I want to create an H-bridge using MOSFETS (IRL620SPBF). I am trying to run a 6V motor which requires max current of 1.2A . The transistors Q111 and Q112 provide level conversion as the 'signal' comes from a Dev board at 3.3V. The DAC signal comes from the Dev board (0 - 3.3V). For prototyping purposes I created an oscillator from 555 timer to act in place of the DAC signal. The frequency ranges from 0 to 1.7KHz. The circuit works but I noticed a few drawbacks. Firstly, the mosfet Q1 gets really hot. These mosfets are rated at 5.2A, 200V which is well within my range. I do not understand why only Q1 gets hot and not Q4. Doesn't the current going through Q1 also goes through Q4 or is there something wrong with the circuit design or is it to do with the switching of the mosfet? I have provided the h-bridge circuit below:

enter image description here

Any comments and suggestions are welcome


New development. Just noticed that when the motor is disconnected from the circuit and the power is connected, mosfet Q3 gets hot. Is this magic?

EDIT 18 March 2014 - Discussion

I took out mosfets Q1 and Q3 and directly connected them to 6V and sending PWM signals to FETS Q2 and Q4. I noticed that the fets get a little warm when driven the motor at full load but they do not get hot at all. When I was using the 4 FET h-bridge I noticed that the top fets would get really hot which was due to improper biasing (the fets weren't switching on fully). The new design is a great improvement. Now I am choosing an N-channel mosfet with much lower Rds(on) compared to the one I am using at the moment which will be STB80NF55L-06T4. Hope that would not consume too much power and dissipate heat. Thank you all

  • \$\begingroup\$ Sounds like Q1 isn't fully saturated. \$\endgroup\$ Commented Mar 16, 2014 at 19:18
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    \$\begingroup\$ Your high side switches Q1 and Q3 are N-mosfets with the Drain connected to 6v, and you are trying to drive the gates with the same 6v. That is what creates the problem. Please refer to this reply \$\endgroup\$
    – alexan_e
    Commented Mar 16, 2014 at 19:25
  • \$\begingroup\$ I was thinking to replace resistor R104 which is a 1K resistor with a smaller value. I read somewhere that the internal capacitor can sometimes not fully charge because of high gate resistance. Am i right? \$\endgroup\$ Commented Mar 16, 2014 at 19:27
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    \$\begingroup\$ You apply 6v to both the Gate and Drain so the Source becomes 6v (at least tries) when the upper mosfets conducts. That reduces Vgs and makes the device work in a half open state where the resistance and voltage drop (across Drain-Source) is increased and that is the cause it gets heated. No resistor change is going to help. You either need a higher voltage source (e.g. 12v) or an isolated voltage source (a separate isolated source to drive the Gates). Of course you can also use P-mosfets for the high side which would work fine. \$\endgroup\$
    – alexan_e
    Commented Mar 16, 2014 at 19:37
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    \$\begingroup\$ Success with modifying the circuit. I suggest you take P FETs for the top positions. OR: if you include a relay for reversing the direction, you need only 1 FET! \$\endgroup\$ Commented Mar 16, 2014 at 20:06

1 Answer 1


To turn on any of the MOSFETs properly in your circuit, the gate voltage has to be significantly higher than the source voltage. For the lower position devices Q2 and Q4, the sources are grounded to 0V therefore they can easily turn on into "saturation" by applying a gate voltage of a few volts above ground.

For the devices connected to the top rail, if you want the on-resistance between drain and source to be really low you have to obey the same rule - the drive voltage to the gate has to be several volts above the source. Now for low volt drop you want the source to be switched virtually to 6V - where in the circuit can the gate receive a voltage of maybe 9 or 10 volts?

There isn't one so please consider two options: -

  • Using P channel MOSFETs at the top rail, source to 6V. P MOS is slower than N MOS but in your application it won't make a difference (1.7kHz).
  • Using a drive circuit derived from a supply voltage that is at least 9V and quite possibly higher to get the best saturation from the MOSFETs.

Now the relays. What are you hoping to achieve here? The MOSFET whose gate is disconnected will float to some almost random voltage level and possibly turn on unexpectedly or just get hot. Get rid of the relays unless you have some cunning plan for their use which eludes me.

This is a standard H bridge using P and N MOSFETs: -

enter image description here

Please ask if you need recommendations for devices.

  • \$\begingroup\$ Thanks for your answer, it looks like it will solve these problems. With the Relays I was hoping to avoid a short circuit situation by turning ON both switches on either side. Also I do not have any other supply on my board apart from 6V. Can I use an OPAMP to perhaps boost the voltage to a value above 6V? \$\endgroup\$ Commented Mar 16, 2014 at 21:00
  • \$\begingroup\$ Only if you supply the op amp with more than 6V. \$\endgroup\$ Commented Mar 16, 2014 at 21:01
  • \$\begingroup\$ I'm screwed. I have no other supply apart from 6V \$\endgroup\$ Commented Mar 16, 2014 at 21:02
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    \$\begingroup\$ @DavidNorman you're not screwed. You can use a 555 timer to make a charge pump (switched capacitor) voltage doubler circuit. This will give you approximately 12V from your 6V at about 20-50mA. And once you have a 12V supply you can go back to N mosfets again :) imgur.com/RkvcZHQ \$\endgroup\$
    – Brian Onn
    Commented Mar 16, 2014 at 21:12
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    \$\begingroup\$ @DavidNorman That should be OK - it's the fets at the top you aren't driving correctly. Having said that the IRL620S is a poor choice for this application. You should consider using a much lower Rds(on) device. You don't need 200V rating either, provided you use a fly-back diode on single ended drives (not top FET). Rds(on) is 0.8 ohms and at 1.5A through the device power dissipated is 1.8 watts- try and find a device that is more like 0.05 ohms and sacrifice the 200V rating. \$\endgroup\$
    – Andy aka
    Commented Mar 17, 2014 at 8:26

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