Chapter 2
Diode Circuits or Uncontrolled Rectifier
2.1 IntroductionBecause 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, Idc, The rms value of the output voltage, Vrms,
The rms value of the output current, Irms
The output DC power, Pdc =Vdc*Idc (2.1) The output AC power, Pac =Vrms *Irms (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 2 dc rms ac V V V = − (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.
dc rms V V
FF = / (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.
1 1 2 2 2 2 2 − = − = − = = FF V V V V V V V RF dc rms dc dc rms dc ac (2.6)
Where VSand IS 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.
1 2 1 2 2 1 2 1 2 − = − = S S S S S i I I I I I THD , and 2 1 1 2 2 1 2 1 2 − = − = S S S S S v V V V V V THD (2.8)
where IS1 and VS1 are the fundamental component of the input currentand voltage, IS and VS
respectively.
In general, power factor in non-sinusoidal circuits can be obtained as following:
φ cos s Voltampere Apparent Power R = = = S S I V P eal PF (2.10)
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. Faactor nt Displaceme Factor Distortion I I I V I V I V P PF S S S S S S S S * cos cos 1 1 1 1 = = = = φ φ (2.11)
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, single-phase half-wave diode rectifier has little practical significance.
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 Fig.2.2 as following: =
∫
= π π ω ω π 0 sin 2 1 m m dc V t d t V V (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:
R V R V I dc m dc = =π
The root mean square (rms) value of a load voltage is defined as: 2 sin 2 1 0 2 2 m m rms V t d t V V =
∫
= π ω ω π (2.14)Similarly, the root mean square (rms) value of a load current is defined as:
R V R V Irms rms m 2 = =
It is clear that the rms value of the transformer secondary current,
I
S is the same as that of the load and diode currentsThen R V I IS D m 2 = = (2.15)
Where, ID is the rms value of diode current.
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
V
dc is defiend as:π π π ω ω π π m m m dc V t d t V V V =
∫
= (−cos −cos(0))= 2 ) sin( 2 1 0 Then, R V R V I dc m dc = =π2 ) sin ( 2 1 0 2 m m rms V t V V =
∫
= π ω π , R V I m rms = 2 and, 2 m S V V =The rms value of the transformer secondery current is the same as that of the load: IS =Vm.2R
Then, the efficiency or rectification ratio is:
rms rms dc dc ac dc I V I V P P * * = = η 40.53% 2 * 2 * = = R V V R V V m m m m π π (b) 1.57 2 2 = = = = π πm m dc rms V V V V FF (c) = = FF2−1= 1.572 −1=1.211 V V RF dc ac
(d) It is clear from Fig2.2 that the PIV is Vm.
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 =Vmsin
( )
ωt ,v
s is positive when 0 <ω
t < π, andv
s is negative when π <ω
t <2π. Whenv
s starts becoming positive, the diode starts conducting and the source keeps the diode in conduction tillω
t reaches π radians. At that instant defined byω
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.2.4 Various waveforms for Half wave diode rectifier with R-L load. Assume Z∠φ =R+ jwL Then Z2 = R2 +w2L2, φ cos Z R= , ωL=Zsinφ and R L ω φ = tan
(
)
( )
+ − = − φ ω φ φ ωω ) sin sin tan
( t m t e Z V t i (2.24)
Starting from ωt = π, as ωt increases, the current would keep decreasing. For some value of
t
ω , 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 whenω
t
reaches β. The value of β can be obtained by substituting that i(
ω
t) = 0wt=β into (2.24) we get:(
)
sin( )
0 sin ) ( tan = + − = − φ β φ φ β β e Z V i m (2.25)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).
) cos 1 ( * 2 sin * 2π 0 ω ω π β β − = = m
∫
m dc V t d t V V (2.26) ) 2 sin( 1 ( 5 . 0 * 2 ) sin ( * 2 1 0 2 β β π ω π β − + = =∫
V t dwt Vm Vrms m (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:
R
wL Z
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.2.9 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:
π ω ω π π m m dc V t d t V V 1 sin 2 0 = =
∫
and R V Idc m π 2 = (2.36)(
)
2 sin 1 0 2 m m rms V t d t V V =∫
= π ω ω π and R V Irms m 2 = (2.38)PIV of each diode = 2Vm (2.40)
2
m
S V
V = (2.41)
The rms value of the transformer secondery current is the same as that of the diode:
R V I IS D m 2 = = (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 % 05 . 81 2 * 2 2 * 2 * * = = = = R V V R V V I V I V P P m m m m rms rms dc dc ac dc π π η (b) 1.11 2 2 2 2 = = = = π πm m dc rms V V V V FF (c) = = FF2−1= 1.112 −1=0.483 V V RF dc ac (d) The PIV is 2Vm
2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load
Another alternative in single-phase full wave rectifier is by using four diodes as shown in Fig.2.12 which known as a single-phase full bridge diode rectifier. It is easy to see the operation of these four diodes. The current flows through diodes D1 and D2 during the positive half cycle of input voltage (D3 and D4 are “OFF”). During
the negative one, diodes D3 and D4 conduct (D1 and D2 are “OFF”).
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 Fig.2.12 Single-phase full bridge diode 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 =300V V V t d t V V m m dc 1 sin 2 190.956 0 = = =
∫
π ω ω π π , A R V I m dc =2π =12.7324(
V t)
d t V V V m m rms 212.132 2 sin 1 1/2 0 2 = = =∫
π ω ω π , R A V I m rms 14.142 2 = = (a) = = =81.06 % rms rms dc dc ac dc I V I V P P η (b) = =1.11 dc rms V V FF (c) 2 1 2 1 0.482 2 2 2 = − = − = − = = FF V V V V V V V RF dc rms dc dc rms dc ac (d) The PIV=300V(e) Input power factor = Re = cos =1
S S S S I V I V Power Apperant Power al φ
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 IS =Io
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
[
]
[
cos 0 cos]
4 1,3,5, ... 2 cos 2 sin * 2 0 0 = = − = − = =∫
n for n I n n I t n n I t d t n I b o o o o n π π π ω π ω ω π π π (2.51)Then from Fourier series concepts we can say:
) ... 9 sin 9 1 7 sin 7 1 5 sin 5 1 3 sin 3 1 (sin * 4 ) (t = I t+ t+ t+ t+ t+ i o ω ω ω ω ω π (2.52) % 46 15 1 13 1 11 1 9 1 7 1 5 1 3 1 )) ( ( 2 2 2 2 2 2 2 = + + + + + + = ∴THD Is t or we can obtain )) ( (I t THD s as the following:
From (2.52) we can obtain the value of is
π 2 4 1 o S I I = % 34 . 48 1 4 2 1 2 4 1 )) ( ( 2 2 2 1 = − = − = − = ∴ π π o o S S s I I I I t I THD
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 Idc =30A and Irms= 30 A
(a) = = =90 % rms rms dc dc ac dc I V I V P P η (b) = =1.11 dc rms V V FF (c) 22 1 2 1 0.482 2 2 = − = − = − = = FF V V V V V V V RF dc rms dc dc rms dc ac (d) The PIV=Vm=300V (e) IS Io 27.01A 2 30 * 4 2 4 1 = π = π =
Input Power factor= =
Power Apperant Power al Re Lag I I I V I V S S S S S S *1 0.9 30 01 . 27 cos * cos * 1 1 φ = φ = =
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 Io to −Io (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.
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 (Io). Also in the time ∆t the supply current will change from Io to −Io 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. − = − m o s V I L u cos 1 1 2ω (2.56) o S o S rd L I f L I V 4 2 4 =− − = π ω (2.61) The DC voltage with source inductance tacking into account can be calculated as following:
o s m rd ce induc source without dc actual dc V V V fL I V = tan − = 2 −4 π (2.62) − = 3 2 2I2 u I o s π π (2.64) 2 sin * 2 8 1 u u I I o S = π (2.68)
( )
− = 3 2 sin 2 u u u pf π π (2.69)Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance
mH
Xs =5 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 o s m rd ce induc source without dc actual dc V V V fL I V = tan − = 2 −4 π
V
Vdc actual = 2*15556−4*50*0.005*200=9703
π
(ii) From (2.56) the commutation angle u can be obtained as following: . 285 . 0 15556 200 * 005 . 0 * 50 * * 2 * 2 1 cos 2 1 cos 1 1 rad V I L u m o s = − = − = − ω − π
The input power factor can be obtained from (2.69) as following
( )
(
)
917 . 0 3 285 . 2 285 . 0 285 . 0 sin * 2 3 2 sin * 2 2 cos * 1 = − = − = = π π π π u u u u I I pf S S A u I IS o 193.85 3 285 . 0 2 200 * 2 3 2 2 2 2 = − = − = π π π π A u u I IS o 179.46 2 285 . 0 sin * 285 . 0 * 2 200 * 8 2 sin * 2 8 1 = = = π π % 84 . 40 1 46 . 179 85 . 193 1 2 2 1 = − = − = S S i II THD2.5 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.
Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier. 6
π
6 5π
Fig.2.23 Primary and diode currents.
Fig.2.24 FFT components of primary and diode currents. By considering the interval from
6
π
to 6 5π
in the output voltage we can calculate the average and rms output voltage and current as following:
m m m dc V t d t V V V 0.827 2 3 3 sin 2 3 5 /6 6 / = = =
∫
π ω ω π π π and R V R V Idc m 0.827* m * * 2 3 3 = = π (2.70)(
m)
m m rms V t d t V V V 0.8407 8 3 * 3 2 1 sin 2 3 5 /6 6 / 2 = + = =∫
π ω ω π π π (2.72) R V Irms = 0.8407 m (2.73)Then the diode rms current is equal to secondery current and can be obtaiend as following:
R V R V I Ir S m 0.4854 m 3 08407 = = = (2.74) Primary current Diode current
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.
ThePIV of the diodes is 2VLL = 3Vm (2.75)
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: (a) VS 265.58 V, Vm 265.58* 2 375.59 V 3 460 = = = = m m dc V V V 0.827 2 3 3 = = π , R V R V Idc m 0827 m 2 3 3 = = π m rms V V =0.8407 , R V Irms = 0.8407 m Then, = = =96.767 % rms rms dc dc ac dc I V I V P P η (b) = =101.657% dc rms V V FF (c) 22 1 2 1 18.28 % 2 2 = − = − = − = = FF V V V V V V V RF dc rms dc dc rms dc ac
(e) The PIV= 3Vm=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:
3 2 1 /3 3 / 0 o o I t d I a
∫
− = = π π ω π[
]
harmonics treplean all for n for n I n for n I t n n I dwt t n I a o o o o n 0 17 , 16 , 11 , 10 , 5 , 4 3 * ,.... 14 , 13 , 8 , 7 , 2 , 1 3 * sin cos * 1 /3 3 / 3 / 3 / = = − = = = = = − −∫
π π ω π ω π ππ π π (2.77) + − − + + −− + = sin8 ... 8 1 7 sin 7 1 5 sin 5 1 4 sin 4 1 2 sin 2 1 sin 3 3 ) (t I I t t t t t t I O O s π ω ω ω ω ω ω (2.78) % 24 . 109 0924 . 1 1 9 * 2 1 2 3 3 / 1 )) ( ( 2 2 2 1 = = − = − = − = π π O o S S s t II I I I THD
Fig.2.26 Primary and secondary current waveforms and FFT components of three-phase half wave rectifier with dc load current
Example 8 Solve example 7 if the load current is 100 A pure DC Solution: (a) VS 265.58 V, Vm 265.58* 2 375.59 V 3 460 = = = = V V V Vdc m 0.827 m 310.613 2 3 3 = = = π , Idc =100A V V Vrms =0.8407 m =315.759 , Irms =100A % 37 . 98 100 * 759 . 315 100 * 613 . 310 = = = = rms rms dc dc ac dc I V I V P P η (b) = =101.657% dc rms V V FF (c) 2 1 2 1 18.28 % 2 2 2 = − = − = − = = FF V V V V V V V RF dc rms dc dc rms dc ac (d) The PIV= 3Vm=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.
1 3 5 4 6 2 b c IL VL Is Ip a
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 :- LL m LL m m dc V t d t V V V V V 3 3 sin 3 3 3 2 1.654 1.3505 3 / 2 3 / = = = = =
∫
π π ω ω π π π (2.91) R V R V R V R V Idc =3 3 m =1.654 m = 3 2 LL =1.3505 LL π π (2.92)(
m)
m m LL rms V t d t V V V V 1.6554 1.3516 4 3 * 9 2 3 sin 3 32 /3 3 / 2 = = + = =∫
π ω ω π π π (2.93) R V Irms =1.6554 m (2.94)Then the diode rms current is R V R V Ir m 0.9667 m 3 6554 . 1 = = (2.95) R V IS =0.9667 2 m (2.96)
Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load voltages.
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 .
Solution: (a) VS 265.58 V, Vm 265.58* 2 375.59 V 3 460 = = = = V V V Vdc =3 3 m =1.654 m =621.226 π , R A V R V Idc =3 3 m =1.654 m = 31.0613 π V V V Vrms m 1.6554 m 621.752 4 3 * 9 2 3+ = = = π , R A V Irms =1.6554 m =31.0876
% 83 . 99 = = = rms rms dc dc ac dc I V I V P P η and (b) = =100.08% dc rms V V FF (c) 22 1 2 1 4 % 2 2 = − = − = − = = FF V V V V V V V RF dc rms dc dc rms dc
ac and (d) The PIV= 3V
m=650.54V
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
[
]
.... ,... 15 , 14 , 12 , 10 , 9 , 8 , 6 , 4 , 3 , 2 , 0 ),... 3 ( 13 2 ), 3 ( 11 2 ), 3 ( 7 2 ), 3 ( 5 2 , 3 2 cos 2 sin * 2 13 11 7 5 1 6 / 5 6 / 6 / 5 6 / = = = = − = − = = − = =∫
n for b I b I b I b I b I b t n n I t d t n I b n o o o o o o o n π π π π π ω π ω ω π π π π π (2.97) − − + + = I t v t t t t t Is o ω ω ω ω ω π 13sin13 1 11 sin 11 1 7 sin 7 1 5 sin 5 1 sin 3 2 ) ( (2.98) % 31 25 1 23 1 19 1 17 1 13 1 11 1 7 1 5 1 )) ( ( 2 2 2 2 2 2 2 2 = + + + + + + + = t I THD sAlso THD(Is(t)) can be obtained as following:
o S I I 3 2 = , IS Io π 3 * 2 1 = then, 1 31.01% / 3 * 2 3 / 2 1 )) ( ( 2 2 1 = − = − = π S S s t II I THD Power Factor = S S S S I I I I 1 1 ) 0 cos( * =
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 three-phase full wave rectifier with source inductance.
Fig.2.37 Three-phase full wave rectifier with source inductance
Commutation angle is , − = − LL o S V I L u cos 1 1 2ω (2.109)
The DC voltage reduction due to source inductance is :Vrd LIo 6fLIo
2 6 = = π ω (2.114) The DC voltage without source inductance tacking into account can be calculated as following:
d LL rd ce induc source without dc actual dc V V V fLI V = tan − =1.35 −6 (2.115) − = 6 3 2I2 u IS o π π and = 2 sin 6 2 1 u u I IS o π
The power factor can be calculated from the following equation:
( )
− = − = = 6 3 sin * 3 2 cos 6 3 2 2 sin 6 2 2 cos 2 1 u u u u u I u u I u I I pf o o S S π π π π πExample 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, Id =300A, L=0.008H , 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:
d LL rd ce induc source without dc actual dc V V V fLI V = tan − =1.35 −6 V Vdcactual=1.35*33000−6*50*.008*300=43830 (iii) the power factor can be obtained from (2.121) then
( )
(
)
0.9644 6 2549 . 0 3 * 2549 . 0 2549 . 0 sin 3 6 3 sin * 3 = − = − = π π π π u u u pf Lagging(iv) The rms value of supply current can be obtained as following:
A u I Is d 239.929 6 2549 . 0 3 * 300 * 2 6 3 2 2 2 = − = − = π π π π
The rms value of fundamental component of supply current can be obtained from (2.120) as
following: u A u I IS o 233.28 2 2549 . 0 sin * 2 * 2549 . 0 * 300 * 3 4 3 2 * 2 sin 2 3 4 1 = = = π π 9644 . 0 2 2549 . 0 cos * 929 . 239 28 . 233 2 cos * 1 = = = u I I pf s S Lagging. % 05 . 24 1 28 . 233 929 . 239 1 2 2 1 = − = − = S S i I I THD
