M.CINDY
D.C.TIBUJI(M.Sc electrical engineering)
All right reserved. No part of this publication should be produced or transmitted in any
form or by any means without written permission of the author.
D.C.TIBUJI
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To Babyscho and Maria Cindy
My brilliant, beautiful wife and baby I say thank you for always being there to comforts
and consoles ask nothing and endures all.
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CONTENTS
PREFACE
ACKNOWLEDGMENT
Chapter 1 WAVEFORM ANALYSIS
1-1. Terminology
Chapter 2 SEMICONDUCTOR DEVICE
2.1 Diode
2.2 Diode characteristic
2.3 Thyristor
2.3 Thyristor characteristic
2.4 Power Transistor 2.5 Transistor characteristic
2.6 The unijunction transistor 2.7 Principle of operation 2.8 Programmable unijunction
transistor
2.9 Diac and Triac
Chapter 3 RECTIFYING CIRCUITS
3.1 types of rectifying circuit 3.2 single-phase half-wave uncontrolled rectification
3.3 single-phase full- wave rectification 3.4 three-phase non-controlled rectification 3.7
single-phase controlled rectification 3.8 single-phase half-controlled rectification 3.9
three-phase controlled rectification.
Chapter 4 CONVERTERS
4.1 different types of converters 4.2 application of some converters.
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PREFACE
This power electronics book is designed to give students who are preparing for the
advance level certificate and baccalaureate technique examination in English the whole
syllabus and this is also as a result of difficulties in getting English text books in power
electronics, so students who have prepared the whole syllabus are having a wide choice
but students who have left out sections of the syllabus will be restricted as to the
questions they my attempt but sufficient practice in answering questions will help you to
develop the necessary skills within your power electronic course and for your
examinations.
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ACKNOWLEDGMENT
I will first like to give thanks to God almighty for granting me the strength to undertake
this idea.
I would like to thank everyone especially; the instructors and students that are
undertaking this course for their ideas that gave me great insight into what I have done.
That not withstanding, I want to acknowledge and thank each of the personalities who
made significant contributions to this book:
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WAVE FORM ANALYSIS
In the analysis of performance with chopped voltage waveform, it is necessary to
determine some features such as average, effective values of current, voltage and power.
Furthermore, knowledge of harmonic content of waveform is often needed, if losses are
to be calculated or if comparism of waveforms is to be made. Application of this
calculation is use throughout project.
TERMINOLOGY
DIRECT CURRENT: it is a unidirectional current, that is a current that flows always in
one sense with a constant value as seen in the figure I below;
i
I
t
ALTERNATING CURRENT: A current changes it sense periodically and whose average
value is zero.
FREQUENCY: the frequency of a sinusoidal current is equal to the number of times that
current repeat with in a second, the unit is Hz.
PERIOD: the period denoted with T = 1/f.
i
Instantaneous value
i-- -- -----------Positive half cycle
π
2π
t
0
Negative half cycle
T
INSTANTANEOUS VALUE: it is the value of current or voltage at a given time.
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Example: I = Imax Sin ωt and
U = Umax Sin ωt
Where u and I are the instantaneous values of voltage and current and
Umax and Imax are the peak or maximum values of voltage and current
ω = 2πf = 2π / T (rad/s)
PEAK-TO-PEAK VALUE: the peak-to-peak value is equal to two peak values.
AVERAGE VALUES
The determination of the average value or direct current (d.c) value across the load is
important. If the average values of the voltage V (t) is periodic with period T, then, the
Average value is given by
t
Vav =1 / T ∫ V(t) dt
0
It is convenient to transform time (t) to radian φ by simply φ = ωt
Since the above function is a Sine function, the average value of the voltage will be given
by
π
Vav =1/T ∫Vmax Sinφ dφ
0
π
=>Vav = 1/2π ∫ VmaxSinφ dφ
0
π
=> Vmax / 2π [-Cosφ]
0
=> Vmax / 2π [-Cosπ]-[-Cos 0]
=>Vmax / 2π [1+1]
Therefore, the average voltage Vav = Vmax / π
Vav = Vmax / π
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THE AVERAGE VALUE OF THE CURRENT (Iav)
π
Iav = 1 / T ∫Imax Sinφ dφ
0
π
 1 / π ∫Imax Sinφ dφ
0
π
 = Imax / π ∫ Sin φ dφ
0
π
 Iav = Imax / π [-Cos φ ]
0
 => Imax / π [-Cos (π) ] - [-Cos (0) ]
=> Imax / π [1+1] = Imax / π X 2
therefore Iav = 2Imax / π
Iav = 2Imax / π
ROOT MEAN SQUARE VALUE OR THE EFFECTIVE VALUE
The root mean square value (R.M.S) of the current or voltage that varies
periodically in time, is the effective value that is equivalent to a constant direct current in
terms of heating .the periodic current and the direct current produce the same average
power in an element. if the instantaneous value of current is given by:
t
Ir.m.s or Ieff = 1 / T ∫ i² (t) dt
0
The root mean square value is too small to determine the value of average power in A.C
load but the voltage and current are not constant values. Hence, the effective value of
voltage in this case is;
π
V²eff = 1 / T ∫ (Vmax Sin φ) ²dφ
0
π
= 1 / 2π ∫V²max Sin²φ dφ
0
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π
=> V²max / 2π ∫Sin²φ dφ
0
But knowing that Sin²φ = 1- Cos 2φ /2
π
 V²max / 2π ∫1 / 2 (1- Cos 2φ) dφ
0
π
 V²max / 2π *1/ 2 [φ – 1/ 2 Sin 2φ]
0
 V²max / 4π [π – 1/ 2 Sin 2π] - [0 – 1/ 2 Sin 2(0)]
=> V²max / 4π [π - 0] - [0 -0] = V²max / 4π [ π ]
 Veff = Vmax / 2
 (V²eff )½ = ( V²max / 4 )½
 Veff = Vmax / 2
Therefore
 Veff = Vmax / 2
THE EFFECTIVE CURRENT
π
I²eff = 1/ T ∫ (Imax Sin φ) ²dφ
0
π
I²eff =1/ π ∫ I²max Sin²φ dφ
0
π
 I²max / π ∫ Sin²φ dφ
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0
π
 I²max / π ∫ 1/ 2(1 – Cos φ ) dφ
0
π
 I²max / 2π [ φ – 1/ 2 Sin 2φ ]
0
 I²max / 2π [ π – 1/ 2 Sin 2π ] - [ 0 – 1/ 2 Sin 2(0) ]
=> I²max / 2π [ π – 0 ] - [ 0 ] = I²max / 2π [ π ]
I²eff =I²max / 2
Therefore
Ieff = Imax / √2
FORM FACTOR (F)
The form factor is applicable to ac waveforms. It provides indication of both distortion
waveforms. It is the ratio of the effective value to the average value.
FORM FACTOR:
F = EFFECTIVE VALUE
AVERAGE VALUE
RIPPLE FACTOR (r)
Although it is the role of a rectifier to convert ac to dc ,the rectifier doesn’t achieve this
goal, the out put voltage or current is unidirectional but still, the periodical fluctuation,
the input voltage, the ripple voltage or current is measure in terms of ripple factor which
is;
r = EFFECTIVE VALUE OF COMPONENT
DC VALUE OF WAVE FORM
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r = √F2 - 1
UTILIZATION FACTOR OF A TRANSFORMER
The utilization factor of a transformer is the ratio of the dc power dissipated in the load
by the ac power in the secondary of the transformer.
UFT = Iav x Vav
Ieff x Veff
CHAPTER: TWO SEMI CONDUCTOR DEVICE
A semi conductor is an element with four electrons on the outer most shell and the
number of electrons in the valence orbit is the key to electrical conductivity.
Semiconductor devices, which are also use as rectifying devices can be, define as a
device, which permits current flow in one direction only, and have the capability to
prevent voltage in the absence of current flow in the opposite direction. The main
devices, semi conductor’s devices are the diode, thyristor, transistor, diac and triac.
DIODE
A diode usually called a junction diode is a semi conductor device that is compose by
putting together two different types of semi conductor materials which are the N and P
type material such that they can permit the flow of electric current during the forward
bias and block the flow of current during the reverse bias.
As said earlier, that the diode is compose of two semi conductor materials, these
materials is being form from silicon or germanium but nowadays germanium is scarcely
used ,so a silicon is a semi conducting material which lies between the insulating and
conducting materials, its current increases with respect to the temperature rise.
In the beginning of this chapter, it was said that silicon is having four electrons in the
outer most shell. if an element with five electrons on the outer most shell is added on to
it, then a free electron will be present and the presence of these free electron causes an
increase in conduction and as electrons are negatively charged, the material is known as
an N-type semi conductor.
Impurity atoms is now added to silicon, the impurity atom is having three electrons on the
outer most shell, then the presence of holes will be notice which can accept an electron.
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The presence of a hole provide a positively charge carrier which will increase conduction
and this material is known as a P-type semiconductor.
The silicon atom in its pure state without any impurity atoms to alter its conductivity is
known as INSTRINSIC semi conductor, when a pure silicon atom is doped or says
impurity atoms added to it, it is called EXTRINSIC semi conductor.
The figure 2.1 below shows the composition of a diode.
When the P-type and the N-type material are put together, what happen, at the junction,
the free electrons of the N-type material and the free holes of the P-type material
combine, so when these union between P-type and N-type materials takes place, the P
side is being left with negative charge while the N-side is left with positive charge.
Therefore, a potential barrier exists across the junction having a value of 0.6 V
THE DIODE CHARACTERISTIC AND SYMBOL
A
I
F. Volt-drop
I
Forward region
V
Leakage I
V
K
Reverse blocking
Reverse breakdown
When the diode is forward bias, the anode of the diode (P-type) is connected to the
positive terminal of the voltage source and the cathode (N-type) is connected to the
negative terminal of a voltage. these will result in current flow once the potential barrier
of 0.6 V is overcome, giving an overall forward voltage drop of 0.7 V at its rated current.,
While the reverse bias is by reversing the polarity of the voltage source from that of the
case with forward bias. Because of this reverse voltage, it will take away the mobile
carriers of holes and electrons away from the junction in both the P and N sides, thus
preventing current flow and giving room for the junction to withstand the supply voltage
without functioning. The junction experiences a high electric field gradient and hence can
be consider as having capacitance. Moreover, because of heat, thermal agitation does
rupture some of the bonds in the crystal, resulting in minority carriers that permit a small
reverse current flow shown as a leakage current. as the reverse voltage increases, it will
lead to an increase in the accelerate rate of the minority carriers across the junction. With
the sufficient energy that the minority carriers are having, they are able to remove others
by collision and when the junction is broken down, its gives the reverse break down
characteristic.
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THYRISTOR
The THYRISTOR is a four layer, three terminal silicon control switch used primary for
ac and dc power control.
STRUCTURE
The silicon control rectifier has four layers of semi conductor materials arranged in a
PNPN structure as seen in fig 2.3 below.
The anode and the cathode terminals are connected to the outer P and N layers. a gate
terminal is connected to the inner P layer and serves to trigger the silicon control rectifier
in to conduction.
Anode
P
J1
N
J2
P
Gate
N
Gate
J3
Cathode
CONSTRUCTION OF SILICON CONTROL RECTIFIER (THYRISTOR)
The diode equivalent circuit can be view in the figure 2.3 above. It consists of two
forward connected diodes formed by PN junctions J1 and J3. A reversed biased diode
formed by NP junction, J2 is connected between J1 and J3.the gate terminal connected to
the common anode of J2 and J3.the construction of a thyristor can be stimulated using
programmable unijunction transistor.
OPERATION
When a forward voltage is apply between the anode and the cathode,
The silicon control rectifier cannot conduct because of the reverse connection of J2
junction applying sufficient positive voltage to the gate.
Causing the J2 junction to break down in the reverse direction, and allowing the J1 and J3
junctions to conduct forward current.
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Once forward current starts flowing, it magnitude is essentially limited only by an
external circuit and the gate no longer needed.
If the anode to cathode source drops to a low value or reverse in polarity, the silicon
control rectifier drops out of conduction due to the bias requirement across the J1 and J3
junctions.
It should be noted that, if the maximum forward and reverse voltages of the silicon
control rectifier were exceeded, the silicon control rectifier would break down regardless
of the gate connection,
THYRISTOR CHARACTERISTIC AND
SYMBOL
I
Forward volt-drop
holding current
latching
current
Reverse leakage current
forward breakdown
Reverse breakdown
Anode
Gate
Cathode
Fig 2.4
The thyristor can be consider as three diodes in series, as in the figure 2.3 above.
Considering the effect of increasing the voltage applied across the thyristor with the
anode positive relative to the cathode. Firstly, the forward leakage current reaches
saturation value due to the action of junction J2 .furthermore a break over value is reach
and the resistance of the thyristor instantly falls to a very low value as shown in the
characteristic curve above. The forward voltage drop is of the order of 1 – 2 volts and
remains nearly constant over a wide variation of current’s
THYRISTOR PROPERTIES
To turn on a thyristor positively in the shortest time, it is desirable to have a gate current
with a fast rise time up to the maximum permitted value.
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The anode must be more positive than the cathode that is, the thyristor being forward
bias.
There must be a current Ig in the gate for a few second. When the conduction is started
the gate losses its control on the thyristor and the conduction ends when the anode is not
more positive than the cathode. A reverse gate current must be prevented to avoid
excessive gate power dissipation. Injection of gate current when the thyristor is reversebiased will increase the leakage current and is best to avoid.
APPLICATION
They are use to control power in a rectifier circuit, dc motors, heaters, lighting systems
and more.
POWER TRANSISTOR
The bipolar power transistor is a three-layer P-NP or N-P-N device having 2 P-N
junctions. bipolar power transistor can also be define as a semi conductor device
constructed or manufactured by putting together two power diode, connected FACE TO
FACE or BACK TO BACK as shown on the figure 2.5 below.
The behavior of a bipolar power transistor within the working interval is marked out in
this manner;
The collector current IC is in function with the base current IB,so a change in base
current will give a corresponding amplified change in the collector-emitter voltage VCE
STRUCTURE AND SYMBOL
N
P
P
N
N
P
collector
C
P
N
B
base
P
E
emitter
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collector
C
P
N
B
Base
P
E
emitter
Fig 2.5
TRANSISTOR CHARACTERISTIC
The transistor characteristic is shown in figure 2.6 below. in a similar way to the other
devices, a break down level is obtain with an increasing voltage .reversal or change of the
collector –emitter voltage will break down the base emitter junction at a low voltage, say
10V. Therefore, the transistor is not operating in the reverse mode.
The power dissipated in the transistor is a function of the product of collector-emitter
voltage with the collector current.
The P-N-P transistor behave like the N-P-N transistor but for the exception that the
current and the voltage directions being reversed.
CHARACTERISTIC CURVE (common-emitter)
IC IB increasing
Saturation
Voltage
Break over voltage
IB = 0
leakage current
VCE
Reverse
Breakdown
Fig 2.6 Characteristic curve for N-P-N transistor
In practice, for power applications, the transistor is operated as a switch.
With or when the base current IB = 0, it is effectively an open-circuit condition. That is,
an open switch mode, as shown in the figure below. The presence of the base current IB
will take the device in to saturation,. That is a closed switch mode. The fact that a
transistor is a controlling device, it is essential to profile the base current to the collector
current. In order to retain the control process, when it is in the saturated state, in order to
avoid excessive base charge, the base current should be just sufficient to maintain
saturation. at start or tun-on,initially. The base current should be high to give a fast turnon. Any adjustment in the collector current must be match by an adjustment also in the
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base current. Turning off the device, the base current should be reduced, at a rate which
the collector current can flow, so as to avoid secondary breakdown,. When the device is
off, a small reverse current is maintained to avoid spurious collector current. When the
transistor is operating as a switch, the transistor power losses are small, being due to the
small leakage current in the open position and the saturation voltage with the collector
current when in the closed position. Typically, the saturation voltage is 1.1 V for a silicon
power transistor.
THE UNIJUNCTION TRANSISTOR (U J T)
A UJT has two extrinsic regions with three external leads. This implies, it is having a
single PN junction similar to the diode. It is having one emitter and two bases. These
contacts are designated as the emitter (E), base 1 (B1) and base 2 (B2).an aluminum wire
is alloyed to the silicon bar summing the emitter lead of the P-material of the PN
junction.
STRUCTURE AND SYMBOL
Base 2
A)
P
Emitter
n
Base 1
B)
B2
E
B1
Figure 2.7 A is the structure and B is the symbol
An equivalent circuit of the unijunction transistor is shown in figure 2.8 below.
B2
R2
E
D
R1
B1
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RB1 and RB2 represent the ohmic resistance of the silicon bar between B1 and B2 on
either side of the PN junction. The total interbase resistance
RBB = RB1 + RB2
The diode shown in the equivalent circuit of a UJT is connected to the junction of RB1
and RB2.its represents the PN junction of the UJT.the supply voltage is applied between
B2 positive and B1 negative. The input signal is applied between the emitter and the base
1 B1
PRINCIPLE OF OPERATION
When VBB is inclined between B2 and B1 current flows through RBB and sets up the
voltage drop at point J.
This makes the cathode of the PN junction more positive than the B1.
With no voltage applied to the emitter, the PN junction is reverse bias.
The amount of reverse bias junction voltage is
VJ =
VBB___ x RB1
RB1 + RB2 VJ=
OR
For the UJT to operate, appositive voltage must be applied to the emitter terminal VE.
VE = VD + VJ
In order for it to conduct, emitter current Ie,at the reverse bias created by VJ at the PN
junction as well as the diode drop Vg,must be overcome at;
VE = VD + VJ
The PN junction becomes forward biased and the UJT fires. According to semi conductor
theorem, holes from the P-materials are injected to the N-region because of the polarity
of VBB across the N-region. The injected silicon holes migrate toward the negative B1
terminal. This increase in free electrons between emitter and B1 causes an effective
production in RB1.as RB1 decreases, reverse bias VJ increases, effectively, increasing
the diode forward bias.
This allowed more holes to be injected from the emitter into the silicon bar and RB1
decreases further. This process is regenerative. This gives the UJT a unique negative
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resistance of the characteristic that is IB increases, RB1 decreases and the voltage drop
across decreases. This is contrary to ohms law of conventional circuit.
CHARACTERISTIC
Vp
Cut-off
Region
Negative resistance
region
saturation
region
IE
Fig 2.9
The negative resistance effects is shown on the emitter characteristics curve in figure 2.9
above, beyond the value of points of the curve RB increases linearly
I – the emitter voltage and resultant becomes positive again.
One of the important parameter of UJT is the INTRINSIC STAND OFF RATIO (η) s.this
refered to the resistance voltage divider ratio
η = __ RB1___
RB1 + RB2
= RB1
RBB
The intrinsic stand off ratio (η)
ranges from 0.4 to 0.1 depending on
the type of UJT.
Vp =VJ + VD = ηVBB + VD
Example;
An oscillator made from a unijunction transistor circuit is supply with 20V and having
Rb2 = 3KΩ a capacitor of 1µf a total resistance through which the capacitor is charge
having a value of 10KΩ
Solution
Data
RT = 10kΩ,Rb2 = 3kΩ, C =1µf ,VBB =20V and
η =0.6
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VBB = 20v
RT=10K
R=3K
E
1µF
B1
Vp = η VBB + Vd
Vp = 0.6 x 20 + 0.7 =>
Vp =12.7 V
Vp = 1 – e-t/T
VBB
e-t/T = 1 – Vp
VBB
ln(e-t/T) = ln(1-Vp/VBB)
-t = ln VBB-VP
T
VBB
-t = T ln VBB-Vp
VBB
t = -T ln VBB-Vp
VBB
T = RC where T is the time constant
=10 x 10-3x 1 x 10-6
T = 10 x 10-3
t = T ln VBB
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VBB-Vp
10 x 10-3 ln 20___
20-12.7
= 10 x 10-3 x 1
f =100HZ
f = 100HZ
t =10-2 = 0.01 sec
PROGRAMMABLE UNIJUNCTION TRANSISTOR
A
G
K
Where; A, is the anode
K, is the cathode
G, is the gate
Figure 2.10
The difference between the UJT and PUT is at the level of the intrinsic stand off ratio.
For the UJT, the stand off ratio is an intrinsic value and given to you by the manufacturer.
While for the PUT, the stand off ratio is programmable by the user. Hence, the stand off
ratio (η) is a value designated by the user. the gate terminal always serves as the reference
voltage and the PUT can come on only when the gate voltage is equal to the anode
voltage,Va =Vg.
The circuit diagram, waveform and structure can be seen below in figure 2.11
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VBB
RT
R1
A
G
CT
R2
VA
VG
VR
STRUCTURE OF P.U.T
G
A
K
P
N
P
N
Example: calculate the frequency of the circuit below;
VBB = 20v
RT =100kΩ
A
R1=15kΩ
G
CT
= 0.47µf
Solution
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R2 = 20kΩ
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Data,
VBB = 20V, R1 = 15KΩ, R2 = 20K, CT = 0.47µf
RT = 100KΩ
VG = R2 X VBB
R2 + R1
VG = 20 X 20
20 + 15
=> VG = 11.4V
T = RC => T = 100 X 103 X 0.47 X 10-6
T = 0.047
t1 = T ln VBB___
VBB - VP
Where VP = VG
t1 = 0.047 ln 20___
20 – 11.4
t1 = 3.9 X 10-2 sec
f = 1_ => f = 1______
T
3.9 X 10-2 sec
f = 25.2 HZ
DIAC
DESCRIPTION: the diac is a four-layer semi conductor diode connected in parallel as
shown in figure 2.12 below.
The diac is an ac switch used primary as a triggering device for most ac circuit.
The diac is non conducting until the current across it tries to excite the break over voltage
in either direction. It is a bi-directional component.
SYMBOLS
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OPERATION
Current flow is initiated when the voltage across the diac reaches the break over voltage
in either directions, once conduction starts current increase rapidly and the voltage across
the diac decrease very fast.
The current pulse is produce when the diac changes from a non-conducting state to a
conducting state, which is use for the triggering purposes. The break over voltage of the
diac is symmetrical in both directions.
For instance, if you have a polarity of positive or negative as indicated in the diagram
above, the left diode conducts when you try to exceed the break over voltage. In this case,
the left latch closes as shown in figure 2.13 above.
On the other hand, if the polarity is opposite to that of A the right latch closes and when
you try to exceed the break over voltage once the diac is conducting. The only way to
open it is by low current drop. This means reducing the rated current below the rated
holding current of the device. Figure 2.14 below shows the schematic symbol of the diac.
CHARACTERISTIC OF A DIAC
I
IH
VBO
VBO
Fig 2.14
POINT TO NOTE
A diac is a two terminal bi-directional diode used as an ac switch.
A diac exhibit negative resistance when the break over voltage is exceeded in either
direction.
APPLICATION
Diac are used as (or in) ac switches pulse generator, relaxation oscillator and firing
circuit.
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Example: as relaxation oscillator circuit
V
R1
R2
X
Y
C
THE TRIAC
DESCRIPTION: a triac is a three terminal semi conductor switch. it s operates much like
two inverse parallel thyristors as such it is capable of conducting in either polarity of the
terminal voltage and can be triggered by either polarity of the gate. A simplification of a
triac and its characteristic curve is shown in figure 2.15 below.
T2
G
Simplify triac
Circuit
T1
T2
G
T1
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RECTIFYING CIRCUITS
A circuit that links an ac supply to a dc load is call a rectifier circuit. These ac supplies
are converted to a direct voltage to feed the load. The direct voltage obtained is not
normally level, as the case of a battery but contain an alternating ripple component
superimposed on the average (dc) level.
The various circuit connections described, although all giving a dc output, but differ
about the ac ripple in the output, the average voltage level, efficiency and their loading
effects on the ac supply system.
TYPES OF RECTIFYING CIRCUIT
Rectifying circuits divide broadly into two groups, namely the HALF-WAVE AND
FULL-WAVE connections.
The expression half-wave describes the fact that the current in each ac supply line is
unidirectional. The half-wave is having another name which I s single-way in describing
the circuits. The control characteristics of the various circuits can be described broadly
into one of the three categories: UNCONTROLLED, FULLY CONTROLLED, and
HALF-CONTROLLED.
SINGLE PHASE HALF –WAVE UNCONTROLLED RECTIFICATION (resistive
load)
The word uncontrolled rectifier circuits contain only diodes giving a dc load voltage
fixed in magnitude relative to the ac supply voltage magnitude.
CIRCUIT DIAGRAM
vd
Vs
Fig 3.1
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vl
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Let’s look at the circuit in two directions, the positive and negative alternance.
The voltage at the terminal of the diode, the loop equation given as:
VS – VD – VL = 0
 VS = VD + VL
At the positive alternance the diode conduct and when the diode conducts, the diode
voltage (VD) = O.
Furthermore, at the negative half cycle the diode will not conduct and the current I = 0
=> VR = I R = O
=> VD =VS
WAVEFORMS
vs
π
2π
t
vl
t
vd
t
Fig 3.2
NUMERICAL APPLICATION OF NON-CONTROLLED SINGLE WAY
RECTIFICATION.
The half wave rectifier is supply from an ac network of 110 V, 50 HZ with a resistive
load of 100 Ω .the transformers is having a transformation ratio of 1/5.
WORK REQUIRED
1234-
Draw the circuit diagram and waveforms of the voltage at the load and diode.
Derive the expression of average voltage and effective voltage.
Calculate the maximum reverse voltage at the diode.
What is the maximum, average and effective current through the load
D.C.TIBUJI
27
M.CINDY
SOLUTION.
1-The circuit diagram
vd
Vs
vl
WAVEFORMS
vs
π
2π
t
vl
t
vd
t
2-Derivation of the expression of average voltage and effective voltage.
π
Vav =1/T ∫Vmax Sinφ dφ
0
π
=>Vav = 1/2π ∫ VmaxSinφ dφ
0
π
=> Vmax / 2π [-Cosφ]
0
28 D.C.TIBUJI
M.CINDY
=> Vmax / 2π [-Cosπ]-[-Cos 0]
=>Vmax / 2π [1+1]
Therefore, the average voltage Vav = Vmax / π
Vav = Vmax / π
For average current
π
Iav = 1 / T ∫Imax Sinφ dφ
0
π
 1 / π ∫Imax Sinφ dφ
0
π
 = Imax / π ∫ Sin φ dφ
0
π
 Iav = Imax / π [-Cos φ ]
0
 => Imax / π [-Cos (π) ] - [-Cos (0) ]
=> Imax / π [1+1] = Imax / π X 2
therefore Iav = 2Imax / π
Iav = 2Imax / π
The effective voltage
π
V²eff = 1 / T ∫ (Vmax Sin φ) ²dφ
0
π
= 1 / 2π ∫V²max Sin²φ dφ
0
D.C.TIBUJI
29
M.CINDY
π
=> V²max / 2π ∫Sin²φ dφ
0
But knowing that Sin²φ = 1- Cos 2φ /2
π
 V²max / 2π ∫1 / 2 (1- Cos 2φ) dφ
0
π
 V²max / 2π *1/ 2 [φ – 1/ 2 Sin 2φ]
0
 V²max / 4π [π – 1/ 2 Sin 2π] - [0 – 1/ 2 Sin 2(0)]
=> V²max / 4π [π - 0] - [0 -0] = V²max / 4π [ π ]
 Veff = Vmax / 2
 √(V²eff ) = (√ V²max /√ 4 )
 Veff = Vmax / 2
Therefore
 Veff = Vmax / 2
3 –the maximum current
Imax = Emax
R
Vs = 1 => Vs = Vp = 110
Vp 5
5
5
Eeff = 22V
30 D.C.TIBUJI
M.CINDY
Eeff = Emax => Emax = Eeff√2
√2
Emax = 22√2 = 31.11V
Imax = 31.11
100
Imax = 0.3111A
The average current
Iav = Eav but Eav = Emax
R
π
Eav = 31.11 = 9.9V
3.14
Iav = 9.9 =0.099A
100
Iav = 0.099A
The effective current
Ieff = Eeff but Eeff = Emax = 31.11
R
2
2
Ieff = 15.55 = 0.1555A
100
Ieff = 0.1555A
4 –the maximum reverse voltage at the load.
PIV = Emax = 31.11V
CONCLUSION: these problem was solved using the sine waveform but its can also be
solved using the cosine waveform, there only difference in the solving is at the level of
the waveform.
Using the cosine waveform to solved the very problem that has been solved using the
sine waveform.
D.C.TIBUJI
31
M.CINDY
1- waveform
vs
π/2
-π/2
vl
vd
2 – Deriving the expression of the average and effective voltage using the cosine
waveform.
π/2
Iav = 1 ⌠
Imax Cos φ dφ
T ⌡
-π/2
where T = π
π/2
π/2
 Iav = 1 ⌠ Imax Cosφdφ = Imax⌠Cosφdφ
π ⌡
π ⌡
- π/2
-π/2
32 D.C.TIBUJI
M.CINDY
π/2
=> Iav = Imax [Sinφ]
π
-π/2
 Iav = Imax [Sin(π/2) – Sin(-π/2)]
π
Iav = Imax [1 + 1]
π
Iav = 2 Imax
π
π
Deriving an expression for the average voltage.
Therefore
π/2
Vav = 1 ⌠ Vmax Cosφ dφ
T⌡
-π/2
where T = 2π
π/2
==> Vav = Vmax ⌠ Cos φ dφ
2π ⌡
-π/2
π/2
=> Vav = Vmax [Sin φ]
2π
-π/2
 Vav = Vmax [Sin(π/2)] - [-Sin(π/2)]
2π
 Vav = Vmax [1+1]
2π
Therefore
Vav = Vmax
π
D.C.TIBUJI
33
M.CINDY
Deriving the expression of the effective voltage is as follows;
π/2
V²eff = 1 ⌠ (Vmax Cosφ) ²dφ
T⌡
-π/2
π/2
=> V²eff = 1 ⌠ V²max Cos²φ dφ
2π ⌡
-π/2
π/2
=> V²eff = V²max ⌠ Cos²φ dφ
2π ⌡
-π/2
But Cos²φ need to be transform in to compound angle Cos 2φ
Cos(a+b) = Cos a Cos b – Sin a Sin b
Cos 2a = Cos a Cos a – Sin a Sin a
Cos 2a = Cos² a - Sin² a
Cos² a + Sin² a = 1
But Sin² a = 1 - Cos² a
Now Cos 2a = Cos² a - Sin² a
Substituting Sin² a
 Cos 2a = Cos² a – (1 - Cos² a)
 Cos 2a = 2 Cos² a – 1
 Cos² a = Cos 2a + 1
2
Therefore
;
Cos²φ
34 D.C.TIBUJI
=
Cos 2φ + 1
2
M.CINDY
π/2
V²eff. = V²max ⌠ Cos² φ dφ
2π ⌡
-π/2
π/2
=> V²eff = V²max ⌠ Cos 2φ + 1 dφ
2π ⌡
2
-π/2
π/2
=> V²eff = V²max
4π
[½ Sin 2φ + φ]
-π/2
=> V²eff = V²max [½ Sin 2(π/2) + π/2] -[½ Sin 2
4π
(-π/2) + (-π/2)]
=> V²eff = Vmax [0 + π/2] - [0 – π/2]
4π
 V²eff = V²max [π/2 + π/2 ]
4π
 V²eff = V²max [ π]
4π
 √V²eff =√ V²max = Veff = Vmax
√4
2
Therefore;
Veff = Vmax
2
This is just to prove to you that either no matter the method, you used the sine or the
cosine, the result is the same.
D.C.TIBUJI
35
M.CINDY
HALF-WAVE RECTIFICATION (R-L) LOAD
(NON-CONTROLLED)
vd
Resistive
load with
Inductance
Fig 3.6
Almost all dc loads contain some inductance. In the waveforms seen below in figure 3.7
is the waveform for inductive load. The circuit in figure 3.6 is an equivalent circuit of an
inductive load.
Current flow will commence directly as the supply voltage goes positive but the presence
of the inductance will delay the current change
, the current still flowing at the end of the half-cycle
The diode remains on and the load sees the negative supply voltage until the current
drops to zero.
WAVE FORM
vs
π
2π
t
vl
α
t
vd
t
36 D.C.TIBUJI
M.CINDY
Fig 3.7
In order for the load not to see the negative supply voltage, current need to stop flowing
at the end of the half cycle.
a diode is connected in parallel with the load to stop this problem of the load,
experiencing the negative supply voltage.
When a diode is connected in parallel with the load as shown in the figure 3.8 below, that
diode is called a commutating diode. More on commutating diode will be discuss in
further chapters.
Commutating diode
Figure 3.8
From figure 3.7, the average voltage of the load can be derived
α
Vav = 1 ⌠ Vmax Sin φ dφ
T⌡
0
where T = 2π
α
=> Vav = Vmax ⌠ Sin φ dφ
2π
⌡
0
α
D.C.TIBUJI
37
M.CINDY
=> Vav = Vmax [ - Cos φ ]
2π
0
=> Vav = Vmax [ - Cos α ] - [- Cos 0 ]
2π
 Vav = Vmax [ - Cos α + 1 ]
2π
Therefore the average voltage Vav is;
Vav = Vmax [1 – Cos α]
2π
The effective voltage of the load
α
V²eff = 1 ⌠ (Vmax Sin φ) ² dφ
T⌡
0
where T = 2π
α
=> V²eff = 1 ⌠ V²max Sin² φ dφ
T ⌡
0
α
=> V²eff = V²max ⌠ Sin²φ dφ
2π ⌡
0
But Sin²φ need to be transform in to compound angle
Cos(a+b) = Cos a Cos b – Sin a Sin b
Cos 2a = Cos a Cos a – Sin a Sin a
Cos 2a = Cos² a - Sin² a
38 D.C.TIBUJI
M.CINDY
Cos² a + Sin² a = 1
But Cos² a = 1 - Sin² a
Now Cos 2a = Cos² a - Sin² a
Substituting Cos² a
 Cos 2a =1 - Sin² a - Sin² a

 Cos 2a =1 – 2 Sin² a

 Sin² a = Cos 2a - 1
-2
Sin² a = 1 – Cos 2a
2
 Sin² φ = 1 – Cos 2φ
2
So
α
V²eff = V²max ⌠ ½ (1 – Cos 2φ) dφ
2π
⌡
0
α
=> V²eff = V²max [φ -½ Sin 2φ]
4π
0
V²max [α - ½ Sin 2α] - [0 - ½ Sin 2(0)]
4π
 V²eff = V²max [α - ½ Sin 2α]
4π
 √V²eff = √ V²max [ α/π – ½π Sin 2α]
√4
 Veff = Vmax √ α/π - ½π Sin 2α
2
D.C.TIBUJI
39
M.CINDY
Therefore the effective voltage Veff is
 Veff = Vmax √ α/π - ½π Sin 2α
2
 Veff = Vmax √ α/π - ½π Sin 2α
2
For the waveform of figure 3.8. the effective value of the current and voltage and also the
average values of voltage and current are the same as that of a resistive load,
Example.
vd
127v
Resistive
load with
Inductance
Fig 3.9 circuit diagram
A single-way uncontrolled rectifier with ( R-L ) load is supply from an ac network of 127
V and a resistance 10Ω with an angular velocity ω = 100π rad/s.the discharge by the
inductance ends at the instance t = 15 ms.
WORK REQUIRED
-draw the waveform across the load.
- derive the expression of the average and the effective voltage and calculate.
SOLUTION
40 D.C.TIBUJI
M.CINDY
Data
V = 127V
R = 10
ω = 100 rad/s
t = 15ms
vs
π
2π
t
vl
α
t
vd
t
Fig 3.10 waveform
Deriving the expression of The average value of the voltage (Vav).
α
Vav = 1 ⌠ Vmax Sin φ dφ
where T = 2π
T⌡
0
α
=> Vav = Vmax ⌠ Sin φ dφ
2π
⌡
0
α
=> Vav = Vmax [ - Cos φ ]
2π
0
D.C.TIBUJI
41
M.CINDY
=> Vav = Vmax [ - Cos α ] - [- Cos 0 ]
2π
 Vav = Vmax [ - Cos α + 1 ]
2π
Therefore the average voltage Vav is;
Vav = Vmax [1 – Cos α]
2π
=> Vav =Vmax [1 – Cos α] but α = ω t
ω= 100π rad/s and t = 15ms
Then, α = 100π x 15 x 10 ¯ ³
 α = 4.7 rad
but we have to change radians to degree Celsius in other to match up with the
formula for average voltage
 α = 360 x 4.7
2 x 3.14
= 270˚
α = 270˚
Vmax =127 √ 2
 Vav = 127 √ 2 [1 – Cos 270 ] = 28.6 V
2π
Vav = 28.6 V
Deriving the expression for the effective value of the voltage is;
42 D.C.TIBUJI
M.CINDY
α
V²eff = 1 ⌠ ( Vmax Sin φ)² dφ
T⌡
0
where T = 2π
α
=> V²eff = 1 ⌠ V²max Sin² φ dφ
T ⌡
0
α
=> V²eff = V²max ⌠ Sin²φ dφ
2π ⌡
0
But Sin²φ need to be transform in to compound angle
Cos(a+b) = Cos a Cos b – Sin a Sin b
Cos 2a = Cos a Cos a – Sin a Sin a
Cos 2a = Cos² a - Sin² a
Cos² a + Sin² a = 1
But Cos² a = 1 - Sin² a
Now Cos 2a = Cos² a - Sin² a
Substituting Cos² a
 Cos 2a =1 - Sin² a - Sin² a
 Cos 2a =1 – 2 Sin² a
 Sin² a = Cos 2a - 1
-2
Sin² a = 1 – Cos 2a
2
 Sin² φ = 1 – Cos 2φ
2
D.C.TIBUJI
43
M.CINDY
So
α
V²eff = V²max ⌠ ½ (1 – Cos 2φ) dφ
2π
⌡
0
α
=> V²eff = V²max [φ -½ Sin 2φ]
4π
0
V²max [α - ½ Sin 2α] - [0 - ½ Sin 2(0)]
4π
 V²eff = V²max [α - ½ Sin 2α]
4π
 √V²eff = √ V²max [ α/π – ½π Sin 2α]
√4
 Veff = Vmax √ α/π - ½π Sin 2α
2
Therefore the effective voltage Veff is
 Veff = Vmax √ α/π - ½π Sin 2α
2
 Veff = Vmax √ α/π - ½π Sin 2α
2
 Veff = 127 √ 2 √[270/π - ½π Sin 2(270)
2
 127 √ 2 √270/180
2
 127 √2 x 1.22 = 110 V
2
44 D.C.TIBUJI
Veff = 110 V
=>
M.CINDY
SINGLE-WAY UNCONTROLLED RECTIFICATION (R-E LOAD)
R
E
Fig 3.11 circuit diagram
1
The single-way or half-wave rectification using R-L load is mostly applicable in the
charging of battery, the function of the circuit can be analysis as follows;
During the positive half cycle, the diode is reverse bias by the battery E and forward bias
by the supply voltage at the secondary of the transformer VS.
Furthermore, during the negative half cycle the diode is reverse biased, then it cannot
conduct.
When the voltage of the battery is greater than the secondary voltage, the diode is
blocked but when the battery voltage E is less than the secondary voltage VS, the diode
will conduct.
WAVEFORM
Vmax
Vs
Voltage across the load
= Vmax - Eb
Eb
α
θ1
π
2π
θ2
Vd
D.C.TIBUJI
45
M.CINDY
Fig 3.12
WAVEFORM ANALYSIS
Vmax– E
is equal to the voltage across the load
Taking a loop in the circuit above will give;
VS – VD – VR – E = O
When the diode is conducting, it is assume to be a short circuit and when there is a short
circuit, the voltage is assume to be zero, so the diode voltage VD = 0
When the diode is block, considering it as a switch, we say the switch is open and when
the switch is open, no conduction, meaning that the diode is block.
That is VR = I R = O => VD = VS – E
θ1 – is the angle at which conduction begins and .its correspond to the moment when VS
=E
Vmax Sinθ1 = E
Sinθ1 = E___
Vmax
θ1 = Sin−¹ E__
Vmax
θ2 – is the angle at which conduction ends
θ2 = π – θ1
46 D.C.TIBUJI
M.CINDY
The angle of conduction α
α = θ2 – θ1
But θ2 = π – θ1
=> α = π – θ1 – θ1
α = π - 2θ1
Therefore the angle of conduction (α)
α = π - 2θ1
Vmax = E + Imax R
 Imax = Vmax – E
R
AVERAGE VALUE OF VOLTAGE (Vav)
θ2
Vav = 1 ⌠ Vmax Sinθ – E dθ
T ⌡
θ1
θ2
θ2
=> 1 ⌠ Vmax Sinθ dθ – 1 ⌠ E dθ
2π ⌡
2π ⌡
θ1
θ1
θ2
θ2
=> Vmax [- Cos θ ] - E [ θ ]
2π
θ1 2π
θ1
Vmax [- Cos θ2 + Cos θ1] - E [ θ2 - θ1 ]
2π
2π
But cos θ2 need to be transform to Cos θ1 so that they can cancel out but
D.C.TIBUJI
θ2 = π - θ1
47
M.CINDY
Cos θ2 ≡≡ Cos (π - θ1)
= Cos π Cos θ1 – Sin π Sin θ1 = - Cos θ1
Vmax [Cos θ1 + Cos θ1] – E [ θ2 - θ1 ]
2π
2π
Vmax [2 Cos θ1] – E α
2π
2π
Vmax Cos θ1 – E α
2π
2π
Therefore the average value of voltage Vav is
Vav = Vmax Cos θ1 – E α
2π
2π
NB: if α is in degree you replace π by 180˚ or you convert α in to radians.
The peak inverse voltage is
PIV = Vmax + E
The charging time t
t= C
Iav
Where – C = capacity of the battery in
(A.H)
The effective value of the voltage
θ2
V²eff = 1 ⌠ ( Vmax Sin θ – E)² dθ
48 D.C.TIBUJI
M.CINDY
T
⌡
θ1
θ2
1 ⌠ V²max Sin²θ - E² dθ
T⌡
θ1
θ2
V²max ⌠ Sin²θ - E² dθ
2π ⌡
θ1
θ2
V²max - E²⌠ Sin² dθ
2π
⌡
θ1
θ2
V²max - E² ⌠ ½ ( 1 – Cos 2θ ) dθ
2π
⌡
θ1
θ2
V²max - E² [θ - ½ Sin θ]
4π
θ1
V²max - E² [ θ2 -½ Sin2(θ2)]-[θ1-½ Sin2(θ1)]
4π
V²max - E² [ θ2 -½ Sin2θ2-θ1+ ½ Sin2θ1]
4π
V²max - E² [ θ2 -θ1+ Sin2(θ2) + Sin2(θ1)]
4π
V²max - E² [ α + Sin2(θ2) + Sin2(θ1)]
4π
But Sin 2θ2 need to be transform to Sin 2θ1.so that they can cancel out .θ2 = π - θ1
So Sin 2θ2 = Sin2 ( π - θ1)
D.C.TIBUJI
49
M.CINDY
Sin (2π - 2θ1) = Sin2πCos2θ1-Cos2πSin2θ1
Sin (2π - 2θ1) = - Sin 2θ1
V²max - E² [ α - Sin2(θ1) + Sin2(θ1)]
4π
V²max - E² [ α ]
4π
V²max - E² [α ]
4π
Veff = √V²max - E² √ [α ]
√4π
Veff = Vmax - E √ [α]
√4π
let T = 2π
Therefore the effective value of voltage is
Veff = Vmax - E √ [α ]
√2T
Example.
Mariah Cindy N.C wants to charge the battery of her phone, needed a universal battery
charge with the following characteristics, voltage of 46 V, and has a capacity of 16.5 A.H
using a single-way rectification circuit which has a secondary voltage of 64 V, 50 HZ
from the transformer. The variable resistor is chosen in the manner that, the maximum
current is 15A.
WORK REQUIRED.
I – Draw the circuit diagram
ii- Draw the waveform across the load
iii –compute:
The angle and duration of conduction in one period
- The value of the variable resistor R
- The value of the average current after deriving the expression of the average
voltage.
50 D.C.TIBUJI
M.CINDY
IV – what is the peak inverse voltage of the diode given that, the characteristics of the
diodes, having a security factor of 70 percent for current and 100 percent for the voltage?
V- Compute the charging time.
SOLUTION
R
E
Fig 3.13 circuit diagram
Fig 3.14 WAVEFORM
Vmax
Vs
Voltage across the load
= Vmax - Eb
Eb
α
θ1
π
2π
θ2
Vd
III –the angle and duration of conduction
α = π - 2θ1
but θ1 =?
Vmax Sinθ1 = E
D.C.TIBUJI
51
M.CINDY
=> θ1 = Sin−¹ E__ = 46
Vmax 64√2
θ1 = π/6 = 30˚
α = π – 2(π/6) = 2π
3
= 120˚
The conduction angle α = 120˚
Duration of conduction t = α
ω
t = 0.0066 sec. or 6.67 ms
t = 2π___
3 x 2π x 50
The value of R
Taking a loop, we are going to have:
Vmax – E – Imax R =0
R = Vmax – E = 64 √2 – 45
Imax
15
R = 3.034 Ω
Deriving of average value of voltage
θ2
Vav = 1 ⌠ Vmax Sinθ – E dθ
T ⌡
θ1
θ2
θ2
=> 1 ⌠ Vmax Sinθ dθ – 1 ⌠ E dθ
2π ⌡
2π ⌡
θ1
θ1
52 D.C.TIBUJI
M.CINDY
θ2
θ2
=> Vmax [- Cos θ ] - E [ θ ]
2π
θ1 2π
θ1
Vmax [- Cos θ2 + Cos θ1] - E [ θ2 - θ1 ]
2π
2π
But cos θ2 need to be transform to Cos θ1 so that they can cancel out but
θ2 = π - θ1
Cos θ2 ≡≡ Cos (π - θ1)
= Cos π Cos θ1 – Sin π Sin θ1 = - Cos θ1
Vmax [Cos θ1 + Cos θ1] – E [ θ2 - θ1 ]
2π
2π
Vmax [2 Cos θ1] – E α
2π
2π
Vmax Cos θ1 – E α
2π
2π
Therefore the average value of voltage Vav is
Vav = Vmax Cos θ1 – E α
2π
2π
But Iav = Vav
R
Vav = Vmax Cos θ1 – E α
2π
2π
= 64 √2 Cos 30 – 46 x 2 x3.14
3.14
2 x3.14 x 3
D.C.TIBUJI
53
M.CINDY
Vav = 9.66 V
So, Iav = Vav
R
9.66
3.034
= 3.1839 A
Therefore, the average value of voltage
Iav = 3.1839 A
The peak inverse voltage PIV
PIV = Vmax + E = 64 √2 + 46
PIV = 136.5 V
Characteristic of diode
Iav = Iav + Iav x 70%
3.1839 +3.1839 x 70 / 100
Iav = 4.45 A
Imax = Imax + Imax x 70%
15 + 15 x 0.7 = 21 A
Imax = 21 A
PIV = PIV +PIV x 100%
54 D.C.TIBUJI
M.CINDY
136.5 + 136.5 * 1 = 273 V
PIV = 273 V
The charging time (t)
C = Iav t => t = C
Iav
t = 16.45
3.1839
t =5.17 H
FULL WAVE RECTIFICATION
BRIDGE RECTIFIER
Vs
Load
Vs is the voltage at the secondary of the transformer
Fig 3.14 circuit diagram
The single-phase bridge connection can be arranged in three diagrammatic layout which
the figure 3.14 above represent one amongst the three.
D.C.TIBUJI
55
M.CINDY
Another diagrammatic layout can be seen below in figure 3.15 which is the simplest and
whilst almost self-explanatory and widely used in electronic circuits.
+
LOAD
Vs
Vs is the voltage at the secondary of the transformer
Figure 3.15-bridge rectifier circuit diagram.
The same circuit is draw to a different diagrammatic layout again, as can be seen in the
figure 3.16 below. its shows clearly the concept of two half-wave circuits in series
making a full-wave connection, two diodes with common cathode feeding into the load,
two diodes with common anodes returning the load current to the other supply line
,however, this layout is rather cumbersome and is used for power applications.
D1
D3
Vs
load
D2
D4
Vs is the voltage at the secondary of the transformer
Figure 3.16
A single-way rectifier is not giving out the out put voltage that is to have in certain
applications because it is not making used of the other half of the supply wave. This
56 D.C.TIBUJI
M.CINDY
problem can be solved by using the full wave rectifications, were both the positive and
the negative half cycles are used.
WAVEFORMS
VS
VL
VD1
ID1
Figure 3.17
Deriving the expression for the average value of voltage across the load..
π
Vav = 1 ⌠ Vmax Sin θ dθ
but T = π
D.C.TIBUJI
57
M.CINDY
T ⌡
0
π
π
Vmax ⌠ Sin θ dθ = Vmax [- Cos θ ]
π ⌡
0
0
Vmax [-Cos π] - [-Cos 0 ]
π
Vmax [1+1] = 2Vmax
π
π
Therefore the average value of the voltage is
Vav = 2Vmax
π
The effective value of the voltage across the load
π
V²eff = 1⌠ V²max Sin² θ dθ
π⌡
0
π
V²max ⌠ Sin² θ dθ
π ⌡
0
But Sin² θ = ½ ( 1 – Cos 2θ )
π
V²max ⌠ ½ (1- Cos 2θ ) dθ
π ⌡
0
58 D.C.TIBUJI
M.CINDY
π
V²max [θ-½ Sin 2θ]
2π
0
V²max [π-½ Sin 2π]-[0-½ Sin 2(0)]
V²eff = V²max
2
=> Veff = Vmax
√2
Therefore the effective voltage is
Veff = Vmax
√2
Example.
A two single way rectifier have been combine to make used of both cycles of the supply
and they comprises of four ideal diodes,suppliy from a sinusoidal voltage source of
210√2 Cos θ,50HZ.
The rectifier feed a 1.5Ω purely resistive load. the voltage drop across each forward
biased is express as VD = 0.7+0.002 Id .the room temperature being 25˚C.
WORK REQUIRED
Calculate for each diode
The peak inverse voltage
The maximum forward current
2) Knowing that the thermal conductance
λ =5w/k.calculate the junction temperature
Solution.
Circuit diagram
D.C.TIBUJI
59
M.CINDY
D1
D3
Vs
load
D2
D4
1) the peak inverse voltage across a diode, diode when reverse biased is subjected to
a maximum voltage of ;
210√2 = 297 V
The maximum forward current
Imax-f = Vmax = 210√2
R
1.5
Imax-f = 198 A
The average current
Each diode conducting during a half-cycle be it positive or negative, it is behaving as a
half wave circuit. Therefore
Iav = Imax-f = 198 = 63.1 A
π
3.14
 Iav = 63.1 A
2) each diode dissipate an average power of:
P =VI or I²R or V²/R
But V =0.7 + 0.002 I²
P = 0.7 Iav + 0.002 I²
60 D.C.TIBUJI
M.CINDY
But Iav = Imax =198 = 63.1 A
π
3.14
I² = Imax-f = (198)²
2
( 2)²
I² =9801 A
Then, Pd = 0.7 x 63.1 + 0.002 x 9801
The power dissipated Pd is 64 w
Pd = λ (Tj – Ta)
Where; PD –is the power dissipated
λ -thermal conductance
Tj absolute temperature of junction and
Ta absolute temperature of air
64 = 5 (Tj – 25) => 64 = 5 (Tj – 25)
5
5
Tj = 37.8˚C
FULL-WAVE RECTIFICATION WITH (E-R) AS LOAD
il
D1
D3
R
VS
E
D2
D4
Figure 3.18
D.C.TIBUJI
61
M.CINDY
WAVEFORMS
vs
voltage across
The load
θo π/2
-π/2 -θo
θo+π/2
il
From the circuit diagram of figure 3.18 above, the functioning of the full-wave rectifier
can be analyze as follows
The diode will conduct only when
Vmax Cos θ1 =E
=> θ1 = Cos−¹ E__
Vmax
The diode start conducting when the angle its sustain is -θ1 and
When the diodes attain an angle of θ1, the diode stop conducting..
The conduction angle of the diode is 2θ1
Then;
V = E + iR => I =V – E
R
I = 1 (Vmax Cos θ1 – E)
R
The current or the diode is open within the interval:
(-π/2, -θ1) and (θ1, π/2)
62 D.C.TIBUJI
M.CINDY
The current flow within this interval
(-θ1, θ1) => i =1 (Vmax Cosθ – E)
R
Since Vmax Cos θ1 = E we have
I= Vmax (Cosθ - Cosθ1)
R
The average value of current across the E-R load
θ1
Iav = 1 ⌠ Imax Sin dθ
T⌡
-θ1
θ1
Iav = 1⌠ Vmax (Cosθ - Cosθ1) dθ
π⌡
R
-θ1
θ1
Iav = Vmax [Sinθ-θ Cosθ1]
πR
-θ1
Vmax [Sinθ1-θ1Cosθ]-[Sin(-θ1)-(-θ1)Cosθ]
πR
Vmax [2Sinθ1-2θ1Cosθ]
πR
Therefore, the average value of current is
Iav = 2Vmax [Sinθ1-θ1Cosθ]
πR
Deriving the expression for the effective value of current
θ1
D.C.TIBUJI
63
M.CINDY
I²eff = 1 ⌠(Vmax Cosθ-Vmax Cosθ1)² dθ
T⌡
R
-θ1
But Vmax Cosθ1 = E
θ1
 1⌠ (Vmax Cosθ – E)² dθ
T⌡
R
-θ1
= V²eff
θ1
V²max-E²⌠ Cos²θ dθ
πR
⌡
-θ1
Cos²θ = ½ (1 + Cos 2θ)
θ1
V²max-E²⌠ ½ (1 + Cos 2θ) dθ = V²eff
πR
⌡
-θ1
θ1
=> V²max-E² [ θ + ½ Sin 2θ ] = V²eff
2πR
-θ1
V²max-E² [ θ1+ ½ Sin 2θ1]-[-θ1+½ Sin 2(-θ1)]
2πR
=> V²max-E² [ Sin 2θ1 ] = V²eff
2πR
Veff = Vmax – E √Sin 2θ1
√ 2πR
Let T = 2π
Then,
Veff = Vmax – E √Sin 2θ1
√TR
64 D.C.TIBUJI
M.CINDY
SINE WAVEFORM FOR E-R LOAD
WAVEFORM ANALYSIS
Vmax– E
is equal to the voltage across the load
θ1 – is the angle at which conduction begins and .its correspond to the moment when VS
=E
Vmax Sinθ1 = E
Sinθ1 = E
Vmax
θ1 = Sin−¹ E__
Vmax
θ2 – is the angle at which conduction ends
θ2 = π – θ1
The angle of conduction α
α = θ2 – θ1
But θ2 = π – θ1
=> α = π – θ1 – θ1
α = π - 2θ1
Therefore the angle of conduction (α)
α = π - 2θ1
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65
M.CINDY
Vmax = E + Imax R
 Imax = Vmax – E
R
AVERAGE VALUE OF VOLTAGE (Vav)
Figure 3.19 waveform
θ2
Vav = 1 ⌠ Vmax Sinθ – E dθ
T ⌡
θ1
but T = π
θ2
θ2
=> 1 ⌠ Vmax Sinθ dθ – 1 ⌠ E dθ
π ⌡
π⌡
θ1
θ1
θ2
θ2
=> Vmax [- Cos θ ] - E [ θ ]
π
θ1 2π
θ1
Vmax [- Cos θ2 + Cos θ1] - E [ θ2 - θ1 ]
π
π
But cos θ2 need to be transform to Cos θ1 so that they can cancel out but
66 D.C.TIBUJI
θ2 = π - θ1
M.CINDY
Cos θ2 ≡≡ Cos (π - θ1)
= Cos π Cos θ1 – Sin π Sin θ1 = - Cos θ1
Vmax [Cos θ1 + Cos θ1] – E [ θ2 - θ1 ]
π
π
Vmax [2 Cos θ1] – E α
π
π
2Vmax Cos θ1 – E α
π
π
Therefore the average value of voltage Vav is
Vav = 2Vmax Cos θ1 – E α
π
π
NB: if α is in degree you replace π by 180˚ or you convert α in to radians.
The peak inverse voltage is
PIV = Vmax + E
The charging time t
t= C
Iav
Where – C = capacity of the battery in
(A.H)
The effective value of the voltage
θ2
V²eff = 1 ⌠ (Vmax Sin θ – E)² dθ
D.C.TIBUJI
67
M.CINDY
T
⌡
θ1
θ2
1 ⌠ V²max Sin²θ - E² dθ
T⌡
θ1
θ2
V²max ⌠ Sin²θ - E² dθ
π ⌡
θ1
θ2
V²max - E²⌠ Sin² dθ
π
⌡
θ1
θ2
V²max - E² ⌠ ½ ( 1 – Cos 2θ ) dθ
π
⌡
θ1
θ2
V²max - E² [θ - ½ Sin θ]
2π
θ1
V²max - E² [ θ2 -½ Sin2(θ2)]-[θ1-½ Sin2(θ1)]
2π
V²max - E² [ θ2 -½ Sin2θ2-θ1+ ½ Sin2θ1]
2π
V²max - E² [ θ2 -θ1+ Sin2(θ2) + Sin2(θ1)]
2π
V²max - E² [ α + Sin2(θ2) + Sin2(θ1)]
2π
But Sin 2θ2 need to be transform to Sin 2θ1.so that they can cancel out .θ2 = π - θ1
So Sin 2θ2 = Sin2 ( π - θ1)
68 D.C.TIBUJI
M.CINDY
Sin (2π - 2θ1) = Sin2πCos2θ1-Cos2πSin2θ1
Sin (2π - 2θ1) = - Sin 2θ1
V²max - E² [ α - Sin2(θ1) + Sin2(θ1)]
2π
V²max - E² [ α ]
2π
V²max - E² [α ]
2π
Veff = √V²max - E² √ [α ]
√2π
Veff = Vmax - E √ [α]
√2π
let T = 2π
Therefore the effective value of voltage is
Veff = Vmax - E √ [α ]
√T
NUMERICAL APPLICATION OF BATTERY CHARGER.
1) Mariah Cindy N.C. battery is to be charge, a battery charger is needed .a battery
charger supply by an iron core transformer with the primary rated 110 V.50HZ. a
battery rated 40V neglecting its internal resistance. the battery charger is compose
of 4 perfect diodes forming a bridge with a protective R.the transformer is also
perfect.
WORK TO BE DONE
a) The interval of conduction of current is equal to 1/3 of the period of the supply.
You are asked to calculate the transformation ratio of the transformer, taking into
consideration that the secondary voltage is of the form
D.C.TIBUJI
69
M.CINDY
V = Vmax Cosθ
b) Calculate the protective resistance R in other to stabilize a peak current of 20A
c) Design the waveform and give the expression of the charging current in function of
θ, I = (θ)
d) Compute the average current of the load.
e) Compute the interval of charging putting in mind that the capacity of the battery is
C = 500AH.
f) Compute the effective current of the load
g) Come out with the efficiency of the charger.
h) Compute the peak inverse voltage of the diode having a safety factor of 5%
j) Produce the waveform of the current available at the secondary of the transformer.
Solution
Data
T = 2π , V = 110V, 50HZ, α= 1/3 x T
Emf = 40V, Imax =20A
a) the transformation ratio (a)
As stated, the interval of conduction is 1/3 the period of the supply
α= 1/3 x 2π => α= 2π/3 but α = 2θ1
2π/3 = 2θ1 => 2π = 6θ1
θ1 = 2π / 6 = π /3
Vmax Cosθ1 = E => Vmax = E / Cosθ1
Vmax = 40 / Cos 60 = 40 / 0.5
Vmax = 80V
70 D.C.TIBUJI
M.CINDY
The transformation ratio a
= Vmax / √2 = 80
110
110 √2
a= VS
VP
a = 0.5
b) the protective resistance R
Vmax = E + Imax R
R = Vmax – E
Imax
R = 80 – 40 = 2Ω =>
20
R = 2Ω
c) the charging current I =f(θ)
The charging current begins at -π /3 and ends at π /3 => θ Œ [-π/3, π/3]
V =E + RI BUT V= Vmax Cosθ
I = V – E = Vmax Cos θ – E
R
R

 I = Vmax (Cos θ – E / Vmax)
R
I = 80 (Cosθ – 40/80) = 40(Cosθ – 0.5)
2
I = 40(Cosθ - 0.5) A
WAVEFORM
D.C.TIBUJI
71
M.CINDY
vs
voltage across
The load
θo π/2
-π/2 -θo
θo+π/2
il
d) calculation of the average value of current
θ1
Iav = 1⌠ I dθ
T⌡
-θ1
π/3
= 2⌠ 40(Cosθ – 0.5) dθ
π⌡
0
π/3
80⌠ Cosθ – 0.5 dθ = Iav
π⌡
0
π/3
Iav = 80 [Sinθ – 0.5θ]
π
0
Iav = 80 [√3/2 –π/6] = 8.7 A
π
Iav = 8.7 A
e) The interval of charging
C = Iav t => t = C / Iav = 500/ 8.7
t = 57.47 H = 57 H 28.2 min
72 D.C.TIBUJI
M.CINDY
f) the effective value of the current
θ1
I²eff = 1⌠ I² dθ
T⌡
-θ1
θ1
= 2⌠ 40²(Cos θ – 0.5)²dθ
π⌡
0
π/3
I²eff = 2⌠ 40²(Cos²θ – Cosθ + ¼)dθ
π⌡
0
π/3
2 (40) ²⌠ 1 + Cos2θ – Cosθ + 1 dθ
π ⌡
2
4
0
π/3
I²eff = 2(40)² [½ (θ + ½ Sin2θ – Sinθ + θ/4]
π
0
Ieff = 25.8 A
g) the efficiency η
η = Pa
Pt
=
 η=
____40 x 8.7____
40 x 8.7 + (25.8)² x 2
E x Iav___
E x Iav + I²R
η = 21%
The maximum reverse voltage across each diode is
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M.CINDY
Vmax = 80 V
When adopting a safety factor of 5% the peak inverse voltage of each diode will be
PIV = PIV + PIV x Sf /100%
PIV = 80 + 80 x 50 / 100 = 120 V
PIV = 120 V
THREE PHASE NON-CONTROLLED RECTIFICATION.
Diodes are use in polyphase systems to give an out put voltage, which has theoretically a
unique directional form. With the three phase system, the time intervals between the
repetitions in the dc load waveforms are not longer but are shorter for a single phase
layout
Practically speaking,, three phase systems or supply are used or are supplying big or
heavy loads with a high value of inductance, as a result, the ripple content of the load
current will be less and it is reasonable to assume the current to be continues and level,
that is. it has negligible ripple.
WAVEFORM OF A THREE-PHASE NETWORK
Vs
Figure 3.20
THREE PHASE SINGLE-WAY RECTIFICATION WITH RESISTIVE LOAD
74 D.C.TIBUJI
M.CINDY
Circuit diagram
VD1
i1
L1
L2
i2
il
i3
L3
Load
VL
N
Figure 3.21
The connection of the polyphase single-way circuit can be seen above in figure 3.21;
each phase of the supply is being connected to the load through the diode.
The load current is return to the supply neutral, as is the case of all half-wave
connections.
The circuit functions in the following manner:
-just one diode is conducting at any given time.
-the diode which is connected to the phase having the highest instantaneous value
-a lead voltage is experience, which is represented on the waveform at the top of the
successive phase voltages.
-V1 is the most positive phase, diode D1 conducts and directly V2 becomes more
positive than V1, the load current is transfer from D1 to diode D2 and the process
continuous.
WAVEFORM
D.C.TIBUJI
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M.CINDY
Load
voltage
Vs
V1
V2
V3
VL
Vav
π/6
5π/6
3π/2
iL
i3
VD1
√3 Vmax
Figure 3.22 representation of 3-phase half-wave, waveforms.
THE AVERAGE VALUE OF THE VOLTAGE.
The period T can be gotten by considering the output waveform.
From the waveform T = 5π – π
6 6.
T = 2π
3
5π/6
Vav = 1⌠ Vmax Sinθ dθ but T = 2π/3
T⌡
π/6
76 D.C.TIBUJI
M.CINDY
5π/6
Vav = 3Vmax⌠ Sinθ dθ
2π
⌡
π/6
5π/6
Vav = 3Vmax [ Cosθ]
2π
π/6
Vav = 3Vmax [ -Cos 5π/6 + Cos π/6 ]
2π
Vav = 3Vmax [ √3 /2 + √3/2 ]
2π
Vav = 3Vmax [2√3/2]
2π
Vav = 3√3 Vmax
2π
Expression of the effective value of voltage
5π/6
V²eff = 1⌠ (Vmax Sinθ )²dθ but T = 2π/3
T⌡
π/6
5π/6
V²eff= 3⌠ V²max Sin²θ dθ
2 π⌡
π/6
D.C.TIBUJI
77
M.CINDY
5π/6
V²eff = 3V²max⌠ Sin²θ dθ
2π
⌡
π/6
But Sin²θ = ½ (1 – Cos2θ)
5π/6
V²eff = 3V²max⌠ ½ (1 – Cos2θ) dθ
2π
⌡
π/6
5π/6
V²eff = 3V²max [θ - ½ Sin2θ]
4π
π/6
= 3V²max [5π/6 - ½ Sin2(5π/6)]-[π/6-½Sin2(π/6)]
4π
= 3V²max [5π/6 - ½ Sin 5π/3-π/6+ ½ Sin π/3]
4π
= 3V²max [2π/3 - ½ Sin 5π/3+ ½ Sin π/3]
4π
Let λ= π/3
= V²max [2λ - ½ Sin 5λ + ½ Sin λ]
2 x 2λ
= V²max [2λ - ½ (Sin 5λ - ½ Sin λ)]
2 x 2λ
But Sin 5λ – Sin λ = 2 Sin2λCos3λ
= V²max [2λ - ½ (2 Sin2λCos3λ)]
2 x 2λ
But Cos 3λ = Cos 3π/3 = -1
78 D.C.TIBUJI
M.CINDY
So,
V²max [2λ - ½ (2 Sin2λ x -1)]
2 x 2λ
= V²max [ 1 + Sin2λ/2λ]
2
Veff = Vmax √1 + Sin 2λ/2λ
√2
Veff = Vmax √1 + Sin 2λ/2λ
√2
Example
A polyphase single-way rectifier of secondary voltage 240V,60HZ per phase having a
load resistance R = 20Ω while diodes are consider ideal. Compute
1- The average voltage across the load.
2-the average current per diode.
3-the effective or root mean square value of voltage across the load.
4-the effective or root meant square value of current per diode and per phase.
5-plot the output waveform
Solution
Data
Vs =240V, 60HZ
R = 20Ω
Veff = Vmax => Vmax =Veff √2
√2
Vmax = 240√2 = 339.4V
1) Vav = 3√3 Vmax
2π
Vav = 3√3 x 339.4
2 x 3.14
Vav = 281V
D.C.TIBUJI
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M.CINDY
2) The average current per diode
Iav per diode = Iav/3 but Iav = Vav/R
Iav = 281/20 => Iav = 14.05A
Iav/D =14.05/3
Iav/D = 4.7 A
2) Veff = Vmax √1 + Sin2λ/2λ
√2
339.4 √1 + .0866/2.09
√2
Veff = 284V
4) Ieff = Veff /R = 284/20 = 14.2A
Ieff = 14.2A
Ieff/D = Ieff/√3 = 14.2/√3 = 8.2 A
Ieff/D = 8.2A
80 D.C.TIBUJI
M.CINDY
THREE PHASE FULL WAVE RECTIFICATION
For the full wave rectification, the load is fed through a three-phase half-wave connection
and the return current path being through an other half-wave connection to one of the
three lines, no neutral is needed from the waveform in the figure below.
Firstly, the load voltage is the addition of the two, three phase half-wave voltages relative
to the supply neutral appearing at the positive and negative sides of the load respectively
Circuit diagram
i1
i3
i5
VL
i4
i6
i2
Figure 3.22-circuit diagram three-phase full-wave diagram
The average value of the three phase full wave can be gotten by multiplying the average
value of half-wave by 2.
The average value for voltage for half-wave is
Vav = 3√3 Vmax
2π
The average value for voltage for full-wave is
D.C.TIBUJI
81
M.CINDY
Vav = 2 x 3√3 Vp max
2π
Vav = 3 VL max
π
We are going to see how the equation is being derived starting from the period to the
finish but we need to study the waveform first then we can proceed to the deriving proper
WAVEFORMS
Vs
V1
V2
V3
VP
θ
π/6
5π/6
3π/2
VL
iL
i5
i6
VD1
VLmax
Figure 3.23 waveform of three-phase full wave rectification.
Deriving the expression for the average voltage
The period T can be investigated by considering the output waveform.
From the waveform T = π – π
2 6.
82 D.C.TIBUJI
M.CINDY
T = 2π or π
6
3
Taking in to consideration that, the full wave is the addition of two half-wave connection,
we can multiply the half-wave load voltage by 2
5π/6
Vav = 1⌠2VmaxSinθ dθ but T = 2π/3 for half-wave
T⌡
π/6
Vav = 2
2π/3
5π/6
⌠ VmaxSinθ dθ
⌡
π/6
Vav = 3 x 2
2π
5π/6
⌠ VmaxSinθ dθ
⌡
π/6
Vav = 6Vmax
2π
5π/6
⌠ Sinθ dθ
⌡
π/6
5π/6
Vav = 6Vmax [- Cosθ]
2π
π/6
Vav = 6Vmax [ -Cos 5π/6 + Cos π/6 ]
2π
D.C.TIBUJI
83
M.CINDY
Vav = 6Vmax [ √3 /2 + √3/2 ]
2π
Vav = 6Vmax [2√3/2]
2π
Vav = 6Vmax [√3]
2π
Vav = 3Vpmax [√3] where VP is the phase voltage
π
Vav = 3√3Vpmax
π
Expression of the effective value of voltage
5π/6
V²eff = 1⌠ 2(Vmax Sinθ) ²dθ but T = 2π/3
T⌡
π/6
5π/6
V²eff= 6⌠ V²max Sin²θ dθ
2 π⌡
π/6
5π/6
V²eff = 6V²max⌠ Sin²θ dθ
2π ⌡
π/6
But Sin²θ = ½ (1 – Cos2θ)
84 D.C.TIBUJI
M.CINDY
5π/6
V²eff = 6V²max⌠ ½ (1 – Cos2θ) dθ
2π ⌡
π/6
5π/6
V²eff = 6V²max [θ - ½ Sin2θ]
4π
π/6
= 6V²max [5π/6 - ½ Sin2(5π/6)]-[π/6-½Sin2(π/6)]
4π
= 6V²max [5π/6 - ½ Sin 5π/3-π/6+ ½ Sin π/3]
4π
= 6V²max [2π/3 - ½ Sin 5π/3+ ½ Sin π/3]
4π
Let λ = π/3 but our period T = 2π/3
= 2V²max [2λ - ½ Sin 5λ + ½ Sin λ]
2 x 2λ
= 2V²max [2λ - ½ (Sin 5λ - ½ Sin λ)]
2 x 2λ
But Sin 5λ – Sin λ = 2 Sin2λCos3λ
= 2V²max [2λ - ½ (2 Sin2λCos3λ)]
2 x 2λ
But Cos 3λ = Cos 3π/3 = -1
So,
2V²max [2λ - ½ (2 Sin2λ x -1)]
2 x 2λ
= 2V²max [ 1 + Sin2λ/2λ]
2
D.C.TIBUJI
85
M.CINDY
Veff = 2Vmax √1 + Sin 2λ/2λ
√2
Veff = 2Vmax √1 + Sin 2λ/2λ
√2
SINGLE PHASE CONTROLLED RECTIFICATION
We have been talking about non-controlled or uncontrolled rectification using diode, now
we want to see how a rectification can be control. It can be be control by the use of
thyristors and triac
HALF-WAVE CONTROLLED RECTIFICATION USING (R-L) LOAD
Almost all dc loads are having inductance. The inductance will cause a delay in the rise
and fall of current .consequently; the thyristor will not turn off until the current exceed
the discharge angle produce by the inductance (λ).
The firing angle (φ) is express with respect to the supply voltage from zero. Once the
thyristor is, fire the load current start rising in value. The energy from the supply absorbs
and stored in the inductor after the current start falling in value, at the point where the
supply is turning to the negative half-cycle, the inductor changes its polarity and restitutes
in the circuit through the supply.
Circuit diagram and waveform
iL
VTh
Load of resistance
With inductance
Vs
86 D.C.TIBUJI
Vl
M.CINDY
Vs
φ
ig
VL
λ
VTh
Figure 2.24 circuit diagram and waveforms of an RL load
Deriving the expression for the average value of voltage of the load.
λ
Vav = 1⌠ Vmax Sinθ dθ
T⌡
φ
λ
Vmax [-Cosθ]
2π
φ
Vav = Vmax [-Cosλ]-[-Cosφ]
2π
Vav = Vmax [Cosφ - Cosλ]
2π
D.C.TIBUJI
87
M.CINDY
Effective value of voltage across the load
λ
V²eff = 1 ⌠ (Vmax Sinθ) ²dθ
T⌡
φ
λ
V²eff = 1 ⌠ V²max Sin²dθ
T⌡
φ
λ
V²eff = V²max ⌠ Sin²θ dθ
2π ⌡
φ
Sin²θ = ½(1 - Cos 2θ)
λ
V²eff = V²max ⌠ ½(1 - Cos 2θ) dθ
2π ⌡
φ
λ
V²eff = V²max [θ - ½ Sin 2θ]
4π
φ
V²eff = V²max [λ - ½ Sin2λ]-[φ - ½ Sin2λ]
4π
V²eff = V²max [λ - ½ Sin2λ - φ + ½ Sin2λ]
4π
=> V²max [λ/π -φ/π + ½π Sin2φ + ½π Sin2λ]
4
Veff = Vmax √ [λ/π -φ/π + ½π Sin2φ - ½π Sin2λ]
2
88 D.C.TIBUJI
M.CINDY
Including a diode in the circuit connected in parallel with the load, this diode is call a
commutating diode or a free wheel diode and the aim or purpose of this diode in this
circuit is to eliminate the discharge angle produce by the inductor.
When the commutating diode is connected, the average and effective values of the circuit
changes
The average value of voltage across the load for the case of commutating diode
π
Vav = 1⌠ Vmax Sinθ dθ
T⌡
φ
π
Vmax [-Cosθ]
2π
φ
Vav = Vmax [-Cos π]-[-Cosφ]
2π
Vav = Vmax [ 1 + Cosφ]
2π
Effective value of voltage across the load
π
V²eff = 1 ⌠ (Vmax Sinθ) ²dθ
T⌡
φ
π
V²eff = 1 ⌠ V²max Sin²dθ
T⌡
φ
π
V²eff = V²max ⌠ Sin²θ dθ
2π ⌡
D.C.TIBUJI
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M.CINDY
φ
Sin²θ = ½(1 - Cos 2θ)
π
V²eff = V²max ⌠ ½(1 - Cos 2θ) dθ
2π ⌡
φ
π
V²eff = V²max [θ - ½ Sin 2θ]
4π
φ
V²eff = V²max [π - ½ Sin2π]-[φ - ½ Sin2φ]
4π
V²eff = V²max [π - ½ Sin2π - φ + ½ Sin2φ]
4π
=> V²max [π/π -φ/π + ½π Sin2φ + ½π Sin2π]
4
Veff = Vmax √ [1 - φ/π + ½π Sin2φ]
2
With the presence of the commutating or free wheel diode in the resistive-inductive load,
the circuit is transform in to a resistive load circuit.
FULL-WAVE CONTROL RECTIFICATION
The full-wave control rectification using a resistive-inductive (R-L) load circuit contains
four thyristors. Conduction cannot take place until the thyristors are fire.
It is only possible for current to flow when the two thyristors working together are fire
simultaneously
The circuit has been design such that current can only flow through the load when pair of
thyristors gate is fire during a half-cycle of the waveform.
90 D.C.TIBUJI
M.CINDY
The other pair of thyristors will be fire on the other half-cycle, for each pair of thyristors
fire; one is serving as the supply and the other as a return. The two pairs are
complimentary in operation.
CIRCUIT DIAGRAM AND WAVEFORMS
Load of resistance
With inductance
Vs
ig
1
ig
3
VL
iL
i1
i3
VT
D.C.TIBUJI
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M.CINDY
Figure 3.25 circuit diagram and waveforms
Average voltage across the load
π+φ
Vav = 1⌠ Vmax Sinθ dθ
T⌡
φ
π+φ
Vav = Vmax ⌠ Sinθ dθ
π ⌡
φ
π+φ
Vav = Vmax [- Cosθ]
π
φ
Vav = Vmax [-Cos (π + φ)]-[-Cosφ]
π
Vav = Vmax [-Cos (π + φ) + Cosφ]
π
Vav = Vmax [Cosφ + Cosφ]
π
Vav = 2Vmax Cosφ
π
92 D.C.TIBUJI
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THE ROOT MEAN SQUARE VALUE OF THE VOLTAGE ACROSS THE LOAD
π+φ
V²r.m.s = 1⌠ (Vmax Sinθ) ² dθ
T⌡
φ
π+φ
V²r.m.s = 1 ⌠ V²max Sin²θ dθ
π⌡
φ
π+φ
V²r.m.s = V²max ⌠ Sin²θ dθ
π ⌡
φ
Sin²θ = ½(1 - Cos 2θ)
π+φ
V²r.m.s = V²max ⌠ ½(1 - Cos 2θ) dθ
π ⌡
φ
π+φ
V²r.m.s = V²max [φ – ½ Sin2θ]
2π
φ
= V²max [π+φ–½ Sin2 (π+φ)]-[φ-½Sin2φ]
2π
= V²max [π –½ Sin2 (π+φ) + ½Sin2φ]
2π
D.C.TIBUJI
93
M.CINDY
= V²max [π]
2π
V²r.m.s = V²max
2
Vr.m.s = Vmax
√2
HALF-CONTROLLED RECTIFICATION
The d.c. load voltage can be control not just using four thyristors, it is possible to control
using two thyristors and two diodes.
When a pair of thyristors and a diode are to function like in this case, the thyristors serve
as the supply to the load while the diodes provide the return path.
The half-controlled circuit is some times call a mixed bridge circuit and this mixed bridge
circuit is being classified in to two, the asymmetrical and the symmetrical bridge circuit.
In the asymmetrical bridge circuit, the two (2) diodes will conduct during the period of
zero load voltage, carrying the freewheeling load current.
The presence of a commutating diode obviously prevents a negative load voltage and
hence enables the thyristors to turn off and regain its blocking state.
To distinguish between the fully-controlled and the half-controlled both technically and
economically,
The half-controlled circuit is cheaper but the alternating current supply is more distort
due to its zero periods
WAVEFORM AND CIRCUIT DIAGRAM
i1
VS
i2
(a)
94 D.C.TIBUJI
(b)
M.CINDY
VS
Vav
Φ
π
i1
i2
vt1
Figure 3.26 waveforms and circuit diagram (a) is an asymmetrical bridge while (b) is a
symmetrical bridge
THE AVERAGE VALUE OF THE LOAD VOLTAGE
π
Vav = 1⌠ Vmax Sinθ dθ
T⌡
φ
π
Vmax [-Cosθ]
π
φ
Vav = Vmax [-Cos π]-[-Cosφ]
π
Vav = Vmax [ 1 + Cosφ]
π
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M.CINDY
Effective value of voltage across the load
π
V²eff = 1 ⌠ (Vmax Sinθ) ²dθ
T⌡
φ
π
V²eff = 1 ⌠ V²max Sin²dθ
T⌡
φ
π
V²eff = V²max ⌠ Sin²θ dθ
π ⌡
φ
Sin²θ = ½(1 - Cos 2θ)
π
V²eff = V²max ⌠ ½(1 - Cos 2θ) dθ
π ⌡
φ
π
V²eff = V²max [θ - ½ Sin 2θ]
2π
φ
V²eff = V²max [π - ½ Sin2π]-[φ - ½ Sin2φ]
2π
V²eff = V²max [π - ½ Sin2π - φ + ½ Sin2φ]
2π
=> V²max [π/π -φ/π + ½π Sin2φ + ½π Sin2π]
2
Veff = Vmax √ [1 - φ/π + ½π Sin2φ]
√2
THREE-PHASE HALF-WAVE CONTROLLED RECTIFICATION
The three phase single-way rectification is the simplest in most of the polyphase rectifier
circuits and is being limited because certain features which is no in the single-way
connection.
The circuit diagram and waveform is in figure 3.27 below.
96 D.C.TIBUJI
M.CINDY
The circuit is controllable; the average value of the load voltage can be control by
adjusting the firing angle.
Waveforms and circuit diagram
VD1
i1
L1
L2
i2
il
i3
L3
Load
VL
N
Load
voltage
Vs
V1
V2
V3
VL
Vav
πs/6
5π/6
3π/2
ig1
ig2
ig3
iL
i3
Vt1
√3 Vmax
Figure 3.27
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M.CINDY
THE AVERAGE VALUE OF VOLTAGE ACROSS THE LOAD.
5π/6+φ
Vav = 1 ⌠
T⌡
Vmax Sinθ dθ
π/6+φ
But T = 2π/3
5π/6+φ
Vav = Vmax⌠
2π/3 ⌡
Sinθ dθ
π/6+φ
5π/6+φ
Vav = Vmax[- Cosθ]
2π/3
π/6+φ
Vav = 3Vmax [- Cos5π/6+φ + Cos π/6+φ]
2π
Vav = 3√3 Vmax Cos φ
2π
98 D.C.TIBUJI
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Chapter 4: CONVERTERS
There are two forms of electrical energy. The direct Current D.C. and the alternating
current a.c.
The second chapter has been dealing with the basic characteristics of all the different
types of rectifying circuit.
The two forms of energy the rectifying circuits has been dealing with, is the alternating
and direct current. Furthermore, the word rectification implies conversion or converting
energy from an a.c.source to a d.c.load.Under certain conditions, practically an example
of this nature, you are having an a.c. load but what is available as a supply is a d.c source.
Under this condition power, flow will be reverse converting from D.C. to a.c.
When a circuit is operating in this mode, it is operating in the inverting mode, that is, as
an inverter.
The word rectifier and inverter is replace by the word convert, when operating in their
various modes.
There are two types of converters in power electronics, the direct or autonomous and the
indirect or non-autonomous converter.
SOME COMMONLY USED CONVERTERS
AC/DC CONVERTER: called a RECTIFIER
DC/DC CONVERTER: called a CHOPPER
DC/AC CONVERTER: called an INVERTER
AC/AC CONVERTER: called a GRADATOR OR A CYCLOCONVERTER
For the DC/AC CONVERTER, fully controlled converter could be used.
In order for the thyristors to conduct, the converter must be connected to a large a.c.
synchronous system. example of such systems are the public supply, the reason for this
condition is to permit the a.c. energy fed back into the a.c system can be absorb by the
many other loads on the system.
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M.CINDY
APPLICATIONS OF SOME CONVERTERS
THE AC/AC CONVERTER
Use for the control of power in a.c.with the used of thyristors GRADATOR.
A GRADATOR can be define as a device, which permits the control of an alternating
power furnished to a resistive or inductive load. Resistive loads are electric ovens and
lighting systems just to name a few. For inductive loads we talk of universal motors and
induction motors e.t.c
COMPOSITION OF A GRADATOR
The gradator comprises of two identical thyristors connected in parallel and the thyristors
can be replaced by a TRIAC, when the power required is low for the application to
control.
At the initial condition, the load current is zero within the intervals (0 , φ) and (π, φ + π)
The thyristors is fire and conduction starts within the intervals (φ , π) and (φ + π , 2π)
The load current = Vmax Sinθ
R
The load current is alternating current but not sinusoidal, it effective value depends on the
firing angle φ
CIRCUIT DIAGRAM AND WAVEFORM OFA RESISTIVE AND INDUCTIVE
LOADS
Th1
Th2
UL
U
100D.C.TIBUJI
Z
M.CINDY
U
π
π +φ
2π
θ
φ
Figure 3.29
ANALYSIS FOR POWER TRANSFER
For a resistive load on controlled rectification the
Veff = Vmax √1 – φ/π + Sin 2φ/2π
√2
When φ = 0
Ieff = Veff = Vmax/√2 = Vmax
R
R
R√2
BUT P = I²R => V²mx R
R²√2²
P = V²max
2R
When the conduction angle is different from zero
P = V²max √1 – φ/π + Sin 2φ/2π
2R
D.C.TIBUJI
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THE CASE OF AN INDUCTIVE LOAD
When φ = θ and L the inductance of the inductor
VL = L di φ = ωt => θ = ωt
dθ
 dθ = ω dt
 dt = dθ
ω
VL = L di = L di
dt
dθ/ω
= Lω di
dθ
V =Vmax Sinθ = Lω di
dθ
=> Vmax Sinθ = di
Lω
dθ
=> ⌠Vmax Sinθ = ⌠di
⌡ Lω
⌡ dθ
=> - Vmax Cosθ + B = I + A
Lω
Where (A,B ΠR)
=> I = - Vmax Cosθ + C
Lω
To have value of C use the initial condition
θ = φ and I = 0
=> Vmax Cosφ = 0
102D.C.TIBUJI
M.CINDY
Lω
C = Vmax Cosφ
Lω
I = - Vmax Cosθ + Vmax Cosφ
Lω
Lω
I = - Vmax ( Cosθ - Cosφ)
Lω
For low power, application the two thyristors is replaced by a single component called a
TRIAC.
The essential characteristic of a TRIAC is that, it can be fire through a gate signal in both
directions of the current flow.
The gate pulses can be either negative or positive. A detail about the Triac has been
explained in the first chapter.
CIRCUIT DIAGRAMS AND WAVEFORM
LOAD
RP
TRIAC
DIAC
C
D.C.TIBUJI
103
M.CINDY
R
Figure 3.30 different circuits of triac and waveform
The circuit in the figure above that comprises of a triac and diac.is a simple circuit to
show how controlled is obtained using the triac.the diac is a gateless triac designed to
break down at low voltage. When the charge on capacitor has reached the diac break
down level, it will be discharge in to the triac gate. The variable resistance in the circuit
help to determine the charging rate of the capacitor C and hence the conduction angle.
THE DC/DC CONVERTER: CHOPPERS
A chopper is a static commutator, constituted of (thyristors or transistor) which permits
the direct transformation of a constant d.c. input into a variable d.c.output voltage. The
principle of functioning of a chopper consists of firing and blocking periodically one or
several thyristors, which functions in force commutation. The input voltage is chops to
give a variable output voltage. This output voltage is as a result of a series of impulse
whose duration is fixed or variable, separated by an interval of time, which can be fixed
or variable.
The average value of the continues output voltage depends on the conduction time and
blocking time of the thyristors and the conduction time and blocking time determine the
period of chopping(Tc).
Let us consider the conduction time t1 and the blocking time t2
The period of chopping Tc
Tc = t1 + t2
The frequency of chopping fc
Fc = 1
Tc
104D.C.TIBUJI
M.CINDY
The duty ratio
δ = t1__
t1 + t2
The average voltage Vav
Vav = VS
t1__
t1 + t2
Where VS is the supply voltage
This equation is for the case of an ideal chopper
CIRCUIT DIAGRAMS AND WAVEFORMS
IH
Ia
H
M
U
ID
B) The curves
IH
Ia
αT
T
t
ID
Ia
t
αT
D.C.TIBUJI
T
105
M.CINDY
QUESTIONS AND ANSWERS.
Q1) A DC/DC converter, which supplies a separately excited dc motor. The e.m.f of the
motor is equal to the voltage at the two terminals of the converter. this e.m.f is related to
the rotational speed by the relation E = KN with K = 0.06 Volt/r.m.s.
The armature current is constant and equal to 40A.the series DC converter is energized
from a 180 V supply.
WORK REQUIRED
a) Give the schematic diagram of the circuit described above
b) Give the curves of ID and IH according to time t
c) According to the duty ratio α, express the mean value of the current
i)
ID in the free wheel diode.
ii) IH in the electronic switch H
N.B at the beginning H is open
d) According to the duty ratio α,express the r.m.s value of the current
i)
ID in the free wheel diode.
ii)
IH in the electronic switch H
e) Give the minimum and maximum values of e.m.f
SOLUTION
a) circuit diagram
IH
Ia
H
M
U
ID
106D.C.TIBUJI
M.CINDY
B) The curves
IH
Ia
αT
T
t
ID
Ia
t
αT
T
c) The expression of the mean value
for the case of the free wheel diode ID from the curve
αT
ID = 1 ⌠ Ia dt
T⌡
0
αT
ID = [Ia t/T ] = IaαT – Ia 0 => ID = Ia α
0
T
T
 ID = 40α
ii)
T
IH = 1 ⌠ Ia dt
T⌡
αT
T
D.C.TIBUJI
107
M.CINDY
ID = [Ia t/T]
αt
Ia T - IaαT
T
T
=> Ia – Ia α= IH
IH = Ia(1- α)
IH = 40(1-α)
T
e) I2H = 1 ⌠ I²a dt
T⌡
αt
T
[Ia²t/T]
= I²H
αt
= I²a (1 – α) => IH = Ia √1 - α
IH = Ia √1 – α
IH = 40 √ 1 - α
d) ii
αT
I²D = 1⌠ I²a dt
T⌡
0
108D.C.TIBUJI
M.CINDY
αT
= [I²a t / T]
0
= I²a α
ID = Ia √α
ID = 40√α
e) The e.m.f is minimum when α = 0
E=αU
E=0
E is maximum when α = 1
E = U = 180 V
Q2) The AC/AC converter of the figure below
-Th1 and Th2 are two ideal thyristors
-U = Umax Sinθ where θ = ωt
α = firing angle of the thyristors
Z = the impedance of the load
Circuit diagram
Th1
Th2
UL
Z
U
a) Briefly describe the operation of this converter.
D.C.TIBUJI
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M.CINDY
b) If the load is considered a purely resistive:
1-Draw the waveform of the voltage across the load as a function of time.
2-Determine the expression of the r.m.s value of current (i) in terms of
resistance(R ) ,firing angle α and r.m.s value of the voltage.
3-deduce the expression of the power transmitted in the load(R)
C) The load is now consider a pure inductance (L)
1- Show that the current through the load can be written as I = Um Cosθ + A where A
is an intergrat
Lw
Ion constant
2-give the expression of A as a function of Um, LW and α
3-briefly explain why the firing angle of the thyristors could not be chosen in the interval
[0,π/2]
4-give the shape of the curve I = f(θ) for θ Œ[0,4π]
Solution
a) when the circuit is supply, during the positive half cycle, at the instant when θ = φ
an impulse is send into the gate of TH1 then, current will circulate through the
thyristor TH1, since the anode-cathode voltage (VAK) of TH1 is positive with
respect to the cathode, current will flow through the load.
When θ = π, the thyristor TH1 is blocked.
Secondly, when the supply frequency is changing from the positive half cycle
to the negative half cycle. the negative half cycle, at the instance when θ = π + φ,
an impulse is send into the gate of thyristor TH2,then since the cathode-anode
voltage is negative with respect to the anode voltage, current will flow through the
thyristor TH2 to the load. When θ = 2π, the thyristor TH2 is blocked.
b) i the waveform of the circuit
U
π
φ
110D.C.TIBUJI
π +φ
2π
θ
M.CINDY
π
I2r.m.s = 1⌠ U2max Sin2θ dθ
π⌡ R2
φ
2)
=>
π
I r.m.s = U max ⌠ Sin2θ dθ
R2 π
⌡
φ
2
=>
2
π
I r.m.s = U max ⌠ (1-Cos 2θ) dθ
2R2π ⌡
φ
2
=>
2
π
I2r.m.s = U2max [θ - ½ Sin 2θ]
2R2π
φ
=>
I2r.m.s = U2max [ ½ - φ/2π +Sin 2φ/4π]
R2
But Ueff = Vmax => Vmax = √2 Ueff
√2
=>
I2r.m.s = (√2 Ueff)2 [ 1 - φ/π +Sin 2φ/2π]
2 R2
Ir.m.s =
3)
Ueff √ [ 1 - φ/π +Sin 2φ/2π]
R
Since the load is a purely resistive;
P = R I2
D.C.TIBUJI
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M.CINDY
P = U2eff [ 1 - φ/π +Sin 2φ/2π]
R
C) U = L di/ dt = L ω di / dθ
U dθ = di ie di = u_ dθ
Lω
Lω
Intergrading both sides we are going to have…
⌠ di =⌠u_ dθ
⌡
⌡Lω
I = 1 ⌠ U dθ but U =Umax Sin θ
Lω⌡
I = 1 ⌠Umax Sin θ dθ
Lω⌡
=> i = - Umax Cos θ + A
Lω
2) To determine A, we take the initial condition
i = 0 when θ = φ
we have to substitute the above condition into the equation below
112D.C.TIBUJI
M.CINDY
=> 0 = - Umax Cos φ + A
Lω
=> A = - Umax Cos φ
Lω
Replacing A gives;
=> i = - Umax Cos θ + Umax Cos φ
Lω
Lω
 Umax [ Cos φ – Cos θ]
Lω
But Imax = Umax
Lω
I = Imax(Cos φ – Cosθ)
3) the firing angle of the thyristors are not chosen within the interval [0, π/2] for the
reason being that , when firing or at the time of firing the second thyristors TH2 at
the interval π + φ, thyristors TH1 will still functional or say conducting which
justify the reason for the second thyristors TH2 not to be fired.
4)i
π/2
π
3π/2 2π 5π/2
3π 7π/2 4π
Q3) a bridge rectifier comprises of four ideal diodes and is fed by a sinusoidal voltage u
= 220√2 Cos θ, θ = ωt = 100πt.
The bridge supplies a 1kΩ resistive load.in other to filter the load voltage a(r>c)
D.C.TIBUJI
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M.CINDY
1234-
sketch a circuit for the installation
Calculate the conduction angle 2θ0, knowing that across the load, Vav = 200v.
Compute for the suitable value for the protective resistance r.
calculate the capacitance C of the capacitor in other to have U/Vav = 1/10
Solution
1) circuit diagram
r
D1
D2
V
C
D3
D4
2) The conduction angle 2θ0
Vav = Vmax Cosθ0
Cosθ0 = Vav/ Vmax = 200/220√2
θ0 = 50°C
2θ0 = 100°C
3) Calculation of r
With the full-wave rectifier we have
tg(θ0) – θ0 = π r
2R
r = 2R (tg 50 – π x 50/180)
π
r = 200 Ω
114D.C.TIBUJI
R
U
M.CINDY
4) calculation of C
in full-wave rectifier we have;
δV
Vav
= 1 (π/2 – θ0)
RCω
=> C = 1 ( π/2 – θ0) Vav
Rω
δV
C=
1 (π/2 – π x 50/180) x 10 = 222 x 10-6
103 x 100π
C = 22.2µf
Q4) A battery charger of 12V is supply from a 220V- 50HZ network. The following will
be required to construct the charge;
- a centre-tapped ideal transformer whose secondary delivers a voltage of 2 x 24
volts.
- Two perfect diodes
- A protective resistance
4.1) draw the circuit diagram
4.2) the effective value of voltage at the secondary of the transformer.
4.3) the period of conduction per half cycle is 5ms, what is the opening time t1 and
closing time t2.
4.4) draw the waveform of the rectified voltage.
4.5) the secondary turns of the transformer has 120 turns, determine the number of
primary turns.
4.6) provide an expression for the average value of the rectified current.
4.7) determine the charging time in hours, if the capacity of the battery is 55AH and
average charging current is 5A.
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M.CINDY
Solution
D1
E
RP
D2
4.2) the effective voltage at the secondary of the transformer,
Veff. = 2x 24 = 48V
4.3) the conduction time λ = 5ms
t1 + t2 = T/2 --------- (1)
But f = 50HZ
Therefore T = 1/50 = 20ms
t2 – t1 = λ ---------- (2)
The equations 1 and 2 can be add up and solved simultaneously.
t1 + t2 = 10 -------------- (1)
t2 – t1 = 5 ---------------- (2)
Adding these two equations gives
2t2 = 15ms => t2 = 7.5ms
t2 = 7.5ms
From the equation 1, t1 = 10 – 7.5
Therefore;
116D.C.TIBUJI
M.CINDY
t1 = 2.5ms
UR
Vm-E
t1
t2 T/2 t3
t4
T
4.5) the primary turns N1
N2 = 120
V1 = 220, V2 = 48V, N1 =?
N1 = V1 = N1 => N2 X V1
N2 V2
V2
=> N1 = 120 X 220/48 = 550 turns
N1 = 550 turns
4.6) expression for the average current
Iav = 2Vmax Cosθ1 – E(λ)
R
π
π
4.7) Q = Iav x t => t = Q/ Iav = 55/5
t = 11 H
Q5) a certain installation is in need of power control device or circuit and a circuit like
the one in the figure below, is given to you.
D.C.TIBUJI
117
M.CINDY
RL
LOAD
RP
II
V
I
C
CIRCUIT ANALYSIS
The aim of this circuit is to control the power consumed in the load RL using the
component II.
The component II has a negligible resistance when it is conducting.
The control circuit, which comprises of the component I, sends pulse or signal in the
element II.
The first pulse is send to the component II at the instant (to) while the second pulse is at (
to + π/2).
The voltage is sinusoidal with effective value of 220V.the frequency is 50HZ and period
T
WORK DEMAND
1) Identify the components I and II
2) What is the role RP in the circuit?
3) Which value can you attribute to RL in order to limit the load current to 3.1A?
4)an ammeter is connected in series with RL to get the reading of the effective value of
the load current of which type is the said ammeter.
5) The first pulse is now taken at the instance
(to = T/4) sketch the following waveforms V(t),VL(t).
6) For to = T/4
-what is the reading of the ammeter of question (4)
-what is the new power consumed by the load
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SOLUTION
1) The components (I) and (II) in the circuit above are DIAC and TRIAC respectively.
2) The main role of R in the circuit is to vary the firing angle of the triac.
3) iLmax = Vmax
RL
RL = Vmax
iLmax
= √2 x Veff
3.1
RL = √2 X 220 = RL = 100.36Ω
3.1
RL = 100.36Ω
4)The type of ammeter is the iron core type or any digital ammeter with an R.S.M scale,
for d.c is the moving core ammeter.
- when to = 0, then, maximum deflection of the ammeter.
- For to = 0, the triac will be conducting at each instant of the period, that is, the ammeter
will display the maximum value.
-the reading of the ammeter is ;
il = V = 220
RL 100
il = 2.2A
- The power consumed P
P = RL X I2L => P = 100 X (2.2)2 = 484W
P = 484W
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5) The waveforms of V(t) and VL
V
t
VL
π
to + T/2
2π
to
6) il = VL
RL
T/2
T
2
But VL = 1⌠ ( VmaxSinθ) dθ +1⌠V2Sin2θdθ
T⌡
T⌡
to
to + T/2
2
T/2
T
But VL2 = 1⌠ 2V2maxSin2θdθ +1⌠2V2Sin2θdθ
T⌡
T⌡
T/4
3 T/4
T/2
T
VL = 2V ⌠ ( 1-Cos2θ)dθ +1⌠(1-Cos2θ)dθ
T ⌡
2
T⌡
2
T/4
3T/4
2
2
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T/2
T
But VL = 2V [ θ-Sin2θ] + [(θ-Sin2θ/2]
T
T/4
3T/4
2
2
VL = V => Ieff = V = V
√2
RL
RL√2
Ieff = 220 = 1.5A
100√2
Ieff = 1.56A
- Power consumed P
P = RL X Ieff2
P = 100 x (1.56)2 = 243W
P = 243W
Q6)
B
H IH
IL
E
VC
ID
VC
Vc =750V, B = inductance, L = 1.3mh, R= 0.112Ω
A train motor is represented by it E.M.F, E in series with it total resistance R = 0.112Ω.
its supplied by a chopper H form by a perfect thyristor. The diode is a free wheel diode
considered perfect.
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STUDY OF THE FIRING CIRCUIT
D
VC
R2
VS
R1
V1
V2
5 Vc
t2
t
t1
-5
The circuit above is an ASTABLE MULTI –VIBRATOR and the waveform. This circuit
is realize with the aid of an OP-AMP mounted in the form of a multi-vibrator for the
purpose of controlling the Thyristor of the chopper
Work required
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1) Come out with a relationship between R1, R2, V1 and V2.
2) At the instance when Vc>V1,
V2 = -Vsat and when
Vc < V1, V2 = +Vsat
From the conditions above, come out the value of V1 suppose R1 = R2 =1KΩ
3) the voltage across the capacitor Vc is given in the waveform above you
are ask to draw on the same waveform, the waveform of V2
4) give the purpose of the diode D in the circuit above and drawn the
waveform of Vs = f(t)
5) Calculate the period of the signal Vs (t) knowing that, the expression of T1
= 2.2 RC with R = 7.6KΩ and C = 0.1µf.determine the frequency of the
supply.
N.B. the OP-AMP is idea and Vsat = ±10V
ANALYSIS ON THE POWER CIRCUIT
The D.C converter is controlled by a system and periodically with frequency f =
600Hz
From 0 to λT, the electronic switch H is close and
From λT to T, the electronic switch H is open
With T = 1/f and λ being the duty ratio is equal to 2/3
1) deduce the value of the load voltage within the interval [0,λ] and [λT,T]
2) produce graphically Vc = f(t)
3) calculate the average value of Vc
4) With the understanding that, in a motor Imax = 447A, Imin =233A and Vav =
340A. calculate the back e.m.f of the motor and plot the following graph; Ic, IH
and Id
Solution
Part 1; control circuit
1) lets use the voltage divider role to analyze these problem
V1 = R1 x V2
R1 + R2
OR
V2 = 1 – R2 x V1
R1
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2) firstly; when Vc > V1, V2 = -Vsat = - 10V
V1 = -10 x 1__
1+1
V1 = -5 V
For VC < V1, V2 = +Vsat = + 10V
V1 = 10 x 1/2 = + 5V
V1 = +5V
3) THE GRAPH OF V2
10V
V2
V1
5V
t
t1
t2
-5V
-10V
4) the purpose of the diode D is to prevent the load from receiving negative pulses
on V2, in other to obtain VS whose waveform is seen below
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Vs
10V
t
t1
to
5) calculation of the period
T = 2.2RC
T = 2.2 x 7.6 x 103 x 0.1 x 106
T = 1.672ms
f = 1/ T = 1/ 1.672 x 10-3
f = 598HZ
Part two. The power circuit
1) Calculation of the load voltage
VC = λV
VC = 2/3 x 750 = 500V
VC = 500V
2) The graphs of Vc, Ic, and IH as a function of time t
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The graphs is seen below
Calculation of e.m.f E of the motor.
VC (av) =E + r Ic
So, E = VC (av) – r Ic
E = 500 – 0.112 x 340 = 462V
E = 462V
VC
750
500
VavC
t
2/3
Ic
447
233
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IH
t
Id
t
2T/3
T
Q7) A battery on charge, within this process of charging, the elm of the battery rises from
E1 = 192(cell discharged) to 220V (normal charge). They wish the current to be constant
and equal to Io = 40A. The charging is done using a Thyristor bridge.
Th1 and Th2 are fired together while Th3 and Th4 are fired T/2 later. The internal
resistance of the battery is 0.1Ω. An inductance (L) for smoothing the charging current Ic
Work required
1 plot the graph across the load and draw the circuit diagram as described above.
2 Express the mean value of voltage V in term of Io, E, and r.
3 calculate the average values of voltage V1 and V2 of V(t) across the load
corresponding respectively to E1 and E2.
4 Deduce the two values θ1 and θ2 corresponding to the firing angle θo
Note: the average value V = 2/π V2 √2 Cos θo
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5 calculate the charging duration if the capacity of the battery is 400AH
Solution
Circuit diagram
L
V2 (t)
r
E
Plotting of V(t)
V(t)
E
θ1
θ2
T/2
t
2) Expressing the average value of the voltage in
Terms of Io, E, and r
Vav = E + Io r
4) data
E1 = 192 V, r = 0.1 Ω, Io = 40A
V1 = E + Io r = 192 + 40 x 0.1
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V1 = 196V
For E2 = 220V
V2 = 220 + 40 x 0.1
V2 = 224V
5) calculation of the firing angles θ1 and θ2
V1 = 2/π x V2√2 Cos θ1
Cos θ1 = 196____
2/π x 224√2
Cos θ1 =0.966
θ1 = 14.8°
θ2 = θ1 + T/2 = 14.8 + 180
θ2 = 194.8°
6) Charging duration (t)
Q = Io x t => t = Q/Io
t = 400/40
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t = 10 hrs
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Q8) Maria Cindy N.C the daughter of the engineer, she saw on one of his father
documents that, SINGLE PHASE FULL-WAVE NON CONTROL rectification using a
resistive load and considering all the diodes perfect.
Maria Cindy now comes to you as a student studying power electronic and wants to know
the following;
Work required
1 the circuit diagram and the meaning of NON CONTROL.
2 calculate the effective value of the voltage, knowing that the average value of the
voltage is 20V
3 When the resistance R = 10Ω, what is the average value of current?
4 The peak value of the diode current
5 the root mean square value of current in the load
6 The average value of the diode current
7 The effective value of the diode current
If the resistive load is now replace with a battery and a resistor, E = 24V and R = 4Ω
1 Sketch the corresponding circuit and calculate the conduction angle θ of one diode
while considering the effective voltage to be 33.4V
2 Calculate the peak value of current in the load
3 Calculate the average value of current in the load
4 The capacity of the battery is 30AH, what is the charging duration of the battery
SOLUTION
1)
Circuit diagram
D1
D3
V
R
D4
D2
NON CONTROL simply means the output voltage cannot be control unlike the control
rectification that the thyristors are being fired which permit the output voltage to be
control.
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2) Vav = 2Vmax
π
 Vav x π = 2Vmax
Vmax = Vav x π
2
But Veff = Vmax
√2
Then Vmax = Veff x √2
So
Vav x π = Veff x 2
2
Veff = Vav x π
2√2
Veff = 20 x 3.14
2√2
Veff = 22.2 V
3) Iav = Vav
R
= 20
10
Iav = 2A
4) Imax = Id = Vmax
R
Imax = 22.2 x √2
10
=> Id = 3.12A
Id = 3.12A
5) Ieff = Veff = 22.2
R
10
Ieff = 2.22A
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6) Id = Iav = 2
2
2
Id = 1A
7) Ieff = Id = 3.12 = 2.2A
√2
√2
Ieff = 2.2A
EXERCISE
Q7) A battery on charge, within this process of charging, the elm of the battery rises from
E1 = 192(cell discharged) to 220V (normal charge). They wish the current to be constant
and equal to Io = 40A. The charging is done using a Thyristor bridge.
Th1 and Th2 are fired together while Th3 and Th4 are fired T/2 later. The internal
resistance of the battery is 0.1Ω. An inductance (L) for smoothing the charging current Ic
Work required
1 plot the graph across the load and draw the circuit diagram as described above.
2 Express the mean value of voltage V in term of Io, E, and r.
3 calculate the average values of voltage V1 and V2 of V(t) across the load
corresponding respectively to E1 and E2.
4 Deduce the two values θ1 and θ2 corresponding to the firing angle θo
Note: the average value V = 2/π V2 √2 Cos θo
5 calculate the charging duration if the capacity of the battery is 400AH.
ANSWERS(V1= 196V, V2 = 224V,θ1 = 14.8,θ2 = 194.8, t= 10hrs)
Q8) Maria Cindy N.C the daughter of the engineer, she saw on one of his father
documents that, SINGLE PHASE FULL-WAVE NON CONTROL rectification using a
resistive load and considering all the diodes perfect.
Maria Cindy now comes to you as a student studying power electronic and wants to know
the following;
Work required
1 Drawn the circuit diagram and the meaning of NON CONTROL.
2 calculate the effective value of the voltage, knowing that the average value of the
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voltage is 20V
3 When the resistance R = 10Ω, what is the average value of current?
4 The peak value of the diode current
5 the root mean square value of current in the load
6 The average value of the diode current
7 The effective value of the diode current
If the resistive load is now replace with a battery and a resistor, E = 24V and R = 4Ω
1 Sketch the corresponding circuit and calculate the conduction angle θ of one diode
while considering the effective voltage to be 33.4V
2 Calculate the peak value of current in the load
3 Calculate the average value of current in the load
4 The capacity of the battery is 30AH, what is the charging duration of the battery
9) THREE PHASE RECTIFICATION
Considering the circuit, if the diodes are being replace with Thyristors and V1, V2 and
V3, which are the phase voltages are jointly connected to the ground at a point N.
Take the firing angle of the Thyristor to be 30°.
WORK REQUIRED
Draw the waveform of the voltages VDN and VDC.
Draw the waveform of the voltage VR
Draw the waveform of the voltage VT1
Indicate the interval of conduction of all the thyristors.
Determine the average value of VR and deduce the average value of current
What is the maximum value of the voltage across R
10)
R
D1
D2
B
VL
D3
D4
The circuit above is supplying a circuit that requires a constant supply.
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The diodes D1,D2,D3,D4 and B are assumed to be ideal. The component B is having a
voltage of 15V, V is a voltage source of root mean square value, 21.2V– 50Hz
A resistive load is connected to the circuit, you are asked to;
identify the component B and give its role in that circuit.
Analyze the operation of the circuit and draw the graphs of V(t) and VL(t)
Knowing that R the protective resistor has a value of 15Ω, determine the maximum
current in this resistor
For one complete cycle determine the duration of conduction of the component B.
What would be the reading of an electromagnetic voltmeter connected between the load.
11)
IH
VL
I
E = 220V
M
ID
Vc
VA
The circuit diagram above is that of a DC/DC converter and a DC motor functioning in
the following conditions.
the current flows through the converter only in one direction and open at a frequency of
1000Hz and closes at the same frequency
The supply is a DC voltage of E = 220V
The duty or cyclic ration α is variable within the interval 0≤ α ≤ 1
The diode is ideal
The load is made up of the following
Perfect inductance L of negligeable resistance and a considerably large enough so as to
maintain a constant current through the load.
A DC motor M which functions at a constant torque
Work required
For a cyclic ratio α = 0.5, give the chronograms of E(t), i(t) and iD(t)
- for a variable cyclic ratio, express the average value of Vc as a funct
show that the average voltage across the inductor VL = 0
134D.C.TIBUJI
ion of E and α
M.CINDY
In connection with the question above established a simple relationship between E(t) and
Va(t)
12)
Ic
E = 12V
R = 0.2Ω
Vc
D
TH
P
C
D
The circuit diagram above represent the circuit diagram of a 12V-50Hz battery intended
to be supply to the AES-SONEL network through a transformer.
the supply voltage is v = 220√2 sin φ
At the secondary of the transformer is 48√2 sinφ
- work required
- state the functioning of D and explain the operation of this circuit
- the thyristor goes on with delay
- plot the chronograms of the voltage across the battery Vc and that of the current Ic
through it
- determine the mean values of the voltages across the battery and the current
- determine the delay angle of the thyristor if the mean value of the current is half of that
of the question above.
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13)
N
L
D4
D6
D3
M
D1
D2
D5
- The figure above is a three phase half control rectification supplied by a transformer
couple in star-delta and the half control circuit is controlling a separately excited DC
motor having a constant field and armature current of 1A and 42A respectively. In other
to vary the speed from 0 to 1500 rpm , the armature voltage is varied through the bridge
circuit. The secondary of the transformer is displaced as follows:
- VR = Vm sinφ
- VS = Vm (sin φ— 2π/3)
- VT = Vm (sin φ— 4π/3)
- on the name plate of the transformer ,the following information can be gotten: V1 =
380 V , N1 = 262, N2 = 90, f = 50Hz.
Work required
- represent as a function of time on the same plane
The load voltage, the diode 1 voltage, the diode 1 current and secondary line current of
the transformer
- calculate the average voltage of the load
- the average voltage at the secondary of the transformer
- the average and effective currents through diode
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14) - the average and effective currents through the secondary of the transformer
D1
A
D2
V1
B
V2
R
D3
C
V
V3
The figure above is a three phase non controlled rectification using a resistive load. The
diodes are all assumed to be ideal.
The circuit is receiving it supply from the AES-SONEL network with the following
characteristics:
V1= Vm cosφ
- V2 = Vm (cos φ— 2π/3)
- V3 = Vm (cosφ— 4π/3)
Where Vm = 220√2V, φ = 2πf, f = 50Hz and R = 500Ω
WORK REQUIRED
- design on the same plane the chronograms of the voltages V1, V2, V3 and V across
the load , indicating all the remarkable points on the X-Y plane
investigate the duration of conduction for each of the diodes in terms of T( the period)
Calculate the average and the root mean square values of the voltages across the load,and
also deduce the power absorbed by the load
Let the diodes of circuit diagram above be replaced by three thyristors, assumed to be
ideal and this thyristors are having a firing angle (φ)
Where 0 < φ < 2π/3
Draw the wave forms of the voltages V1, V2, V3 and V across the load
Prove that the expression of the average voltage across the load is :
Vav = 3√3 Vm cos φ/2π
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15)
D1
U1
R
u
U2
D2
The supply voltage of the circuit below is 100 sinθ and the transformer is having 200
turns on the primary and on the 2 turns on the secondary
WORK REQUIRED
explained the functioning of the circuit above with the aid of graphs
Calculate: u1 and u2
What does T stands for
16)
Rv = 30Ω
D1
Rs
V
U = 64 v
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M.CINDY
the maximum current is considered to be 15A then
Calculate:
- the conduction angle and duration
- give the value of the variable resistor Rs
- the peak inverse voltage
- the charging current
- the charging duration when the capacity of the battery is 30AH
17)
I
Vs = 120v
Ic
RL= 8Ω
The figure above is the circuit diagram for DC/DC converter:
1) what is a chopper
2) which are the electronic switches that are usually used as a chopper
3) give one application of each of them
4) what is the significant of the two vertical lines in the symbol of the chopper.
5) draw the waveform of VL(t)
18)
Maria Cindy decided to construct a bridge rectifier circuit containing a resistive load of
value 50 Ω which is supply from an AC single phase network.
Maria Cindy Njang Chofor wants to find out from the students of (ET -7) FULL
GOSPEL COMPREHENSIVE HIGH SCHOOL MUYUKA, if the father is saying the
truth that his students are very powerful. She starts by asking;
Draw the circuit diagram showing the direction of currents in each branch
Draw the waveform of; a) the voltage across the load and that of each diode and b)
the current through the load and diodes.
Investigate the peak inverse voltage at each diode
The average value of current through the load
The power dissipated in the load
Consider the current to be constant in the case when the supply is cut off and a battery
of e.m.f, 10V with internal resistance r of 10 mkΩ is connected in parallel with
the load;- a) calculate the equivalent thevenin voltage and resistance seen from the
terminals of the load, - b) if a resistance for protection is connected in series with
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the thevenin resistance and the thevenin voltage so as to limit the current to 20 A,
- c) draw the new circuit diagram and calculate the value of the resistance for
protection, - d) considering the new circuit diagram the value of value of the
resistor for protection to be 10Ω.draw the waveform of the voltage and current
across the load.
From the results of question 6, - a) calculate the instant that conduction begins, - b)
the instant when conduction ends and – c) the duration of conduction.
19)
D
D1
D2
R
B
A
D3
A
D4
C
V(t)
The expression of the supply network for the circuit above is ;
V(t) Vm sin θ. All the diodes and thyristor are assumed to be ideal
The resistance R = 35 Ω .the ammeter is of the ferromagnetic type.
WORK REQUIRED
What is the name given to the circuit
Explain the functioning of the circuit and indicate the duration of current through the
resistance R
IF the thyristor is fired at t = 0 and t = T/2 and the indication of the ammeter is 4A
Draw the waveform of the voltage across the resistanceR
Calculate the root mean aquare of the voltage across the load resistance and the
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maximum value of V(t)
If the thyristor is now fired at t = T/4 and t = 3T/4
Draw the waveform of the voltage across the load resistance
Calculate the current indicated by the ammeter
The thyristor is now fired only at t = T/4
Draw the waveform of the voltage across the load
Calculate the current indicated by the ammeter
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