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One might wish to switch DC through a transformer with an H-bridge. I'm leaving out antiparallel diodes and snubber caps, for simplicity.

schematic

simulate this circuit – Schematic created using CircuitLab

One issue with this approach is that the upper FETs need isolated gate drives, referenced to the windings of the transformer. This can increase the cost and complexity of the system substantially. It occurs to me that one might use SCRs in place of the upper transistors.

schematic

simulate this circuit

SCRs will have higher losses than FETs, and they consume more drive current, but they're substantially easier to drive and harder to kill, and cheaper at high voltages and currents. At high voltages and currents, one might use IGBTs instead of FETs, which would have losses more comparable to those of SCRs. SCRs would require separate antiparallel diodes.

Will this work? Is it done? Am I missing some critical flaw?

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  • \$\begingroup\$ What's wrong with using chips commonly available for driving N channel FETs in the "high" position? They are cheap too but one man's "cheap" may not be the same as someone else's. \$\endgroup\$
    – Andy aka
    May 27, 2015 at 15:15
  • \$\begingroup\$ Why not make the entire bridge out of SCR's? That is what basic DC Motor controls are. You don't need the FET's unless you are PWMing something. \$\endgroup\$
    – R Drast
    May 27, 2015 at 15:16
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    \$\begingroup\$ @RDrast How, exactly, in a DC H-bridge, do you propose the SCRs will turn off, other than turning all on and shorting out the supply, effectively causing a full supply hiccup? DC motor controls with SCRs all run, AFAIK, on AC transformers. \$\endgroup\$
    – Asmyldof
    May 27, 2015 at 15:21
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    \$\begingroup\$ @Asmyldof True that... I wasn't really thinking of the DC H-Bridge feed. Not enough coffee :) \$\endgroup\$
    – R Drast
    May 27, 2015 at 15:45
  • \$\begingroup\$ Since you would have to wait for the SCR to turn off (current to zero), fast-decay PWM would not be feasible. I believe slow-decay PWM would still be feasible by leaving the SCR on and modulating the opposite MOSFET. The decay could be sped-up by putting TVS diodes in series with the diode clamps to the positive rail causing them to clamp at a higher voltage. This is something that cannot be done with MOSFETs on the high-side due to their body diodes. \$\endgroup\$
    – Tut
    May 27, 2015 at 18:19

1 Answer 1

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A reason SCRs aren't used for push-pull schemes involving high frequencies, as far as I am aware is the abominable turn-off time of the things.

Gate trigger to conductance can be in the micro-second range for most types, but for many types that I am aware of turn off times are in the 10 to 100us range, which is pretty much the death of any efficient use of a square wave or PWM signal for power conversion.

There may be whole series of SCRs that are super fast in both regards, but I haven't heard of them.

EDIT: (In SCR's turn-off time is not the transition of stopping to conduct, which would not make much sense anyway, but the time the device is expected to be "relaxed" before a voltage is again applied that might induce current over its holding current. This is the time needed for the blocking diode barrier to regain its withstand voltage. So some tweaking may be possible by using an SCR rated for 200V at 40V, but it won't be an order of magnitude.)

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  • \$\begingroup\$ If SCR2 is on and M4 off, and these are instantly switched, SCR2 will still be conducting as M4 conducts, forming a short through both, no? \$\endgroup\$
    – rdtsc
    May 27, 2015 at 16:34
  • \$\begingroup\$ @rdtsc Yes, that's why, when SCR2 is conducting through M2, you turn off M2 and then wait for SCR2 to recover, before you trigger M4. This is why the slowness of SCRs is super annoying in a set-up like this. \$\endgroup\$
    – Asmyldof
    May 27, 2015 at 16:36
  • \$\begingroup\$ In addition, you must be absolutely sure that SCR2 has turned off before turning on M4. Otherwise the current through M4 will cause SCR2 to remain on. Even worse than the spike caused by shoot-through in an all MOSFET design. This turn-off time would also be dependant on the inductance of the load. \$\endgroup\$
    – Tut
    May 27, 2015 at 17:24

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