I have a PWM bridge driving a higher than 1k load from 200V supply. The simplified, equivalent schematic is this:schematic

The drivers I'm using are slow turn-on/fast turn-off and, in the schematic, it's a 50% duty-rati.,The parasitics modeled here also involve some capacitance across the filtering inductor. These will cause some spikes, sure, but they pale compared to the switching problem and, while I can get away with 100-200mA peak, what I have here is a bit scary: spikes

The driving, as seen here, is a much nicer version to what will be in reality. The problem is that the mid-point voltage, V(mid), doesn't fall with the ending of the commanding pulse, only with the rising of the complementary pulse, which makes it overlap a bit and cause a spike, in addition to the inevitable parasitic. Here's a zoomed-in portion: enter image description here

You can see that the upper drive, V(gh,mid), is turned off fast and has plenty of dead-time, while V(mid) keeps on going until V(gl) starts.

I know there are circuits trying to circumvent the parasitic capacity, but they only affect the response (slow zero as opposed to a notch), because the current as seen from the bridge is still there, which is what is the problem.

My question(s): is there any way of making V(mid) fall with the end of the command pulse, actually having a dead-time in there? Another winding with a forced current, maybe? Some clever snubbers? Anything? Or is this "unobtainium"?

Update: This is a close-up with Id(M3): zoom2

The dead-time is 25n but the rise time is imposed to be 25n, too. Tfall=1n but, apparently it doesn't affect the response as I'd want. Increasing the dead-time to 50n (keeping tr) results in this: dt=50n

  • \$\begingroup\$ What is a "higher than 1k load"? \$\endgroup\$
    – jippie
    Commented Dec 29, 2015 at 15:28
  • \$\begingroup\$ What happens if you extend the dead time in the driver section? It is only like 40ns now. \$\endgroup\$
    – jippie
    Commented Dec 29, 2015 at 15:35
  • 1
    \$\begingroup\$ Can you plot the current in M3 drain too? \$\endgroup\$
    – jippie
    Commented Dec 29, 2015 at 15:48
  • 1
    \$\begingroup\$ You might want to keep your time scales in mind. Sure, you've got a big current spike, but the half-amplitude width is only about 10 nsec. I think you're worrying unnecessarily. \$\endgroup\$ Commented Dec 29, 2015 at 16:42
  • 1
    \$\begingroup\$ Why do I not see ANY load current other than your "spikes" of concern ? \$\endgroup\$
    – Marla
    Commented Dec 29, 2015 at 16:57

1 Answer 1


Just because M1 turns off (and M3 isn't on yet) doesn't mean that v(Mid) will go low -- what happens depends on the load current (and load inductance).

However in your circuit, that's not what is causing the shoot through current -- notice that in your 2nd waveform, V(gh,mid) rises slightly at around 6.05 us. This is because V(mid) is falling (by M3 turning on), but M1 (which was off) turns back on again. M1 turns of because of drain-gate capacitance which couples the increasing VDS of M1 to the gate of M1.

You can improve this by reducing the value of the 12 ohm, and/or slowing down the turn on of M3. Basically in these situations you need to ensure the turn-off 'strength' is stronger than the turn-on capability.

In more technical terms -- when turning on M3, the dV/dt of v(mid) is limited by the Miller capacitance CDG of M3 to dV/dt = I_gate/Cdg. However, similarly the gate current induced in M1 is also I = Cdg*dV/dt, which is a similar amount. To avoid M1 turning on, you need to have a higher magnitude of gate turn-off current than turn-on current.

  • \$\begingroup\$ V(gh,mid)@6.05us=~160mV, is that enough to turn on M1? Still, I changed the gate drive to be a 100Ohm anti-parallel with a diode and the spikes lowered a bit. Increasing R to 1kOhm drastically reduced the spikes and dissipated power, but the gate drive is weak and I fear for low pulse widths. I'll have to concoct some sort of constant current with two slopes, slow before Miller and fast afterwards, while keeping the turn-off diode, but I suspect complications (both electronic and money). Thank you for the explanation. I'll still wait a while to see what other answers come up. \$\endgroup\$ Commented Dec 29, 2015 at 17:21
  • \$\begingroup\$ No, 160 mV shouldn't really turn on the FET. Fundamentally, you need stronger turn-off then turn-on strength. Try putting the GND end of your VCCS's for gate drive to ~ 3 V. This will turn on with less current, and drive VGS negative for off. Then 'just' come up with a circuit that does something similar. \$\endgroup\$
    – jp314
    Commented Dec 29, 2015 at 18:08
  • \$\begingroup\$ If I do that, then I need to do it for M4, too, which has fixed gound reference, and then for M3+M2 when the negative voltage comes around. It's clear that this is the cause so the driving must be changed, and it will be costly, no way around it. Maybe with the slower turn on the dead-time can be reduced, too. Thank you for the "light", I marked this answer as the answer. \$\endgroup\$ Commented Dec 29, 2015 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.