Three phase semi converter

it is the brief introduction about three phase semi converter. it will helpfull for the students of power electronics subject.

1. Three phase semi converter Power electronics - I PRESENTED BY:- Rami nirav (161310109048) Arpit raval (161310109050) Sem.:-5 Electrical engineering

2. Fig.:- three phase semi converter with RL load

3. the circuit of three phase semi converter consists three SCRs, three diodes and load of RL type. The firing angle should be measured from ω t = 60°, scr1 is fired at ω t = 60 °+0°= 60 ° and for α = 60 °, scr1 is fired at ω t = 60 °+60 °= 120°. Before turning off the scr next scr will not be fired so scr id fired after turning off the previous scr. For example for α =60° if scr is fired it start conducting from ω t=120° and it conduct from ω t=120°to ω t=240° Now next scr is fired at ω t=240° and at same instant previous scr gets turned off. scr will be conducting from ω t=240°to ω t=360°. Next scr is fired at ω t=360°and at the same instant the previous scr will get turned off.

4. Scr3 is already conducting through D 2 . scr1 conduct for 120° and it fired at ω t=60° The output voltage v ab and v ac are obtained when scr1 is conducting. Similarly v bc and v ba are obtained when scr2 is conducting. Again scr2 conducts for 120°and it is fired at ω t=180°. At ω t=300°, when scr3 is fired, voltage v ca and v cb are obtained. Scr3 conduct for 120° from ω t=300°to ω t=420° the above cycle repeat with scr1.

5. The output voltage and current waveforms for α =60° scr1 is fired at ω t=120° and scr3 is already conducting through d 2 .scr1 and d 3 conduct simultaneously for 120° and voltahe v ac is obtained. Scr2 is fired at ω t=240° and it conducts through diode d 1 . In this case voltage v ba is obtained. Similarly the scr3 conducts and v cb is obtained. here it should be noted that freewheeling diode does not come into play for α≤ 60° and voltage v ab , v bc , v ca do not appear in the output voltage waveform for α =60°.

6. The phase voltage are ; v an = v m sin ω t v bn = v m sin( ω t-2∏/3) v cn = v m sin( ω t+2 ∏/3 ) And corresponding line to line voltage are v ac = v an – v cn = √3 v m sin( ω t-∏/6) v cb = v cn – v bn = √3v m sin( ω t+∏ /6 ) v ab = v an – v bn = √3v m sin( ω t+∏ /6)

7. For α≤ 60° ( continues output voltage);the average voltage is given by

8. Thank you