltage is at its maximum, which is
consistent with the fact that the current through an inductor lags the voltage across
it. During this time, energy is transferred from the ac source and is stored in the
magnetic field of the inductor.
For the interval π/2 and π
The source voltage decreases from its positive maximum to zero. The induced
voltage in the iductor reverses polarity and oposes the associated decrease in
current, thereby aiding the diode foward current. Therefore, the current starts
decreasing gradually at a delayed time, becoming zero when all the energy stored
by then inductor is released to the circuit. Again this is consistent with the fact that
current lags voltage in an inductive circuit. Hence, even after the source voltage
has dropped past zero volts, there is still load current, which exists a little more
than half a cycle.
For the interval greater than π
At π, the source voltage reverses and starts to increase to its negative maximum.
However, the voltage induced across the inductor is still positive and will sustain
forward conduction of the diode until this induced voltage decreases to zero. When
this induced voltage falls to zero, the diode will now be reversed biased, but would
have conducted forward current for an angle θ, where θ = π + σ. σ is the extended
angle of current conduction due to the energy stored in the magnetic field being
returned to the source.
The instantaneous supply voltage vs is given by
vs = vR + vL
From figure 3.4, for angles less than ωt2 the inductor is storing energy in its
magnetic field from the source and the inductor voltage would be such as to oppose
the growth of current and the supply voltage. For angles greater than ωt2 the
inductor voltage would have reversed and would aid the supply voltage to prevent
the fall of current. Hence, the average inductor voltage is zero.
The average output voltage is given by
The average load current is given by
The addition of a freewheeling diode
The average dc voltage varies proportionately to [1 - cos(π + σ)]. This can be
made to be a maximum, thereby increasing the average dc voltage, by making
cos(π + σ) a maximum. The maximum value that this can take is given by cos(π +
σ) = 1, which can be obtained if σ = 0. We can make σ = 0 with the addition of a
freewheeling diode given by Dm as shown with the dotted line.
When the supply voltage goes to zero, the current from D1 is transferred across to
diode Dm. This is called commutation of diodes. The result is the charge in the
inductor will be used to keep diode Dm on, instead of previously forcing D1 to
remain in its foward state. This would reduce the value of the extended angle of
conduction of the diode D1, σ to zero.
Intuitively we can see that if the value of the inductance is high, it will store more
charge and therefore be able to keep diode Dm on for a longer time. In other
words, if the circuit has a long time constant given by
then the inductor would be able to keep diode Dm on for the entire duration of the
negative half-cycle, and by so doing, maintain a continuous load current.
Example 3.2
Section 3 - Diode Circuits And Rectifiers
Example 3.2
Consider the circuit shown below, where the battery voltage E = 12V and its
capacity is 100 W-h. The average charging current should be Idc = 5A. The primary
input voltage is Vp = 120 V, 60 Hz and the transformer has a turns ratio of n =
2:1.
Calculate
a. the conduction angle σ of the diode
b. the current-limiting resistor R
c. the power rating PR
d. the charging time, h, in hours
e. the rectifier efficiency η and
f. the peak inverse voltage PIV of the diode
The waveforms of the voltage and load current are shown below.
Solution
E = 12 V
Vp = 120 V
Vm = √2 Vs = √2(60) = 84.85 V
a.
for Vs > E the diode D1 conducts.
The angle α when the diode starts to conduct is found from:
Vm sin α = E
α = 8.13°
β = 180° - α = 180° - 8.13° = 171.87°
conduction angle δ= β- α= 171.87° - 8.13°
δ = 163.74°
b. The average charging surrent Idc is given by
β=π-α
R = 4.26 Ω
c. The rms battery current Irms is given by
= 67.4
Irms = √67.4 = 8.2A
Power rating of resistor =
= 286.4 W
d. Power delivered to the battery = Pdc
Pdc = IdcE = 5 x 12 = 60 W
But WH rating of battery = 100 WH
∴ 100 WH = 60 x h
e. Rectifier efficiency η is given by
f. The PIV of the diode is
PIV = Vm + E = 84.85 + 12 = 96.85
Electrical Home
Section 3 - Diode Circuits And Rectifiers
Ronald De Four
Single Phase Full Wave Rectifiers
Two types of full wave rectifiers exist. They are:
1. That formed by using a center-tapped transformer and two diodes and
2. That formed by using a transformer and four diodes also known as a bridge
rectifier.
Full-Wave Rectifier With Center-Tapped Transformer (R Load)
The circuit and associated waveforms are shown in figure 3.5. Each half of the
transformer with its associated diode acts as a half-wave rectifier. On the positive
half cycle of supply voltage, diode D1 is forward biased while diode D2 is reversed
biased. On the negative half cycle of supply voltage, diode D2 is forward biased
while diode D1 is reversed biased.
Figure 3.5 - Full-Wave Rectifier With Center-Tapped Transformer
Each diode conducts on alternate half cycles of supply voltage producing a fullwave output voltage across the load. Dc saturation of the transformer core does
not exist here since there is no dc current flowing through the transformer.
The average output voltage is given by
The peak inverse voltage of the diodes is 2Vm
Comparison Between Half-Wave & Full-Wave Center-Tapped Rectifiers R
load
Table 3.1
Performance Parameter
Half-Wave
Rectifier (R Load)
Full-Wave Center-Tapped
Transformer Rectifier (R
Load)
Efficiency (η)
40.5%
81%
Form Factor (FF)
157%
111%
Ripple Factor (RF)
121%
48.2%
Transformer Utilization Factor
(TUF)
28.6%
57.32%
Peak Inverse Voltage (PIV)
Vm
2Vm
Crest Factor (CF)
2
1.414
Table 3.1 shows that the full-wave center-tapped rectifier has improved
performance over the half-wave rectifier.
Example 3.3
Section 3 - Diode Circuits And Rectifiers
Example 3.3
The rectifier shown in the figure below has a purely resistive load. Determine
a. the efficiency
b. the form factor
c. the ripple factor
d. the transformer utilization factor
e. the peak inverse voltage of the diode D1 and
f. the crest factor of the input current
Solution
The average output voltage is given by
The average load current is given by
The rms value of the output voltage is given by
a.
b. Form factor
c. Ripple factor
d. Transformer utilization factor
Rms value of transformer secondary voltage = vs
This is the value of the transformer secondary voltage for ½ of the secondary
winding.
Rms value of transformer secondary voltage associated with this ½ winding
secondary voltage is the same as the (rms) load current for this ½ winding.
= rms value of load current for this ½ winding
The Volt-Ampere rating of the transformer
e. PIV = 2Vm
f.
Section 3 - Diode Circuits And Rectifiers
Single Phase Full Wave Rectifiers
Full-Wave Bridge Rectifier R Load
As mentioned earlier, one can also implement a single-phase full-wave rectifier
using four diodes. The diagram of the full-wave bridge rectifier and associated
waveforms are shown in figure 3.6.
On the positive half cycle of transformer secondary supply voltage, diodes D1 and
D2 conduct, supplying this voltage to the load. On the negative half cycle of supply
voltage, diodes D3 and D4 conduct supplying this voltage to the load.
It can be seen from the waveforms that the peak inverse voltage of the diodes is
only Vm The average output voltage is the same as that for the center-tapped
transformer full-wave rectifier.
Figure 3.6 - Full-Wave Bridge Rectifier With R load and Associated Waveforms
Single Phase Full-Wave Bridge Rectifier With RL Load
With a resistive load, the load current is identical in shape to the output voltage.
Most loads are inductive and the load current with these loads depends on the
value of load resistance and load inductance.
Figure 3.7 is a circuit of a bridge rectifier with an inductive load and a battery.
Figure 3.7 - Bridge Rectifier With RL Load & Battery
The input voltage is given by:
and the load current can be obtained from the solution of the differential equation
.............(1)
Recall
The above equation is a non-exact first order linear differential
equation, hence by definition it must be of the form
A(x) dy/dx + B(x)y = C(x)
.............(LE.1)
If an equation is not exact, it is natural to make it exact by the
introduction of an appropriate factor, which is then called an
integrating factor. By dividing each member of equation LE.1 by A(x)
we obtain
dy/dx + P(x)y = Q(x)
.............(LE.2)
For the moment, let us assume that there exists for equation LE.2 a
positive integrating factor v(x) > 0, a function of x alone. Then
v(x)[dy/dx + P(x)y] = v(x)Q(x)
.............(LE.3)
must be an exact equation. But equation LE.3 can be easily put into
the form
[v(x)P(x)y - v(x)Q(x)]dx + v(x)dy = 0
.............(LE.4)
which is of the form of an exact first order differential equation.
Mdx + Ndy = 0
where M = vPy - vQ
of x alone.
and
.............(LE.5)
N = v, and v, P, and Q are functions
Recall
Say we have an equation of the form
M(x,y)dx + N(x,y)dy = 0
.............(E.1)
in which seperation of variables is not possible. Suppose,
also, that a function f(x,y) can be found such that it has
for its differential the expression Mdx + Ndy = 0; that is
dF = Mdx + Ndy
.............(E.2)
Then certainly
F(x,y) = c
.............(E.3)
following from the fact that dF = 0. Now we know that if
an equation E.1 is exact then a function F exists which will
satisfy E.2. But from calculus
.............(E.4)
so
.............(E.5)
These two equations lead to
.............(E.6)
therefore
.............(E.7)
Therefore if equation LE.5 is to be exact it follows from the
requirement of equation E.7 that v must satisfy the equation
vP = dv/dx
.............(LE.6)
Solving LE.6 we can readily get v
.............(LE.7)
In other words, we have shown that if equation LE.2 has a positive
integrating factor independant of y, then that factor must be given by
equation LE.7. Let us multiply equation LE.2 by the integrating factor,
obtaining
.............(LE.8)
Now the left member of LE.8 is the derivative of the product
.............(LE.9)
and the right side is a function of x only. Hence equation LE.8 is exact,
which is what we wanted to show. Now let us proceed to solve and
obtain a general solution for y. Integrating equation LE.8 with respect
to x is done as shown below
.............(LE.10)
Solving for y yields a general solution as shown in equation LE.11
(where r = Q, and h =
).
.............(LE.11)
Now applying this solution of a first order linear differential equation to equation
(1) which is
.............(1)
which can be rewritten as
.............(16)
where
yields the solution of equation (2) as
.............(3)
.............(4)
Now
where
Therefore
and
.............(5)
where Z is the load impedance given by
Continuous Load Current
Examination of the load current waveform for this circuit reveals the following:
1. at ωt = π,
iL = I1
2. at ωt = 0,
iL = I1
hence the value of the constant C in equation (5) and the value of I1 can be
determined by imposing these conditions in the load current equation.
Imposing the condition at ωt = π,
on the load current equation
iL = I1
yields
.............(6)
Substituting for C in equation (5) yields
.............(7)
Under steady-state conditions
Applying this condition to equation (7) yields
where
.............(8)
Now substituting for I1 in equation (7) yields
.............(9)
for
and
Equation (9) is the equation for the continuous load current of a bridge rectifier
with an inductive load and a battery in series with the load.
The rms diode current for diodes D1 and D2 is given by
.............(10)
and the rms output current can be obtained by combining the rms diode current of
diodes D1 and D2 with that of diodes D3 and D4 and is given by
.............(11)
The average diode current is given by
.............(12)
Discontinuous Load Current
Under these conditions, the load current flows only for the interval given by
The diode starts to conduct at ωt = α where α is given by
At ωt = α,
given by
and the general solution of the load current equation which is
yields
Substituting for C in the load current equation yields
At ωt = β, the load current falls to zero and
.
Substituting this condition in the load current equation yields
The value of β can be determined from this transcendental equation by an iterative
method of solution. Starting with β = 0, and increasing its value by a very small
amount until the left hand side of the equation becomes zero.
Knowing α, β and the load current equation, the rms current through diodes D1 and
D2 is obtained from
and the average diode current from
Section 3 - Diode Circuits And Rectifiers
Three-Phase Bridge Rectifier
The diagram for a three-phase bridge rectifier is shown in figure 3.8. together with
the supply voltage, output voltage and diode current waveforms.
Figure 3.8 - Three-Phase Bridge Rectifier & Associated Waveforms
The diodes are numbered in order of conduction sequences where the conduction
sequence for the diodes is 12, 23, 34, 45, 56 and 61. In one cycle, each pair of
diodes conducts for 60° and each diode conducts for 120°. The pair of diodes
connected to the supply lines with the highest instantaneous line-to-line voltage
will conduct. Since there are six commutations taking place in one cycle this
rectifier is also called a six-pulse converter.
Three-Phase Bridge Rectifier R Load
Taking the line voltage waveform as a cosine wave, since each pair of diodes is on
for 60° then they begin conduction at -30° and end at +30°.
If Vm is the peak phase voltage of the transformer secondary winding, and there
are six pulses in one cycle, then the average output voltage is given by
The rms output voltage is given by
The peak current through the diode is given by Im where
The rms diode current is obtained by integrating over 60° and multiplying the
integral by 2 since each diode conducts for 60° in a cycle. This current is given by
From the line current waveform, it is clear that the current in line-a flows for four
sixty degrees in one cycle. Hence, the rms value of transformer secondary current
is given by
Example 3.4
Performance Parameters of Half-Wave, Full-Wave Center-Tapped & 3Phase Bridge Rectifiers R load
Table 3.2
Performance
Parameter
Single
PhaseHalfWave Rectifier
(R Load)
Single PhaseFullWave CenterTapped
Transformer
Rectifier
(R Load)
Three-Phase
BridgeRectifier
(R Load)
Efficiency (η)
40.5%
81%
99.83%
Form Factor (FF)
157%
111%
100.08%
Ripple Factor (RF)
121%
48.2%
4%
Transformer
Utilization Factor
(TUF)
28.6%
57.32%
95.42%
Peak Inverse
Voltage (PIV)
Vm
2Vm
√3Vm
Crest Factor (CF)
2
1.414
From the above table, it is clear that the three-phase bridge rectifier is far superior
with respect to performance parameters in comparison to the single phase half
wave and single phase full wave center-tapped rectifiers.
Three-Phase Bridge Rectifier RL Load
The instantaneous output voltage of this rectifier is given by
for
.............(1)
The load current can be obtained from the solution of the equation
.............(2)
The solution is of the form
.............(3)
where Z is the load impedance given by
and
Applying the condition at
,
IL = I1
to equation (3) the constant C can be determined. Application of this condition
yields
Substituting for C in equation (3) yields
Under steady-state conditions
Applying this condition to the above equation gives the value of I1 as
Substituting for I1 in the equation of instantaneous load current yields
for
Since each diode conducts for 120°, the rms diode current is twice the rms current
over the 60° interval in a cycle. This is given by
The rms output current is obtained by combining the rms current of each diode
over a cycle. This is given by
The average diode current is given by
Section 3 - Diode Circuits And Rectifiers
Example 3.4
A three phase rectifier has a purely resistive load of R. Determine
a. the efficiency
b. the form factor
c. the ripple factor
d. the transformer utilization factor
e. the peak inverse voltage of each diode and
f. the peak current through a diode.
The rectifier delivers Idc = 60 A at an output voltage of Vdc = 280.7V and the
source frequency is 60 Hz.
Solution
a. Efficiency,
Now
Pdc = VdcIdc
where Vdc= average output voltage
Idc= average output (load) current
Pac = VrmsIrms
where Vrms= rms value of output voltage
Irms= rms value of output (load) current
Since
where Vm= peak voltage of transformer secondary
then
and since
Vrms = 1.6554Vm
then
b. Form factor
c. Ripple factor
Now, the output voltage is comprised of two components
a. a dc value
b. an ac or ripple component
The rms output voltage is given by
d.
where Vs = rms voltage of transformer secondary
Is = rms current of transformer secondary
Vs = 0.707Vm
Is = current in one line of transformer = 0.7804Im
where Im = peak secondary line current
But
∴ VA rating of transformer = 3VsIs
e. Vm = peak line to neutral voltage
But Vdc = 1.654Vm = 280.7 V
PIV= peak value of secondary line to line voltage = 169.7√3 = 293.9 V
f. The average diode current Idc is given by
If the average load current is Idcand each diode is on for 120° of a cycle of
360°
then average diode current = average load current
Section 3 - Diode Circuits And Rectifiers
Rectifier Filter Circuits
●
It has been observed that the output of rectifiers contain harmonics.
●
DC filters are used to smooth out the dc output voltage of rectifiers.
●
DC filters are usually L type, C type or LC type, these are shown in figure 3.9.
●
●
●
Due to the rectification action, the input current of a rectifier contain
harmonics.
An ac filter of LC type is used to reduce the harmonic content of the supply
current. An ac filter is shown in figure 3.9.
The examples given below would examine the steps involved when designing
ac and dc filters for rectifier circuits.
Figure 3.9 - AC & DC Filters
Example 3.5
Example 3.6
Example 3.7
Example 3.8
Electrical Home
Section 3 - Diode Circuits And Rectifiers
Example 3.6
A single phase bridge rectifier is supplied from a 120 V 60 Hz source. The load
resistance is
R = 500 ohms.
a. Design a C filter so that the ripple factor of the output voltage is less than 5%
b. With the value of the capacitor C in part (a), calculate the average load
voltage Vdc
Solution
i. Charging of capacitor
When the instantaneous voltage vs is higher than the capacitor voltage
vc, diodes D1 and D2 or D3 and D4 conduct supply voltage thereby
charging the capacitor.
ii. Discharging of capacitor
When the instantaneous supply voltage vs falls below the instantaneous
capacitor voltage vc diodes D1 and D2 or D3 and D4 are reverse biased
and the capacitor discharges through the load resistor.
iii. The capacitor voltage varies between Vcmin and Vcmax and this is known
as the output ripple voltage.
iv. The output voltage waveforms and equivalent circuit on charging and
discharging are shown below.
At t = t1, the capacitor is charged at the peak supply voltage Vm.
The equation for discharge of the capacitor is given by
differentiating we get
Integrating bith sides yields
Let ek = K
Applying the initial condition of discharge:
at t = 0,
vc = Vm
and
Hence the output or capacitor voltage during the discharge period is given by
The peak to peak ripple voltage Vr(pp) is given by
Vr(pp) = vL(t=0) - vL(t=t2)
Now e-x ≈ 1 - x
For large R, t1 « t2
hence
but
The average load voltage Vdc is given by
The rms output ripple voltage Vac can be approximated to
Ripple factor RF is given by
b.
Section 3 - Diode Circuits And Rectifiers
Example 3.7
Reduction of harmonics in output voltage with the use of an LC filter.
An LC filter shown below is used to reduc the ripple content of the output voltage
for a single phase full wave rectifier. The load resistance is R = 40 ohms, load
inductance L = 10 mH and source frequency is 60 Hz.
a. Determine the values of Le and Ce so that the ripple factor of the output
voltage is 10%.
Solution
To make it easier for the nth harmonic ripple current to pass through the capacitor,
the load impedance must be 10 times greater than the capacitor reactance, ie.,
The rms value of the nth harmonic component appearing in the output is obtained
using
where Vn = rms value of the nth harmonic
The total rms ripple voltage due to all harmonics is given by
The output voltage of a bridge rectifier can be described by a power series given by
where
∴ bn = 0
The computation can be simplified by only considering the dominant harmonics.
This can be seen as the second harmonic whose rms value is given by
Ce can be obtained using
∴ Le = 30.83 mH
Section 3 - Diode Circuits And Rectifiers
Example 3.8
This example examines an input ac filter design in order to keep the harmonic line
current within a specified value.
An LC filter shown below is used to reduce the input current harmonics in a single
phase full wave rectifier given below. The load current is ripple free and its average
value is Ia If the supply frequency is f = 60 Hz determine
a. the resonant frequency of the filter so that the total input harmonic current is
reduced to 1% of the fundamental component.
b. the HF of the input current
c. and input PF in that filter.
Solution
a. The equivalent circuit of the nth harmonic current is shown above where:
Isn = rms value of the nth harmonic current appearing in the supply
In = rms value of the nth harmonic current produced by the rectifier switching
action
NOTE: It is Isn that must be reduced to a specified value relative to the
fundamental current so that:
a. the supply would have a small harmonic content and
b. the remainder of the harmonic current would flow through the capacitor
and diodes
The potential difference across Li and Ci due to the current source In(ω) is
given by
The total rms harmonic current flowing in the supply is given by Inwhere
The harmonic factor r is given by
Now the input current expressed in a fourier series is given by i1(t) where
where
The rms value of the fundamental current is given by Is1where
and the rms value of the nth harmonic is given by Inwhere
If we only consider the third harmonic
Now the filter circuit would be in resonance when
NOTE: At this frequency 30.72 Hz the filter would ensure that the total input
harmonic current is reduced to 1% of the fundamental component.
b. Harmonic factor
The rms value of the input current is given by
HF = 48.43 %
c. Power factor
From the waveform, the displacement angle φ is zero,
hence
displacement factor DF = cos φ = cos 0 = 1
PF = 0.9 lagging
Section 3 - Diode Circuits And Rectifiers
Effects of Source Inductance
●
●
●
The effects of source inductance were ignored in the previous derivations of
output voltages produced by rectifiers.
The effect of source inductance is to reduce the rectifier output voltage.
This effect can be observed with the use of the three-phase bridge rectifier
and associated waveforms in figure 3.10.
Figure 3.10 - Three-phase Bridge Rectifier With Source Inductance
The effect of the source inductances are as follows:
●
In this circuit, the diodes with the most positive voltage will conduct.
●
●
●
●
●
●
Circuit operation at the point where ωt = π will now be considered.
Just before ωt = π, Vac is more positive than the other two line voltages and
as a result, diodes D1 and D2 supply load current.
At the point just after ωt = π, line voltage Vbc is the most positive and diode
D1 is turning off while diode D3 is turning on to supply load current along with
diode D2.
But at ωt = π, line voltages Vac and Vbc are both equal, and load current is still
flowing through diode D1.
Due to the source inductance L1, the load current flowing through diode D1
cannot fall to zero immediately for load current to be transferred to diode D3.
As the current through diode D1 given as id1 decreases, a voltage +vL1 is
induced across inductor L1. The resulting output voltage is given by vL, where
vL = vac + vL1
●
At the same time, id3 the current through diode D3 increases from zero,
inducing an equal voltage
-vL2 across inductor L2. The resulting output voltage becomes
vL = vbc - vL2
●
●
●
The result is that the anode voltages of diodes D1 and D3 are equal and both
diodes conduct for a specified period called the commutation or overlap angle
µ.
The transfer of current from one diode to another is called commutation and
the reactance corresponding to the inductance causing the overlap is called
the commutating reactance.
The effect of this overlap is to reduce the average output voltage of the
rectifier. The reduction in output voltage is determined using the following
method:
The voltage across the inductor L2 is given by
Assuming a linear rise of current from zero to Idc , we can write
This process of transferring current from one diode to another occurs six
times for a three-phase bridge rectifier, hence the average reduction in
voltage due to the presence of commutating inductances is given by Vx
where
If all the inductances are equal then
Electrical Home
Section 4 - Thyristors
Objectives
●
Understand the different types of thyristors and their characteristics.
●
Examine the methods of turning on and turning off thyristors.
●
Demonstrate the techniques for di/dt and dv/dt protection of thyristors.
●
●
Identify the conditions and techniques for series and parallel operation of
thyristors.
Examine thyristor firing circuits.
Introduction
●
●
The thyristor is one of the most important power semiconductor devices.
They are operated as bistable switches, operating from nonconducting to
conducting state.
Section 4 - Thyristors
Thyristor Characteristics
A thyristor is a four layer pnpn semiconductor device consisting of three pn
junctions. It has three terminals: an anode a cathode and a gate. Figure 4.1
shows the thyristor symbol and a sectional view of the three pn junctions.
Figure 4.1 Thyristor Symbol & pn Junctions
When the anode voltage is made positive with respect to the cathode, junctions J1
and J3 are forward biased and junction J2 is reverse biased. The thyristor is said
to be in the forward blocking or off-state condition. A small leakage current flows
from anode to cathode and is called the off-state current.
If the anode voltage VAK is increased to a sufficiently large value, the reverse
biased junction J2 would breakdown. This is known as avalanche breakdown and
the corresponding voltage is called the forward breakdown voltage VBO. Since the
other two junctions J1 and J3 are already forward biased, there will be free
movement of carriers across all three junctions. This results in a large forward
current. The device is now said to be in a conducting or on-state. The voltage drop
across the device in the on-state is due to the ohmic drop in the four layers and is
very small (in the region of 1 V). In the on-state the anode current is limited by
an external impedance or resistance as shown in figure 4.2(a).
V-I Characteristics of Thyristor
Figure 4.2 shows the V-I characteristics and the circuit used to obtain these
characteristics.
Figure 4.2 Thyristor Circuit & V-I Characteristics
The important points on this characteristic are :
●
Latching Current IL
This is the minimum anode current required to maintain the thyristor in the
on-state immediately after a thyristor has been turned on and the gate signal
has been removed.
If a gate current greater than the threshold gate current is applied until the
anode current is greater than the latching current IL then the thyristor will be
turned on or triggered.
●
Holding Current IH
This is the minimum anode current required to maintain the thyristor in the
on-state.
To turn off a thyristor, the forward anode current must be reduced below its
holding current for a sufficient time for mobile charge carriers to vacate the
junction. If the anode current is not maintained below IH for long enough,
the thyristor will not have returned to the fully blocking state by the time the
anode-to-cathode voltage rises again. It might then return to the conducting
state without an externally-applied gate current.
●
Reverse Current IR
When the cathode voltage is positive with respect to the anode, the junction
J2 is forward biased but junctions J1 and J3 are reverse biased. The thyristor
is said to be in the reverse blocking state and a reverse leakage current
known as reverse current IR will flow through the device.
●
Forward Breakover Voltage VBO
If the forward voltage VAK is increased beyond VBO , the thyristor can be
turned on. But such a turn-on could be destructive. In practice the forward
voltage is maintained below VBO and the thyristor is turned on by applying a
positive gate signal between gate and cathode.
●
Once the thyristor is turned on by a gate signal and its anode current is
greater than the holding current, the device continues to conduct due to
positive feedback even if the gate signal is removed. This is because the
thyristor is a latching device and it has been latched to the on-state.
Section 4 - Thyristors
Two - Transistor Model of Thyristor
This model is used to demonstrate the regenerative or latching action due to
positive feedback in the thyristor. A thyristor can be considered as two
complementary transistors. One being pnp and the other npn. The two-transistor
model is shown in figure 4.3 below.
Figure 4.3 Two-Transistor Model of Thyristor
The collector current IC of a transistor is related to the emitter current IE and the
leakage current of the collector base junction ICBO as
.............(1)
The emitter current of transistor Q1 is the anode current IA of the thyristor and
collector current IC1 is given by
.............(2)
where α1 and ICBO1 are the current gain and leakage current respectively for
transistor Q1.
Similarly, the collector current for transistor Q2 is IC2 where
.............(3)
where α2 and ICBO2 are the current gain and leakage current respectively for
transistor Q2.
Combining the two collector currents IC1 and IC2 yields
.............(4)
When a gate current IG is applied to the thyristor
.............(5)
Solving for anode current IA in equation 5 yields
.............(6)
The current gain α1 varies with emitter current IE1 which is equal to IA; and α2
varies with emitter current IE2 which is equal to Ik.
A typical variation of current gain α with emitter current IE is shown in figure 4.4.
Figure 4.4 Typical Variation of Current Gain With Emitter Current
If the gate current IG is increased from zero to some positive value, this will
increase the anode current IA as shown by equation 6. An increase of IA which is
an increase of IE1 would increase α1 as shown in figure 4.4 and also α2 since
. The increase in values of both α1 and α2 would further increase the
value of anode current IA which is a regenerative or positive feedback effect.
If α1 and α2 approach unity, the denominator of equation 6 approaches zero and a
large value of anode current is produced causing the thyristor to turn on as a result
of the application of a small gate current.
The capacitance of the pn junctions are shown in figure 4.5 below.
Figure 4.5 Two-transistor Transient Model of Thyristor
Under transient conditions, the capacitances of the pn junctions influence the
characteristics of the thyristor.
If a thyristor is in the blocking state and a rapidly rising voltage is applied to the
device, high currents would flow through the junction capacitors. The current
through capacitor Cj2 can be expressed as
where
Cj2 = capacitance of junction j2
Vj2 = voltage of junction j2
qj2 = charge in junction j2
If the rate of rise of voltage dv/dt is large, then ij2 would be large, which would
result in increased leakage currents ICBO1 and ICBO2. High enough values of ICBO1
and ICBO2 may cause α1 and α2 to approach unity, resulting in undesirable turn on
of the thyristor.
It must be noted that a large current through the junction capacitors may
cause damage to the device.
Section 4 - Thyristors
Thyristor Turn-on
A thyristor is turned on by increasing the anode current. This can be accomplished
in the following ways.
Thermals
If the temperature of a thyristor is high, there will be an increase in the number of
electron-hole pairs. This would increase the leakage current. This increase in
leakage current causes the anode current to increase and as a result causes α1 and
α2 to increase. Due to the regenerative action, the sum α1 + α2 may tend to unity
and the thyristor may be turned on. This type of turn-on may cause thermal
runaway and should be avoided.
Light
If light is allowed to strike the junction of a thyristor, the electron-hole pairs will
increase and this may cause the thyristor to be turned on. This is the principle of
operation of light activated thyristors.
High Voltage
If the forward anode to cathode voltage VAK is increased beyond the forward
breakdown voltage VBO , high enough leakage currents will flow, causing
regenerative turn-on. This type of turn-on is destructive and should be avoided.
dv/dt
From equation 6, if the rate or rise of the anode to cathode voltage is high, (for
example, when there is a voltage spike), the charging current of the capacitive
junctions may be high enough to turn on the thyristor. A high value of charging
current may cause damage to the thyristor and must be avoided. Hence, thyristors
must be protected against high dv/dt and must be operated within the
manufacturer's dv/dt specifications.
Gate Current
The injection of gate current into a forward biased thyristor would turn-on the
device. As the gate current is increased, the forward voltage required to turn-on
the device decreases. This is shown in figure 4.6.
Figure 4.6 Effects of Gate Current on Forward Blocking Voltage
Figure 4.7 shows the waveform of anode current as a function of time, following
the application of a gate signal. The following time delays can be described from
the waveforms:
Figure 4.7 Thyristor Turn-on Characteristics
Turn-on Time ton
The turn-on time ton is defined as the time interval between 10% of steadystate gate current and 90% of steady-state thyristor on-state current.
Delay Time td
The delay time td is defined as the time interval between 10% of gate current
and 10% of thyristor on-state current.
Rise Time tr
The rise time is defined as the time required for the anode current to rise from
10% of the on-state current to 90% of the on-state current.
Note:
Gate Control Circuit Design
Consideration must be given to the following points when designing gate control
circuits.
●
●
●
The gate signal should be removed after the thyristor has been turned on. A
continuous gate signal will increase the power loss in the gate junction.
No gate signal should be applied when the thyristor is reversed biased. If a
gate signal is applied under these conditions, the thyristor may fail due to an
increased leakage current.
The width of the gate pulse must be greater than the time required for the
anode current to rise to the holding current. In practice, the gate pulse width
is made wider than the turn-on time of the thyristor.
Electrical Home
Section 4 - Thyristors
Di/dt Protection
A minimum time is required for the thyristor to spread the current conduction
uniformly throughout the junctions. If this time is not allotted and the rate of rise
of anode current is very high compared to the spreading velocity at turn-on, then
this could lead to localised "hot-spot" heating and the device may fail as a result of
excessive heating.
Protection against di/dt is necessary and an example is shown in figure 4.8 below.
The circuit analysis is as follows:
Figure 4.8
For an inductive load, when thyristor T1 is turned off, free-wheeling diode Dm
conducts load current. If thyristor T1 is fired when diode Dm is still conducting,
di/dt can be very high. In order to reduce the high di/dt a series inductor Ls is
added to the circuit as shown. The forward di/dt is given as
dv/dt Protection
As seen earlier, a high dv/dt may cause damage to a thyristor. In order to protect a
thyristor from high dv/dt, the circuits shown in figure 4.9 below could be used.
Figure 4.9 dv/dt Protection
If the switch S1 in figure 4.9(a) is closed at time t = 0, a step voltage will be
applied across thyristor T1 and dv/dt may be high enough to turn on the thyristor.
dv/dt can be limited by connecting capacitor Cs across the thyristor as shown in
figure 4.9(b). Since
then
and the rate of rise of voltage is limited by the value of the capacitor used. In order
to limit the capacitor discharge current when the thyristor is turned on, a resistor
Rs is inserted in series with the capacitor as shown in figure 4.9(c). This resistor
capacitor arrangement is known as a snubber circuit.
For figure 4.9(c), when switch S1 is closed at time t = 0, the voltage across the
capacitor is given by
and this charging capacitor voltage is seen by the thyristor anode to cathode
terminals as VAK. This is depicted by the waveform of figure 4.9(d). The rate of rise
of voltage across the thyristor can be represented by
where 0.632VS is one time constant.
The value of the snubber time constant RSCS can be found for a known dv/dt. And
for a known discharge current ITD , the value of resistor RS can be found using
It is sometimes necessary to use one resistor for dv/dt and another for limiting the
discharge current of the snubber capacitor. This arrangement is shown in figure
4.9(e). In this circuit, R1 and CS are used for dv/dt protection, while R1 + R2 is
used for limiting the capacitor discharge current.
The load can also be placed in series with the snubber components as shown in
figure 4.9(f).
Example 4.1
Section 4 - Thyristors
Example 4.1
The input voltage to the circuit shown below is Vs = 200 V with load resistance R =
5 ohms. The load and stray inductances are negligible and the thyristor is operated
at a frequency of 2 kHz. If the required dv/dt is 100 V/µs and the discharge current
is to be limited to 100 A, determine
a. the values of Rs and Cs
b. the snubber losses and
c. the power rating if the snubber resistor.
Solution
ITD = 100 A
R=5Ω
Vs = 200 V
(a) Using Kirchoffs voltage law
at t = 0, vc = 0
differentiating
Therefore integrating each side yields
Recall that 102 = 100
Therefore log10 100 = 2
So if we let i =
Section 4 - Thyristors
Thyristor Turn-Off
A thyristor which is in the on-state can be turned off or commutated by reducing
the anode current to a level below the holding current and keeping the anode
current below this level for a sufficiently long time so that the excess carriers in the
four layers are swept out or recombined.
Thyristors can either be line commutated or forced commutated. Figure 4.10
displays the turn-off characteristics of a line commutated thyristor.
The two outer pn junctions exhibit torn-off characteristics similar to that of a diode.
The reverse recovery time trr and peak reverse recovery current IRR are given by
trr = ta + tb
For this line commutated thyristor, a reverse voltage appears across the thyristor
immediately after the forward current goes through the zero value. This reverse
voltage will accelerate the turn-off process by sweeping out excess carriers in
junctions J1 and J3.
Figure 4.10 Turn-Off Characteristics of Line Commutated Thyristor
The inner junction J2 will require a time known as the recombination time trc to
recombine the excess carriers and the negative reverse voltage reduces this
recombination time.
●
Turn-off Time tq
This is the sum of the reverse recovery time trr and the recombination time
trc. It is defined as the time interval between the instant when the on-state
current has decreased to zero and the instant when the thyristor is capable of
withstanding forward voltage without turning on. It depends on the peak value
of on-state current and the instantaneous on-state voltage.
●
Reverse Recovery Charge QRR
This is defined as the amount of charge which has to be recovered during the
turn-off process. Its magnitude is determined by the area enclosed by the path
of the reverse recovery current and depends on:
1. The rate of fall of on-state current and
2. The peak value of on-state current before turn-off
Section 4 - Thyristors
Thyristor Types
Thyristors are manufactured almost exclusively by diffusion. Thyristors can be
classified into nine categories depending on the physical construction, turn-on time
and turn-off time. These categories are:
1. Phase Control Thyristors (SCRs)
2. Fast Switching Thyristors (SCRs)
3. Gate Turn-off Thyristors (GTOs)
4. Bidirectional Triode Thyristors (TRIACs)
5. Reverse Conducting Thyristors (RCTs)
6. Static Induction Thyristors (SITHs)
7. Light Activated Silicon Controlled Rectifiers (LASCRs)
8. FET Controlled Thyristors (FET-CTHs)
9. MOS Controlled Thyristors (MCTs)
Phase Control Thyristors
This type of thyristor operates at line frequency and is turned-off by natural
commutation. The turn-off time is of the order of 50 to 100 µs. Since the thyristor
is made of silicon and it is a controlled device, it is called a silicon control rectifier
(SCR). This type is used in low speed switching applications and is also known as a
converter thyristor. Modern thyristors use an amplifying gate circuit to turn-on the
main thyristor. In figure 4.11 an auxiliary thyristor TA is gated on by an external
gate signal. The amplified output of this auxiliary thyristor is used to supply gate
signal to the main thyristor TM.
Figure 4.11 Amplifying Gate Thyristor
Fast Switching Thyristor
This type of thyristor is used in high speed switching applications and are forced
commutated. They have fast turn-off times in the range of 5 to 50 µs. The on-state
voltage drop varies approximately as an inverse function of the turn-off time tq.
This type is known as an inverter thyristor.
Gate Turn-off Thyristor
A gate turn-off thyristor is turned on by applying a positive gate signal and turned
off with the application of a negative gate signal. GTOs have several advantages
over SCRs. These advantages are:
1. Elimination of commutating components resulting in a reduction in cost,
weight and volume.
2. Reduction in acoustic and electromagnetic noise due to the elimination of
commutation chokes.
3. Faster turn-off permitting high switching frequency applications.
4. Improved efficiency of converters.
GTOs, when used in low power applications have several advantages over bipolar
transistors. These advantages are:
1. Higher blocking voltage capability.
2. A high ratio of peak controllable current to average current.
3. A high ratio of peak surge current to average current.
4. A high on-state gain i.e. anode current to gate current.
5. A pulsed gate signal of shorter duration.
Controllable Peak On-state Current ITGQ
This is defined as the peak value of on-state current which can be turned off by
gate control.
Bi-directional Triode Thyristor
The bi-directional thyristor or TRIAC conducts current in both directions and is
normally used in ac phase control circuits. It can be considered as two SCRs
connected in anti-parallel with a common gate connection. The TRIAC symbol and VI characteristics are shown in figure 4.12.
Figure 4.12 TRIAC Characteristics
A bi-directional device cannot have an anode and a cathode, hence the terminals
are labeled gate, MT1 and MT2. If terminal MT1 is positive with respect to MT2, the
triac can be turned on by applying a positive gate signal between gate and MT1. If
terminal MT2 is negative with respect to MT1, the triac can be turned on by
applying a negative gate signal between gate and MT1. It is not necessary to have
both polarities of gate signals to turn on a thyristor. If operated in quadrant 1, it
needs a positive gate voltage and current, while if operated in quadrant III it needs
a negative gate voltage and current.
Reverse Conducting Thyristor
An RCT is a thyristor with a built in anti-parallel diode as shown in figure 4.13.
Figure 4.13 Reverse Conducting Thyristor
This thyristor is used in chopper and inverter circuits to allow reverse current due
to inductive loads to flow through the antiparallel diode. The reverse blocking
voltage of an RCT is very low typically in the range of 30 to 40 volts.
Static Induction Thyristor
A SITH is a minority carrier device. It is turned on by applying a positive gate
voltage and turned off by applying a negative gate voltage. It has fast switching
speed in the range of 1 to 6 µs and high dv/dt and di/dt capabilities.
Light Activated Silicon Controlled Thyristor
This device is turned on when light is incident on the silicon wafer. Electron-hole
pairs which are produced by the incident light trigger the device when an electric
field is applied. LASCRs are used in high voltage high current applications like
HVDC transmission. These devices offer complete electrical isolation between the
light triggering source and the switching device. LASCRs cannot be turned off by
the gate.
FET Controlled Thyristor
The FET controlled thyristor consists of a FET in parallel with a thyristor as shown in
figure 4.14. The thyristor is triggered when a gate voltage is applied to the
MOSFET. These devices cannot be turned off by gate control.
Figure 4.14 FET Controlled Thyristor
MOS Controlled Thyristor
These devices combines the features of a regenerative four layer thyristor and a
MOS gate structure. The equivalent circuit and symbol for this device is shown in
figure 4.15. The features of a MCT are:
1. Low forward voltage drop during conduction.
2. Fast turn-on and turn-off times.
3. Low switching losses.
4. Low reverse voltage blocking capability.
5. High gate input impedance.
This type is used in low speed switching applications and is also known as a
converter thyristor. Modern thyristors use an amplifying gate circuit to turn-on the
main thyristor. In figure 4.11 an auxiliary thyristor TA is gated on by an external
gate signal. The amplified output of this auxiliary thyristor is used to supply gare
signal to the main thyristor TM.
Figure 4.15 MOS Controlled Thyristor
Section 4 - Thyristors
Series Operation of Thyristors
Thyristors are connected in series to improve their overall voltage rating. The
characteristics of thyristors of the same type are not the same, as shown in figure
4.16 and hence auxiliary components must be added to thyristors connected in
series to ensure proper operation.
Figure 4.16 Off-State Characteristics of Two Thyristors of Same Type
In figure 4.16 it is clearly seen that for the same off-state current the off-state
voltages differ. Voltage sharing networks are required for both reverse and offstate conditions. Resistors placed in parallel with the thyristors are used to
accomplish voltage sharing between thyristors placed in series. The voltage sharing
resistors for n thyristors in series are shown in figure 4.17.
Figure 4.17 Three Series Connected Thyristors
For equal voltage sharing, the off-state currents differ as shown in figure 4.18.
Figure 4.18 Forward Leakage Currents For Equal Voltage Sharing
Let the off-state current of thyristor T1 be represented by ID1. For ns thyristors in
the string, and the other ns - 1 thyristors having the same off-state current, then
ID2 = ID3 = IDn
and
ID1 < ID2
Note: Since thyristor T1 has the least off-state current, then this thyristor
will share the highest voltage because the voltage drop across the resistor
R in parallel with this thyristor will be larger than that across the resistor R
for the other thyristors.
If I1 is the current through resistor R which is connected across thyristor T1, then
the current through the other resistors are equal and given by
ID2 = ID3 = IDn
The off-state current spread is given by ∆ID where
∆ID = ID2 - ID1
∆ID = (IT - I2) - (IT - I1)
∆ID = I1 - I2
therefore
I2 = I1 - ∆ID
The voltage drop across thyristor T1 is given by VD1 where,
VD1 = RI1
Using Kirchhoff's voltage law where supply voltage is given by Vs yields
Vs = VD1 + (ns - 1)I2R = VD1 + (ns - 1)(I1 - ∆ID)R
Vs = VD1 + (ns - 1)I1R - (ns - 1)∆IDR
Vs = nsVD1 - (ns - 1)∆IDR
Solving the above equation for VD1 the voltage across thyristor T1 yields
The voltage across the thyristor T1 will be a maximum when the off-state current
spread ∆ID is maximum. The worse case steady-state voltage across the thyristor
T1 is seen when no current flows through thyristor T1. Under these conditions,
ID1 = 0
and
∆ID = ID2
The voltage across thyristor T1 is now given as VDS(max), where
During turn-off the differences in forward leakage currents causes differences in
stored charge which in turn causes differences in reverse voltage sharing. This is
shown in figure 4.19. The thyristor with the least recovery charge or with the
lowest reverse recovery time will see the highest transient voltage. The junctions
capacitances which control the transient voltage distribution will be inadequate for
this process and an external capacitor C1 should be used as shown in figure 4.17.
Resistor R1 limits the capacitor discharge current. The same components R1 and C1
are used for both transient voltage sharing and dv/dt protection.
Figure 4.19 Reverse Recovery Time & Voltage Sharing
The voltage difference between thyristors T1 and the other thyristors is given by
where Q1 is the charge stored by thyristor T1 and Q2 is the charge stored by the
other thyristors such that
Q2 = Q3 = Qn
and
Q1 < Q2
The transient voltage across thyristor T1 is given by
The worse case transient voltage occurs when Q1 = 0, hence ∆Q = Q2 and is given
by
The transient voltage derating factor which is normally used to increase the
reliability of the string is given by
Example 4.2
Section 4 - Thyristors
Parallel Operation of Thyristors
When the load current exceeds the rating of a single thyristor, thyristors are
connected in parallel to increase the overall current capability. Contrary to what
might be expected, the load current is not shared equally between the thyristors,
because thyristors are not perfectly matched. Figure 4.20 shows the V-I
characteristics of two thyristors T1 and T2 which are connected in parallel.
Figure 4.20 V-I Characteristics of Two Parallel Thyristors
The thyristor carrying the higher current would dissipate more power which in turn
will increase the junction temperature and hence decrease the internal resistance.
This in turn will increase its current sharing capcity and maybe damage the
thyristor. This process is termed thermal runawy and is not common only to
thyristors. Thermal runaway may be prevented by using one common heat sink to
ensure that both thyristors are operating at the same temperature. Equal current
sharing could be accomplished with the use of a small resistor or inductor in series
with each thyristor as shown in figure 4.21.
Figure 4.21 Current Sharing of Thyristors in Parallel
When a resistor is used to produce equal current sharing, the losses in the series
resistor is very high and may be unacceptable. When magnetically coupled
inductors are used for the purpose of current sharing, if thyristor T1 current
increases, a voltage of opposite polarity to that of the coil in series with T1 will be
induced in the coil in series with thyristor T2. The polarity of this voltage is as such
to increase the anode potential of thyristor T2, thereby increasing the current flow
through this thyristor.
Section 4 - Thyristors
Thyristor Firing Circuits
In thyristor converters high ac voltages exists between anode and cathode of the
thyristor, while low voltage level pulses are placed between gate and cathode.
Isolation is necessary between the gate-cathode circuit and the anode-cathode
circuit. This isolation is accomplished with the use of:
1. Optocouplers and
2. Pulse transformers.
In the case of optocoupler isolation, the low voltage gate drive circuit is optically
isolated from the high voltage anode-cathode circuit as shown in figure 4.22.
Figure 4.22 Optical Isolation Using Photo SCR
In the above circuit, the gate drive circuit is connected to the light emitting diode
D1 via a current limiting resistor R1. Pulses sent to the light emitting diode D1 turns
on the photo SCR T1 which in turn triggers the power thyristor TL. Hence the gate
drive circuit is optically isolated from the output circuit. Instead of using an
optocoupler, a pulse transformer could be used to magnetically isolate the gate
drive circuit from the anode-cathode circuit.
Section 5 - Controlled Rectifiers
Objectives
●
To examine the principles of operation of single-phase and three-phase
controlled rectifiers to convert a fixed ac voltage into a variable dc voltage.
●
Describe the application of thyristors in controlled rectification
●
Analyse and evaluate the performance of controlled rectifiers.
●
●
Examine the problems of poor power factor and harmonic distortion of supply
current due to controlled rectifiers.
Determine the filtering requirements for controlled rectifiers.
Introduction
●
●
●
●
●
As was observed earlier, diode rectifiers produce a fixed output voltage. In
order to control the output voltage of a rectifier, phase control thyristors must
be used instead of diodes.
The output voltages of these rectifiers are varied by varying the delay or firing
angle of the thyristor.
Phase controlled thyristors are turned on by the application of a short pulse to
the gate and they are turned off by the process of natural or line
commutation.
Phase controlled rectifiers are also called ac-dc converters and are widely used
in industrial applications.
There are two types of phase control converters:
1. Single phase converters.
2. Three-phase converters
●
Each type of converter can be divided into three categories:
1. Semiconverter
2. Full converter
3. Dual converter
●
●
●
A semiconverter is a one quadrant converter, having one polarity of voltage
and current.
A full converter is a two quadrant device whose output voltage polarity can be
either positive or negative but whose output current has one polarity.
A dual converter is a four quadrant device whose output voltage and current
can be of either positive of negative polarity.
Electrical Home
Section 5 - Controlled Rectifiers
Single Phase Half-Wave Thyristor Converter With R Load
Figure 5.1 below shows a single phase half-wave thyristor converter with a resistive
load. For the positive half cycle of input voltage, the thyristor T1 is forward biased
and when the thyristor is fired at ωt = α it conducts and the input voltage appears
across the load. When the input voltage goes negative at ωt = π, the thyristor is
reversed biased and it is turned off. The delay angle α, is defined as the time the
input voltage starts to go positive to the time the thyristor is fired.
Figure 5.1 - Single Phase Thyristor Converter With R Load
The average output voltage Vdc is given by
The output voltage Vdc can be varied from Vm/π to zero as the firing angle α varies
from zero to π.
The rms output voltage is given by
Example 5.1
Section 5 - Controlled Rectifiers
Example 5.1
The converter shown below has a purely resistive load R and a delay angle
Determine
a.
b.
c.
d.
e.
the rectification efficiency
the form factor (FF)
the ripple factor (RF)
the transformer utilization factor (TUF)
and the peak inverse voltage of the thyristor
Solution
delay angle,
Vdc = 0.1592 Vm
.
a.
b. Form Factor (FF) =
c. Ripple Factor (RF)
d. Vs= rms voltage of transformer secondary winding
Is= rms value of transformer secondary current
= rms value of load current
e. Peak Inverse Voltage (PIV) = Vm
Section 5 - Controlled Rectifiers
Single Phase Semiconverters
Figure 5.2 shows a single phase semiconverter with a highly inductive load, such
that the load current is assumed continuous and ripple free.
Figure 5.2 - Single-Phase Semiconverter
Click the stop/play button to stop or play the animation or click the step button to
step forward through the animated sequence above.
During the positive half cycle of input voltage, thyristor T1 is forward biased.
During the interval 0 ≤ ωt ≤ α, no supply current can flow through the thyristors
and the load current freewheels through the freewheeling diode Dm. When T1 is
fired at ωt = α, the load is connected to the input supply via T1 and D2 for the
period α ≤ ωt ≤ π.
For the period π ≤ ωt ≤ (π + α), the input voltage is negative and the thyristor T1
and diode D2 will be reversed biased. Under these conditions, freewheeling diode
Dm is forward biased and provides a path for the flow of load current. During the
negative half cycle of input voltage, thyristor T2 is forward biased and when a firing
pulse occurs at ωt = π + α, diode Dm will be reversed biased and the load will be
connected to the supply via thyristor T2 and diode D1.
The average output voltage is found from
The output voltage can be varied from a maximum of 2Vm/π to a minimum of zero
as the firing angle α varies from zero to π.
The rms output voltage is given by
Section 5 - Controlled Rectifiers
Single-Phase Semiconverter With RL Load
The load current of a semiconverter is dependent on the load resistance,
inductance and the battery voltage E in series with the load. The load current of a
semiconductor is made up of two components:
1. That component in the range 0 ≤ ωt ≤ α when the freewheeling diode is
forward biased and load current is provided by the energy stored in the
magnetic field.
2. The component in the range α ≤ ωt ≤ π when load current is provided by the
supply.
Load Current For Interval 0 ≤ ωt ≤ α
In this time range, the freewheeling diode Dm is forward biased and load current is
supplied by the energy stored in the magnetic field of the inductor. The load
current is given by iL1 and the voltage equation for the circuit is given by
.............(1)
.............(2)
where α is given as
Equation (2) is of the form
whose solution is given as
where
Therefore the solution for equation 2 can be written as
Applying the initial conditions
yields
Hence,
At the end of this interval i.e. when ωt = α, the load current is given by IL1, where
Load Current For Interval α ≤ ωt ≤ π
During this interval the thyristor conducts and supply voltage is connected to the
load via thyristor T1 and diode D2.
The load current for this interval is given by iL2 and the voltage equation can be
written as
Solving the above equation for the load current iL2 yields
Now
where
Therefore
where Z is the load impedance given by
The constant C can be determined from the initial conditions
Therefore the load current is given by
When thyristor T1 is turned off at ωt = π, the load current iL2(t) goes back to its
original value IL0 where,
The rms current flowing through one thyristor is given by
The average current flowing through one thyristor is given by
The rms value of load current is given by
The average load current is given by
Section 5 - Controlled Rectifiers
Single-Phase Full Converter
Figure 5.3 shows a single-phase full converter with a highly inductive load so that
the load current is continuous and ripple free. Thyristors T1 and T2 are forward
biased during the positive half cycle of supply voltage. When these two thyristors
are fired simultaneously at ωt = α, the load is connected to the supply via
thyristors T1 and T2. Thyristors T1 and T2 will continue to conduct beyond ωt = π
as a result of the presence of an inductive load. During the negative half cycle,
thyristors T1 and T2 are forward biased and at ωt = π + α, these thyristors are
fired into conduction causing thyristors T1 and T2 to be reversed biased and be
turned off due to line commutation. For the period α ≤ ωt ≤ π, the input voltage and
current are positive and power flows from supply to load and the converter is said
to be in rectification mode.
Figure 5.3 - Single-Phase Full Converter
Click the stop/play button to stop or play the animation or click the step button to
step through the animated sequence above.
For the period π ≤ ωt ≤ π + α, the input voltage is negative, input current is positive
and power flows from the load to the supply. The converter is said to be operated
in the inversion mode. This converter supplies two quadrant operation since the
output voltage can either be positive or negative depending on the value of the
firing angle.
The average output voltage is given as
The output voltage can be varied from 2Vm/π to -2Vm/π when a varies from 0 to π.
The rms output voltage is given by
Section 5 - Controlled Rectifiers
Single-Phase Full Converter with RL load
The output load current of the converter comprises of two components per cycle.
One component flows when thyristors T1 and T2 are fired and connects the supply
voltage to the load and the other component flows when thyristors T3 and T4 are
turned on again connecting supply voltage to the load. Since both components of
current are identical, only one component will be studied. For the interval
α ≤ ωt ≤ (α + π)
the load current is given by
The solution for iL takes the form
at ωt = α, iL = IL0 hence
Hence the load current during this interval is given by
The current magnitude at the end of the first conponent of load current is the same
as that at the beginning of the second component of load current, that is at
ωt = α + α, iL = IL0 and the current IL0 can be obtained by substituting this
condition in the above equation which yields
The value of firing angle a at which current IL0 = 0 can be obtained for known
values of the parameters in the above equation, using an iterative method.
The rms value of thyristor current is given by
The rms output current is given by
The average current of one thyristor is given by
The average output current is given by
Idc = IA + IA = 2IA
Section 5 - Controlled Rectifiers
Three-Phase Half-Wave Converter
Three-phase converters provide higher average output voltage than their single
phase counterparts. In addition, the output voltage ripple frequency is higher in
three-phase converters than in single phase converters. Figure 5.4 shows a threephase half wave converter and the associated waveforms.
Thyristor T1 is fired at ωt = π/6 + α since the phase voltage van is the most
positive phase voltage for the interval π/6 ≤ ωt ≤ 5π/6. Thyristor T2 is fired at
ωt = 5π/6 + α, since vbn is the most positive phase voltage for the next 120°.
When thyristor T2 is fired on, T1 will be turned off due to the line to line voltage
Vab is now negative. Thyristor T3 is fired at ωt = 3π/2 + α. When this happens,
thyristor T2 will be turned off.
Figure 5.4 Three-Phase Half-Wave Converter
Click the stop/play button to stop or play the animation or click the step button to
step forward through the animated sequence above.
In figure 5.4 the load current is continuous since the load is highly inductive. For a
purely resistive load and firing angle α > π/6, the load current would be
discontinuous and each thyristor would be commutated when the polarity of the
phase voltage is reversed. The frequency of the output ripple voltage is 3fs, where
fs is the frequency of the supply. This converter is not normally implemented in
practical circuits because the supply current contain dc components.
For a continuous load current, the average output voltage is given by:
The rms output voltage is obtained from
For a resistive load and α ≥ π/6
And the rms output voltage for a resistive load is given by
Section 5 - Controlled Rectifiers
Three-Phase Semiconverter
Figure 5.5 Three-Phase Semiconverter with Highly Inductive Load
Figure 5.6 Three-Phase Semiconverter for α < π/3
Figures 5.5 and 5.6 show a three-phase semiconverter with a highly inductive load.
For delay angle α ≥ π/3
The delay angle α can be varied between 0 and π. For the period π/6 ≤ ωt ≤ 7π/6,
thyristor T1 is forward biased and if it is fired at ωt = π/6 + α, thyristor T1 and
diode D1 conducts the supply line voltage vac across the load. At ωt = 7π/6, the
line voltage vac starts to go negative. This causes freewheeling diode Dm to
conduct load current causing thyristor T1 and diode D1 to turn off. Each thyristor
conducts for a duration of π - α which is less than 2π/3.
For delay angle α ≤ π/3
Under these conditions, each thyristor conducts along with two diodes (one diode
and a thyristor at any one time) for the interval 2π/3.
Output Voltage of Semiconverter
The three line to neutral voltages are given by:
The line to line voltages are given by:
where Vm is the peak phase voltage.
For delay angle α ≥ π/3
Under these conditions the output voltage is discontinuous and the average output
is given by
The maximum output voltage occurs when α = 0 and is given by
The rms output voltage is given by
For delay angle α ≤ π/3
Under these conditions the output voltage is continuous and the output voltage is
given by
The rms output voltage is given by
Section 5 - Controlled Rectifiers
Three-Phase Full Converter
Figure 5.7 Three-Phase Full Converter
A diagram of a three-phase full converter with a highly inductive load is shown in
figure 5.7 together with the associated voltage and current waveforms. This
converter provides two quadrant operation and thyristors are fired at an interval of
π/3 degrees. Since thyristors are fired every 60°, the frequency of the output ripple
voltage is six times the frequency of the supply voltage. At ωt = π/6 + α, thyristor
T6 is already conducting and thyristor T1 is turned on. For the interval
π/6 ≤ ωt ≤ π/2 thyristors T1 and T6 conduct, and line to line voltage vab appears
across the load. At ωt = π/2 + α, thyristor T2 is turned on and thyristor T6 is
turned off due to natural commutation. This occurs because when thyristor T2 is
turned on the line to line voltage across thyristor T6 is the positive voltage vbc from
cathode to anode which reverse biases thyristor T6. During the interval
(π/2 + α) ≤ ωt ≤ (5π/6 + α), thyristors T1 and T2 conduct and line to line voltage
appears across the load. The firing sequence of the thyristors is: 12, 23, 34, 45, 56
and 61.
Determination of average output voltage and rms value of output voltage.
The line to neutral voltages are defined as:
The corresponding line to line voltages are:
The average output voltage is given by
The maximum output voltage is obtained when α = 0 and is given by
The rms value of output voltage is given by:
Fourier Transform - Revision
Introduction
Periodic phenomena occur quite frequently in Electrical Engineering, indeed the source of electrical
power is a periodic signal so it stands to reason that manipulation of this a.c. source will result in
mostly periodic signals. It is an important practical problem to represent the corresponding periodic
functions in terms of simple periodic functions such as sine and cosine.
Fourier series are series of sine and cosine terms and arise in the important task of representing
complex periodic functions. The theory of Fourier series is quite involved, but the application of these
series is simple.
Recall A function f(x) is called periodic if it is defined for all real x and if there is some positive
number p such that
f(x + p) = f(x)
for all x
................F.1
This number p is called the period of f(x). The graph of such a function is obtained by
periodic repetition of its graph in any interval of length p.
Note that f = c = const is also a periodic function in the sense of the definition,
because it satisfies (F.1) for every positive p.
From (F.1) we have
f(x + 2p) = f[(x + p) + p]
= f(x + p)
= f(x)
So we can say
f(x + np) = f(x)
for all x
................F.2
Furthermore, if f(x) and g(x) have period p, then the function
h(x) = af(x) = bg(x)
(a,b = const)
also has the period p.
The Fourier Coefficients
The Fourier Series, as already mentioned allows us to represent a periodic function as a series of sine
and cosine terms. Say we have such a periodic funtion f(x). Then we can reprensent f(x) as shown
below.
f(x) = a0 + a1cosx + b1sinx + a2cos2x + b2sin2x + .....
Grouping the sine and cosine terms yields,
................F.3
That is we assume that the series converges. We see that each term of the series has the period 2π.
Hence, if the series converges, its sum will be a function of period 2π.
Given such a function we want to find the coefficients a0, an and bn of the corresponding series (F.3).
We determine a0. Integrating on both sides of (F.3) from -π to π, we get
Hence our first result is
................F.4
We now determine a1, a2,... by a similar procedure. We multiply (F.3) by cos mx, where m is any
fixed positive integer, and integrate from -π to π.
................F.5
Recall
The first integral in (F.5) is zero. By applying (1) and (2) to (F.5), we obtain
Integration shows that all the terms on the right of equations (F.6) and (F.7) are zero, except for the
last term in (F.6) which equals π when n = m. Since in (F.5) this term is multiplied by am, the right
side in (F.5) equals amπ. Our second result is (if we write n in place of m).
................F.8
We finally determine b1,b2,..., in (F.3). If we multiply (F.3) by sin mx, where m is any fixed positive
integer, and then integrate from integrate from -π to π.
................F.9
The first integral in is zero. The next integral is of the kind considered before, and is zero for all n =
1,2,... For the last integral, using (3), we obtain
................F.10
The last term is zero. The first term is also zero, except at n = m when it is equal to π. Since in (F.9)
this term is multiplied by bm, the right side in (F.9) equals bmπ. Our final result is (if we write n in
place of m).
................F.10
Summary
So in summary, if we have a function f(x) of period 2π, we can represent it as a sum of sin and
cosine terms given by the fourier series
where
n = 1,2,...
n = 1,2,...
Section 6 - AC Voltage Controllers
Objectives
Introduction
●
●
An ac voltage controller is defined as a circuit consisting of thyristors between
the supply and the load for the purpose of varying the rms value of ac load
voltage.
AC voltage controllers are used in:
a. Industrial heating
b. No-load transformer tap changing
c. Light controls and
d. Speed control of induction motors
●
Two types of control strategy are used to control power flow between supply
and load:
a. On-off control
b. Phase angle control
●
●
●
In on-off control, thyristors connect the load to the ac source for a few cycles
of input voltage and then disconnect it for another few cycles.
In phase angle control, thyristors connect the load to the ac source for a
portion of each cycle of input voltage.
AC voltage controllers are classified as follows:
1. Single phase controllers
a. Unidirectional or half wave control
b. Bidirectional or full-wave control
2. Three-Phase Controllers
a. Unidirectional or half wave control
b. Bidirectional or full-wave control
Section 6 - AC Voltage Controllers
On-off Control
A single-phase full-wave controller is shown in figure 6.1 to explain the principle of
on-off control.
Figure 6.1 - On-off Control
●
●
●
●
●
The thyristor switch which consists of thyristors T1 and T2 connect the ac
supply to the load for a time tn where tn consists of an integral number of
cycles.
It must be noted that the thyristors are turned on at the zero-voltage crossing
of input voltage.
This type of control is applied in applications which have a high mechanical
inertia and high thermal time constant e.g. industrial heating and speed
control of motors.
The harmonics generated by the switching action is reduced due to zerovoltage and zero-current switching.
If a sinusoidal voltage of the form
is connected to the load for n cycles and disconnected for m cycles the rms
output voltage is given by
where
and k is called the duty cycle
●
The input power factor is given by
●
The following must be noted when using an on-off ac voltage controller:
1. The power factor and output voltage varies with the square root of the
duty cycle.
2. If T is the period of the input voltage, then (m+n)T is the period of the onoff control. (m+n)T must be less than the mechanical or thermal time
constant of the load.
3. If m and n are in days, then the square root of k would give erroneous
results if used to compute the power factor.
Section 6 - AC Voltage Controllers
Phase Control
Single-Phase Unidirectional Controller
A single-phase unidirectional phase control ac voltage controller is shown in figure
6.2 below. The power flow to the load is controlled by delaying the firing angle of
thyristor T1 over the positive half cycle of voltage waveform.
Figure 6.2 - Single-Phase Half-Wave Controller
The rms output voltage is given by
The average value of output voltage is given by
It should be noted that as the delay angle is varied between 0 and π, the rms
output voltage varies between Vm/√2 and Vm/2 , while the average output voltage
varies between 0 and -Vm/π.
The output voltage and input current are asymmetrical and contain a dc
component. This may produce a saturation problem if an input transformer exists.
Section 6 - AC Voltage Controllers
Phase Control
Single-Phase Bidirectional Controller With R Load
A single-phase full-wave rectifier with resistive load is shown in figure 6.3 below.
Figure 6.3 - Single-Phase Full-Wave Controller
The problem of dc input current could be prevented with the use of the bidirectional
controller.
Thyristor T1 controls power flow during the positive half-cycle of input voltage,
while thyristor T2 provides control during the negative half cycle of input voltage by
varying the delay angle.
It must be noted that the firing pulses of thyristors T1 and T2 are kept 180 degrees
apart.
If the delay angle of thyristors T1 and T2 are equal and given by α, then the rms
output voltage is given by
Section 6 - AC Voltage Controllers
Phase Control
Single-Phase Bidirectional Controller With RL Load
Figure 6.4 shows a single-phase full-wave controller with an inductive load.
Figure 6.4 Single-Phase Full-Wave Controller with RL Load
If thyristor T1 is fired on the positive half cycle and carries the load current, then,
at ωt = π, when the input voltage starts to go negative, the load current does not
fall to zero due to the load inductance.
The thyristor T1 will continue to conduct until its current falls to zero at ωt = β.
The conduction angle of thyristor T1 is given by δ, where
δ=β-α
The angle δ depends on
1. delay angle α
2. power factor angle of load θ.
The instantaneous thyristor current can be found from the equation
The solution of the above equation is of the form
where load impedance
and
The constant A can be determined from the initial condition:
At ωt = α, i1 = 0 hence A can be given as
therefore the current can be given as
When the thyristor T1 is turned off at ωt = β, the current i1 falls to zero.
Substituting this condition in the above equation yields
from which the angle β which is known as the extinction angle can be determined
by an iterative method.
The rms output voltage is given by
The rms thyristor current is given by
and the rms output current is obtained by combining the rms current of each
thyristor and given as Io where
For resistive loads, the gating signals for a controller could be short pulses.
For inductive loads, short pulses should not be used as gating signals. This can be
explained as follows:
When thyristor T2 is fired at ωt = π + α, thyristor T1 is still conducting due to load
inductance. When thyristor T1 current falls to zero at time ωt = β, the gate pulses
to thyristor T2 has already ceased and hence thyristor T2 will not turn on. This can
be corrected by using continuous gating signals of duration π - α. The disadvantage
of continuous gating signal is an increase in switching losses and hence a larger
isolation transformer is required for the gate circuit.
Section 6 - AC Voltage Controllers
Single-Phase Transformer Tap Changers
The diagram of a single phase transformer tap changer is shown in figure 6.5
below.
Figure 6.5 - Single-Phase Transformer Tap Changer
The waveforms for the transformer tap changer is shown in figure 6.6.
Figure 6.6 Waveforms of Single-Phase Transformer Tap Changer
Three-Phase Half-Wave Controllers
The circuit diagram of a three-phase half-wave controller with a star connected
resistive load is shown in figure 6.7 below.
The current flow to the load is controlled by thyristors T1, T3 and T5 while the three
diodes provide the return current path. The thyristor firing sequence is T1, T3, T5
and current flow through the controller is only possible if at least one thyristor is
conducting.
For rms input phase voltage of Vs, the instantaneous input phase voltages are
given by
and the input line voltages are given by
The input and output waveforms for α = 60o and α = 150o are shown in figure 6.8.
Three conduction ranges would be described for this controller.
For 0° ≤ α ≤ 90°
For this range, two or three devices could conduct simultaneously and the
possible combinations are:
a. two thyristors and one diode
b. one thyristor and one diode
c. one thyristor and two diodes
If three devices conduct, then normal three-phase operation is in effect and the
output phase voltage is the same as the input phase voltage i.e.
van = vAN = √2 Vs sin ωt
If two devices conduct at the same time, current flows only through two lines
and the third line can be considered to be open circuited and the line to line
voltage would appear across the load. The output phase voltage would be one
half the line voltage. Example if c is open circuited, then
Figure 6.8 Waveforms of Three-Phase Half-Wave Controller & Y-Connected
Load
The waveforms of an output phase voltage can be drawn directly from the input
phase and line voltages by noting the output phase voltage corresponds to the
input phase voltage if three devices are conducting, the output phase voltage
corresponds to one-half the input line voltage if two devices are conducting and
the output phase voltage corresponds to zero if all thyristors are turned off.
For this range of firing angle, one thyristor is conducting at any time and the
return path is shared by either one or two diodes.
The rms output phase voltage is given by
Output phase voltage van is made up of the following parts:
1. At ωt = α, thyristor T1 is fired and
❍
line voltage vAB supplies load current via thyristor T1 and diode D6
❍
line voltage vAC supplies load current via thyristor T1 and diode D2
Hence, three devices are conducting and the output phase voltage van is
equal to the input phase voltage vAN.
These three devices conduct until line voltage vBC becomes more positive
than line voltage vAB and this reverse biases diode D6, resulting in only
line voltage vAC supplying load current thereafter.
The three devices conduct for the interval ωt = α to ωt = 2π/3, hence the
output phase voltage van for this interval is given by the input phase
voltage VAN:
2. At ωt = 2π/3 diode D6 commutates and:
line voltage vAC supplies load current via thyristor T1 and diode D2.
Hence, only two devices are conducting and the output phase voltage van
is given as one-half the line voltage vAC.
❍
These two devices conduct until thyristor T3 is fired at ωt = 2π/3 + α for
the vBC waveform which reverse biases thyristor T1 and causes it to
commutate. Now the output sees vAC for the interval ωt = 2π/3 to ωt =
2π/3 + α. If the vAC waveform starts from zero on the ωt axis, then the
output sees vAC from ωt = 2π/3 to ωt = 2π/3 + α, where the vAC
waveform is given by:
vAC = √6Vssinωt
The output phase voltage van for this interval is given by:
3. At ωt = 2π/3 + α thyristor T3 fires and:
line voltage vBA supplies load current via thyristor T3 and diode D4
line voltage vBC supplies load current via thyristor T3 and diode D2.
Hence, three devices are conducting and the output phase voltage van is
equal to the input phase voltage vAN.
❍
❍
These three devices conduct until line voltage vCA becomes more positive
than line voltage vBC and this reverse biases diode D2, resulting in only
line voltage vBA supplying load current thereafter. The three devices
conduct for the interval ωt = 2π/3 + α to ωt = 4π/3, hence the output
phase voltage van for this interval is given by the input phase voltage vAN:
4. At ωt = 4π/3 diode D2 commutates and:
line voltage vBA supplies load current via thyristor T3 and diode D4.
Hence, only two devices are conducting and the output phase voltage van
is given as one-half the line voltage vBA.
❍
These two devices conduct until thyristor T5 is fired at ωt = 4π/3 + α for
the vCA waveform which reverse biases thyristor T3 and causes it to
commutate. Now the output sees vBA for the interval ωt = 4π/3 to ωt =
4π/3 + α. If the vBA waveform starts from zero on the ωt axis, then the
output sees vBA from ωt = 3π/2 to ωt = 3π/2 + α, where the vBA
waveform is given by:
vBA = √6Vssinωt
The output phase voltage van for this interval is given by:
5. At ωt = 4π/3 + α thyristor T5 fires and:
line voltage vCA supplies load current via thyristor T5 and diode D4
❍
line voltage vCB supplies load current via thyristor T5 and diode D6.
Hence, three devices are conducting and the output phase voltage van is
equal to the input phase voltage vAN.
❍
These three devices conduct until line voltage vAB becomes more positive
than line voltage vCA and this reverse biases diode D4, resulting in only
line voltage vCB supplying load current thereafter. The three devices
conduct for the interval ωt = 4π/3 + α to ωt = 6π/3, hence the output
phase voltage van for this interval is given by the input phase voltage vAN:
The full rms output phase voltage over an interval of 2π is given by:
For 90° ≤ α ≤ 120°
When the thyristor anode to cathode voltage goes negative, the thyristor will
commutate, hence the range of the line voltages will change as follows:
1. vAC will be placed across the load for the interval ωt = π/2 and ωt = π
since the thyristor in question will commutate when its anode voltage goes
negative.
2. vBA will be placed across the load for the interval ωt = 3π/2 and ωt = 2π,
since the thyristor in question will commutate when its anode voltage goes
negative.
Hence the rms output voltage for this interval is given by:
For 120° ≤ α ≤ 210°
In this interval, only one thyristor and one diode conduct at the same time
since:
1. with α>=120o, a return diode is reversed biased and only one line voltage
supply appears across the load.
2. The supply voltage vAC is connected to the load for the interval ωt = α -
π/6 and π, since for ωt >= π, vAC is negative and the thyristor in question
commutates.
3. Line voltage vBA will be placed across the load for the interval ωt = 3π/2 2π/3 + α and ωt = 2π, since the thyristor in question will commutate when
its anode voltage goes negative.
Hence the rms output voltage for this interval is given by:
Section 6 - AC Voltage Controllers
Three-Phase Full-Wave Controller
The diagram of a three-phase full-wave controller is shown in figure 6.9.
Figure 6.9 Three-Phase Bidirectional Controller
The unidirectional controller contain dc input current and high harmonic content
due to the asymmetrical nature of the output voltage waveform.
The firing sequence of thyristors is T1, T2, T3, T4, T5and T6.
For rms input phase voltage of Vs, the instantaneous input phase voltages are
given by
and the input line voltages are given by
The waveforms of output phase voltage and input voltages are shown in figure
6.10.
Figure 6.10 Waveforms of 3-Phase Bidirectional Controller
For the range 0° ≤ α ≤ 60°
Two and three thyristors conduct at one time, and the rms output phase
voltage is given by:
For the range 60° ≤ α ≤ 90°
Two thyristors conduct at one time, and the rms output phase voltage is given
by:
For the range 90° ≤ α ≤ 150°
Two thyristors conduct at one time, and the rms output phase voltage is given
by:
Basic Circuit Theory
Basic Definitions
Electron: an indivisible particle of negative charge. The amount of charge is measured in
coulombs (C). The magnitude of the charge associated with an electron is 1.602x10-l9 C.
Current: charge in motion (electrons). Current is measured in units of amperes, or more simply
amp.
Voltage: an electric potential difference that causes electron flow. It is also called electromotive
force (EMF). An analogy often used to describe current and voltage is water in a pipe. Current is
analogous to the flow of water, while voltage is analogous to the pressure.
Conductor: a material that allows a continuous current to pass through it under the action of a
fixed voltage. An example of a good conductor is copper or aluminum which is used in homes and
offices for all electrical connections.
Insulator: the opposite of a conductor, it does not allow a continuous current to pass though it
under the action of a fixed voltage. An example of an insulator is the plastic on electrical cords.
Using our water analogy, a conductor can be envisioned as the region inside a pipe, while an
insulator can be envisioned as the actual material of the pipe which contains the water flow.
Switch: used to control the flow of electrons, or current as it is commonly called. Ideally, a switch
turns on or off instantly, and has no voltage across it while it is conducting. In our water analogy,
an ideal switch would cut the flow immediately, from completely on to completely off in an instant.
Common Passive Circuit Elements
All circuit elements can be separated into two groups: active and passive. The electrical definition
is very similar to the common definition: active circuit elements are capable of delivering power,
while passive elements are capable of receiving, and possibly storing, power. In our water
analogy, a pump would be an active element. A narrow section of pipe that restricts the flow, a
tank, and a water wheel would all be examples of passive elements.
Resistors: circuit elements that literally "resist" current flow. Voltage is higher on the end of the
resistor that sees the current first. Figure 1 shows two schematic representations of a resistor. In
our water analogy, a resistor would be a narrow section of pipe that restricts the flow.
Figure 1. Schematic representations of a resistor
The on-resistance (RDS(on)) of our HEXFET® power MOSFETs is usually one of two parameters
critical to the designer. The other is breakdown voltage (V(BR)DSS) or how much voltage the device
can block when it is off. On-resistance is merely the resistance from drain to source of the power
MOSFET in the "on" state. In the "off" state, the resistance is extremely high, but instead of RDS(off),
we measure it as leakage current, or IDSS.
Capacitors: circuit elements that store electrons. In many instances, they are used as a
rechargeable battery, providing a stable voltage reference far from the input power point. They
have many different uses in electrical circuits in addition to simply storing electrons. There are
many different types of capacitors, including aluminum electrolytic, tantalum electrolytic, ceramic
disk, mica, polycarbonate, polypropylene, and polystyrene.
Two important considerations in the selection of capacitors are equivalent series inductance
(ESL), and equivalent series resistance (ESR). Ideally, these two parameters should be as close
to zero as possible, especially as frequency increases. The capacitors above are mentioned
approximately in order of decreasing ESL and ESR. Aluminum electrolytic capacitors have
extremely high capacitive values, but also high ESL and ESR. This makes them good for dc
applications, such as the capacitors on the output of a bridge rectifier, to provide the dc supply to
the rest of the circuit. Polypropylene and polystyrene capacitors have very low capacitive values,
but also extremely low ESR and ESL values making them good for extremely high frequency
applications.
Stray capacitance exists in all circuits to some extent. While usually to ground, it can occur
between any two points with different potentials. All semiconductor devices have capacitance
between their external terminals, and are specified on the data sheets. Figure 2 shows several
different schematic representations of capacitors. In our water analogy, a capacitor would be a
tank storing water for later use.
Figure 2.Schematic Representations of Capacitors
Stray capacitance is also responsible for electro-static discharge (ESD). ESD is responsible for the shock you
receive in the winter after walking across a carpeted room and touching the doorknob. ESD is particularly
dangerous to MOS-gated semiconductors. The amount of static required to cause damage is so small, that a
person can damage a device without knowing it. This is why anyone who handles MOS-gated semiconductors
must follow strict ESD prevention procedures. Following proper procedures is essential as devices can be
damaged, reducing their lifetime, with no perceivable effects at the time of damage.
Figure 3. Schematic representations of inductors
Inductors: circuit elements that resist change. If, after a period of current flow, an attempt is made to interrupt
the current flow, the inductor will continue to force current. Figure 3 shows the schematic representations of
two different inductors. In our water analogy, an inductor would be a water wheel - it is difficult to start
spinning, but once it is spinning, it is difficult to stop.
Figure 4.Toroidal Inductor
Inductors are typically manufactured by winding wire in a toroidal (donut) shape shown in Figure 4. If the
inductor is wound around a non-ferromagnetic material such as plastic, ceramic, cardboard, or merely air, the
inductance per unit volume is considerably less than if the inductor is wound on a ferromagnetic core. The
upper inductor in Figure 4 depicts an air-cored inductor, while the lower inductor depicts a ferromagnetic cored
inductor. Ferromagnetic refers to magnetic materials, whose characteristics greatly vary.
Figure 5. B-H Characteristics for a Magnetic Material.
Figure S shows the B-H characteristics for a ferromagnetic material where B is the magnetic flux density, and
H is the magnetic field. Operation follows the line, in the direction indicated by the arrow. Although the
explanation of this figure is beyond the scope of this module, some important concepts can be observed
without a thorough understanding of the plot. During operation, the operating point slides along the curve in the
direction of the arrows. If the positive magnetic flux density (B) is not offset by an equal negative magnetic flux
density, the operation curve will slowly creep up, until the material saturates (magnetic flux density (B) is at a
maximum and cannot further increase).
At saturation the inductance drops to the value of an equivalent air-cored inductor, and the current through it is
merely limited by the core's internal resistance which is usually quite low. This is seen at the top of the above
curve where the lines flatten, and further increases in flux density (B) are not allowed. Saturation can be
caused by one of two mechanisms. First, if the magnetic material is underdesigned, and the flux generated by
the current in the winding is greater than the core can handle, the material will saturate. In the above figure,
this would place the operating point at the top of the B-H curve.
The second method applies if the magnetic material is not allowed to reset between consecutive pulses.
Sufficient time between pulses is necessary to allow the energy stored in the magnetic element to go to zero,
or reset. If the design does not allow this to occur, the flux in the magnetic element will build up, or staircase,
with each consecutive pulse until the device saturates. This results in a large current which usually destroys
the semiconductors in its path. This phenomenon also affects transformers which are merely special cases of
the inductor.
Figure 6. Schematic Representation of a Transformer
The final circuit element is the transformer. Figure 6 shows the schematic representation of a transformer. A
transformer could be thought of as a ferromagnetic-cored inductor with two or more sets of wires wound on it.
Saturation is also a problem in transformers. Thus transformers and inductors are sometimes lumped together
and simply called magnetics.
Transformers are most commonly used for one of two purposes. The first is isolation, which is typically needed
between two sections of a system which have different ground levels. The second is to change voltage levels.
A familiar example is the large ac adapter wall plug supplied with most portable equipment for home use. The
adaptor box contains a transformer which steps the voltage down from the line voltage, usually to around 12V,
which is then further conditioned by two diodes, and finally supplied to the equipment.
Leakage inductance is a critical parameter for transformers, generators, and motors. Leakage inductance is
the difference between the self-inductance and the mutual inductance of the primary and secondary windings.
Its value is typically quite small, but very important in determining the characteristics and operation of the
circuit. It is of particular interest as the switching device may be asked to dissipate the energy stored in the
leakage inductance. The leakage inductance contributes to a turn-off voltage spike seen by the switching
device. If the energy and/or voltage is sufficient, a snubber may need to be added to the circuit to protect the
switching device from damage due to this spike. IR specifies the amount of energy HEXFET® power
MOSFETs can dissipate in this mode and are tested as shown in Figure 7, the unclamped inductive test
circuit.
Figure 7. Unclamped Inductive Test Circuit
Basic Electrical Definitions
Power is defined as current multiplied by voltage:
P=Vx I
where P is the power measured in watts (W) (also joules per second), V is the steady state voltage measured
in volts (V), and I is the steady state current measured in amps (A).
Energy is defined as current multiplied by voltage, multiplied by time:
E=IxVxT
where "E" is the energy measured in joules (also watt-seconds), "V" is the instantaneous voltage measured in
volts, "i" is the instantaneous current measured in amps, and "T" is the time period measured in seconds.
To calculate power, given energy and frequency, multiply energy by the frequency. For example, if an IGBT
has a total switching energy loss of 1.4mJ under a given set of operating conditions, and is operated at 20kHz,
the total power loss due to switching will be 28W.
E (1.4mJ) x f (20kHz) = P (28W)
ac versus dc
Direct current (dc) has a constant magnitude. In contrast, alternating current (ac) has a magnitude dependent
on time. it follows a sinusoidal waveform, shown below. ac is generated by moving a copper winding through a
magnetic field. This causes a voltage to be developed on the winding. Generators in the United States operate
at 60Hz, but many places in the world, 50Hz is the standard. Hz is the abbreviation for Hertz, which is the unit
of measure for frequency. Frequency is only defined for regular waveforms that repeat indefinitely. Frequency
is how many times per second the same position on the waveform occurs. Thus, in the figure below, sixty
peaks will pass in one second if the frequency is 60Hz. T is the period, while 1/T is the frequency.
Figure 8. 60Hz Sine Wave.
Nearly all current starts off as ac, which is generated through an electromechanical process, and is then
converted to dc. It is difficult to generate dc directly, as it requires either a dynamo or a chemical reaction such
as the one within in a solar cell which converts sunlight into dc voltage. In applications where dc is present,
there is usually a nearby ac source. For example, in your automobile the battery that drives the lights, all the
electronics, and all the motors are typically 12 volts dc. This battery is charged by the alternator which is
basically a small generator driven by the engine. A three-phase diode bridge is responsible for converting the
ac output of the alternator to be compatible with the dc battery.
The last important concept is the role of frequency on magnetics. It is beyond the scope of this training module
to explain why, but as the operating frequency of a circuit increases, the physical size of the magnetics
(remember this means both inductors and transformers) shrinks. This is one of the reasons designers are
constantly increasing the frequency of their designs. In the power supply world, one of the benchmarks of a
design is how many watts per cubic inch the power supply delivers. One way to substantially increase this
number is by moving to higher frequency, and hence, physically smaller magnetic components. The tradeoff of
higher frequency operation is increased switching losses in the semiconductor devices, whether it be a diode,
IGBT, or power MOSFET.
Basic Semiconductor Theory
Semiconductor is an appropriate name for the device because it perfectly describes the material
from which it's made -- not quite a conductor, and not quite an insulator. To produce a useful
device, semiconductor material is processed into MOSFET, IGBT, SCR, or diode devices, etc.
All IR semiconductor products originate from silicon wafers. Silicon is one of the most common
elements on earth. It is the basis of sand and glass in the form of silicon dioxide (SiO2). After being
refined, silicon is supplied as amorphous silicon which means that the atoms are randomly
arranged in the material. Under the proper conditions, silicon can be manufactured into epitaxial
chunks which basically means a single crystal. Most gems are examples of single crystals.
Diamonds, for example, are merely carbon atoms arranged in a particular 3-D or lattice structure
shown in Figure 9. We make use of silicon in a similar lattice form to manufacture semiconductors.
Figure 9. Silicon Lattice Structure.
A pure silicon lattice is a good insulator because the atoms are arranged so that all electrons are
bonded to a silicon nucleus. In order to change the resistivity of the silicon, it is necessary to
introduce impurities, which changes the number of electrons in the lattice structure. This is called
doping, and may result in extra electrons (called n-type), or missing electrons (called p-type) in the
lattice. Typical dopants include boron and phosphorous.
Missing electrons are also called holes. There is no physical basis for this nomenclature, but it is
easy to understand how it came about. Assume that you have six paper cups and five balls. Line
the cups in a row, and put balls in the five right cups. Now move the ball in the second cup to the
first, the ball in the third to the second cup and so on. It appears that the empty cup is moving to
the right when in reality, the balls are merely shifting to the left. This movement is possible
because of the different number of cups and balls. The significance of this is that electrons can
move approximately two times faster than holes, and as such, n-type material is much preferred to
p-type material.
Semiconductor devices can also be divided into two groups based on how current is conducted
through the material. A device in which the current is conducted by the charges dominant in the
lattice is called a majority carrier device (e.g., electrons in n-type material, or holes in p-type
material. If the current is conducted by charges not dominant in the lattice, the resulting device is
called a minority carrier device (e.g., electrons in p-type material, or holes in retype material). This
very important distinction has a large bearing on the device's operation, especially the recovery
characteristics. A fish tank can be used to illustrate how current is conducted through silicon. If
you put an air hose onto the bottom of the fish tank, it takes some time for the air to bubble out of
the tank. This is analogous to a minority carrier device (current carrier is different from the bulk
media). On the other hand, if you put a water hose onto the bottom of the tank, it doesn't take any
time for the water to reach equilibrium. This is analogous to a majority carrier device (current
carrier is the same as the bulk media).
Basic Semiconductor Processing
Wafers are sliced from a single silicon crystal which has to be "grown." This is done by melting
silicon in a crucible. Pure silicon occurs in two forms - either as a single crystal, or as a collection
of atoms with no particular arrangement, called polysilicon. A "seed" or a small silicon crystal is
inserted into the crucible holding the molten polysilicon. As the seed is slowly drawn out, the
molten silicon aligns with the crystal lattice in the seed. As it cools, the molten silicon expands on
this crystal lattice forming an ingot as shown in Figure 10. The entire ingot is drawn out as a single
crystal made up of many silicon atoms. This ingot is then sliced into thin wafers, and each wafer is
polished to a mirror-like finish. The mirror-like finish of the silicon wafer needs to have a pattern
etched into it to make a useful circuit, or circuit element (discrete).
Figure 10. Monocrystalline Silicon Ingot.
Figure 11 depicts a mask used to transfer the desired pattern onto the silicon wafer. The mask
pattern (either positive or negative) is then projected onto the wafer by one of several different
methods. A particular device, or design, requires a number of different masks - this collection is
known as a mask set. The fewer masks in the mask set. the lower the processing costs.
Figure 11. Mask.
The mask contains one image repeated numerous times. The masks are used in the photolithography process
to transfer the patterns to the silicon wafer. A photo-sensitive material is applied to the wafer, and the mask
exposes certain areas of the wafer. This causes a chemical change in the photo-sensitive material. A chemical
is then used to etch portions away, leaving a pattern on the wafer. This is repeated a number of times, with
some or all of the following intermediate steps occurring between mask steps.
Diffusion is the process by which dopants are added to the wafer. By using the appropriate mask, a certain
pattern is diffused into the wafer. Diffusion is usually followed by a drive-in step by heating the wafer for a
particular amount of time. Controlling the time and temperature defines the profile diffused into the wafer.
The diffusion process must be tightly controlled. With advances in IC processing, a method called ion
implantation has been developed. Instead of controlling the time and temperature of a diffusion furnance, ion
implantation makes use of an extremely high voltage electron gun which accelerates the dopants, and "shoots"
them into the wafer. By adjusting the high voltage, the implant depth is controlled. It is possible to get very
precise profiles by using this method.
The "wires" of the integrated circuit world are constructed using either metal, or polysilicon, which may be
heavily doped to reduce its resistance. Either of these materials can be used to cover the entire surface, or to
make tightly controlled patterns via the aforementioned processes. Sometimes problems happen with "step
coverage" of the metal as shown in Figure 12., when the metal has to cover what looks like a single stair step.
The metal thickness tends to thin at the outer most portion of the step, and can lead to failure if the metal
becomes discontinuous across this step. Silicon dioxide is commonly used to insulate the metal from
contacting other layers.
Figure 12. Metal Step Coverage.
Device Cross Section
The following sections show the cross sectional views and describes the operation of six different
devices. More detailed information is available in each device's specific Training Module. The goal of
this section is to compare and contrast the various devices: pn diode, Schottky diode, SCR, MOSFET,
IGBT, and Control IC.
pn Diode
As shown in Figure 13, the top metal is the anode, while the bottom is the cathode. The action occurs
at the interface, called the junction, between the implanted p-type and n-type materials. When a
positive voltage is applied between the anode and cathode, current will flow through the diode,
provided the voltage is greater than "a diode drop" which, for standard pn diodes, is usually around
0.7V. As the forward current (IF) increases, the voltage drop (VF) will also increase. However, most of
the voltage drop is the initial 0.7V drop which occurs when any amount of current flows through the
diode.
Figure 13. PN Diode
If a negative voltage is applied across the pn junction (anode to cathode), the device exhibits very high
resistance to current flow, and the small amount of current that does conduct is called the leakage
current (IRM).
When a diode is conducting in the forward direction and is asked to block in the reverse direction, it
"forgets" it is a diode for a period of time and allows current to conduct. After this short period of time,
called the reverse recovery time, or trr, the diode "remembers" it is a diode and begins blocking
current. However, during this recovery time, a large current conducts through the diode, called reverse
recovery current, or Irr. The shape of the waveforms during this period are critical to the operation of
the rest of the circuit, which is why IR has developed the HEXFRED® diode, an ultra-fast, but ultra-soft
diode unlike snappier diodes from our competitors that usually cause excessive voltage ringing in the
circuit. As temperature increases, the forward voltage decreases, while the reverse recovery current
and charge increase.
Schottky Diode
As shown in Figure 14, the Schottky diode is very similar to a standard pn diode, but instead of having an
implanted p-layer, the action occurs at the interface between the barrier metal and the silicon. The guard rings are
used to make the device's reverse breakdown characteristics more rugged. Since both metal and the silicon are ntype materials, the conduction occurs through majority carriers only, with no minority carrier injection, storage, or
recombination . This explains the Schottky diode's lack of reverse recovery, making it ideal for high frequency
applications.
Figure 14. Schottky Diode
The barrier metal also is responsible for the Schottky diode's low forward voltage drop, making it ideal for use in
low voltage systems. Of course, the tradeoff is the reverse leakage current, which is many times that seen in pnjunction diodes. In some applications, and especially during burn-in, this leakage current may cause the device to
exceed its rated junction temperature. It needs to be included in any junction temperature calculations. As
temperature increases, the forward drop decreases, while the reverse leakage current greatly increases.
Figure 15. SCR
SCR
The SCR (silicon controlled rectifier), or thyristor, is one of the original high power semiconductor switching
technologies. As shown in Figure 15, the SCR is a four layer device, npnp from top to bottom (or cathode to
anode). It is a latching device; once it is turned on, or "fired," it remains on until the current is removed. For this
reason, its primary application is phase-control of ac signals. Figure 16 shows that by controlling where on the
cycle the SCR is turned on, the output power level is controlled. SCRs designed for these line frequency (50-60
Hz) applications are called phase control SCRs.
Figure 16. Phase Control of ac Waveform
The second family of SCRs is the inverter type. These are used in pulsed power applications involving higher
frequencies. The main difference between the two families is the turn-off time (tq). A device's tq is measured as the
time required for the device to be in the "off" state before voltage is reapplied. Inverter SCRs typically have a tq of
less than 30 microseconds (ms). Similar phase control SCRs have tq ratings of several hundred ms.
Like with the pn diode, as the temperature increases, the voltage drop decreases. Perhaps more importantly, as
temperature increases, the current required to fire the SCR decreases. At low temperatures, the gate triggering
circuitry must supply enough current to ensure the device fires, while at high temperatures, the SCR is susceptible
to spurious firing due to noise. The device's gate triggering circuitry must ensure this does not happen.
Figure 17. HEXFET power MOSFET
MOSFET
As shown in Figure 17 above, the HEXFET® power MOSFET is named for the hexagonal shape of its individual
cells. Current flows from the source metallization down through the device, and out through the drain contact.
Vertical current flow is the reason the HEXFET is also called a vertical MOSFET. Nearly all power MOSFETs on
the market employ this vertical structure.
The MOSFET (Metal Oxide Semiconductor Field Effect Transistor) is used primarily in medium-power circuits
where switching speed is critical. This power device is extremely easy to drive as it requires voltage, not current on
the gate. And due to its wide acceptance, the MOSFET market is growing at a rapid pace.
It is important to select the proper voltage MOSFET as the RDS(on) increases exponentially with increasing
breakdown voltage. Also, as the device heats up due to power dissipation, its RDS(on) increases. Thus, in most
applications, the 25 degC RDS(on) rating is not accurate. The actual RDS(on) rating could be twice as high. This is not
always a negative attribute. It's what allows power MOSFETs to easily be used in parallel.
IGBT
The IGBT (Insulated Gate Bipolar Transistor) represents the union of a power MOSFET and a power bipolar (BJT)
transistor, incorporating the best features of each. But while the MOSFET can be used in applications exceeding 1
MHz, the fastest IGBTs are limited to only a fraction of that. Therefore, the only real drawback of the IGBT is its
switching speed. Yet the conduction characteristics of the IGBT really outshine those of the power MOSFET,
especially at voltages greater than about 200V. If you have a midfrequency, high voltage design, look at IGBTs.
Increasing the operating temperature of an IGBT causes its switching losses to increase substantially, thus
decreasing the maximum operating frequency. The conduction voltage of IGBTs is not strongly affected by
temperature, and can even decrease with temperature at certain current densities. This causes concern for
designers with regards to the IGBT's parallelibility. Still, IR believes that by following a few simple guidelines,
IGBTs can be successfully paralleled. (See Design Tip 94-6 for more information on parallel operation of IGBTs.)
Control IC
In 1987, IR introduced the IR2110, a half-bridge, high voltage, MOS Gate Driver, or Control IC . Since that time,
many variations on the IR2110 have been introduced. These devices are unique in that they simplify the drive
circuitry required to drive a high-side, n-channel MOSFET at voltages up to 600V. These devices allow designers
to reduce the parts count and design time for the drive circuitry. Figure 18 shows a cross-section of the die. One of
the most successful devices in this family is the IR2151, a self-oscillating, half-bridge driver designed primarily for
electronic ballast applications.
Future 18. Control IC
Modes of Operation
The switching devices (MOSFET, IGBT, and SCR) can be operated in one of several modes. The SCR is unique in
this group in that it is a self-commutating device, which means the user is required to turn on, or "fire" the SCR. But
at the zero crossing, the SCR switches off. This makes the SCR extremely useful in ac applications, or where the
current decays to zero at the point when the SCR should be turned off. One limitation of the SCR is that it cannot
easily be used to switch dc loads.
By far the largest number of IR devices are used in hard switching as depicted in Figure 19. The switching
waveforms on data sheets are for hard switching. The conditions are very difficult on the switching device, i.e., high
current must be switched off to high voltage. Considerable power is dissipated in the switching interval due to these
conditions.
Figure 19. Hard Switching Waveforms
Pulse-Width Modulation (PWM) is a special case of hard switching. In many applications, it is desirable to replicate
a sine wave as shown in Figure 20. One way to accomplish this is to approximate the sine wave with narrow
square pulses of varying duty cycle. After this waveform is smoothed, typically by an inductor, it appears very
similar to the desired sine wave. The frequency of the desired sine wave is called the carrier frequency, while the
frequency at which the switch operates is called the modulation frequency. To make the replicated waveform
closely match the desired waveform, the modulation frequency is usually at least ten times the carrier frequency.
Some applications employ resonant mode switching. In these applications, the current and/or voltage is a sine
wave as opposed to the square waves common in PWM techniques. Operation in the resonant mode has lower
switching losses, and is used with devices that have high switching losses, or to push operating frequency higher.
Figure 20. Sine Wave Generation through PWM.
Some applications operate in the linear mode. As applied to switching devices, linear mode means that discrete
changes in the control signal result in proportional discrete changes in the output. When a circuit is operated in the
linear mode, the switching element limits the current in the circuit, while normally the circuit itself, rather than the
switching element, limits the current. This limitation leads to high power dissipation in the switching device.
Parallel operation of semiconductors requires extra effort on the part of the design engineer. When operating
semiconductors in parallel, the critical parameter is the temperature coefficient. The temperature coefficient reflects
how the semiconductor's voltage drop responds to changes in temperature. In general, semiconductors have a
negative temperature coefficient, with the one notable exception of the power MOSFET in that its temperature
coefficient is positive. A negative temperature coefficient means that the voltage drop across the semiconductor
decreases as the temperature increases. This causes problems such as current hogging, thermal runaway, and
hotshots. As the device heats up due to normal power dissipation, the voltage drop decreases. This allows more
current to flow, generates more heat, and further reduces the voltage drop, creating a regenerative effect. When
paralleling any IR device (other than the MOSFET) special design considerations are necessary to prevent
potential current sharing problems.
The following documents provide more information on this subject: IR Design Tip 94-6A, "Parallel Operation of
IGBTs"; IR Application Note AN-990, "Application Characterization of IGBTs"; and "Paralleling of Power MOSFETs
for Higher Power Output," by James B. Forsythe (PowerCon '81).
Packaging
Many IR devices are available either in plastic or hermetically sealed metal packages. In the past, the deciding
which package to use was divided between commercial versus military. Today the division is not as clear. Military
does not automatically mean hermetic, nor does hermetic automatically mean more reliable. However, most space
applications still require hermetic packages. In applications where the device is exposed to high temperature, and
high humidity, a hermetically packaged device will improve its reliability. The TO-3, once the power transistor
package, is no longer competitive with the new package styles. The TO-3 is difficult to heat sink, and must be
isolated externally. New packages, such as the TO-254 (M-Pak) are tab mounted, and isolated, making assembly
much easier.
Most heat sinkable plastic packages have a metal tab connected to the heat sink by the user. The die is mounted
directly on this tab, usually the positive terminal of the device. In some applications, it is desireable to have the heat
sink grounded, while in other applications it is easier to insulate the heat sink from the rest of the system. In cases
with a grounded heat sink, it is necessary to isolate the device from the heat sink. For this purpose, IR
manufactures special versions of the TO-220 and TO-247 packages called Full-Paks that have a very thin plastic
coating on the exposed metal of the device. The plastic provides up to 2500Vrms isolation voltage, while being thin
enough to only moderately increase the thermal resistance of the mounting system. As an alternative to buying
these Full-Pak devices from IR, the user can isolate the back of the package from the heat sink using a thermally
conductive, electrically isolating material as shown in Figure 22. Some common types are mica, pressed ceramic
wafer, polyimide, and elastomeric insulators, with the latter gaining more and more popularity. The screw hole in
the TO-247ac package is already isolated, so only the back of the package needs to be isolated. The screw hole of
the TO-220AB is not isolated, so the user must isolate it. One way is by using a nylon shoulder washer with a
standard steel screw. The steel screw is typically 4-40 whereas a 6-32 screw is normally used when isolation is of
no concern.
The extremely broad R product line is packaged in anything from a tiny,3-lead surface mount package (barely
visible to the naked eye) to a huge "hockey puk" package (greater than 4" in diameter!). IR application note AN-995
discusses IR's various surface mount packages, and how to mount them. Most high power products (SCRs, and
diodes in both discrete and module packages) are simple to mount. These devices are typically large, and
connnected using large bolts, or are stud mounted. Remember, however, not to exceed the torque or force
specified on the data sheet. For the hockey puk packages, a suitable mounting clamp must be used as shown in
Figure 22 below. In fact, the puk will appear open-circuited if pressure is not applied since this is a compression
bonded device.
Figure 22. Isolation System for Standard TO-220AB and TO-247AC
Figure 23. Mounting Hockey Puk SCRs and Diodes
Mounting Techniques
Most mounting questions concern the TO-220AB and TO-247AC package styles used to house
diodes, MOSFETs, and IGBTs, and the three methods used to make electrical connections to the
die within these packages: wire bonding, soldering, and compression mounting. Each method has
its own advantages and disadvantages.
Figure 23. Wire Bonding
Wire bonding, shown in Figure 23, uses a small diameter (typically <=20 mils) wire that is ultrasonically
bonded (melted) at each connection point. Advantages: wire bonding is quick and easy, and can be completely
automated. Disadvantages: increased voltage drop due to the small wires, low fusing current, expensive
equipment required.
Solder mounting shown in Figure 24 below is typically used in smaller diodes (< 10A), mostly the familar axialleaded diode.
Figure 24. Solder Mounting
Some smaller IR Schottky diodes (SMB and SMC) also use this technique. Advantages: both the voltage drop,
and fusing current are improved. For example, the 30BQ015 Schottky diode is rated at 3 amps and 15 volts,
but the surge rating is 650 amps because the die is soldered directly to the leadframe. Disadvantage: both
sides of the die must be solderable.
Figure 25. Compression Bonding
Compression bonding (Figure 25 above) is used in devices where power cycling capability is required, typically
high power diodes and SCRs. Advantage: in these applications, compression bonding makes a much better
connection because there is no fixed (soldered) interface to fatigue, and the fusing current is very high.
Disadvantage: compression bonding is more expensive, and requires physically rugged die. All IR hockey
puks are compression bonded. Some stud mounted diodes and SCRs, and some diode and SCR modules are
also compression bonded.
Topical Applications
International Rectifier manufactures components used in the efficient control of energy. High
efficiency is a major requirement because of today's rising energy costs and the need for
battery-powered systems to be able to run extended periods of time.
In Figure 26, the line power or raw energy source may be an electric utility, automobile
alternator (ac), battery (dc), or a power supply. IR components take this raw power and
condition it into more useful energy. Some examples are switching off unneeded circuits in
portable electronics, variable ac to control motors, and variable dc to control motors, electronic
lamp ballasts, etc. All IR components fit into the basic power conversion function sub-blocks:
Figure 26. The Basic Power Conversion Functions
These blocks are similar to the USDA's Basic Food Groups wherein each meal should include
one item from each group. Similarly, designs will often need one or more items from each
group. The following sections discuss each block in detail, the associated IR product, and how
to use them.
Input Rectification
Input rectification is the first stage in most electronic devices using ac power. The design uses
four diodes/rectifiers arranged in a bridge configuration for single-phase inputs, or six diodes
arranged in a bridge for three-phase inputs. Standard recovery diodes are specified since the
speed of the diode is not important. Standard recovery diodes have excellent forward voltage
drop, and have lower relative costs than other families of diodes.
Figure 27. Bridge Rectifier
A bridge rectifier (both single- and three-phase shown in Fig. 27 above) converts the ac waveform on the left to
the dc level on the right. IR sells both single- and three-phase bridges in plastic isolated-base modules. Diode
bridges can also comprise discrete diodes, or doubler diode modules (two diodes in series). The choice
depends on the desired mounting method, and the current requirement.
In the above example, the dc voltage will equal the peak of the ac input voltage due to the capacitor on the
output of the bridge rectifier. Since ac voltage is measured in volts-rms, peak voltage is equal to 1.414 times
the rms voltage. Select the rectifiers based on this peak voltage, allowing extra margin for high line conditions.
SCRs may be used to limit current, or output voltage. In some cases, designers require SCRs in their input
bridges. Phase control SCRs are used in smart bridge configurations, so called because they can be used to
limit inrush current.
A problem with passive rectification is that current is only drawn from the line when the line voltage exceeds
the capacitor voltage. This results in a current waveform shown on the left in the above diagram. The
mismatch between the current and voltage waveforms is called power factor, and results in inefficient
utilization of the power source. Additionally, some government regulations require new designs to meet a
specific power factor.
Typical Applications
A boost converter (shown below) is the most common circuit used to improve the power factor. As shown on
the left, the voltage and current waveforms are very similar. This results in near-perfect power factoring.
Figure 28. Boost Converter
Figure 29. DC Chopper
DC Chopper
The dc output voltage is in the 400 to 500 volt range, and calls for a 450 to 600 volt rating on both the power
switch and the diode. IR has focused on this application with its low gate charge HEXFET® power MOSFETs
and HEXFRED® ultrafast diodes. This configuration is one most commonly used in power electronics circuits.
Depending on the application, a transformer may be used to provide voltage isolation or to change voltage
levels according to the turns ratio between the primary and secondary sides. This single switch, single ended
configuration is also known as a forward converter in the power supply world. In power supply applications, the
inductance and the freewheeling diode are on the secondary side of the transformer.
Much emphasis is placed on the selection of the switch (either a MOSFET or IGBT), but the selection of the
diode is equally important. The diode characteristics affect the operation of the switch itself. For low voltage
applications, a Schottky diode is ideal. For higher voltage applications, IR's HEXFRED is an excellent choice.
Figure 30. Half Bridge
Half Bridge
The half-bridge is a higher power version of the previous circuit. It is the workhorse of the power electronics
industry, finding use in power supplies, motor controls, and lighting ballasts. In a power supply, the inductor is
actually a transformer. In a motor drive, the inductor is the motor windings. In a lighting ballast, the inductor is
in series with the lamp, which is in parallel with the lower capacitor.
The greatest design problem with the half-bridge configuration is driving the upper switch, which can be either
an IGBT or a HEXFET. To properly drive a MOS-gated transistor, the gate voltage must be greater than the
emitter/source voltage by approximately 15V. When the emitter/source terminal is connected to a fixed voltage
reference (i.e., ground in the case of the bottom switch), this is a simple task. However, the emitter/source of
the high-side switch swings between ground (when the lower switch is conducting), and nearly the positive rail,
which requires the gate voltage to be above the positive rail. This is typically a problem, since the positive rail
is usually the highest voltage available in the system.
Several methods are used to solve the problem of driving the upper switch (for a list, see AN978A), all of which
are relatively complex, and each has drawbacks. IR produces a range of devices to solve this problem: the
IR2100 Series Control IC drivers. They use a technique called bootstrapping to generate the gate drive signal
for the upper switch. IR manufactures a line of these devices that focus on the electronic ballast market, the
IR215x. They include both control circuitry and gate drive circuitry--all on one chip.
In even higher power circuits, the two capacitors on the right side of the circuit may be replaced with a halfbridge identical to the one on the left. This is called a full bridge. In addition to higher power applications, full
bridges are also used in reversible dc motor drives. The configuration allows voltage to be applied to the load
in both directions.
Typical Applications
Figure 31. Three-Phase Bridge
The three-phase bridge can be thought of as three half-bridges. Three-phase outputs are mainly used for
motor drives or ac inverters. The IR2130 Control IC has been designed specifically for this application.
Figure 32. Push-Pull Configuration
The push-pull configuration is used in power supply and low power UPS systems.
Figure 33. Two Transistor Forward Converter
The two transistor forward converter topology is commonly used in power supply and switched reluctance (SR)
motor designs. Its advantage is that the voltage requirement of the switch and freewheeling diode is half that of
the one transistor forward topology, often used in off-line power supplies requiring the greater capabilitiesof
800 volt HEXFET® power MOSFETs. The increased benefits and lower cost outweigh the complexity of the
design. Typical applications include low power converters with high switching frequencies.
Figure 34. High Side dc Switch
Placing the switch on the high side offers protection against the most common short circuit, a short to the
chassis (ground). If this occurs, the load is connected to ground on both ends, so the ground is unenergized.
However, if the switch and load were reversed, a short to ground would energize the load.
IR manufactures the lR62xx series of high side intelligent switches. They offer overcurrent, overtemperature,
and ESD protection. And because these devices accept ground-referenced logic-level signals as control, the
problem of driving a high-side switch is addressed.
In some applications, high-side intelligent switches can be used to replace electromagnetic relays. However,
applications that require voltage isolation must use IR microelectronic relays. These solid state devices cost
more than equivalent mechanical devices, but feature superior reliability. Additionally, microelectronic relays
are available to control ac loads. This line has found application in telecommunications, instrumentation, and
process control.