I am working on making an autonomous robot. Right now, I am involved in designing an H-Bridge using discrete components for my robot. I am almost done with the H-Bridge design, and want to test it now. However, I am not being able to decide a dead time for the input complementary PWM I will provide as an input to the H-Bridge. I know that to prevent overshoot, you have to delay the turn-on of the low-side FET by at least as much as the turn-off time of the high-side FET. The same goes of course for the other transition, when you switch from low-side to the high-side.

I have searched about finding the dead time online and I haven't found any useful information. The information I found was for H-Bridges who have gate driver circuitry designed discretely, and use formulas to find the actual turn on and turn off times of the MOSFET. Whereas, I am using IR2112 to drive the MOSFETs (IRFZ44N in my case).

I am attaching the schematic of my H-Bridge circuit. Please view it and tell me how should I calculate the dead time. I do know the turn on/off delays of both the gate driver IC and the MOSFET. Should I add the turn off delay of both these to get my dead time?

enter image description here

  • \$\begingroup\$ You know the gate charge of the FETs, and you know the gate current. \$\endgroup\$ Jul 18, 2016 at 11:57
  • \$\begingroup\$ @IgnacioVazquez-Abrams Yes, I know the gate charge of the FETs and the rated outut current of the IR2112. Is the dead time equal to t=Q/I? \$\endgroup\$ Jul 18, 2016 at 12:04
  • \$\begingroup\$ @IgnacioVazquez-Abrams I am sorry I didn't understand what you mean by both halves? \$\endgroup\$ Jul 18, 2016 at 12:07
  • \$\begingroup\$ If you're driving a simple DC motor just go for the highest dead-time your controller supports. It should not matter if dead time is 0.1µs or 1ms in that case. \$\endgroup\$
    – JimmyB
    Jul 18, 2016 at 14:56
  • \$\begingroup\$ @JimmyB Why is that? \$\endgroup\$ Jul 19, 2016 at 12:38

1 Answer 1


Dead time is to prevent shoot through i.e. both MOSFETs conducting appreciable current at the same time.

I do know the turn on/off delays of both the gate driver IC and the MOSFET

That's good because you are almost there.

Should I add the turn off delay of both these to get my dead time?

No, add the turn-on time of one MOSFET/driver to the turn-off time of the other MOSFET/driver to give a conservative estimate. Calculate this for switching high and switching low and choose the highest value of combined times.

If you wanted to reduce the value a fraction it's down to how much shoot-through current you can suffer - I would probably want to simulate this on something like LTSpice.

  • \$\begingroup\$ should I add only the turn on time of the High side MOSFET or Driver with the turn off time of the Low side MOSFET or Driver? or should I add the turn on time of the high side (MOSFET + Driver) and the turn of time of the low side (MOSFET + Driver)? \$\endgroup\$ Jul 18, 2016 at 12:42
  • \$\begingroup\$ and I know the turn on/off times of these ICS from their data sheets. I hope I don't have to calculate the "actual" turn on/off times somehow... \$\endgroup\$ Jul 18, 2016 at 12:43
  • \$\begingroup\$ I am sorry for bothering you, but I don't know why am I getting confused.. \$\endgroup\$ Jul 18, 2016 at 12:44
  • \$\begingroup\$ and shouldn't the value be same for both switching high and low because there are same MOSFET and driver on both sides? \$\endgroup\$ Jul 18, 2016 at 12:46
  • \$\begingroup\$ Switching high isn't usually exactly the same time as switching high. The times have to include driver and mosfet unless you know the driver to be much quicker than the mosfet. A little bit of shoot-through isn't usually a problem providing the peak current about one tenth that of normal load currents when switching. But shoot-through current rises very quickly if the dead time is too low. \$\endgroup\$
    – Andy aka
    Jul 18, 2016 at 13:25

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