EE 435- Electric Drives
Dr. Ali M. Eltamaly
Chapter 2
Diode Circuits or Uncontrolled Rectifier
2.1 Introduction
Because of their ability to conduct current in one direction, diodes are used in rectifier
circuits. The definition of rectification process is “ the process of converting the alternating
voltages and currents to direct currents and the device is known as rectifier” It is extensively
used in charging batteries; supply DC motors, electrochemical processes and power supply
sections of industrial components.
The most famous diode rectifiers have been analyzed in the following sections. Circuits and
waveforms drawn with the help of PSIM simulation program.
There are two different types of uncontrolled rectifiers or diode rectifiers, half wave and full
wave rectifiers. Full-wave rectifiers has better performance than half wave rectifiers. But the
main advantage of half wave rectifier is its need to less number of diodes than full wave
rectifiers. The main disadvantages of half wave rectifier are:
1- High ripple factor,
2- Low rectification efficiency,
3- Low transformer utilization factor, and,
4- DC saturation of transformer secondary winding.
2.2 Performance Parameters
In most rectifier applications, the power input is sine-wave voltage provided by the electric
utility that is converted to a DC voltage and AC components. The AC components are
undesirable and must be kept away from the load. Filter circuits or any other harmonic reduction
technique should be installed between the electric utility and the rectifier and between the
rectifier output and the load that filters out the undesired component and allows useful
components to go through. So, careful analysis has to be done before building the rectifier. The
analysis requires define the following terms:
The average value of the output voltage, Vdc ,
The average value of the output current, I dc ,
The rms value of the output voltage, Vrms ,
The rms value of the output current, I rms
The output DC power, Pdc = Vdc * I dc
(2.1)
The output AC power, Pac = Vrms * I rms
(2.2)
The effeciency or rectification ratio is defiend as η = Pdc / Pac
(2.3)
The output voltage can be considered as being composed of two components (1) the DC
component and (2) the AC component or ripple. The effective (rms) value of the AC component
of output voltage is defined as:2
Vac = Vrms
− Vdc2
(2.4)
The form factor, which is the measure of the shape of output voltage, is defiend as shown in
equation (2.5). Form factor should be greater than or equal to one. The shape of output voltage
waveform is neare to be DC as the form factor tends to unity.
FF = Vrms / Vdc
(2.5)
The ripple factor which is a measure of the ripple content, is defiend as shown in (2.6). Ripple
factor should be greater than or equal to zero. The shape of output voltage waveform is neare to
be DC as the ripple factor tends to zero.
9
Diode Circuits or Uncontrolled Rectifier
2
2
Vrms
− Vdc2
Vac
Vrms
RF =
=
=
− 1 = FF 2 − 1
(2.6)
2
Vdc
Vdc
Vdc
Where VS and I S are the rms voltage and rms current of the transformer secondery respectively.
Total Harmonic Distortion (THD) measures the shape of supply current or voltage. THD
should be grearter than or equal to zero. The shape of supply current or voltage waveform is near
to be sinewave as THD tends to be zero. THD of input current and voltage are defiend as shown
in (2.8.a) and (2.8.b) respectively.
THDi =
I S2 − I S21
I S2
VS2 − VS21
VS2
= 2 − 1 , and
=
−1
(2.8)
THDv =
I S21
I S1
VS21
VS21
where I S1 and VS1 are the fundamental component of the input current and voltage, I S and VS
respectively.
In general, power factor in non-sinusoidal circuits can be obtained as following:
Real Power
P
=
= cos φ
(2.10)
PF =
Apparent Voltamperes VS I S
Where, φ is the angle between the current and voltage. Definition is true irrespective for any
sinusoidal waveform. But, in case of sinusoidal voltage (at supply) but non-sinusoidal current, the
power factor can be calculated as the following:
Average power is obtained by combining in-phase voltage and current components of the same
frequency.
P
V I cos φ1 I S1
PF =
= S S1
=
cos φ1 = Distortion Factor * Displacement Faactor
(2.11)
VS I S
VS I S
IS
Where φ1 is the angle between the fundamental component of current and supply voltage.
Distortion Factor = 1 for sinusoidal operation and displacement factor is a measure of
displacement between v(ωt ) and i (ωt ) .
Single-Phase Half Wave Diode Rectifier With Resistive Load
Fig.2.1 shows a single-phase half-wave diode rectifier with pure resistive load. Assuming
sinusoidal voltage source, VS the diode beings to conduct when its anode voltage is greater than
its cathode voltage as a result, the load current flows. So, the diode will be in “ON” state in
positive voltage half cycle and in “OFF” state in negative voltage half cycle. Fig.2.2 shows
various current and voltage waveforms of half wave diode rectifier with resistive load. These
waveforms show that both the load voltage and current have high ripples. For this reason, singlephase half-wave diode rectifier has little practical significance.
Fig.2.1 Single-phase half-wave diode rectifier with resistive load.
10
Chapter Two
Fig.2.2 Various waveforms for half wave diode rectifier with resistive load.
The average or DC output voltage can be obtained by considering the waveforms shown in
π
Fig.2.2 as following: Vdc
Vm
1
V
t
d
t
ω
ω
=
sin
=
m
π
2π ∫0
(2.12)
Where, Vm is the maximum value of supply voltage.
Because the load is resistor, the average or DC component of load current is: I dc =
Vdc Vm
=
R π R
The root mean square (rms) value of a load voltage is defined as:
π
Vrms
V
1
=
Vm2 sin 2 ωt dωt = m
∫
2π 0
2
(2.14)
Vrms Vm
=
R
2R
It is clear that the rms value of the transformer secondary current, I S is the same as that of the
load and diode currents
V
Then I S = I D = m
(2.15)
2R
Where, I D is the rms value of diode current.
Similarly, the root mean square (rms) value of a load current is defined as: I rms =
Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The
efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1.
Solution: From Fig.2.2, the average output voltage Vdc is defiend as:
π
Vdc
V
V
1
=
Vm sin(ωt ) dωt = m (− cos π − cos(0)) = m
π
2π
2π
∫
0
Then, I dc =
Vdc Vm
=
R πR
Diode Circuits or Uncontrolled Rectifier
π
Vrms
V
1
=
(Vm sin ωt ) 2 = m ,
2π
2
∫
0
I rms =
11
Vm
V
and, VS = m
2R
2
The rms value of the transformer secondery current is the same as that of the load: I S = Vm .2 R
Then, the efficiency or rectification ratio is:
Vm Vm
*
Pdc
Vdc * I dc
π
πR
η=
=
= 40.53%
=
V m Vm
Pac Vrms * I rms
*
2 2R
Vm
Vrms
π
(b) FF =
= 2 = = 1.57
Vm 2
Vdc
π
Vac
= FF 2 − 1 = 1.57 2 − 1 = 1.211
Vdc
(d) It is clear from Fig2.2 that the PIV is Vm .
(c) RF =
2.3.2 Half Wave Diode Rectifier With R-L Load
In case of RL load as shown in Fig.2.3, The voltage source, if vs = Vm sin (ωt ) , vs is positive
when 0 < ω t < π, and vs is negative when π < ω t <2π. When vs starts becoming positive, the
diode starts conducting and the source keeps the diode in conduction till ω t reaches π radians. At
that instant defined by ω t =π radians, the current through the circuit is not zero and there is some
energy stored in the inductor. The voltage across an inductor is positive when the current through
it is increasing and it becomes negative when the current through it tends to fall. When the
voltage across the inductor is negative, it is in such a direction as to forward-bias the diode. The
polarity of voltage across the inductor is as shown in the waveforms shown in Fig.2.4.
When vs changes from a positive to a negative value, the voltage across the diode changes its
direction and there is current through the load at the instant ω t = π radians and the diode
continues to conduct till the energy stored in the inductor becomes zero. After that, the current
tends to flow in the reverse direction and the diode blocks conduction. The entire applied voltage
now appears across the diode as reverse bias voltage. An expression for the current through the
diode can be obtained by solving the deferential equation representing the circuit.
Fig.2.3 Half Wave Diode Rectifier With R-L Load
Chapter Two
12
Fig.2.4 Various waveforms for Half wave diode rectifier with R-L load.
Assume Z∠φ = R + j wL Then Z 2 = R 2 + w 2 L2 ,
ωL
Z
R = Z cos φ , ωL = Z sin φ and tan φ =
wL
R
t
ω


−
Φ
Vm 
tan φ 
(2.24)
i (ωt ) =
sin (ωt − φ ) + sin (φ )e


R
Z 



Starting from ω t = π, as ωt increases, the current would keep decreasing. For some value of
ωt , say β, the current would be zero. If ω t > β, the current would evaluate to a negative value.
Since the diode blocks current in the reverse direction, the diode stops conducting when ωt
reaches β. The value of β can be obtained by substituting that i (ωt ) = 0 wt = β into (2.24) we get:
β 
−
Vm 
tan φ 
i(β ) =
sin (β − φ ) + sin (φ )e
(2.25)
=0
Z 


The value of β can be obtained from the above equation by using the methods of numerical
analysis. This average value can be obtained as shown in (2.26). The rms output voltage in this
case is shown in equation (2.27).
β
Vdc
V
V
= m * ∫ sin ωt dωt = m * (1 − cos β )
2π 0
2π
(2.26)
β
Vrms
Vm
1
=
* ∫ (Vm sin ωt ) 2 dwt =
* β + 0.5(1 − sin( 2 β )
2π 0
2 π
(2.27)
2.4 Single-Phase Full-Wave Diode Rectifier
The full wave diode rectifier can be designed with a center-taped transformer as shown in Fig.2.8,
where each half of the transformer with its associated diode acts as half wave rectifier or as a
bridge diode rectifier as shown in Fig. 2.12. The advantage and disadvantage of center-tap diode
rectifier is shown below:
13
Diode Circuits or Uncontrolled Rectifier
Advantages
• The need for center-tapped transformer is eliminated,
• The output is twice that of the center tapped circuit for the same secondary voltage, and,
• The peak inverse voltage is one half of the center-tap circuit.
Disadvantages
• It requires four diodes instead of two, in full wave circuit, and,
• There are always two diodes in series are conducting. Therefore, total voltage drop in the
internal resistance of the diodes and losses are increased.
The following sections explain and analyze these rectifiers.
2.4.1 Center-Tap Diode Rectifier With Resistive Load
In the center tap full wave rectifier, current flows through the
load in the same direction for both half cycles of input AC
voltage. The circuit shown in Fig.2.8 has two diodes D1 and D2
and a center tapped transformer. The diode D1 is forward bias
“ON” and diode D2 is reverse bias “OFF” in the positive half
cycle of input voltage and current flows from point a to point b.
Whereas in the negative half cycle the diode D1 is reverse bias
“OFF” and diode D2 is forward bias “ON” and again current
flows from point a to point b. Hence DC output is obtained
across the load.
Fig.2.8 Center-tap diode rectifier
Fig.2.9 Various current and voltage waveforms for center-tap diode rectifier with resistive load.
[
In case of pure resistive load, Fig.2.9 shows various current and voltage waveform for
converter in Fig.2.8. The average and rms output voltage and current can be obtained from the
waveforms shown in Fig.2.9 as shown in the following:
Vdc =
1
π
π
∫
0
Vm sin ωt dωt =
2 Vm
π
and
I dc =
2 Vm
π R
(2.36)
14
Chapter Two
Vrms =
1
π
π
2
∫ (Vm sin ωt ) dωt =
0
Vm
2
and
I rms =
Vm
2 R
PIV of each diode = 2Vm
V
VS = m
2
The rms value of the transformer secondery current is the same as that of the diode:
V
IS = ID = m
2R
(2.38)
(2.40)
(2.41)
(2.41)
Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency,
(b) Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of diode D1.
Solution:- The efficiency or rectification ratio is
2 Vm 2 Vm
*
Pdc
Vdc * I dc
π
πR
=
=
= 81.05%
η=
Vm
Vm
Pac Vrms * I rms
*
2
2R
Vm
V
2 = π = 1.11
(b) FF = rms =
2 Vm 2 2
Vdc
π
Vac
= FF 2 − 1 = 1.112 − 1 = 0.483
Vdc
(d) The PIV is 2Vm
(c) RF =
2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load
Another alternative in single-phase
full wave rectifier is by using four
diodes as shown in Fig.2.12 which
known as a single-phase full bridge
diode rectifier. It is easy to see the
operation of these four diodes. The
current flows through diodes D1 and
D2 during the positive half cycle of
Fig.2.12 Single-phase full bridge diode rectifier.
input voltage (D3 and D4 are “OFF”). During
the negative one, diodes D3 and D4 conduct (D1 and D2 are “OFF”).
In positive half cycle the supply voltage forces diodes D1 and D2 to be "ON". In same time it
forces diodes D3 and D4 to be "OFF". So, the current moves from positive point of the supply
voltage across D1 to the point a of the load then from point b to the negative marked point of the
supply voltage through diode D2. In the negative voltage half cycle, the supply voltage forces the
diodes D1 and D2 to be "OFF". In same time it forces diodes D3 and D4 to be "ON". So, the
current moves from negative marked point of the supply voltage across D3 to the point a of the
load then from point b to the positive marked point of the supply voltage through diode D4. So, it
15
Diode Circuits or Uncontrolled Rectifier
is clear that the load currents moves from point a to point b in both positive and negative half
cycles of supply voltage. So, a DC output current can be obtained at the load in both positive and
negative halves cycles of the supply voltage. The complete waveforms for this rectifier is shown
in Fig.2.13.
Fig.2.13 Various current and voltage waveforms of Full bridge single-phase diode rectifier.
Example 4 The rectifier shown in Fig.2.12 has a purely resistive load of R=15 Ω and, VS=300 sin
314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor,
(d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.
Solution: Vm = 300 V
Vdc =
1
π
π
∫ Vm sin ωt dωt =
2 Vm
π
0
1/ 2
Vrms
1 π

=  ∫ (Vm sin ωt )2 dωt 
 π 0

=
= 190.956 V , I dc =
2 Vm
= 12.7324 A
π R
V
Vm
= 212.132 V , I rms = m = 14.142 A
2R
2
Pdc
V I
= dc dc = 81.06 %
Pac Vrms I rms
V
(b) FF = rms = 1.11
Vdc
(a) η =
2
2
Vrms
− Vdc2
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 0.482
2
Vdc
Vdc
Vdc
(d) The PIV=300V
V I cos φ
Re al Power
(e) Input power factor =
= S S
=1
Apperant Power
VS I S
2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load Current
The full bridge single-phase diode rectifier with DC load current is shown in Fig.2.14. In this
circuit the load current is pure DC and it is assumed here that the source inductances is negligible.
In this case, the circuit works as explained before in resistive load but the current waveform in the
supply will be as shown in Fig.2.15. And the rms value of the input current is I S = I o
16
Chapter Two
Fig.2.14 Full bridge single-phase diode rectifier with DC load current.
Fig.2.15 Various current & voltage waveforms for single-phase diode bridge rectifier for pure DC load.
The supply current in case of pure DC load current is shown in Fig.2.15, as we see it is odd
function, then an coefficients of Fourier series equal zero, an = 0 , and
bn =
2
π
I o * sin nωt dωt
π∫
=
0
=
2 Io
[− cos nωt ]π0
nπ
(2.51)
2 Io
[cos 0 − cos nπ ] = 4 I o for n = 1, 3, 5, .............
nπ
nπ
Then from Fourier series concepts we can say:
i (t ) =
4 Io
π
1
1
1
1
* (sin ωt + sin 3ωt + sin 5ωt + sin 7ωt + sin 9ωt + ..........)
3
5
7
9
2
2
2
2
2
2
(2.52)
2
1 1  1 1  1   1   1 
∴THD( I s (t )) =   +   +   +   +   +   +   = 46% or we can obtain
 3   5   7   9   11   13   15 
THD ( I s (t )) as the following:
From (2.52) we can obtain the value of is I S1 =


2
 IS 
I
 − 1 =  o
∴THD ( I s (t )) = 

4 Io
 I S1 

 2π
4 Io
2π
2


 −1 =



 2π

 4

2

 − 1 = 48.34%


Diode Circuits or Uncontrolled Rectifier
17
Example 5 solve Example 4 if the load is 30 A pure DC
Solution: From example 4 Vdc= 190.986 V, Vrms=212.132 V I dc = 30 A and I rms = 30 A
P
V I
V
(a) η = dc = dc dc = 90 % (b) FF = rms = 1.11
Pac Vrms I rms
Vdc
2
2
2
Vrms
− Vdc
Vac
Vrms
(c) RF =
=
=
− 1 = FF 2 − 1 = 0.482
2
Vdc
Vdc
Vdc
(d) The PIV=Vm=300V
4 Io
4 * 30
=
= 27.01A
(e) I S1 =
2π
2π
Input Power factor=
V I * cos φ
I * cos φ
27.01
Re al Power
= S S1
= S1
=
*1 = 0.9 Lag
Apperant Power
VS I S
IS
30
2.4.5 Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier.
Fig.2.15 Shows the single-phase diode bridge rectifier with source inductance. Due to the value
of LS the transitions of the AC side current iS from a value of I o to − I o (or vice versa) will not
be instantaneous. The finite time interval required for such a transition is called commutation
time. And this process is called current commutation process. Various voltage and current
waveforms of single-phase diode bridge rectifier with source inductance are shown in Fig.2.16.
Fig.2.15 Single-phase diode bridge rectifier with source inductance.
Fig.2.16 Various current & voltage waveforms for single-phase diode bridge rectifier with source inductance.
18
Chapter Two
Let us study the commutation time starts at t=10 ms as indicated in Fig.2.16. At this time the
supply voltage starts to be negative, so diodes D1 and D2 have to switch OFF and diodes D3 and
D4 have to switch ON as explained in the previous case without source inductance. But due to the
source inductance it will prevent that to happen instantaneously. So, it will take time ∆t to
completely turn OFF D1 and D2 and to make D3 and D4 carry the entire load current ( I o ). Also
in the time ∆t the supply current will change from I o to − I o which is very clear in Fig.2.16.
Fig.2.17 shows the equivalent circuit of the diode bridge at time ∆t .
Fig.2.17 The equivalent circuit of the diode bridge at commutation time ∆t .
 2ωLs I o 

u = cos −1 1 −
(2.56)
V
m


− 4ω LS I o
Vrd =
= −4 f LS I o
(2.61)
2π
The DC voltage with source inductance tacking into account can be calculated as following:
2V
Vdc actual = Vdc without sourceinduc tan ce − Vrd = m − 4 fLs I o
(2.62)
π
Is =
I S1 =
pf =
2 I o2
π u 
−
π  2 3 
(2.64)
8I o
u
* sin
2
2 πu
2 sin (u )
(2.68)
(2.69)
π u 
u π − 
 2 3
Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance
X s = 5 mH supply to feed 200 A pure DC load, find: (i) Average DC output voltage, (ii) Power
factor. And (iii) Determine the THD of the utility line current.
Solution: (i) From (2.62), Vm = 11000 * 2 = 15556V
2V
Vdc actual = Vdc without sourceinduc tan ce − Vrd = m − 4 fLs I o
π
19
Diode Circuits or Uncontrolled Rectifier
Vdc actual =
2 *15556
π
− 4 * 50 * 0.005 * 200 = 9703V
(ii) From (2.56) the commutation angle u can be obtained as following:
2ωLs I o 
2 * 2 * π * 50 * 0.005 * 200 
−1 
−1 
u = cos 1 −

Vm
 = cos 1 −


15556
 = 0.285 rad .

The input power factor can be obtained from (2.69) as following
I
2 * sin (u )
2 * sin (0.285)
u
=
= 0.917
pf = S1 * cos  =
IS
2
π u 
 π .285 
u π  −  0.285 π  −
3 
 2 3
2
IS =
2 I o2  π u 
−
=
π  2 3 
I S1 =
8I o
u
* sin =
2
2 πu
2 * 200 2  π 0.285 
 2 − 3  = 193.85 A
π
8 * 200
 0.285 
* sin 
 = 179.46 A
2 π * 0.285
 2 
2
2
 IS 
193.85 

 − 1 = 
THDi = 
 − 1 = 40.84%
179
.
46
I


 S1 
2.5 Three Phase Diode Rectifiers
2.5.1 Three-Phase Half Wave Rectifier
Fig.2.21 shows a half wave three-phase diode
rectifier circuit with delta star three-phase
transformer. In this circuit, the diode with highest
potential with respect to the neutral of the transformer
conducts. As the potential of another diode becomes
the highest, load current is transferred to that diode,
and the previously conduct diode is reverse biased
“OFF case”.
For the rectifier shown in Fig.2.21 the load voltage, primary diode currents and its FFT
components are shown in Fig.2.22, Fig.2.23 and Fig.2.24 respectively.
π
6
5π
6
Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier.
20
Chapter Two
Fig.2.23 Primary and diode currents.
Primary current
Diode current
Fig.2.24 FFT components of primary and diode currents.
5π
in the output voltage we can calculate the average
6
6
and rms output voltage and current as following:
By considering the interval from
Vdc
3
=
2π
Vrms =
5π / 6
∫ Vm sin ωt dωt =
π /6
5π / 6
3
2π
∫ (Vm sin ωt )
π /6
2
π
to
3 3 Vm 0.827 * Vm
3 3 Vm
=
= 0.827Vm and I dc =
R
2 *π * R
2π
dωt =
1 3* 3
+
Vm = 0.8407 Vm
2 8π
(2.70)
(2.72)
0.8407 Vm
(2.73)
R
Then the diode rms current is equal to secondery current and can be obtaiend as following:
08407 Vm
Vm
(2.74)
= 0.4854
Ir = IS =
R
R 3
I rms =
Diode Circuits or Uncontrolled Rectifier
21
Note :the rms value of diode current has been obtained from the rms value of load current divided
by 3 because the diode current has one third pulse of similar three pulses in load current.
(2.75)
ThePIV of the diodes is 2 VLL = 3 Vm
Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side and
the load resistance is R=20 Ω. If the source inductance is negligible, determine (a) Rectification
efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse
voltage (PIV) of each diode.
Solution:
460
= 265.58 V , Vm = 265.58 * 2 = 375.59 V
(a) VS =
3
3 3 Vm 0827 Vm
3 3 Vm
Vdc =
= 0.827 Vm , I dc =
=
2π R
R
2π
0.8407 Vm
P
V I
Vrms = 0.8407 Vm ,
Then, η = dc = dc dc = 96.767 %
I rms =
R
Pac Vrms I rms
V
(b) FF = rms = 101.657 %
Vdc
(c) RF =
2
2
2
Vrms
− Vdc
Vac
Vrms
=
=
− 1 = FF 2 − 1 = 18.28 %
2
Vdc
Vdc
Vdc
(e) The PIV= 3 Vm=650.54V
2.5.2 Three-Phase Half Wave Rectifier With
DC Load Current and zero source inductance
In case of pure DC load current as shown in
Fig.2.25, the diode current and primary current
are shown in Fig.2.26.
To calculate Fourier transform of the diode
current of Fig.2.26, it is better to move y axis to
make the function as odd or even to cancel one
coefficient an or bn respectively. If we put Y-axis
at point ωt = 30o then we can deal with the secondary current as even functions. Then, bn = 0 of
secondary current. Values of an can be calculated as following:
1
a0 =
2π
an =
1
π
π /3
∫ I o dωt =
−π / 3
π /3
Io
3
∫ I o * cos nωt dwt
−π / 3
=
Io
[sin nωt ]−ππ//33
nπ
Io
* 3 for n = 1,2,7,8,13,14,....
nπ
I
= − o * 3 for n = 4,5,10,11,16,17
nπ
= 0 for all treplean harmonics
=
(2.77)
22
Chapter Two
I s (t ) =
IO
3I O 
1
1
1
1
1

+
 sin ωt + sin 2ωt − sin 4ωt − sin 5ωt + sin 7ωt + sin 8ωt − −...
π 
3
2
4
5
7
8

(2.78)
2
2
2 *π 2
− 1 = 1.0924 = 109.24%
9
New axis
I 
THD( I s (t )) =  S 
 I S1 




 Io / 3 
−1 = 
 −1 =
 3I O 
 π 2 


Fig.2.26 Primary and secondary current waveforms and FFT components of three-phase half wave
rectifier with dc load current
23
Diode Circuits or Uncontrolled Rectifier
Example 8 Solve example 7 if the load current is 100 A pure DC
460
Solution: (a) VS =
= 265.58 V , Vm = 265.58 * 2 = 375.59 V
3
3 3 Vm
Vdc =
= 0.827 Vm = 310.613V , I dc = 100 A
2π
Vrms = 0.8407 Vm = 315.759 V , I rms = 100 A
P
V I
310.613 * 100
η = dc = dc dc =
= 98.37 %
Pac Vrms I rms 315.759 *100
V
(b) FF = rms = 101.657 %
Vdc
2
2
2
Vrms
− Vdc
Vac
Vrms
=
=
− 1 = FF 2 − 1 = 18.28 %
(c) RF =
2
Vdc
Vdc
Vdc
(d) The PIV= 3 Vm=650.54V
2.5 Three-Phase Full Wave Diode Rectifier
2.5.1 Three-Phase Full Wave Rectifier With Resistive Load
In the circuit of Fig.2.30, the AC side inductance LS is neglected and the load current is pure
resistance. Fig.2.31 shows complete waveforms for phase and line to line input voltages and
output DC load voltages. Fig.2.32 shows diode currents and Fig.2.33 shows the secondary and
primary currents and PIV of D1. Fig.2.34 shows Fourier Transform components of output DC
voltage, diode current secondary current and Primary current respectively.
IL
Ip
Is
3
1
5
VL
a
b
c
4
6
2
Fig.2.30 Three-phase full wave diode bridge rectifier.
For the rectifier shown in Fig.2.30 the waveforms is as shown in Fig.2.31. The average output
voltage is :Vdc =
3
π
I dc =
Vrms =
I rms
2π / 3
∫
3 Vm sin ωt dωt =
π /3
3 3 Vm
π
=
3 2 VLL
π
= 1.654Vm = 1.3505VLL
3 3 Vm 1.654Vm 3 2 VLL 1.3505VLL
=
=
=
π R
R
πR
R
3
π
2π / 3
∫ (
3 Vm sin ωt
π /3
1.6554 Vm
=
R
)2 dωt =
3 9* 3
+
Vm = 1.6554 Vm = 1.3516VLL
2
4π
(2.91)
(2.92)
(2.93)
(2.94)
24
Chapter Two
Then the diode rms current is
Ir =
1.6554 Vm
Vm
= 0.9667
R
R 3
I S = 0.9667
2
(2.95)
Vm
R
(2.96)
Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load
voltages.
Fig.2.32 Diode currents.
Diode Circuits or Uncontrolled Rectifier
25
Fig.2.33 Secondary and primary currents and PIV of D1.
Fig.2.34 Fourier Transform components of output DC voltage, diode current secondary current and
Primary current respectively of three-phase full wave diode bridge rectifier.
Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load
resistance is R=20 Ω. If the source inductance is negligible, determine (a) The efficiency, (b)
Form factor (c) Ripple factor (d) Peak inverse voltage (PIV) of each diode .
460
Solution: (a) VS =
= 265.58 V ,
Vm = 265.58 * 2 = 375.59 V
3
3 3 Vm
3 3 Vm 1.654Vm
Vdc =
= 1.654Vm = 621.226 V , I dc =
=
= 31.0613 A
R
π
π R
Vrms =
1.6554 Vm
3 9* 3
+
Vm = 1.6554 Vm = 621.752 V , I rms =
= 31.0876 A
2
4π
R
26
Chapter Two
η=
Pdc
V I
V
= dc dc = 99.83 % and (b) FF = rms = 100.08 %
Pac Vrms I rms
Vdc
(c) RF =
2
2
2
Vrms
− Vdc
Vac
Vrms
=
=
− 1 = FF 2 − 1 = 4 % and (d) The PIV= 3 Vm=650.54V
2
Vdc
Vdc
Vdc
2.5.2 Three-Phase Full Wave Rectifier With DC Load Current
The supply current in case of pure DC load current is shown in Fig.2.35. Fast Fourier
Transform of Secondary and primary currents respectively is shown in Fig2.36.
As we see it is odd function, then an=0, and
bn =
2
π
5π / 6
∫ I o * sin nωt dωt
π /6
=
2 Io
[− cos nωt ]π5π/ 6/ 6
nπ
2 Io
2 Io
2 Io
2 Io
( 3 ),......(2.97)
( 3 ), b13 =
(− 3 ), b11 =
(− 3 ), b7 =
π
13π
11π
7π
5π
bn = 0, for n = 2,3,4,6,8,9,10,12,14,15,.............
b1 =
2 Io
I s (t ) =
3 , b5 =
2 3I o 
1
1
1
1

 sin ωt − sin 5v ωt − sin 7ωt + sin 11ωt + sin 13ωt  (2.98)
π 
13
11
7
5

2
2
2
2
2
2
2
1 1  1   1   1   1   1   1 
THD ( I s (t )) =   +   +   +   +   +   +   +  
 5   7   11   13   17   19   23   25 
2
= 31%
Also THD ( I s (t )) can be obtained as following:
2
2* 3
IS =
I o , I S1 =
I o then, THD ( I s (t )) =
3
π
Power Factor =
2
 IS 

 − 1 =
I
 S1 
2/3
2*3/π 2
− 1 = 31.01%
I S1
I
* cos(0) = S1
IS
IS
Fig.2.35 The D1 and D2 currents, secondary and primary currents.
Diode Circuits or Uncontrolled Rectifier
27
Fig2.36 Fast Fourier Transform of Secondary and primary currents respectively.
2.5.4 Three-Phase Full Wave Diode Rectifier With Source Inductance
The source inductance in three-phase diode bridge rectifier Fig.2.37 will change the shape of
the output voltage than the ideal case (without source inductance) as shown in Fig.2.31. The DC
component of the output voltage is reduced. Fig.2.38 shows The output DC voltage of threephase full wave rectifier with source inductance.
Fig.2.37 Three-phase full wave rectifier with source inductance
Fig.2.38 The output DC voltage of three-phase full wave rectifier with source inductance
28
Chapter Two

2ω LS I o 
Commutation angle is , u = cos −1 1 −

VLL 

(2.109)
6ω LI o
= 6 fLI o
(2.114)
2π
The DC voltage without source inductance tacking into account can be calculated as following:
(2.115)
Vdc actual = Vdc without sourceinduc tan ce − Vrd = 1.35VLL − 6 fLI d
The DC voltage reduction due to source inductance is : Vrd =
2 6 Io  u 
2 I o2  π u 
I
and
=
sin  
−
S
1
π  3 6 
πu
2
The power factor can be calculated from the following equation:
IS =
pf =
I S1
u
cos   =
IS
2
2 6 Io  u 
sin  
πu
2
3 * sin (u )
u
cos   =
2
2I π u 
π u 
u π − 
− 

π  3 6
 3 6
2
o
Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has
8 mH source inductance to feed 300A pure DC load current Find;
(i)
Commutation time and commutation angle.
(ii)
DC output voltage.
(iii) Power factor.
(iv) Total harmonic distortion of line current.
Solution: (i) By substituting for ω = 2 * π * 50 , I d = 300 A , L = 0.008 H , VLL = 33000V in
(2.109), then u = 0.2549 rad . = 14.61 o
(ii) The the actual DC voltage can be obtained from (2.115) as following:
Vdc actual = Vdc without sourceinduc tan ce − Vrd = 1.35VLL − 6 fLI d
Vdcactual = 1.35 * 33000 − 6 * 50 * .008 * 300 = 43830V
(iii) the power factor can be obtained from (2.121) then
3 * sin (u )
3 sin (0.2549)
pf =
=
= 0.9644 Lagging
π
u
π
0
.
2549




u π  −  0.2549 * π  −
6 
 3 6
3
(iv) The rms value of supply current can be obtained as following:
2 I d2  π u 
2 * 300 2  π 0.2549 
Is =
* −
=
−
 = 239.929 A
6 
π  3 6 
π
3
The rms value of fundamental component of supply current can be obtained from (2.120) as
4 3 Io
4 3 * 300
 0.2549 
u
sin   * 2 3 =
* sin 
following: I S1 =
 = 233.28 A
πu 2
π * 0.2549 * 2
 2 
2
pf =
I S1
 u  233.28
 0.2549 
* cos  =
* cos
 = 0.9644 Lagging.
Is
 2  239.929
 2 
2
2
 IS 
239.929 

 − 1 = 
THDi = 
 − 1 = 24.05%
I
233
.
28


1
S

