Power electronics digital notes b. Tech III year

Figure: 5.39 Circuit diagram of load commutated CSI
The power switching devices used here is the same, i.e. four Thyristors only in a full- bridge 
configuration. The positive direction for load current and voltage is shown in Fig. 5.40 Before t = 0, the 
capacitor voltage is , i.e. the capacitor has left plate negative and right plate positive. At that time, the 
thyristor pair, Th2 & Th4 was conducting. When (at t = 0), the thyristor pair, Th1 & Th3 is triggered by 
the pulses fed at the gates, the conducting thyristor pair, Th2 & Th4 is reverse biased by the capacitor 
voltage C = −Vv 1 , and turns off immediately. The current path is through Th1, load (parallel 
combination of R & C), Th3, and the source. The current in the thyristors is I
Ti
, the output current is 
Iac= I 

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Figure: 5.40 Voltage and current waveforms of load commutated CSI
Numerical Problems 
1.
A single-phase half bridge inverter has a resis load of 2.4 W and the d.c. input voltage of 48 V. 
Determine:- 
(i) RMS output voltage at the fundamental frequency 
(ii) Output power 
P
0 
(iii) Average and peak currents of each transistor 
(iv) Peak blocking voltage of each transistor. 
(v) Total harmonic distortion and distortion factor. 
(vi) Harmonic factor and distortion factor at the lowest order harmonic. 
Solution: 
(i) RMS output voltage of fundamental frequency, 
E
1 = 0.9 ¥ 48 = 43.2 V. 
(ii) RMS output voltage, Eorms = 
E 
= 48 V. 
Output power = 
E 
2
/
R 
= (48)
2
/2.4 = 960 W. 

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(iii) Peak transistor current = 
Ip 
= 
Ed
/
R 
= 48/2.4 = 20 A. 
Average transistor current = 
Ip
/2 = 10 A. 
(iv) Peak reverse blocking voltage, 
VBR 
= 48 V.