Chapter 2
Diode Circuits or Uncontrolled Rectifier
2.1 Introduction
The only way to turn on the diode is when its anode voltage becomes higher than
cathode voltage as explained in the previous chapter. So, there is no control on the
conduction time of the diode which is the main disadvantage of the diode circuits. Despite
of this disadvantage, the diode circuits still in use due to it’s the simplicity, low price,
ruggedness, ….etc.
Because of their ability to conduct current in one direction, diodes are used in rectifier
circuits. The definition of rectification process is “ the process of converting the
alternating voltages and currents to direct currents and the device is known as rectifier” It
is extensively used in charging batteries; supply DC motors, electrochemical processes and
power supply sections of industrial components.
The most famous diode rectifiers have been analyzed in the following sections. Circuits
and waveforms drawn with the help of PSIM simulation program [1].
There are two different types of uncontrolled rectifiers or diode rectifiers, half wave and
full wave rectifiers. Full-wave rectifiers has better performance than half wave rectifiers.
But the main advantage of half wave rectifier is its need to less number of diodes than full
wave rectifiers. The main disadvantages of half wave rectifier are:
1- High ripple factor,
2- Low rectification efficiency,
3- Low transformer utilization factor, and,
4- DC saturation of transformer secondary winding.
2.2 Performance Parameters
In most rectifier applications, the power input is sine-wave voltage provided by the
electric utility that is converted to a DC voltage and AC components. The AC components
are undesirable and must be kept away from the load. Filter circuits or any other harmonic
reduction technique should be installed between the electric utility and the rectifier and
between the rectifier output and the load that filters out the undesired component and
allows useful components to go through. So, careful analysis has to be done before
building the rectifier. The analysis requires define the following terms:
The average value of the output voltage, Vdc ,
The average value of the output current, I dc ,
The rms value of the output voltage, Vrms ,
The rms value of the output current, I rms
The output DC power, Pdc = Vdc * I dc
(2.1)
The output AC power, Pac = Vrms * I rms
(2.2)
P
The effeciency or rectification ratio is defiend as η = dc
(2.3)
Pac
The output voltage can be considered as being composed of two components (1) the DC
component and (2) the AC component or ripple. The effective (rms) value of the AC
component of output voltage is defined as:-
18
Diode Circuits or Uncontrolled Rectifier
2
2
(2.4)
Vac = Vrms
− Vdc
The form factor, which is the measure of the shape of output voltage, is defiend as
shown in equation (2.5). Form factor should be greater than or equal to one. The shape of
output voltage waveform is neare to be DC as the form factor tends to unity.
V
(2.5)
FF = rms
Vdc
The ripple factor which is a measure of the ripple content, is defiend as shown in (2.6).
Ripple factor should be greater than or equal to zero. The shape of output voltage
waveform is neare to be DC as the ripple factor tends to zero.
2
2
2
Vrms
− Vdc
Vac
Vrms
=
=
− 1 = FF 2 − 1
2
Vdc
Vdc
Vdc
The Transformer Utilization Factor (TUF) is defiend as:P
TUF = dc
VS I S
RF =
(2.6)
(2.7)
Where VS and I S are the rms voltage and rms current of the transformer secondery
respectively.
Total Harmonic Distortion (THD) measures the shape of supply current or voltage. THD
should be grearter than or equal to zero. The shape of supply current or voltage waveform
is near to be sinewave as THD tends to be zero. THD of input current and voltage are
defiend as shown in (2.8.a) and (2.8.b) respectively.
THDi =
THDv =
I S2 − I S21
I S21
VS2 − VS21
VS21
=
=
I S2
I S21
−1
VS2
VS21
(2.8.a)
−1
(2.8.b)
where I S1 and VS1 are the fundamental component of the input current and voltage, I S
and VS respectively.
Creast Factor CF, which is a measure of the peak input current IS(peak) as compared to its
rms value IS, is defiend as:I S ( peak )
(2.9)
CF =
IS
In general, power factor in non-sinusoidal circuits can be obtained as following:
P
Real Power
PF =
=
= cos φ
(2.10)
Apparent Voltamperes VS I S
Where, φ is the angle between the current and voltage. Definition is true irrespective for
any sinusoidal waveform. But, in case of sinusoidal voltage (at supply) but non-sinusoidal
current, the power factor can be calculated as the following:
19
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Average power is obtained by combining in-phase voltage and current components of
the same frequency.
P
V I1 cos φ1 I S1
(2.11)
PF =
=
=
cos φ = Distortion Factor * Displaceme nt Faactor
VS I S
VS I S
IS
1
Where φ1 is the angle between the fundamental component of current and supply
voltage.
Distortion Factor = 1 for sinusoidal operation and displacement factor is a measure of
displacement between v(ωt ) and i (ωt ) .
2.3 Single-Phase Half-Wave Diode Rectifier
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 with respect to this high
voltage. It is quite often justifiable to use the ideal diode model. An ideal diode has zero
conduction drops when it is forward-biased ("ON") and has zero current when it is reversebiased ("OFF"). The explanation and the analysis presented below are based on the ideal
diode model.
2.3.1 Single-Phase Half Wave Diode Rectifier With Resistive Load
Fig.2.1 shows a single-phase half-wave diode rectifier with pure resistive load.
Assuming sinusoidal voltage source, VS the diode beings to conduct when its anode
voltage is greater than its cathode voltage as a result, the load current flows. So, the diode
will be in “ON” state in positive voltage half cycle and in “OFF” state in negative voltage
half cycle. Fig.2.2 shows various current and voltage waveforms of half wave diode
rectifier with resistive load. These waveforms show that both the load voltage and current
have high ripples. For this reason, single-phase half-wave diode rectifier has little practical
significance.
The average or DC output voltage can be obtained by considering the waveforms shown
in Fig.2.2 as following:
π
Vdc
V
1
=
Vm sin ωt dωt = m
π
2π
∫
(2.12)
0
Where, Vm is the maximum value of supply voltage.
Because the load is resistor, the average or DC component of load current is:
V
V
I dc = dc = m
R π R
(2.13)
The root mean square (rms) value of a load voltage is defined as:
π
Vrms =
V
1
Vm2 sin 2 ωt dωt = m
2π
2
∫
(2.14)
0
Similarly, the root mean square (rms) value of a load current is defined as:
V
V
I rms = rms = m
R
2R
20
(2.15)
Diode Circuits or Uncontrolled Rectifier
It is clear that the rms value of the transformer secondary current, I S is the same as that
of the load and diode currents
V
Then I S = I D = m
(2.15)
2R
Where, I D is the rms value of diode current.
Fig.2.1 Single-phase half-wave diode rectifier with resistive load.
Fig.2.2 Various waveforms for half wave diode rectifier with resistive load.
21
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The
efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of
diode D1 and (f) Crest factor.
Solution: From Fig.2.2, the average output voltage Vdc is defiend as:
π
Vdc
V
V
1
=
Vm sin(ωt ) dωt = m (− cos π − cos(0)) = m
2π
2π
π
∫
0
Then, I dc =
Vdc Vm
=
R πR
π
Vrms
V
1
=
(Vm sin ωt ) 2 = m ,
2π
2
∫
0
I rms =
Vm
V
and, VS = m
2R
2
The rms value of the transformer secondery current is the same as that of the load:
V
I S = m Then, the efficiency or rectification ratio is:
2R
Vm Vm
*
Pdc
Vdc * I dc
π πR
η=
=
= 40.53%
=
Vm Vm
Pac Vrms * I rms
*
2 2R
Vm
V
π
(b) FF = rms = 2 = = 1.57
Vm 2
Vdc
π
Vac
= FF 2 − 1 = 1.57 2 − 1 = 1.211
Vdc
Vm Vm
P
π π R
(d) TUF = dc =
= 0.286 = 28.6%
Vm Vm
VS I S
2 2R
(e) It is clear from Fig2.2 that the PIV is Vm .
I S ( peak ) Vm / R
(f) Creast Factor CF, CF =
=
=2
IS
Vm / 2 R
(c) RF =
2.3.2 Half Wave Diode Rectifier With R-L Load
In case of RL load as shown in Fig.2.3, The voltage source, VS is an alternating
sinusoidal voltage source. If vs = Vm sin (ωt ) , vs is positive when 0 < ω t < π, and vs is
negative when π < ω t <2π. When vs starts becoming positive, the diode starts conducting
and the source keeps the diode in conduction till ω t reaches π radians. At that instant
defined by ω t =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
22
Diode Circuits or Uncontrolled Rectifier
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 waveforms shown
in Fig.2.4.
When vs changes from a positive to a negative value, the voltage across the diode
changes its direction and there is current through the load at the instant ω t = 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 as reverse bias voltage.
An expression for the current through the diode can be obtained by solving the
deferential equation representing the circuit. It is assumed that the current flows for 0 < ω t
< β, where β > π ( β is called the conduction angle). 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 < ω t < 2π, the diode blocks current and acts as an open
switch. For this period, there is no equation defining the behavior of the circuit.
For 0 < ω t < β, the following differential equation defines the circuit:
di
(2.17)
L + R * i = Vm sin (ωt ), 0 ≤ ωt ≤ β
dt
Divide the above equation by L we get:
V
di R
(2.18)
+ * i = m sin (ωt ), 0 ≤ ωt ≤ β
dt L
L
The instantaneous value of the current through the load can be obtained from the
solution of the above equation as following:
R
R
⎤
− ∫ dt ⎡
∫ dt Vm
L
L
⎢
i (t ) = e
e
*
sin ωt dt + A⎥
(2.19)
L
⎢
⎥
⎣
⎦
Where A is a constant.
R
R
⎤
t V
− t⎡
(2.20)
Then; i (t ) = e L ⎢ e L * m sin ωt dt + A⎥
L
⎢
⎥
⎣
⎦
By integrating (2.20) we get:
∫
∫
i (t ) =
Vm
R 2 + w 2 L2
(R sin ωt − ωL cosωt ) +
R
− t
Ae L
23
(2.21)
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Fig.2.3 Half Wave Diode Rectifier With R-L Load
Fig.2.4 Various waveforms for Half wave diode rectifier with R-L load.
Assume Z∠φ = R + j wL
Then Z 2 = R 2 + w 2 L2 ,
R = Z cos φ , ωL = Z sin φ and tan φ =
Z
ωL
wL
R
Substitute these values into (2.21) we get the following equation:
R
− t
V
i (t ) = m (cos φ sin ωt − sin φ cosωt ) + Ae L
Z
Φ
R
R
− t
V
Then, i (t ) = m sin (ωt − φ ) + Ae L
Z
(2.22)
24
Diode Circuits or Uncontrolled Rectifier
The above equation can be written in the following form:
ωt
R
−
ωt
−
Vm
Vm
ωL
sin (ωt − φ ) + Ae
=
sin (ωt − φ ) + Ae tan φ
i (t ) =
Z
Z
(2.23)
The value of A can be obtained using the initial condition. Since the diode starts
conducting at ω t = 0 and the current starts building up from zero, i (0 ) = 0 (discontinuous
conduction). The value of A is expressed by the following equation:
V
A = m sin (φ )
Z
Once the value of A is known, the expression for current is known. After evaluating A,
current can be evaluated at different values of ωt .
ωt ⎞
⎛
−
Vm ⎜
tan φ ⎟
(2.24)
i (ωt ) =
sin (ωt − φ ) + sin (φ )e
⎟
Z ⎜⎜
⎟
⎝
⎠
Starting from ω t = π, as ωt increases, the current would keep decreasing. For some
value of ωt , say β, the current would be zero. If ω t > β, the current would evaluate to a
negative value. Since the diode blocks current in the reverse direction, the diode stops
conducting when ωt reaches β. The value of β can be obtained by substituting that
i (ωt ) = 0 wt = β into (2.24) we get:
β
⎛
−
Vm ⎜
i(β ) =
sin (β − φ ) + sin (φ )e tan φ
⎜
Z ⎜
⎝
⎞
⎟
⎟=0
⎟
⎠
(2.25)
The value of β can be obtained from the above equation by using the methods of
numerical analysis. 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 to ground are the same. This
average value can be obtained as shown in (2.26). The rms output voltage in this case is
shown in equation (2.27).
β
Vdc
V
V
= m * sin ωt dωt = m * (1 − cos β )
2π
2π
∫
(2.26)
0
β
Vrms
Vm
1
=
* ∫ (Vm sin ωt ) 2 dwt =
* β + 0.5(1 − sin( 2 β )
2π
2
π
0
25
(2.27)
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
2.3.3 Single-Phase Half-Wave Diode Rectifier With Free Wheeling Diode
Single-phase half-wave diode rectifier with free wheeling diode is shown in Fig.2.5.
This circuit differs from the circuit described above, which had only diode D1. This circuit
shown in Fig.2.5 has another diode, marked D2. This diode is called the free-wheeling
diode.
Fig.2.5 Half wave diode rectifier with free wheeling diode.
Fig.2.6 Various voltage waveforms of diode rectifier with free-wheeling diode.
Fig.2.7 Various current waveforms of diode rectifier with free-wheeling diode.
26
Diode Circuits or Uncontrolled Rectifier
Let the source voltage vs be defined as Vm sin (ωt ) which is positive when 0 < ωt < π
radians and it is negative when π < ω t < 2π radians. When vs is positive, diode D1
conducts and the output voltage, vo become positive. This in turn leads to diode D2 being
reverse-biased during this period. During π < wt < 2π, the voltage vo 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 vo would be zero volts, assuming that the
diode drop is negligible. Additionally when diode D2 conducts, diode D1 remains reversebiased, because the voltage across it is vs which is negative.
When the current through the inductor tends to fall (when the supply voltage become
negative), the voltage across the inductor become negative and its voltage tends to forward
bias diode D2 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 because the current free-wheels
through D2.
Fig.2.6 shows various voltage waveforms of diode rectifier with free-wheeling diode.
Fig.2.7 shows various current waveforms of diode rectifier with free-wheeling diode.
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 < ω t < 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
< ω t < π, the differential equation (2.18) applies. The solution of this equation will be as
obtained before in (2.20) or (2.23).
ωt
⎛
−
Vm ⎜
tan φ
i (ωt ) =
sin (ωt − φ ) + sin (φ ) e
Z ⎜⎜
⎝
⎞
⎟
⎟
⎟
⎠
0 < ωt < π
(2.28)
For the negative half-cycle ( π < ωt < 2π ) of the source voltage D1 is OFF and D2 is
ON. Then the driving voltage is set to zero and the following differential equation
represents the circuit in this case.
L
di
+ R* i = 0
dt
for π < ωt < 2π
(2.29)
The solution of (2.29) is given by the following equation:
i (ωt ) = B e
−
ωt − π
tan φ
(2.30)
The constant B can be obtained from the boundary condition where i (π ) = B is the
starting value of the current in π < ωt < 2π and can be obtained from equation (2.23) by
substituting ωt = π
27
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
π
−
V
Then, i(π ) = m (sin(π − φ ) + sin (φ ) e tan φ ) = B
Z
The above value of i (π ) can be used as initial condition of equation (2.30). Then the load
current during π < ωt < 2π is shown in the following equation.
π
⎛
−
Vm ⎜
i (ωt ) =
sin (π − φ ) + sin (φ ) e tan φ
Z ⎜⎜
⎝
⎞ − ωt − π
⎟ tan φ
⎟e
⎟
⎠
for π < ωt < 2π
(2.31)
For the period 2π < ωt < 3π the value of i (2π ) from (2.31) can be used as initial
condition for that period. The differential equation representing this period is the same as
equation (2.28) by replacing ω t by ωt − 2π and the solution is given by equation (2.32).
This period ( 2π < ωt < 3π ) differ than the period 0 < wt < π in the way to get the constant
A where in the 0 < ωt < π the initial value was i (0) = 0 but in the case of 2π < ωt < 3π the
initial condition will be i (2π ) that given from (2.31) and is shown in (2.33).
ωt − 2π
−
Vm
(2.32)
i (ωt ) =
sin (ωt − 2π − φ ) + Ae tan φ for 2π < ωt < 3π
Z
The value of i (2π ) can be obtained from (2.31) and (2.32) as shown in (2.33) and
(2.34) respectively.
π
⎛
−
Vm ⎜
sin (π − φ ) + sin (φ ) e tan φ
i (2π ) =
⎜
Z ⎜
⎝
⎞ − π
⎟ tan φ
⎟e
⎟
⎠
(2.33)
Vm
sin (− φ ) + A
(2.34)
Z
By equating (2.33) and (2.34) the constant A in 2π < ωt < 3π can be obtained from the
following equation:
V
A = i (2π ) + m sin (φ )
(2.35)
Z
Then, the general solution for the period 2π < ωt < 3π is given by equation (2.36):
i (2π ) =
V
V
⎛
⎞ −
i (ωt ) = m sin (ωt − 2π − φ ) + ⎜ i(2π ) + m sin (φ )⎟e
Z
Z
⎝
⎠
ωt − 2π
tan φ
2π < ωt < 3π (2.36)
Where i (2π ) can be obtained from equation (2.33).
Example 2 A diode circuit shown in Fig.2.3 with R=10 Ω, L=20mH, and VS=220 2
sin314t.
(a) Determine the expression for the current though the load in the period 0 < ωt < 2π
and determine the conduction angle β .
(b) If we connect free wheeling diode through the load as shown in Fig.2.5 Determine
the expression for the current though the load in the period of 0 < ωt < 3π .
28
Diode Circuits or Uncontrolled Rectifier
Solution: (a) For the period of 0 < ωt < π , the expression of the load current can be
obtained from (2.24) as following:
ωL
314 * 20 *10 −3
φ = tan −1
= tan −1
= 0.561 rad . and tan φ = 0.628343
R
10
Z = R 2 + (ωL) 2 = 10 2 + (314 * 20 *10 − 3 ) 2 = 11.8084Ω
ωt
⎛
−
Vm ⎜
i (ωt ) =
sin (ωt − φ ) + sin (φ ) e tan φ
⎜
Z ⎜
⎝
[
⎞
⎟
⎟
⎟
⎠
220 2
sin (ωt − 0.561) + 0.532 * e −1.5915 ωt
11.8084
i (ωt ) = 26.3479 sin (ωt − 0.561) + 14.0171 * e −1.5915 ωt
=
]
The value of β can be obtained from the above equation by substituting for i ( β ) = 0 .
Then, 0 = 26.3479 sin (β − 0.561) + 14.0171 * e −1.5915 β
By using the numerical analysis we can get the value of β. The simplest method is by
using
the
simple
iteration
technique
by
assuming
−1.5915 β
and substitute different values for β in the
Δ = 26.3479 sin (β − 0.561) + 14.0171 * e
region π < β < 2π till we get the minimum value of Δ then the corresponding value of β is
the required value. The narrow intervals mean an accurate values of β . The following
table shows the relation between β and Δ:
β
1.1 π
1.12 π
1.14 π
1.16 π
1.18 π
1.2 π
Δ
6.49518
4.87278
3.23186
1.57885
-0.079808
-1.73761
It is clear from the above table that β ≅ 1.18 π rad. The current in β < wt < 2π will be
zero due to the diode will block the negative current to flow.
(b) In case of free-wheeling diode as shown in Fig.2.5, we have to divide the operation of
this circuit into three parts. The first one when 0 < ωt < π (D1 “ON”, D2 “OFF”), the
second case when π < ωt < 2π (D1 “OFF” and D2 “ON”) and the last one when
2π < ωt < 3π (D1 “ON”, D2 “OFF”).
¾ In the first part ( 0 < ωt < π ) the expression for the load current can be obtained as
In case (a). Then:
i ( wt ) = 26.3479 sin (ωt − 0.561) + 14.0171 * e −1.5915 wt
for 0 < ωt < π
the current at ωt = π is starting value for the current in the next part. Then
29
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
i (π ) = 26.3479 sin (π − 0.561) + 14.0171 * e −1.5915 π = 14.1124 A
¾ In the second part π < ωt < 2π , the expression for the load current can be obtained
from (2.30) as following:
−
ωt −π
tan φ
i (ωt ) = B e
where B = i (π ) = 14.1124 A
Then i (ωt ) = 14.1124 e −1.5915(ωt −π )
for ( π < ωt < 2π )
The current at ωt = 2π is starting value for the current in the next part. Then
i (2π ) = 0.095103 A
¾ In the last part ( 2π < ωt < 3π ) the expression for the load current can be obtained
from (2.36):
V
V
⎛
⎞ −
i (ωt ) = m sin (ωt − 2π − φ ) + ⎜ i (2π ) + m sin (φ )⎟e
Z
Z
⎝
⎠
ωt − 2π
tan φ
∴ i (ωt ) = 26.3479 sin (ωt − 6.8442) + (0.095103 + 26.3479 * 0.532)e −1.5915(ωt − 2π )
∴ i (ωt ) = 26.3479 sin (ωt − 6.8442) + 14.1131e −1.5915(ωt − 2π ) for ( 2π < ωt < 3π )
2.4 Single-Phase Full-Wave Diode Rectifier
The full wave diode rectifier can be designed with a center-taped transformer as shown in
Fig.2.8, where each half of the transformer with its associated diode acts as half wave
rectifier or as a bridge diode rectifier as shown in Fig. 2.12. The advantage and
disadvantage of center-tap diode rectifier is shown below:
Advantages
• The need for center-tapped transformer is eliminated,
• The output is twice that of the center tapped circuit for the same secondary voltage,
and,
• The peak inverse voltage is one half of the center-tap circuit.
Disadvantages
• It requires four diodes instead of two, in full wave circuit, and,
• There are always two diodes in series are conducting. Therefore, total voltage drop
in the internal resistance of the diodes and losses are increased.
The following sections explain and analyze these rectifiers.
30
Diode Circuits or Uncontrolled Rectifier
2.4.1 Center-Tap Diode Rectifier With Resistive Load
In the center tap full wave rectifier, current flows through the load in the same direction
for both half cycles of input AC voltage. The circuit shown in Fig.2.8 has two diodes D1
and D2 and a center tapped transformer. The diode D1 is forward bias “ON” and diode D2
is reverse bias “OFF” in the positive half cycle of input voltage and current flows from
point a to point b. Whereas in the negative half cycle the diode D1 is reverse bias “OFF”
and diode D2 is forward bias “ON” and again current flows from point a to point b. Hence
DC output is obtained across the load.
Fig.2.8 Center-tap diode rectifier with resistive load.
Fig.2.9 Various current and voltage waveforms for center-tap diode rectifier with resistive load.
In case of pure resistive load, Fig.2.9 shows various current and voltage waveform for
converter in Fig.2.8. The average and rms output voltage and current can be obtained from
the waveforms shown in Fig.2.9 as shown in the following:
31
Chapter Two
Vdc =
1
Dr. Ali M. Eltamaly, King Saud University
π
V sin ωt dωt =
π∫ m
0
I dc =
2 Vm
(2.36)
π
2 Vm
π R
Vrms =
(2.37)
π
(V sin ωt )
π∫ m
1
2
0
dωt =
Vm
2
(2.38)
Vm
(2.39)
2 R
(2.40)
PIV of each diode = 2Vm
V
VS = m
(2.41)
2
The rms value of the transformer secondery current is the same as that of the diode:
V
(2.41)
IS = ID = m
2R
I rms =
Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The
efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of
diode D1 and(f) Crest factor of transformer secondary current.
Solution:- The efficiency or rectification ratio is
2 Vm 2 Vm
*
Pdc
Vdc * I dc
π
πR
=
=
= 81.05%
η=
Vm
Vm
Pac Vrms * I rms
*
2
2R
Vm
V
2 = π = 1.11
(b) FF = rms =
2 Vm 2 2
Vdc
π
Vac
= FF 2 − 1 = 1.112 − 1 = 0.483
Vdc
2 Vm 2 Vm
Pdc
π π R
=
= 0.5732
(d) TUF =
Vm Vm
2 VS I S
2
2 2R
(e) The PIV is 2Vm
(c) RF =
(f) Creast Factor of secondary current, CF =
I S ( peak )
IS
32
Vm
= R =2
Vm
2R
Diode Circuits or Uncontrolled Rectifier
2.4.2 Center-Tap Diode Rectifier With R-L Load
Center-tap full wave rectifier circuit with RL load is shown in Fig.2.10. Various voltage
and current waveforms for Fig.2.10 is shown in Fig.2.11. An expression for load current
can be obtained as shown below:
Fig.2.10 Center-tap diode rectifier with R-L load
Fig.2.11 Various current and voltage waveform for Center-tap diode rectifier with R-L load
33
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
It is assumed that D1 conducts in positive half cycle of VS and D2 conducts in negative
half cycle. So, the deferential equation defines the circuit is shown in (2.43).
di
L
+ R * i = Vm sin(ωt )
(2.43)
dt
The solution of the above equation can be obtained as obtained before in (2.24)
ωt ⎞
⎛
−
Vm ⎜
tan φ ⎟
(2.44)
i (ωt ) =
sin (ωt − φ ) + sin (φ )e
⎟ for 0 < ωt < π
Z ⎜⎜
⎟
⎝
⎠
In the second half cycle the same differential equation (2.43) and the solution of this
equation will be as obtained before in (2.22)
ωt − π
−
Vm
i (ωt ) =
sin (ωt − π − φ ) + Ae tan φ
Z
(2.45)
The value of constant A can be obtained from initial condition. If we assume that
(2.46)
i(π)=i(2π)=i(3π)=……..=Io
Then the value of I o can be obtained from (2.44) by letting ωt = π
π
⎛
−
Vm ⎜
I o = i (π ) =
sin (π − φ ) + sin (φ )e tan φ
⎜
Z ⎜
⎝
⎞
⎟
⎟
⎟
⎠
(2.47)
Then use the value of I o as initial condition for equation (2.45). So we can obtain the
value of constant A as following:
π −π
−
Vm
i (π ) = I o =
sin (π − π − φ ) + Ae tan φ
Z
V
Then; A = I o + m sin (φ )
Z
Substitute (2.48) into (2.45) we get:
(2.48)
ωt −π
Vm
Vm
⎛
⎞ − tan φ
i (ωt ) =
sin (ωt − π − φ ) + ⎜ I o +
sin (φ )⎟e
, then,
Z
Z
⎝
⎠
ωt −π
⎡
−
Vm ⎢
i (ωt ) =
sin (ωt − π − φ ) + sin (φ )e tan φ
Z ⎢
⎣⎢
ωt −π
⎤
−
⎥ + I e tan φ
⎥ o
⎦⎥
(for π < ωt < 2π )
(2.49)
In the next half cycle 2π < ωt < 3π the current will be same as obtained in (2.49) but we
have to take the time shift into account where the new equation will be as shown in the
following:
ωt − 2π
⎡
−
Vm ⎢
sin (wt − 2π − φ ) + sin (φ )e tan φ
i (ωt ) =
Z ⎢
⎣⎢
ωt − 2π
⎤
−
⎥ + I e tan φ
⎥ o
⎦⎥
34
(for 2π < ωt < 3π )
(2.50)
Diode Circuits or Uncontrolled Rectifier
2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load
Another alternative in single-phase full wave rectifier is by using four diodes as shown
in Fig.2.12 which known as a single-phase full bridge diode rectifier. It is easy to see the
operation of these four diodes. The current flows through diodes D1 and D2 during the
positive half cycle of input voltage (D3 and D4 are “OFF”). During the negative one,
diodes D3 and D4 conduct (D1 and D2 are “OFF”).
Fig.2.12 Single-phase full bridge diode rectifier.
Fig.2.13 Various current and voltage waveforms of Full bridge single-phase diode rectifier.
35
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
In positive half cycle the supply voltage forces diodes D1 and D2 to be "ON". In same
time it forces diodes D3 and D4 to be "OFF". So, the current moves from positive point of
the supply voltage across D1 to the point a of the load then from point b to the negative
marked point of the supply voltage through diode D2. In the negative voltage half cycle,
the supply voltage forces the diodes D1 and D2 to be "OFF". In same time it forces diodes
D3 and D4 to be "ON". So, the current moves from negative marked point of the supply
voltage across D3 to the point a of the load then from point b to the positive marked point
of the supply voltage through diode D4. So, it is clear that the load currents moves from
point a to point b in both positive and negative half cycles of supply voltage. So, a DC
output current can be obtained at the load in both positive and negative halves cycles of the
supply voltage. The complete waveforms for this rectifier is shown in Fig.2.13
Example 4 The rectifier shown in Fig.2.12 has a purely resistive load of R=15 Ω and,
VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form
factor, (c) Ripple factor, (d) TUF, (e) The peak inverse voltage, (PIV) of each diode, (f)
Crest factor of input current, and, (g) Input power factor.
Solution: Vm = 300 V
Vdc =
1
π
V sin ωt dωt =
π∫ m
2 Vm
π
0
= 190.956 V , I dc =
2 Vm
= 12.7324 A
π R
1/ 2
Vrms
⎡1 π
⎤
=⎢
(Vm sin ωt )2 dωt ⎥
⎢⎣ π 0
⎥⎦
(a) η =
∫
=
V
Vm
= 212.132 V , I rms = m = 14.142 A
2R
2
Pdc
V I
= dc dc = 81.06 %
Pac Vrms I rms
(b) FF =
Vrms
= 1.11
Vdc
2
2
2
Vrms
− Vdc
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 0.482
2
Vdc
Vdc
Vdc
(d) TUF =
Pdc
190.986 *12.7324
=
= 81 %
212.132 * 14.142
VS I S
(e) The PIV= Vm =300V
(f) CF =
I S ( peak )
IS
=
300 / 15
= 1.414
14.142
(g) Input power factor =
I2 *R
Re al Power
= rms
=1
Apperant Power
VS I S
36
Diode Circuits or Uncontrolled Rectifier
2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load Current
The full bridge single-phase diode rectifier with DC load current is shown in Fig.2.14.
In this circuit the load current is pure DC and it is assumed here that the source inductances
is negligible. In this case, the circuit works as explained before in resistive load but the
current waveform in the supply will be as shown in Fig.2.15.
The rms value of the input current is I S = I o
Fig.2.14 Full bridge single-phase diode rectifier with DC load current.
Fig.2.15 Various current and voltage waveforms for full bridge single-phase diode rectifier with
DC load current.
37
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
The supply current in case of pure DC load current is shown in Fig.2.15, as we see it is
odd function, then an coefficients of Fourier series equal zero, an = 0 , and
bn =
2
π
I o * sin nωt dωt
π∫
=
0
=
2 Io
[− cos nωt ]π0
nπ
(2.51)
2 Io
[cos 0 − cos nπ ] = 4 I o for n = 1, 3, 5, .............
nπ
nπ
Then from Fourier series concepts we can say:
i (t ) =
4 Io
1
1
1
1
* (sin ωt + sin 3ωt + sin 5ωt + sin 7ωt + sin 9ωt + ..........)
π
3
5
7
9
2
2
2
2
2
2
(2.52)
2
⎛1⎞ ⎛1⎞ ⎛ 1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
∴ THD( I s (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 46% or we can obtain
⎝ 3 ⎠ ⎝ 5 ⎠ ⎝ 7 ⎠ ⎝ 9 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 15 ⎠
THD( I s (t )) as the following:
From (2.52) we can obtain the value of is I S1 =
4 Io
2π
2
⎞
⎛
⎟
⎜
2
2
⎛ 2π ⎞
⎛ IS ⎞
Io ⎟
⎜
⎟ − 1 = 48.34%
⎟⎟ − 1 =
∴ THD ( I s (t )) = ⎜⎜
− 1 = ⎜⎜
⎟
⎜ 4 Io ⎟
4
I
⎝ S1 ⎠
⎠
⎝
⎟
⎜
π
2
⎠
⎝
Example 5 solve Example 4 if the load is 30 A pure DC
Solution: From example 4 Vdc= 190.986 V, Vrms=212.132 V
I dc = 30 A and I rms = 30 A
P
V I
(a) η = dc = dc dc = 90 %
Pac Vrms I rms
V
(b) FF = rms = 1.11
Vdc
2
2
2
Vrms
− Vdc
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 0.482
2
Vdc
Vdc
Vdc
Pdc
190.986 *30
=
= 90 %
VS I S 212.132 * 30
(e) The PIV=Vm=300V
(d) TUF =
(f) CF =
I S ( peak )
IS
=
30
=1
30
4 Io
4 * 30
=
= 27.01A
2π
2π
Re al Power
=
Input Power factor=
Apperant Power
(g) I S1 =
=
VS I S1 * cos φ
I * cos φ
27.01
= S1
=
*1 = 0.9 Lag
VS I S
IS
30
38
Diode Circuits or Uncontrolled Rectifier
2.4.5 Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier.
Fig.2.15 Shows the single-phase diode bridge rectifier with source inductance. Due to
the value of LS the transitions of the AC side current iS from a value of I o to − I o (or vice
versa) will not be instantaneous. The finite time interval required for such a transition is
called commutation time. And this process is called current commutation process. Various
voltage and current waveforms of single-phase diode bridge rectifier with source
inductance are shown in Fig.2.16.
Fig.2.15 Single-phase diode bridge rectifier with source inductance.
Fig.2.16 Various current and voltage waveforms for single-phase diode bridge rectifier with source
inductance.
Let us study the commutation time starts at t=10 ms as indicated in Fig.2.16. At this
time the supply voltage starts to be negative, so diodes D1 and D2 have to switch OFF and
39
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
diodes D3 and D4 have to switch ON as explained in the previous case without source
inductance. But due to the source inductance it will prevent that to happen instantaneously.
So, it will take time Δt to completely turn OFF D1 and D2 and to make D3 and D4 carry
the entire load current ( I o ). Also in the time Δt the supply current will change from I o to
− I o which is very clear in Fig.2.16. Fig.2.17 shows the equivalent circuit of the diode
bridge at time Δt .
Fig.2.17 The equivalent circuit of the diode bridge at commutation time Δt .
From Fig.2.17 we can get the following equations
di
VS − Ls S = 0
(2.53)
dt
Multiply the above equation by dωt then,
VS dωt = ωLs diS
(2.54)
Integrate both sides of the above equation during the commutation period ( Δt sec or u
rad.) we get the following:
VS dωt = ωLs diS
π +u
−Io
∫ Vm sin ωt dωt = ωLs ∫ diS
π
(2.55)
Io
Then; Vm [cos π − cos(π + u )] = −2ωLs I o
Then; Vm [− 1 + cos(u )] = −2ωLs I o
2ωLs I o
Then; cos(u ) = 1 −
Vm
⎛ 2ωLs I o ⎞
⎟⎟
Then; u = cos −1 ⎜⎜1 −
(2.56)
V
m
⎝
⎠
⎛ 2ωLs I o ⎞
u 1
⎟⎟
And Δt = = cos −1 ⎜⎜1 −
(2.57)
ω ω
V
m
⎝
⎠
It is clear that the DC voltage reduction due to the source inductance is the drop across
the source inductance.
di
(2.58)
vrd = Ls S
dt
40
Diode Circuits or Uncontrolled Rectifier
π +u
−Io
∫ vrd dω t = ∫ ω LS diS = −2ω LS I o
Then
π
π +u
∫ vrd dω t
(2.59)
Io
is the reduction area in one commutation period Δt . But we have two
π
commutation periods Δt in one period of supply voltage. So the total reduction per period
π +u
is: 2
∫ vrd dω t = −4 ω LS I o
(2.60)
π
To obtain the average reduction in DC output voltage Vrd due to source inductance we
have to divide the above equation by the period time 2π . Then;
− 4ω LS I o
Vrd =
= −4 f LS I o
(2.61)
2π
The DC voltage with source inductance tacking into account can be calculated as
following:
2V
Vdc actual = Vdc without sourceinduc tan ce − Vrd = m − 4 fLs I o
(2.62)
π
To obtain the rms value and Fourier transform of the supply current it is better to move
the vertical axis to make the waveform odd or even this will greatly simplfy the analysis.
So, it is better to move the vertical axis of supply current by u / 2 as shown in Fig.2.18.
Moveing the vertical axis will not change the last results. If you did not bleave me keep
going in the analysis without moveing the axis.
Fig. 2.18 The old axis and new axis for supply currents.
Fig.2.19 shows a symple drawing for the supply current. This drawing help us in getting
the rms valuof the supply current. It is clear from the waveform of supply current shown in
Fig.2.19 that we obtain the rms value for only a quarter of the waveform because all for
quarter will be the same when we squaret the waveform as shown in the following
equation:
41
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
π
Is =
2
π
u/2
[
∫
0
Then; I s =
2
2
⎛ 2I o ⎞
ωt ⎟ dωt + ∫ I o2 dωt ]
⎜
⎠
⎝ u
u/2
2 I o2 ⎡ 4 u 3 π u ⎤
+ − ⎥=
⎢
π ⎢⎣ 3u 2 8 2 2 ⎥⎦
Is
(2.63)
2 I o2 ⎡ π u ⎤
−
π ⎢⎣ 2 3 ⎥⎦
u
Io
−
(2.64)
π
u
2
2π
u
π+
2
π
u
2
2
− Io
π−
u
2
2π −
u
2
Fig.2.19 Supply current waveform
To obtain the Fourier transform for the supply current waveform you can go with the
classic fourier technique. But there is a nice and easy method to obtain Fourier transform
of such complcated waveform known as jump technique [ ]. In this technique we have to
draw the wave form and its drevatives till the last drivative values all zeros. Then record
the jump value and its place for each drivative in a table like the table shown below. Then;
substitute the table values in (2.65) as following:
42
Diode Circuits or Uncontrolled Rectifier
Is
u
Io
−
π
u
2
u
2
π−
− Io
u
2
2π
u
π+
2
2π −
u
2
I s′
2Io
u
u
2
−
π
u
2
π+
π−
−
2I o
u
u
2
u
2
2π −
u
2
Fig.2.20 Supply current and its first derivative.
Table(2.1) Jumb value of supply current and its first derivative.
Js
−
u
2
u
2
0
π−
u
2
0
0
2I o
2I
2I
I s′
− o
− o
u
u
u
It is an odd function, then ao = an = 0
Is
π+
u
2
0
2I o
u
⎤
⎡m
1 m
J s′ sin nωt s ⎥
(2.65)
⎢ J s cos nωt s −
n s =1
⎥⎦
⎢⎣ s =1
1 ⎡ − 1 2I o ⎛
u⎞
u ⎞ ⎞⎤
⎛ u⎞
⎛u⎞
⎛
⎛
bn =
⎜ sin n⎜ − ⎟ − sin n⎜ ⎟ − sin n⎜ π − ⎟ + sin n⎜ π + ⎟ ⎟⎥
⎢ *
nπ ⎣ n
u ⎝
2⎠
2 ⎠ ⎠⎦
⎝ 2⎠
⎝2⎠
⎝
⎝
8I
nu
(2.66)
bn = 2 o * sin
2
n πu
8I
u
b1 = o * sin
(2.67)
πu
2
1
bn =
nπ
∑
∑
43
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
8I o
u
* sin
2
2 πu
8I o
u
* sin
2
I
2 πu
⎛u⎞
⎛u⎞
cos⎜ ⎟
pf = S1 * cos⎜ ⎟ =
IS
⎝2⎠
⎝2⎠
2 I o2 ⎡ π u ⎤
−
π ⎢⎣ 2 3 ⎥⎦
Then; I S1 =
(2.68)
(2.69)
⎛u⎞ ⎛u⎞
4 sin ⎜ ⎟ cos⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠ = 2 sin (u )
=
⎡π u ⎤
⎡π u ⎤
u π⎢ − ⎥
u π⎢ − ⎥
⎣ 2 3⎦
⎣ 2 3⎦
Example 6 Single phase diode bridge rectifier connected to 11 kV,
50 Hz, source inductance X s = 5 mH supply to feed 200 A
pure DC load, find:
i. Average DC output voltage.
ii. Power factor.
iii. Determine the THD of the utility line current.
Solution: (i) From (2.62), Vm = 11000 * 2 = 15556V
2V
Vdc actual = Vdc without sourceinduc tan ce − Vrd = m − 4 fLs I o
π
Vdc actual =
2 *15556
π
− 4 * 50 * 0.005 * 200 = 9703V
(ii) From (2.56) the commutation angle u can be obtained as following:
⎛ 2ωLs I o ⎞
2 * 2 * π * 50 * 0.005 * 200 ⎞
⎟⎟ = cos −1 ⎛⎜1 −
u = cos −1 ⎜⎜1 −
⎟ = 0.285 rad .
Vm ⎠
15556
⎝
⎠
⎝
The input power factor can be obtained from (2.69) as following
2 * sin (u )
pf =
I S1
⎛u⎞
* cos⎜ ⎟ =
IS
⎝2⎠
IS =
2 I o2 ⎡ π u ⎤
=
−
π ⎢⎣ 2 3 ⎥⎦
I S1 =
8I o
u
* sin =
2
2 πu
⎡π
u⎤
u π⎢ − ⎥
⎣ 2 3⎦
=
2 * sin (0.285)
⎡π
.285 ⎤
0.285 π ⎢ −
3 ⎥⎦
⎣2
= 0.917
2 * 200 2 ⎡ π 0.285 ⎤
⎢⎣ 2 − 3 ⎥⎦ = 193.85 A
π
8 * 200
⎛ 0.285 ⎞
* sin ⎜
⎟ = 179.46 A
2 π * 0.285
⎝ 2 ⎠
2
2
⎛I ⎞
⎛ 193.85 ⎞
THDi = ⎜⎜ S ⎟⎟ − 1 = ⎜
⎟ − 1 = 40.84%
179
.
46
I
⎝
⎠
⎝ S1 ⎠
44
Diode Circuits or Uncontrolled Rectifier
2.5 Three Phase Diode Rectifiers
2.5.1 Three-Phase Half Wave Rectifier
Fig.2.21 shows a half wave three-phase diode rectifier circuit with delta star three-phase
transformer. In this circuit, the diode with highest potential with respect to the neutral of
the transformer conducts. As the potential of another diode becomes the highest, load
current is transferred to that diode, and the previously conduct diode is reverse biased
“OFF case”.
Fig.2.21 Half wave three-phase diode rectifier circuit with delta star three-phase transformer.
For the rectifier shown in Fig.2.21 the load voltage, primary diode currents and its FFT
components are shown in Fig.2.22, Fig.2.23 and Fig.2.24 respectively.
π
6
5π
6
Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier.
45
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Fig.2.23 Primary and diode currents.
Primary current
Diode current
Fig.2.24 FFT components of primary and diode currents.
46
Diode Circuits or Uncontrolled Rectifier
5π
in the output voltage we can calculate the
6
6
average and rms output voltage and current as following:
By considering the interval from
Vdc
3
=
2π
I dc =
5π / 6
∫ Vm sin ωt dωt =
π /6
π
to
3 3 Vm
= 0.827Vm
2π
3 3 Vm 0.827 * Vm
=
R
2 *π * R
3
2π
Vrms =
I rms =
5π / 6
2
∫ (Vm sin ωt ) dωt =
π /6
(2.70)
(2.71)
1 3* 3
+
Vm = 0.8407 Vm
2 8π
0.8407 Vm
R
(2.72)
(2.73)
Then the diode rms current is equal to secondery current and can be obtaiend as
following:
Vm
08407 Vm
Ir = IS =
= 0.4854
(2.74)
R
R 3
Note that the rms value of diode current has been obtained from the rms value of load
current divided by 3 because the diode current has one third pulse of similar three pulses
in load current.
(2.75)
ThePIV of the diodes is 2 VLL = 3 Vm
Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side
and the load resistance is R=20 Ω. If the source inductance is negligible, determine (a)
Rectification efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization
factor, (e) Peak inverse voltage (PIV) of each diode and (f) Crest factor of input current.
Solution:
460
(a) VS =
= 265.58 V , Vm = 265.58 * 2 = 375.59 V
3
Vdc =
3 3 Vm
= 0.827 Vm ,
2π
I dc =
3 3 Vm 0827 Vm
=
2π R
R
Vrms = 0.8407 Vm
I rms =
0.8407 Vm
R
47
Chapter Two
η=
Dr. Ali M. Eltamaly, King Saud University
Pdc
V I
= dc dc = 96.767 %
Pac Vrms I rms
(b) FF =
Vrms
= 101.657 %
Vdc
2
2
2
Vrms
− Vdc
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 18.28 %
2
Vdc
Vdc
Vdc
(d) I S =
0.8407 Vm
1
I rms =
3
3 R
Pdc
TUF =
=
3 * VS I S
(0.827Vm ) 2 / R
= 66.424 %
0.8407 Vm
3 * Vm / 2 *
3R
(e) The PIV= 3 Vm=650.54V
(f) CF =
I S ( peak )
IS
=
Vm / R
= 2.06
0.8407 Vm
3R
48
Diode Circuits or Uncontrolled Rectifier
2.5.2 Three-Phase Half Wave Rectifier With DC Load Current and zero source
inductance
In case of pure DC load current as shown in Fig.2.25, the diode current and primary current
are shown in Fig.2.26.
New axis
Fig.2.25 Three-phase half wave rectifier with dc load current
Fig.2.26 Primary and secondary current waveforms and FFT components of three-phase half wave
rectifier with dc load current
49
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
To calculate Fourier transform of the diode current of Fig.2.26, it is better to move y
axis to make the function as odd or even to cancel one coefficient an or bn respectively. If
we put Y-axis at point ωt = 30o then we can deal with the secondary current as even
functions. Then, bn = 0 of secondary current. Values of an can be calculated as following:
1
a0 =
2π
an =
1
π
π /3
I
∫ I o dωt = 3o
(2.76)
−π / 3
π /3
∫ I o * cos nωt dwt
−π / 3
Io
[sin nωt ]−ππ//33 = I o * 3 for n = 1,2,7,8,13,14,....
nπ
nπ
I
= − o * 3 for n = 4,5,10,11,16,17
nπ
= 0 for all treplean harmonics
=
I s (t ) =
IO
3I O
+
π
3
1
1
1
1
1
⎛
⎞
⎜ cos ωt + cos 2ωt − cos 4ωt − cos 5ωt + cos 7ωt + cos 8ωt − −... ⎟
2
4
5
7
8
⎝
⎠
(2.77)
(2.78)
2
⎛I ⎞
THD( I s (t )) = ⎜⎜ S ⎟⎟
⎝ I S1 ⎠
2
⎛
⎞
⎜
⎟
⎜ Io / 3 ⎟
−1 = ⎜
⎟ −1 =
⎜ 3I O ⎟
⎜ π 2 ⎟
⎝
⎠
2 *π 2
− 1 = 1.0924 = 109.24%
9
It is clear that the primary current shown in Fig.2.26 is odd, then, an=0,
bn =
2
π
2π / 3
∫ I o * sin nωt dωt
0
=
2Io
[− cos nωt ] 20π / 3
nπ
3I o
for n = 1,2,4,5,7,8,10,11,13,14,....
nπ
= 0 for all treplean harmonics
=
iP (t ) =
(2.79)
3I O ⎛
1
1
1
1
1
⎞
⎜ sin ωt + sin 2ωt + sin 4ωt + sin 5ωt + sin 7ωt + sin 8ωt − −...⎟
π ⎝
2
4
5
7
8
⎠
The rms value of I P =
⎛I ⎞
THD ( I P (t )) = ⎜⎜ P ⎟⎟
⎝ I P1 ⎠
2
2
Io
3
⎛
⎜
−1 = ⎜
⎜
⎜
⎝
(2.80)
(2.81)
2
2 ⎞
Io ⎟
2
3 ⎟ − 1 = ⎛ 2π ⎞ − 1 = 67.983% (2.82)
⎜
⎟
3I O ⎟
⎝3 3⎠
⎟
π 2 ⎠
50
Diode Circuits or Uncontrolled Rectifier
Example 8 Solve example 7 if the load current is 100 A pure DC
460
Solution: (a) VS =
= 265.58 V , Vm = 265.58 * 2 = 375.59 V
3
3 3 Vm
Vdc =
= 0.827 Vm = 310.613V , I dc = 100 A
2π
Vrms = 0.8407 Vm = 315.759 V , I rms = 100 A
P
V I
310.613 * 100
η = dc = dc dc =
= 98.37 %
Pac Vrms I rms 315.759 *100
V
(b) FF = rms = 101.657 %
Vdc
2
2
2
Vrms
− Vdc
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 18.28 %
2
Vdc
Vdc
Vdc
1
1
I rms =
*100 = 57.735 A
3
3
Pdc
310.613 * 100
=
= 67.52 %
TUF =
3 * VS I S 3 * Vm / 2 * 57.735
(d) I S =
(e) The PIV= 3 Vm=650.54V
(f) CF =
I S ( peak )
IS
=
100
= 1.732
57.735
2.5.3 Three-Phase Half Wave Rectifier With Source Inductance
The source inductance in three-phase half wave diode rectifier Fig.2.27 will change the
shape of the output voltage than the ideal case (without source inductance) as shown in
Fig.2.28. The DC component of the output voltage is reduced due to the voltage drop on
the source inductance. To calculate this reduction we have to discuss Fig.2.27 with
reference to Fig.2.28. As we see in Fig.2.28 when the voltage vb is going to be greater than
the voltage va at time t (at the arrow in Fig.2.28) the diode D1 will try to turn off, in the
same time the diode D2 will try to turn on but the source inductance will slow down this
process and makes it done in time Δt (overlap time or commutation time). The overlap
time will take time Δt to completely turn OFF D1 and to make D2 carry the entire load
current ( I o ). Also in the time Δt the current in Lb will change from zero to I o and the
current in La will change from I o to zero. This is very clear from Fig.2.28. Fig.2.29 shows
the equivalent circuit of three phase half wave diode bridge in commutation period Δt .
51
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Fig.2.27 Three-phase half wave rectifier with load and source inductance.
Fig.2.28 Supply current and output voltage for three-phase half wave rectifier with pure
DC load and source inductance.
Fig.2.29 The equivalent circuit for three-phase half wave diode rectifier in commutation period.
52
Diode Circuits or Uncontrolled Rectifier
From Fig.2.29 we can get the following equations
di
va − La D1 − Vdc = 0
(2.83)
dt
di
(2.84)
vb − Lb D 2 − Vdc = 0
dt
subtract (2.84) from(2.83) we get:
di ⎞
⎛ di
va − vb + L⎜ D 2 − D1 ⎟ = 0
dt ⎠
⎝ dt
Multiply the above equation by dωt the following equation can be obtained:
(va − vb )dωt + ωL(diD 2 − diD1 ) = 0
substitute the voltage waveforms of va and vb into the above equation we get:
2π ⎞ ⎞
⎛
⎛
⎜Vm sin (ωt ) − Vm sin ⎜ ωt −
⎟ ⎟dωt = ωL(diD1 − diD 2 )
3 ⎠⎠
⎝
⎝
π ⎞⎞
⎛
⎛
Then;
⎜ 3 Vm sin ⎜ ωt + ⎟ ⎟dωt = ωL(diD1 − diD 2 )
6 ⎠⎠
⎝
⎝
Integrating both parts of the above equation we get the following:
5π
+u
6
∫
5π
6
Then;
Then;
Io
⎛0
⎞
π⎞
⎛
⎜
⎟
3 Vm sin ⎜ ωt + ⎟dωt = ωL⎜ diD1 − diD 2 ⎟
6⎠
⎝
⎜I
⎟
0
⎝ o
⎠
∫
∫
π ⎞⎞
⎛ ⎛ 5π π ⎞
⎛ 5π
3 Vm ⎜ cos⎜
+ ⎟ − cos⎜
+ u + ⎟ ⎟ = −2ωLI o
6 ⎠⎠
⎝ 6
⎝ ⎝ 6 6⎠
3 Vm (cos(π ) − cos(π + u )) = −2ωLI o
3 Vm (− 1 + cos(u )) = −2ωLI o
2ωLI o
Then; 1 − cos(u ) =
3 Vm
2ωLI o
Then; cos(u ) = 1 −
3 Vm
Then;
⎛
2ωLI o ⎞
⎟
Then u = cos −1 ⎜⎜1 −
(2.85)
⎟
3
V
m⎠
⎝
⎛
2ωLI o ⎞
u 1
⎟
(2.86)
Δt = = cos −1 ⎜⎜1 −
⎟
ω ω
3
V
m⎠
⎝
It is clear that the DC voltage reduction due to the source inductance is equal to the drop
across the source inductance. Then;
di
vrd = L D
dt
53
Chapter Two
5π
+u
6
Dr. Ali M. Eltamaly, King Saud University
Io
∫ vrd dωt = ∫ ωL diD = ωLI o
Then,
5π
6
5π
+u
6
∫ vrd dωt
(2.87)
0
is the reduction area in one commutation period Δt . But, we have three
5π
6
commutation periods, Δt in one period. So, the total reduction per period is:
5π
+u
6
3*
∫ vrd dωt = 3ωLI o
5π
6
To obtain the average reduction in DC output voltage Vrd due to source inductance we
have to divide the total reduction per period by 2π as following:
3ωLI o
(2.88)
= 3 f L Io
Vrd =
2π
Then, the DC component of output voltage due to source inductance is:
Vdc Actual = Vdc without
− 3 f L Io
(2.89)
source
induc tan ce
Vdc
Actual
=
3 3 Vm
− 3 f L Io
2π
(2.90)
Example 9 Three-phase half-wave diode rectifier connected to 66 kV,
50 Hz , 5mH supply to feed a DC load with 500 A DC,
fined the average DC output voltage.
⎛ 66000 ⎞
Solution: vm = ⎜
⎟ * 2 = 53889V
⎝ 3 ⎠
(i) Vdc Actual = Vdc without
− 3 f L Io
source
induc tan ce
Vdc
Actual
=
3 3 Vm
3 * 3 * 53889
− 3 f L Io =
− 3 * 50 * 0.005 * 500 = 44190V
2π
2π
2.5 Three-Phase Full Wave Diode Rectifier
The three phase bridge rectifier is very common in high power applications and is
shown in Fig.2.30. It can work with or without transformer and gives six-pulse ripples on
the output voltage. The diodes are numbered in order of conduction sequences and each
one conduct for 120 degrees. These conduction sequence for diodes is 12, 23, 34, 45, 56,
and, 61. The pair of diodes which are connected between that pair of supply lines having
the highest amount of instantaneous line to line voltage will conduct. Also, we can say that,
the highest positive voltage of any phase the upper diode connected to that phase conduct
54
Diode Circuits or Uncontrolled Rectifier
and the highest negative voltage of any phase the lower diode connected to that phase
conduct.
2.5.1 Three-Phase Full Wave Rectifier With Resistive Load
In the circuit of Fig.2.30, the AC side inductance LS is neglected and the load current is
pure resistance. Fig.2.31 shows complete waveforms for phase and line to line input
voltages and output DC load voltages. Fig.2.32 shows diode currents and Fig.2.33 shows
the secondary and primary currents and PIV of D1. Fig.2.34 shows Fourier Transform
components of output DC voltage, diode current secondary current and Primary current
respectively.
For the rectifier shown in Fig.2.30 the waveforms is as shown in Fig.2.31. The average
output voltage is :Vdc =
3
π
I dc =
2π / 3
∫
3 Vm sin ωt dωt =
3 3 Vm
π /3
π
=
3 2 VLL
π
(2.91)
= 1.654Vm = 1.3505VLL
3 3 Vm 1.654Vm 3 2 VLL 1.3505VLL
=
=
=
π R
R
πR
R
Vrms =
I rms =
3
π
2π / 3
∫ (
3 Vm sin ωt
)2 dωt =
π /3
(2.92)
3 9* 3
+
Vm = 1.6554 Vm = 1.3516VLL (2.93)
2
4π
1.6554 Vm
R
(2.94)
Then the diode rms current is
Ir =
1.6554 Vm
Vm
= 0.9667
R
R 3
I S = 0.9667
2
(2.95)
Vm
R
(2.96)
IL
Ip
Is
3
1
5
VL
a
b
c
4
6
Fig.2.30 Three-phase full wave diode bridge rectifier.
55
2
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load
voltages.
Fig.2.32 Diode currents.
56
Diode Circuits or Uncontrolled Rectifier
Fig.2.33 Secondary and primary currents and PIV of D1.
Fig.2.34 Fourier Transform components of output DC voltage, diode current secondary current and
Primary current respectively of three-phase full wave diode bridge rectifier.
Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the
load resistance is R=20 Ω. If the source inductance is negligible, determine (a) The
efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak
inverse voltage (PIV) of each diode and (f) Crest factor of input current.
57
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Solution: (a) VS =
3 3 Vm
Vdc =
π
460
= 265.58 V ,
3
Vm = 265.58 * 2 = 375.59 V
= 1.654Vm = 621.226 V , I dc =
3 3 Vm 1.654Vm
=
= 31.0613 A
π R
R
1.6554 Vm
3 9* 3
= 31.0876 A
+
Vm = 1.6554 Vm = 621.752 V , I rms =
2
4π
R
Vrms =
Pdc
V I
= dc dc = 99.83 %
Pac Vrms I rms
V
(b) FF = rms = 100.08 %
Vdc
η=
2
2
2
Vrms
− Vdc
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 4 %
2
Vdc
Vdc
Vdc
V
1.6554 Vm
2
I rms = 0.8165 *
= 1.352 m
R
R
3
(d) I S =
TUF =
Pdc
=
3 * VS I S
(1.654Vm ) 2 / R
V
3 * Vm / 2 * 1.352 m
R
= 95.42 %
(e) The PIV= 3 Vm=650.54V
I
3 Vm / R
(f) CF = S ( peak ) =
= 1.281
Vm
IS
1.352
R
2.5.2 Three-Phase Full Wave Rectifier With DC Load Current
The supply current in case of pure DC load current is shown in Fig.2.35. Fast Fourier
Transform of Secondary and primary currents respectively is shown in Fig2.36.
As we see it is odd function, then an=0, and
bn =
2
π
5π / 6
∫ I o * sin nωt dωt
π /6
2 Io
[− cos nωt ]π5π/ 6/ 6
nπ
2 Io
2 Io
2 Io
3 , b5 =
(− 3 ), b7 =
(− 3 )
b1 =
5π
7π
π
2 Io
2 Io
( 3 ), b13 =
( 3 ),.............
b11 =
11π
13π
bn = 0, for n = 2,3,4,6,8,9,10,12,14,15,.............
=
I s (t ) =
(2.97)
2 3I o ⎛
1
1
1
1
⎞
⎜ sin ωt − sin 5v ωt − sin 7ωt + sin 11ωt + sin 13ωt ⎟ (2.98)
π ⎝
13
11
7
5
⎠
58
Diode Circuits or Uncontrolled Rectifier
2
2
2
2
2
2
2
⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
THD ( I s (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟
⎝ 5 ⎠ ⎝ 7 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 17 ⎠ ⎝ 19 ⎠ ⎝ 23 ⎠ ⎝ 25 ⎠
= 31%
2
Also THD( I s (t )) can be obtained as following:
IS =
2
2* 3
I o , I S1 =
Io
3
π
2
⎛I ⎞
THD( I s (t )) = ⎜⎜ S ⎟⎟ − 1 =
⎝ I S1 ⎠
2/3
2*3/π 2
− 1 = 31.01%
Fig.2.35 The D1 and D2 currents, secondary and primary currents.
59
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Fig2.36 Fast Fourier Transform of Secondary and primary currents respectively.
For the primary current if we move the t=0 to be as shown in Fig.2.28, then the function
will be odd then, an = 0 , and
2π / 3
π /3
π
⎞
2 ⎛⎜
bn =
I1 * sin nωt dωt + 2 I1 * sin nωt dωt +
I1 * sin nωt dωt ⎟
⎟
π⎜
2π / 3
π /3
⎝ 0
⎠
π
2I ⎛
2π ⎞
(2.99)
= 1 ⎜1 − cos nπ + cos n − cos n
⎟
nπ ⎝
3
3 ⎠
2 * 3I1
bn =
for n = 1,5,7,11,13,...............
nπ
bn = 0, for n = 2,3,4,6,8,9,10,12,14,15,.............
∫
I P (t ) =
∫
∫
1
1
1
2 * 3I1 ⎛
1
⎞
⎜ sin ωt + sin 5ωt + sin 7ωt + sin 11ωt + sin 13ωt ⎟ (2.100)
π ⎝
5
7
11
13
⎠
2
2
2
2
2
2
2
⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
THD( I P (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟
⎝ 5 ⎠ ⎝ 7 ⎠ ⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 17 ⎠ ⎝ 19 ⎠ ⎝ 23 ⎠ ⎝ 25 ⎠
= 30%
Power Factor =
2
I S1
I
* cos(0) = S1
IS
IS
2.5.4 Three-Phase Full Wave Diode Rectifier With Source Inductance
The source inductance in three-phase diode bridge rectifier Fig.2.37 will change the
shape of the output voltage than the ideal case (without source inductance) as shown in
Fig.2.31. The DC component of the output voltage is reduced. Fig.2.38 shows The output
DC voltage of three-phase full wave rectifier with source inductance.
60
Diode Circuits or Uncontrolled Rectifier
Fig.2.37 Three-phase full wave rectifier with source inductance
Fig.2.38 The output DC voltage of three-phase full wave rectifier with source inductance
Let us study the commutation time starts at t=5ms as shown in Fig.2.39. At this time Vc
starts to be more negative than Vb so diode D6 has to switch OFF and D2 has to switch
ON. But due to the source inductance will prevent that to happen instantaneously. So it will
take time Δt to completely turn OFF D6 and to make D2 carry all the load current ( I o ).
Also in the time Δt the current in Lb will change from I o to zero and the current in Lc
will change from zero to I o . This is very clear from Fig.2.39. The equivalent circuit of the
three phase diode bridge at commutation time Δt at t = 5ms is shown in Fig.2.40 and
Fig.2.41.
61
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Fig.2.39 Waveforms represent the commutation period at time t=5ms.
Fig.2.40 The equivalent circuit of the three phase diode bridge at commutation time Δt at
t = 5ms
62
Diode Circuits or Uncontrolled Rectifier
Fig.2.41 Simple circuit of the equivalent circuit of the three phase diode bridge at commutation
time Δt at t = 5ms
From Fig.2.41 we can get the following defferntial equations:
di
di
Va − La D1 − Vdc − Lb D 6 − Vb = 0
dt
dt
Va − La
diD1
di
− Vdc − Lc D 2 − Vc = 0
dt
dt
(2.102)
diD1
= 0 , substitute this value
dt
and (2.102) we get the following differential equations:
di
Va − Vb − Lb D 6 = Vdc
dt
di
Va − Vc − Lc D 2 = Vdc
dt
By equating the left hand side of equation (2.103) and (2.104) we get the
differential equation:
Note that, during the time Δt , iD1 is constant so
Va − Vb − Lb
(2.101)
diD 6
di
= Va − Vc − Lc D 2
dt
dt
in (2.101)
(2.103)
(2.104)
following
(2.105)
diD 6
di
(2.106)
− Lc D 2 = 0
dt
dt
The above equation can be written in the following manner:
(Vb − Vc )dt + Lb diD 6 − Lc diD 2 = 0
(2.107)
(Vb − Vc )dω t + ω Lb diD 6 − ω Lc diD 2 = 0
(2.108)
Integrate the above equation during the time Δt with the help of Fig.2.39 we can get the
limits of integration as shown in the following:
Vb − Vc + Lb
π / 2+u
0
Io
π /2
π / 2+u
Io
0
∫ (Vb − Vc )dω t + ∫ ω Lb diD6 − ∫ ω Lc diD 2 = 0
∫
π /2
⎛
2π
⎛
⎜Vm sin ⎜ ω t −
3
⎝
⎝
2π
⎞
⎛
⎟ − Vm sin ⎜ ω t +
3
⎠
⎝
⎞⎞
⎟ ⎟dω t + ωLb (− I o ) − ωLc I o = 0
⎠⎠
assume Lb = Lc = LS
63
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
π / 2+u
2π ⎞
2π ⎞⎤
⎡
⎛
⎛
Vm ⎢− cos⎜ ω t −
= 2ω LS I o
⎟
⎟ + cos⎜ ω t +
3 ⎠
3 ⎠⎥⎦ π / 2
⎝
⎝
⎣
⎡
2π ⎞
2π ⎞
⎛π
⎛π
⎛ π 2π
Vm ⎢− cos⎜ + u −
⎟ + cos⎜ + u +
⎟ + cos⎜ −
3 ⎠
3 ⎠
3
⎝2
⎝2
⎝2
⎣
= 2ω LS I o
π⎞
7π ⎞
⎡
⎛
⎛
⎛ −π ⎞
⎛ 7π
Vm ⎢− cos⎜ u − ⎟ + cos⎜ u +
⎟ + cos⎜
⎟ − cos⎜
6⎠
6 ⎠
⎝
⎝
⎝ 6 ⎠
⎝ 6
⎣
⎛ π 2π
⎞
⎟ − cos⎜ +
3
⎠
⎝2
⎞⎤
⎟⎥
⎠⎦
⎞⎤
⎟⎥ = 2ω LS I o
⎠⎦
⎡
3
3⎤
⎛ 7π ⎞
⎛ 7π ⎞
⎛π ⎞
⎛π ⎞
+
⎟+
⎟ − sin (u ) sin ⎜
⎢− cos(u ) cos⎜ ⎟ − sin (u ) sin ⎜ ⎟ + cos(u ) cos⎜
⎥
2 ⎦
⎝ 6 ⎠ 2
⎝ 6 ⎠
⎝6⎠
⎝6⎠
⎣
=
2ω LS I o
Vm
⎤
⎡
2ω LS I o
3
3
cos(u ) − 0.5 sin (u ) −
cos(u ) + 0.5 sin (u ) + 3 ⎥ =
⎢−
Vm
2
⎦
⎣ 2
3[1 − cos(u )] =
cos(u ) = 1 −
2ω LS I o
Vm
2ω LI o
2ω LI o
2 ω LS I o
=1−
=1−
VLL
3 Vm
2 VLL
⎡
2ω LS I o ⎤
u = cos −1 ⎢1 −
(2.109)
⎥
VLL ⎦
⎣
⎡
2ω LS I o ⎤
u 1
(2.110)
Δt = = cos −1 ⎢1 −
⎥
V
ω ω
LL
⎦
⎣
It is clear that the DC voltage reduction due to the source inductance is the drop across the
source inductance.
di
(2.111)
vrd = LS D
dt
Multiply (2.111) by dωt and integrate both sides of the resultant equation we get:
π
2
+u
Io
∫ vrd dω t = ∫ ω LdiD = ω LS I o
π
(2.112)
0
2
π
2
+u
∫ vrd dω t
is the reduction area in one commutation period Δt . But we have six
π
2
commutation periods Δt in one period so the total reduction per period is:
64
Diode Circuits or Uncontrolled Rectifier
π
2
6
+u
∫ vrd dω t = 6ω LS I o
(2.113)
π
2
To obtain the average reduction in DC output voltage Vrd due to source inductance we
have to divide by the period time 2π . Then,
6ω LI o
Vrd =
= 6 fLI o
(2.114)
2π
The DC voltage without source inductance tacking into account can be calculated as
following:
(2.115)
Vdc actual = Vdc without sourceinduc tan ce − Vrd = 1.35VLL − 6 fLI d
Fig.2.42 shows the utility line current with some detailes to help us to calculate its rms
value easly.
Is
u
Io
2π
+u
3
2π u
+
6 2
− Io
2π
3
Fig.2.42 The
utility line current
π u
⎡
⎤
+
u
2
3 2
⎢
⎥
2 ⎛ Io ⎞
⎢ ⎜ ωt ⎟ dωt + I d2 dωt ⎥ =
Is =
π⎢ ⎝u ⎠
⎥
u
0
⎢
⎥
⎣
⎦
∫
∫
2 I o2 ⎡ 1 3 π u
⎤
u
+
+
−
u
⎢
⎥
π ⎣ 3u 2
3 2
⎦
2 I o2 ⎡ π u ⎤
Then I S =
(2.116)
−
π ⎢⎣ 3 6 ⎥⎦
Fig.2.43 shows the utility line currents and its first derivative that help us to obtain the
Fourier transform of supply current easily. From Fig.2.43 we can fill Table(2.2) as
explained before when we study Table (2.1).
65
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
u
Is
Io
7π u
−
6 2
π
− Io
I s′
6
−
5π u
−
6 2
u
2
u
Io
u
5π u
−
6 2
π
I
− o
u
11π u
−
6 2
6
−
7π u
−
6 2
11π u
−
6 2
u
2
Fig.2.43 The utility line currents and its first derivative.
Table(2.2) Jumb value of supply current and its first derivative.
5π u
5π u
−
+
6
6
6 2
6 2
Is
0
0
0
0
Io
Io
− Io
− Io
I s′
u
u
u
u
It is an odd function, then ao = an = 0
Js
π
π
u
2
−
+
bn =
⎤
1 ⎡m
1 m
J
cos
n
ω
t
J s′ sin nωt s ⎥
−
⎢
s
s
nπ ⎣⎢ s =1
n s =1
⎦⎥
bn =
1
nπ
∑
7π u
−
6 2
0
− Io
u
u
2
7π u
+
6 2
0
Io
u
11π u
−
6
2
0
Io
u
∑
11π u
+
6
2
0
− Io
u
(2.117)
⎡ − 1 Io ⎛
⎛π u ⎞
⎛π u ⎞
⎛ 5π u ⎞
⎛ 5π u ⎞
− ⎟ + sin n⎜
+ ⎟
⎢ * ⎜ sin n⎜ − ⎟ − sin n⎜ + ⎟ − sin n⎜
n
u
6
2
6
2
6
2
⎝
⎠
⎝
⎠
⎝
⎠
⎝ 6 2⎠
⎝
⎣
⎛ 7π u ⎞
⎛ 7π u ⎞
⎛ 11π u ⎞
⎛ 11π u ⎞ ⎞⎤
− sin n⎜
− ⎟ + sin n⎜
+ ⎟ + sin n⎜
− ⎟ − sin n⎜
+ ⎟ ⎟⎥
2⎠
2 ⎠ ⎠⎦
⎝ 6 2⎠
⎝ 6 2⎠
⎝ 6
⎝ 6
nu ⎡
nπ
5nπ
7 nπ
11nπ ⎤
−
−
+
cos
cos
cos
cos
(2.118)
2 ⎢⎣
6
6
6
6 ⎥⎦
n 2πu
Then, the utility line current can be obtained as in (2.119).
bn =
2I o
i(ω t ) =
* sin
4 3 ⎡ ⎛u⎞
1
1
⎛ 5u ⎞
⎛ 7u ⎞
sin ⎜ ⎟ sin (ωt ) − 2 sin ⎜ ⎟ sin (5ωt ) − 2 sin ⎜ ⎟ sin (7ωt ) +
⎢
π u ⎣ ⎝2⎠
⎝ 2 ⎠
⎝ 2 ⎠
5
7
+
(2.119)
1
⎤
⎛ 11u ⎞
⎛ 13u ⎞
sin ⎜
⎟ sin (11ωt ) + 2 sin ⎜
⎟ sin (13ωt ) − − + + ⎥
⎝ 2 ⎠
⎝ 2 ⎠
11
13
⎦
1
2
2 6 Io ⎛ u ⎞
sin ⎜ ⎟
πu
⎝2⎠
The power factor can be calculated from the following equation:
Then; I S1 =
66
(2.120)
Diode Circuits or Uncontrolled Rectifier
pf =
I S1
⎛u⎞
cos ⎜ ⎟ =
IS
⎝2⎠
Then; pf =
2 6 Io ⎛ u ⎞
sin ⎜ ⎟
πu
⎝2⎠
⎛u⎞
cos ⎜ ⎟
⎝2⎠
2 I o2 ⎡ π u ⎤
−
π ⎢⎣ 3 6 ⎥⎦
3 * sin (u )
(2.121)
⎡π u ⎤
u π⎢ − ⎥
⎣ 3 6⎦
Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz
supply has 8 mH source inductance to feed 300A pure DC load current Find;
(i)
commutation time and commutation angle.
(ii)
DC output voltage.
(iii) Power factor.
(iv) Total harmonic distortion of line current.
Solution: (i) By substituting for ω = 2 * π * 50 , I d = 300 A , L = 0.008 H , VLL = 33000V
in (2.109), then u = 0.2549 rad . = 14.61 o
u
Then, Δt = = 0.811 ms .
ω
(ii) The the actual DC voltage can be obtained from (2.115) as following:
Vdc actual = Vdc without sourceinduc tan ce − Vrd = 1.35VLL − 6 fLI d
Vdcactual = 1.35 * 33000 − 6 * 50 * .008 * 300 = 43830V
(iii) the power factor can be obtained from (2.121) then
3 * sin (u )
3 sin (0.2549 )
= 0.9644 Lagging
pf =
=
0
.
2549
π
u
π
⎡
⎤
⎡
⎤
u π ⎢ − ⎥ 0.2549 * π ⎢ −
6 ⎥⎦
⎣3
⎣ 3 6⎦
(iv) The rms value of supply current can be obtained from (2.116)as following
Is =
2 I d2 ⎡ π u ⎤
−
=
π ⎢⎣ 3 6 ⎥⎦
2 * 300 2 ⎛ π 0.2549 ⎞
*⎜ −
⎟ = 239.929 A
6 ⎠
π
⎝3
The rms value of fundamental component of supply current can be obtained from (2.120)
as following:
4 3 Io ⎛ u ⎞
4 3 * 300
⎛ 0.2549 ⎞
I S1 =
sin ⎜ ⎟ * 2 3 =
* sin ⎜
⎟ = 233.28 A
πu 2
π * 0.2549 * 2
⎝2⎠
⎝ 2 ⎠
pf =
I S1
⎛ u ⎞ 233.28
⎛ 0.2549 ⎞
* cos⎜ ⎟ =
* cos⎜
⎟ = 0.9644 Lagging.
Is
⎝ 2 ⎠ 239.929
⎝ 2 ⎠
2
2
⎛ IS ⎞
⎛ 239.929 ⎞
⎜
⎟
THDi = ⎜
⎟ − 1 = ⎜⎝ 233.28 ⎟⎠ − 1 = 24.05%
⎝ I S1 ⎠
67
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
2.7 Multi-pulse Diode Rectifier
Twelve-pulse bridge connection is the most widely used in high number of pulses
operation. Twelve-pulse technique is using in most HVDC schemes and in very large
variable speed drives for DC and AC motors as well as in renewable energy system. An
example of twelve-pulse bridge is shown in Fig.2.33. In fact any combination such as this
which gives a 30o-phase shift will form a twelve-pulse converter. In this kind of converters,
each converter will generate all kind of harmonics described above but some will cancel,
being equal in amplitude but 180o out of phase. This happened to 5th and 7th harmonics
along with some of higher order components. An analysis of the waveform shows that the
AC line current can be described by (2.83).
2 3 ⎛
1
1
1
1
⎞
iP (t ) =
I d ⎜ sin (ωt ) −
sin (5ωt ) + sin (13ωt ) − sin (23ωt ) + sin (25ωt ) − +...⎟ (2.83)
π
⎝
11
23
13
2
2
2
⎠
25
2
2
2
⎛1⎞ ⎛1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞
THD ( I P (t )) = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟
⎝ 11 ⎠ ⎝ 13 ⎠ ⎝ 23 ⎠ ⎝ 25 ⎠ ⎝ 35 ⎠ ⎝ 35 ⎠
= 13.5%
As shown in (10) the THDi is about 13.5%. The waveform of utility line current is
shown in Fig.2.34. Higher pulse number like 18-pulse or 24-pulse reduce the THD more
and more but its applications very rare. In all kind of higher pulse number the converter
needs special transformer. Sometimes the transformers required are complex, expensive
and it will not be ready available from manufacturer. It is more economic to connect the
small WTG to utility grid without isolation transformer. The main idea here is to use a sixpulse bridge directly to electric utility without transformer. But the THD must be lower
than the IEEE-519 1992 limits.
2N :1
a
Vd
a1
b1
c
c1
b
2 3 N :1
a2
b2
c2
Fig.2.33 Twelve-pulse converter arrangement
68
Diode Circuits or Uncontrolled Rectifier
(a) Utility input current.
(b) FFT components of utility current.
Fig.2.34 Simulation results of 12.pulse system.
69
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
Problems
1- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz supply to feed
5Ω pure resistor. Draw load voltage and current and diode voltage drop waveforms
along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple
factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse
Voltage (PIV) of the diode. (e) Crest factor of supply current.
2- The load of the rectifier shown in problem 1 is become 5Ω pure resistor and 10 mH
inductor. Draw the resistor, inductor voltage drops, and, load current along with supply
voltage. Then, find an expression for the load current and calculate the conduction
angle, β . Then, calculate the DC and rms value of load voltage.
3- In the rectifier shown in the following figure assume VS = 220V , 50Hz, L = 10mH
and Ed = 170V . Calculate and plot the current an the diode voltage drop along with
supply voltage, vs .
vL vdiode
i
+
+
+
vs
-
Ed
-
Assume there is a freewheeling diode is connected in shunt with the load of the
rectifier shown in problem 2. Calculate the load current during two periods of supply
voltage. Then, draw the inductor, resistor, load voltages and diode currents along with
supply voltage.
5- The voltage v across a load and the current i into the positive polarity terminal are as
follows:
v(ωt ) = Vd + 2 V1 cos(ωt ) + 2 V1 sin (ωt ) + 2 V3 cos(3ωt )
4-
i (ωt ) = I d + 2 I1 cos(ωt ) + 2 I 3 cos(3ωt − φ )
Calculate the following:
(a) The average power supplied to the load.
(b) The rms value of v(t ) and i (t ) .
(c) The power factor at which the load is operating.
6- Center tap diode rectifier is connected to 220 V, 50 Hz supply via unity turns ratio
center-tap transformer to feed 5Ω resistor load. Draw load voltage and currents and
diode currents waveforms along with supply voltage. Then, calculate (a) The
rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization
Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply
current.
7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz supply to feed 5Ω
resistor. Draw the load voltage, diodes currents and calculate (a) The rectfication
effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF)
(d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.
70
Diode Circuits or Uncontrolled Rectifier
8-
910-
11-
12-
1314-
1516-
17-
18-
If the load of rectifier shown in problem 7 is changed to be 5Ω resistor in series with
10mH inductor. Calculate and draw the load current during the first two periods of
supply voltages waveform.
Solve problem 8 if there is a freewheeling diode is connected in shunt with the load.
If the load of problem 7 is changed to be 45 A pure DC. Draw diode diodes currents
and supply currents along with supply voltage. Then, calculate (a) The rectfication
effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF)
(d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. (f)
input power factor.
Single phase diode bridge rectifier is connected to 220V ,50Hz supply. The supply
has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC
current. Draw, output voltage, diode currents and supply current along with the supply
voltage. Then, calculate the DC output voltage, THD of supply current and input
power factor, and, input power factor and THD of the voltage at the point of common
coupling.
Three-phase half-wave diode rectifier is connected to 380 V, 50Hz supply via
380/460 V delta/way transformer to feed the load with 45 A DC current. Assuming
ideal transformer and zero source inductance. Then, draw the output voltage,
secondary and primary currents along with supply voltage. Then, calculate (a)
Rectfication effeciency. (b) Crest factor of secondary current. (c) Transformer
Utilization Factor (TUF). (d) THD of primary current. (e) Input power factor.
Solve problem 12 if the supply has source inductance of 4 mH.
Three-phase full bridge diode rectifier is connected to 380V, 50Hz supply to feed
10Ω resistor. Draw the output voltage, diode currents and supply current of phase a.
Then, calculate: (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c)
Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e)
Crest factor of supply current.
Solve problem 14 if the load is 45A pure DC current. Then find THD of supply
current and input power factor.
If the supply connected to the rectifier shown in problem 14 has a 5 mH source
inductance and the load is 45 A DC. Find, average DC voltage, and THD of input
current.
Single phase diode bridge rectifier is connected to square waveform with amplitude of
200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the
rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply
current along with the supply voltage. Then, calculate the DC output voltage, THD of
supply current and input power factor.
In the single-phase rectifier circuit of the following figure, LS = 1 mH and
Vd = 160V . The input voltage vs has the pulse waveform shown in the following
figure. Plot is and id waveforms and find the average value of I d .
71
Chapter Two
Dr. Ali M. Eltamaly, King Saud University
id
+
iS
Vd
VS
-
f = 50 Hz
200V
ωt
120o
60
o
120
o
60o
60
72
o
120
o