The inductive load means resistance and inductance in the load. Such loads are DC motors. Because of the inductive (R-L) load, the load current shape is changed. Hence operation of the full bridge converter can be discussed into three modes :
i) Continuous load current
ii) Continuous and ripple free current for large inductive load iii) Discontinuous load current
3.4.2.1 Continuous Load Current
In the continuous load current, the load or output current i0 flows continuously. The waveforms are shown in Fig. 3.4.7.
Fig. 3.4.7 W aveform s o f 1(}> fu ll converter fo r inductive load having continuous load current
Power Devices and Machines 3 - 5 4 Phase Controlled Rectifiers (AC/DC Converters) current keeps on reducing.
fro m Ti to ti + cx due to inductance voltage At n+ a, SCRs T3 and T4 are triggered.
The load current starts increasing. The load current remains continuous in the load. The similar operation repeats. The ripple in the load current reduces as the load inductance is increased.
3.4.2.2 Continuous and Ripple Free Current for Large Inductive Load
s,
Answ er follow in g question after reading this topic
1. Draw the circuit diagram o f a single phase fully controlled bridge rectifier and sketch the waveforms o f output voltage, output current, supply current and SCR current for a level (ripple free) load. Marks [5], M ay-2000. 2 0 0 1 ; Marks [10]. D ec.-2004
Power Devices and Machines 3 - 5 5 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.4.9 W aveform s o f 1<J> fu ll converter fo r continuous and ripplefree load cu rre n t in case o f large inductive load
))*► Example 3.4.3 : For the 1<)> full converter having inductive load and continuous load current, obtain the following :
i) Average output voltage V0^av^
ii) RMS output voltage V0^rms^ [Dec.-2004, 3 Marks]
S olution : i) Average o utp ut voltage fo r inductive load The average output voltage is given as,
1 T
Vo(av) = f
J
vo(“0
0
Observe the waveforms of l<f> full converter for inductive load given in Fig. 3.4.7 and Fig. 3.4.9. The output voltage waveform has a period from a to 7t+a ; i.e. n. And vQ (cot) = Vm sin (ot during this period. Hence above equation becomes,
j 7i+a
Vo (a v ) = -
J
Vm sin d(s>ta
r .nTi+a
= — - COS ( 01
K 1 J«
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Power Devices & Machines 3 - 56 Phase Controlled Rectifiers (AC/DC Converters) 2 V„
o(av ) cos a ... (3.4.5)
This is the expression for average load voltage of l<f> full converter for inductive load.
Plot of V0(av) versus firin g angle (a)
Following table lists the values of VQ/av\ with firing angle (a)
a Vo{*v) = Km c o s a
0 2V
- f = 0 637 Vm
30° 0.55 Vm
60° 0.318 Vm
90° 0
120° - 0.318 Vm
150° - 0-55 Vm
180° - 0-637 Vm
Table 3.4.1 VC(av,) w ith respect to a
Observe that VG (av) is positive for a < 9&. Hence it is called rectification. For a > 0, V0 (av) is negative. Hence it is called inverting mode of operation. In inverting mode, output energy is fedback to the source.
Fig. 3.4.10 V ariation o f VD ^ w ith respect to a
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Power Devices & Machines 3 - 5 7 Phase Controlled Rectifiers (AC/DC Converters)
ii) RMS value o f o utp ut voltage fo r inductive load The rms value is given as,
J \ vo (<°0 d(ot
2
0
2
- | V 2 sin2 cof d cof a
1
a
... (3.4.6)
Thus the rms value of load voltage is same as rms value of the AC supply voltage.
»»► Example 3.4.4 : Draw the circuit arrangement o f a single phase full converter feeding a general load comprising o f R, L and E. Sketch the AC supply voltage o/p voltage and the load current waveforms. Assuming continuous load current operation, derive an expression for DC output voltage.
A single phase full converter feeding an RLE load is fed by 230 V, 50 Hz mains.
If R = 0.5 Q, L = 8 rnH and E = 50 volts, assuming that conduction is continuous and firing angle is 4 0 find average value o f load current.
S olution : C ircuit diagram and waveform s
Fig. 3.4.11 shows the circuit diagram of full converter supplying RLE load.
vs = 230 V. 50 Hz
E = 50 R = 0.5 Q
L = 8 mH
Fig. 3.4.11 14> fu ll converter feeding RLE load
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Power Devices and Machines 3 - 58 Phase Controlled Rectifiers (AC/DC Converters) The RLE load is normally motor load. 'R' is the resistance and 'L' is an inductance of armature winding of the motor. 'E' is the induced emf of the motor. When the load current is continuous, then waveforms of this circuit will be similar to that of RL load. Hence with small ripple in output current, the waveforms of this circuit will be similar to those shown in Fig. 3.4.7. Note that 'E' is not reflected in the waveforms as long as output current (i0) is continuous.
If output current (iQ) is constant and ripple free, then the waveforms will be similar to those shown in Fig. 3.4.9.
RMS and average o utp ut voltage
The output voltage waveform remains same with RL load and RLE load when i0 is continuous. Therefore the rms and average values of output voltage will be same as those derived in previous example for RL load, i.e.,
2V Vo(av) = - f - cos a V , v o(rms) v = -2L
v
^ 2 = VsSecond part : To obtain average load current
The ripple in the load current (i0) depends upon values of R, L and E. If load inductance is small, then iG can become discontinuous. In Fig. 3.4.7, observe that iQ repeats at the intervals of ;r . The waveform of i0 remains same whenever Tj-Tj or T3-T4 conducts. Hence in any interval (i.e. a < cof < a or rc+a < cof < 2n + a) the equivalent circuit will be as shown below.
Fig. 3.4.12 Equivalent c irc u it when T yT 2 o r r 3-T4 conduct By applying KVL to above circuit,
Vm sin cof = Ri0 + L + E
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Power Devices and Machines 3 - 6 0 Phase Controlled Rectifiers (AC/DC Converters)
full converter, derive an expression for average load voltage.
S olution : We know that freewheeling action does not take place in 1<J> full converter
Power Devices and Machines 3 - 6 1 Phase Controlled Rectifiers (AC/DC Converters) diode is more forward biased compared to T| and T2. Hence freewheeling diode conducts.
The freewheeling diode is connected across the output V0. Hence Vo =0 during freewheeling. The energy stored in the load inductance is circulated back in the load itself.
Fig. 3.4.14 shows the waveforms of this operation. The output voltage becomes zero in the freewheeling periods. Compare the load voltage waveform of Fig. 3.4.13 with that of l<j>full converter with resistive load (Fig. 3.4.3). They are same. Hence the average load voltage can be obtained from equation 3.4.3. i.e.,
Vo(av) = ^ - ( 1 + coso) ...(3.4.10)
Fig. 3.4.13 Freewheeling diode conducts from k to rc+a due to inductive load
Fig. 3.4.14 W aveform s o f 14> fu ll converter fo r highly inductive load and freew heeling diode across the load
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Power Devices and Machines 3 - 6 2 Phase Controlled Rectifiers (AC/DC Converters)
i) Fourier series for supply current
ii) RMS value o f nth harmonic o f supply current iii) Fundamental component o f supply current iv) RMS value o f supply current
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Power Devices and Machines 3 - 63 Phase Controlled Rectifiers (AC/DC Converters)
Power Devices and Machines 3 - 6 4 Phase Controlled Rectifiers (AC/DC Converters)
Therefore Fourier series is,
4 I
Power Devices and Machines 3 - 65 Phase Controlled Rectifiers (AC/DC Converters) ii) Supply power factor (PF) iii) Harmonic factor (HF)
iv) Current distortion factor (CDF) S olution : i) Displacem ent factor
Power Devices and Machines 3 - 6 6 Phase Controlled Rectifiers (AC/DC Converters)
From result of previous example and equation 3.4.18, 2V2 /o(av)
Power Devices and Machines 3 - 6 7 Phase Controlled Rectifiers (AC/DC Converters)
• Thus half controlled bridge draws 50 % reactive power compared to that of full controlled bridge.
))*► Example 3.4.10 : A single phase full converter is operated front a 120 V, 60 Hz supply.
The load current with an average valve of la is continuous, with negligible ripple current. If the turns ratio of the transformer is unity, if the delay angle is a = Calculate the
i) HF o f input current ii) DF
iii) PF
S olution : Given data
Supply voltage, Vs = 120 Delay angle, a =
i) Harm onic fa c to r (HF)
For continuous load current, the harmonic factor is fixed for full converter. And it is given by equation 3.4.21 as,
Power Devices and Machines 3 - 68 Phase Controlled Rectifiers (AC/DC Converters)
HF = 0.4834 or 48.34 % ii) Displacem ent fa c to r (DF)
For 1 <j> full converter, DF is given by equation 3.4.19 as,
iii) Power facto r (PF)
For 1 § full converter, PF is given by equation 3.4.20 as, nc 2V2
PF = ---cos a
71
= 0.45
This is lagging PF, since current lags the voltage.
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Power Devices and Machines 3 - 69 Phase Controlled Rectifiers (AC/DC Converters)
Example 3.4.11 : A single phase full converter operates with 220 V, 50 Hz ac input and supplies output load consisting of R-L load with very high inductance drawing level load current 10 A and operated at firing angle of 30°. Find
-i) RMS supply current, iii) Input displacement factor, v) Power factor
S olution : Given : 1 <|>FCB
ii) Fundamental component o f input current, iv) Harmonic factor
vi) Output voltage. [M ay-2000,10 M arks]
v: 220 V Vm = 220^/2 = 311.12 V
Power Devices and Machines 3 - 70 Phase Controlled Rectifiers (AC/DC Converters)
vi) O utput voltage
2Vm 2x311.12 n
Vo(av) = ~ ~ ~ ~ cos^ = 171.53 V.
Example 3.4.12 : A single-phase fully controlled bridge converter supplies an inductive load. Assuming that the output current is virtually constant and is equal to ld, determine the following performance measures, if the supply voltage is 230 V and if the firing angle is maintained at (n/6) radians.
i) Average output voltage
ii) Fundamental power factor or displacement factor (DF) iii) Supply power factor (PF)
Power Devices and Machines 3 - 7 1 Phase Controlled Rectifiers (AC/DC Converters)
Fig. 3.4.15 S olution : In the positive half cycle and short circuit of T2, the situation will be as shown in Fig. 3.4.15 (a)
• Here observe that Tj is forward biased but it will start conducting when it is triggered.
• T4 and diode D3 are reverse biased in
Fig. 3.4.15 (a) C ircuit diagram positive half cycle.
Fig. 3.4.16 shows the situation in positive and negative half cycles. In positive half cycle the controlled supply will be applied to load. But in negative half cycle, supply is shorted through T4.
Fig. 3.4.16 W aveform o f c irc u it o f Fig. 3.4.15(a)
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Power Devices and Machines 3 - 7 2 Phase Controlled Rectifiers (AC/DC Converters)
))»► Example 3.4.14 : Draw voltage and current waveform for circuit shown in Fig. 3.4.17 [May-2007, 4 Marks]
Resistive load for a = 30°
Fig. 3.4.17
Solution : Fig. 3.4.18 shows the voltage and current waveforms.
Fig. 3.4.18 Output voltage and current waveforms
• The output voltage for a = 30° is shown in Fig. 3.4.18(b).
• For resistive load, the shape of output voltage and that of output current are same. And,
V.
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Power Devices and Machines 3 - 73 Phase Controlled Rectifiers (AC/DC Converters)