unipolar voltage switching

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[Moved]unipolar voltage switching

dear friends can any one help me? i have designed single phase full bridge inverter with pic18f458 by using ECCP module. there is no need of dead time in this technique. for freeweeling of current antiparallel diode is used, but in conventional unipolar voltage switching antiparalell diode is not used for freeweeling and proper dead time is necessary because in each leg both switches are invert of each other. why we use conventional unipolar voltage switching (spwm) with op amps or with up down counter of dsPIC?

why antiparallel diod is not used for freeweeling in UP DOWN counter mode while dsPIC for SPWM?

is ECCP modules unipolar is sperior because no need of dead time?

thanks in advance
 

I can't agree to your assumption that unipolar pwm scheme doesn't need deadtime. It also involves synchronous commutation between high- and low-side of each bridge leg, but with different timing for both bridge legs.

MOSFET and IGBT bridges incoprorate diodes, in so far there's no need to place separate diodes. Driving a bridge into freewheeling mode by leaving both switches of a bridge leg off for a certain time is generally unwanted in inverters because the output voltage can't be maintained with reactive load.
 


thanks for reply and dear friend i need some description of why output voltage can,nt be controlled with reactive load,(while using ECCP module of PIC18f458 where both switches turned off for a certain time)?

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i think diod perform the same function of freewheeling as IGBT do?
 

For explanation of unipolar pwm scheme, see e.g. https://www.powere.dynamictopway.com/i16.htm

It's said "Switch pairs (S1, S4) and (S2, S3) are complementary (when one switch in a pair is closed, the other is opened)".

Operating the switches complementary assures that the output voltage is independent of current polarity. In contrast, if you open a switch at some time and close the complementary switch later or not at all, the commutation time depends on current polarity. If the switch is carrying forward current, the output commutates immediately, with reverse current polarity, the voltage is maintained until the current drops to zero. The latter can only happen with reactive load. You are losing the capability to control the output voltage waveform in this case, the sine output of an inverter becomes distorted.

If you don't see how it happens, you should perform an inverter simulation with reactive load.
 

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