High and Low side motor driver circuit

The circuit above is used to drive a 48v motor as shown. I am using:

  • IRF3205 MOSFETs (datasheet)
  • IR2110 High and Low side driver (datasheet)
  • 22uF bootstrap capacitor for PWM frequency of 30 KHz.

I managed to find online and here on the website the meaning of bootstrap circuitry and how it provides the extra voltage to overcome the high side MOSFET source voltage. The problem I am facing is the choice of a suitable Vcc value.

I am running the motor on 4 sealed Lead-acid 12V 12Ah batteries in series for a total of 48v. The data sheet of the IR2110 shows a maximum value for Vcc =25 V and I see this makes sense as the maximum voltage that the IRF3205's low side Gate can take= 20V with respect to the Source.

If I choose Vcc= 25V, which I think will already blow up the low side MOSFET's gate, then on the high side I will get total of 50V minus the voltage drop across the bootstrap diode, since the capacitors will be charged with 25V and we have Vcc=25V giving a total of 50V. This is not sufficient to turn on the Gate of the high side MOSFET since its Source will have 48V so Vgs=50-48=2V

How can I achieve a Vgs>=8V with the restriction of Vcc given that I can't use an external supply other than my four 12v sealed Lead acid batteries that are in series?

  • \$\begingroup\$ The circuit is courtesy from : "Syed Tahmid" tahmidmc.blogspot.com.eg/2013/01/… \$\endgroup\$ Dec 28, 2016 at 16:32
  • \$\begingroup\$ Can you get 12V from the first lead-acid battery in your series-connected bank? \$\endgroup\$ Dec 28, 2016 at 16:34
  • \$\begingroup\$ Yes it's possible. But how can I make use of it? \$\endgroup\$ Dec 28, 2016 at 16:35

2 Answers 2


You've got a few options to obtain a suitable voltage for driving MOSFET gates:

  • Take +12V from the first lead-acid battery in your series-connected bank of lead-acid batteries. Even a drained lead-acid battery will supply sufficient voltage for gate driving.
  • Use a step-up converter to make higher voltage from +5V (the VDD). With a capacitive charge pump, you can 2x the voltage. With an inductive boost, you aren't limited to only 2x.
  • Use a step-down regulator (linear or buck) to create the suitable gate driving voltage from +48V.

Either of these methods can be used to create VCC in the range of voltages that you require.

enter image description here source: block diagram is from p.4 of the IR2110 datasheet

see also: related thread about bootstrap gate driving

  • \$\begingroup\$ The problem is: I can take 12V from a battery as input to the driver's Vcc which is enough. If I take 12V through the bootstrap diode;the capacitor will charge to 12V and I will get Vg= 12V+12V=24V. The MOSFET will turn on but its source voltage will immediately rise to:Vs= 48V. so Vgs=24-48=-24V which will result in the MOSFET not turning on. Is this true? If I take 36V through the diode I will get Vg=36+36=72V and Vgs=72-48=24V which will blow the MOSFET gate according to datasheet. similarly with 24V through the diode I will get Vg=24+24=48V and Vgs=48-48=0V. What am I missing here? \$\endgroup\$ Dec 28, 2016 at 18:05
  • 1
    \$\begingroup\$ @Khaled Let's say that in your circuit Vcc=+12V and Vs=+48V. The bootstrap circuit will drive the gate with ≈Vcc+Vs . The voltage will be supplied by a bootstrap capacitor which is charged to ≈Vcc, and which gets connected to Vs. The purpose of the bootstrap gate drive circuit for the high-side N-channel MOSFET is to apply sufficient Vgs on top of Vs, even when Vs > Vcc. [Hope I was able to identify what you're missing.] \$\endgroup\$ Dec 28, 2016 at 18:11
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    \$\begingroup\$ Earlier, I wrote about the bootstrap gate driver principle of operation. That post revolves around FAN7842 bootstrap gate driver. The principle is the same as IR2110, though. \$\endgroup\$ Dec 28, 2016 at 18:19

You will need two power supplies: 48V (to drive the motor) and something around 12V (to drive the MOSFET gates). Here's a simplified arrangement, with the driver details and right half of the H-bridge omitted:


simulate this circuit – Schematic created using CircuitLab

In your more complex situation, you will not be using switches of course, but some kind of MOSFET driver. But I hope this gives you an idea of the power supply arrangement. Many MOSFET driver ICs will require only that you supply C1, and a connection to 12V. The equivalents of D1 and the switches are inside the IC.

Think of the gate voltage of M2 not relative to ground, but relative to M2's source. C1 is essentially a charge pump, and provides 12V between M2's source and gate, regardless of the voltage at M2's source.

Since the current required for the MOSFET driving circuitry is generally low, you could also get the 12V from 48V with a linear regulator.

  • \$\begingroup\$ If I take 12V from a single battery to charge the bootstrap capacitor; the gate voltage Vg=12+12=24V which will turn on the MOSFET. But the MOSFET's source voltage will rise to 48V so Vgs=24-48=-24V so it won't be turned on. How to solve this? \$\endgroup\$ Dec 28, 2016 at 18:15
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    \$\begingroup\$ @KhaledIsmail Remember voltages are differences. The MOSFET doesn't care what the voltage at the gate compared to ground is: it only cares about the gate voltage compared to its source. See edits. \$\endgroup\$
    – Phil Frost
    Dec 28, 2016 at 18:58

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