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I have a FET based H-bridge being controlled by a micro. I was worried about software errors being able to turn on the top and bottom FETs at the same time, so I opted for the control scheme below for each half of the bridge. While this works fine, difference in the switching times of the two FETs means a partial short occurs - I see the voltage on Vcc is pulled down (even though it has quite a lot of capacitance) during switching. The duration of the 'short' is very brief (looks like <10us on my scope).

Is this inevitable with this control scheme, or can I get some improvement? I tried making R1 and R2 bigger in simulation and then strapping diodes across them (pretty sure I had seen this done before somewhere), so that the FETs turn off faster than they turn on. This seems susceptible to variations in threshold voltage, temperature changes etc. Is there any better solution?

I'm aware that there are ICs dedicated to this function, but I'd like to try and keep this simple. If the only way to do it is with more transistors, then so be it, but I wanted to make sure there isn't some simple solution involving R/Cs, diodes etc that I haven't thought of.

enter image description here

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    \$\begingroup\$ fyi 10us is not short. FET will be switching in 10's of ns. \$\endgroup\$
    – user16222
    Jun 21, 2015 at 19:35

3 Answers 3

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You could improve the circuit by reducing the switching time and then adding the diodes.

Add a NPN/PNP emitter follower stage before the MOSFETs (just two cheap transistors) then do the resistor/diode to the gate thing.

Resistors R2/R3 will probably be in the hundreds of ohms to get the shoot-through to a negligible value.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ (FET specific obviously) but I usually find that R3~200R and R2~100R and any shoot through is all but gone \$\endgroup\$
    – user16222
    Jun 21, 2015 at 20:00
  • \$\begingroup\$ @SpehroPefhany couldn't Q2 and Q3 be eliminated? Perhaps by increasing R2 and R3, and reducing R1? \$\endgroup\$
    – Jodes
    Jun 20, 2016 at 15:54
  • \$\begingroup\$ Q2 and Q3 provide the high current drive to make the MOSFETs switch quickly. They reduce the effect of R1 by a factor of \$\beta\$ without increasing power consumption. \$\endgroup\$ Jun 20, 2016 at 17:19
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Make sure to pole M1 the other way around (D to VCC and S to load)! In your circuit the body diode of the p-channel MOSFET is forward biased all the time M2 is turned on.

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Having started with a high speed H-bridge design as my first proper project, it really sounds like not having to use an IC driver is simpler; it's not. At all.

If you want a simple, quick way to make an H-bridge, without having to worry about signal inversion and dead-time, use a driver IC.

IR2183 works just fine and covers everything up to the beefy 7000pF drive gates at like 100kHz. (Then it breaks down a bit, drive speed wise).

If you really don't want to, you must use another transistor set to drive each gate of your H-Bridge transistor. This is a lot harder, but a lot cooler.

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