Following up from a previous question, in the MIC4606 datasheet, it is suggested to protect the "xHS" (called Vs or Vss on most parts I found) pins with a Schottky diode and a resistor, as seen in the following datasheet figure: Negative voltage resistor and diode Source: datasheet

In the datasheet, the following is mentioned:

Adding a series resistor in the switch node limits the peak high-side driver current during turn-off, which affects the switching speed of the high-side driver. The resistor in series with the HO pin may be reduced to help compensate for the extra HS pin resistance.

In my case, I found that Rg has to be 4 ohm (for a total resistance from VDD to an ideal gate capacitor of 7.9 ohm). Less resistance would pull a higher peak current than this driver can handle, and higher resistance would produce excessive switching losses (this value chosen to fit entire temperature range and allow some tolerance). However, this paragraph tells me to reduce the high side resistance. My question is: should I?

Looking at the high side driver circuit below, my intuition is that lowering Rg (say from 4 to 1 ohm) and adding Rhs (3 ohm) would mostly preserve turn off characteristics, but turn on would be at a higher, potentially problematic current. High side driver circuit of MIC4606

I would like to add Rhs to address negative voltage spikes. I also want to add the suggested clamp diode, however this diode is parallel with the body diode of the lower nmos, which means that without this resistor I may have to oversize it to address cases such as an unpowered bridge and a motor being manually spun. With a 3 ohm resistor as suggested, a 0.7V Schottky diode, and considering that my nmos has a max diode forward voltage of 1.1V, the maximum continuous current through the Schottky diode would be 100 mA.

Maybe I can add a 0.3 ohm resistor and pick a diode capable of more than 1A continuously, without altering Rg. Would that be sound?


1 Answer 1


Correct, for an RHS big enough to drop a volt or more at turn-on, the clamp diode will be activated.

This can be compensated by dividing the RG into forward and reverse paths using a parallel diode with a separate resistor. As here or here,
What is the use of the diode connected anti parallel to the gate resistance in the general MOSFET/IGBT gate driver circuit?
Purpose of 2 resistors and a diode in a MOSFET circuit

In general, you can control rise and fall separately this way. Often, turn-off is made shorter, for a number of reasons.

A higher value RHS would seem preferable, as for example 100mA at 1V is 0.1W, or 0.05W average at 50% duty cycle. A SOD-323 or larger diode would be fine here, and likely even a 0402 resistor would do -- give or take gate power dissipation which may dominate over this figure. It would seem silly to need a diode bigger than SOD-123, and resistor larger than 1206.

As RHS decreases, power dissipation increases, also common mode currents flow from the inverter into the control circuitry. Note that this is also true of the bootstrap diode path, particularly for the first low-side pulse when it charges. What effect this has, depends on the layout of the entire circuit, so I can't offer any analysis offhand, but beware that it may cause problems.

  • \$\begingroup\$ Would adding the diode and a second resistor interfere with this IC's adaptive dead time circuit? It senses the voltage on the gate output pin waiting for it to drop to 2.2V at the output. With a 0.7 V diode, lets say you'd have 3V at the gate. My nmos has a Vgs(on)_min of 2.2V. \$\endgroup\$
    – Anas Malas
    Commented Dec 8, 2022 at 6:36
  • \$\begingroup\$ If that point is fine, I came up with this design (decoupling and bootstrap are not yet complete). Splitting the charge and discharge resistors allowed me to stay within the 1A typ of the gate driver throughtout the temp range with less FET losses. Now the charge path only has 3 ohm (for a total of 7.9-9.4 ohms over 0-80C), and the discharge path has 3 ohms and a 0.7V diode (at 0.8A peak, that's equivalent to a total of 8.3-9.6 ohm if at that current Vf is 0.7V. Even at Vf = 0.4V, thats 7.9-9.4V). Diodes are SOD123 so should be fine even at Vf=4 (also 0.1W) \$\endgroup\$
    – Anas Malas
    Commented Dec 8, 2022 at 6:53
  • \$\begingroup\$ The above is true in general, not necessarily for a particular kind of driver. The MIC4606 watches the HS pin, not HO, so it will be fine with this as well. I don't get where you find more than 6 ohms, unless there's something about the driver and MOSFET I don't know. Note that the MIC4606 is in the ballpark of 12 ohms, internally: the 1A typ. peak output current is into a short circuit from 12V supply. So your 3 or 6 ohms will not do very much in comparison. Typically one would use 10-100 ohms to notably affect risetime in this situation (if needed at all). \$\endgroup\$ Commented Dec 8, 2022 at 18:40
  • \$\begingroup\$ I must confess that I don't exactly know how to accurately calculate MOSFET switching losses. I have been using TI's FET losses calculator which is included in their Power Stage Designer 4.0... I tried to feed it the worst case scenario current (say a 99.9% duty cycle) and dragged the resistance slider until driver current was below 1A and total losses were below 5W. This is the result. I got the driver's resistance from this typical graph, which is the only place these values are mentioned. No min or max. Where's 12 ohm from? \$\endgroup\$
    – Anas Malas
    Commented Dec 9, 2022 at 7:45
  • \$\begingroup\$ 12 ohms is the driver's average output resistance, Vcc/Iout(pk). External resistors less than that simply have less impact on rise/fall time because the driver internal resistance or current dominates. \$\endgroup\$ Commented Dec 9, 2022 at 19:38

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