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I'm designing a motor driver circuit which can only operate in one direction - thus is fairly simple:

enter image description here

Bottom most line is sourcing 5V to VDD and the one above is the PWM drive signal. This looks way too simple to me and I just know that's not how things work in electronics. I found this example of an H bridge driver where I saw there are 100nF capacitors on both drivers as well as C1 between the motor and its source voltage.

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source

What is the purpose of those capacitors and are they mandatory in such application?

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3 Answers 3

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The only thing the MOSFET driver does is connect the MOSFET gate to VCC or GND via internal transistors.

This means any impedance in the driver's VCC or GND traces is then in series with the gate.

A decoupling capacitor next to the driver ensures a low power supply impedance for quick current delivery to the FET gate and fast switching. Without the capacitor, VCC will sag, potentially triggering undervoltage protection on the driver (if it has one) or not turning the FET on completely until VCC stabilizes.

However, inductance in series with the gate has another effect: it can make your FETs oscillate. If this occurs in a high current circuit they will usually explode.

The lack of decoupling caps in your design makes me thing you've never heard of this, so you should pay careful attention to the layout and routing of your gate traces, they should be short and low inductance. Add a low value resistor like 33R in series with the gate of each FET to prevent oscillation too. If you don't need it, you can always put a 0R resistor. But if you need it and you don't have the footprint, then... oops.

Likewise the GROUND connection from the driver to the FETs should be low-inductance, ideally a ground plane.

Note you also need a flyback diode that can take the full motor current.

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  • \$\begingroup\$ Thanks a lot for the tips, I really was looking for some advice like that. You're completely right I'm just a hobbyist but this is very interesting to me and I believe that doing is the best way of learning. I had trouble determining the value of the decoupling capacitor - what I found is that I should just put 0.1uF and 10uF in parallel to account for low and high frequency drive. I'm planning to build it on a piece of perfboard where transistors will be next to each other same as the driver ics. Traces will be made from solder on the board - is that enough of a low inductance? I'm not ... \$\endgroup\$
    – php_nub_qq
    Feb 19, 2021 at 15:15
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    \$\begingroup\$ For decoupling caps: minimum 0.1µF ceramic X7R per chip, and an electrolytic somewhere on nearby, value not critical like 100µF, first one that falls off the capacitor bin when you shake it is okay. Low value electrolytics have very high ESR so the 100µF value isn't about the capacitance, it's about the ESR. Besides they're all more or less the same price. \$\endgroup\$
    – bobflux
    Feb 19, 2021 at 15:22
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    \$\begingroup\$ "explode" well, that depends how much energy is available for fireworks, in a H bridge driver when both FETs conduct at the same time due to regrettable timing mistake in the drivers, they basically short the power supply so if it is beefy enough, "accidental separation of MOSFET into several pieces" definitely occurs lol \$\endgroup\$
    – bobflux
    Feb 19, 2021 at 15:27
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    \$\begingroup\$ Also don't forget the flyback diode \$\endgroup\$
    – bobflux
    Feb 19, 2021 at 15:36
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    \$\begingroup\$ Since you asked earlier, here's a nice picture of a MOSFET that didn't have time to melt \$\endgroup\$
    – bobflux
    Feb 20, 2021 at 14:11
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All the capacitors in the schematic you show are power supply decoupling capacitors. While this is a good and standard practice for almost all circuit, those that switch rapidly, like a MOSFET circuit and also drivers like your TC4420s, often will malfunction due to voltage spikes/sags if not properly decoupled.

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What is the purpose of those capacitors and are they mandatory in such application?

To answer this we need to look at the power supply, the mosfets, and the inductance of the motor and its associated cabling.

  1. Some power supplies do not tolerate current flowing backward into them.
  2. Some power supplies will not permit current flowing backward into them.
  3. Some power supplies are at a distance from their load, and the cabling connecting them has an inductance which hinders current suddenly flowing backward into them.

If any of these is the case, the capacitor C1, or some other circuitry is (very likely) mandatory.

The motor (and it's associated cabling if it has any) has inductance. If there is current flowing through the motor, and the mosfets are switched off, the inductance will attempt to keep the current flowing. Current may flow through the fly-back diodes or body-diodes of the MOSFETs and into the capacitor, if the capacitor is present, or into the power supply if the power supply allows it, and can tolerate it and the power supply is not too far from the mosfets.

If the capacitor C1 is missing, and the power supply does not permit current to flow back into it, or the power supply is located far enough away that inductance of the power bus hinders current from suddenly flowing backward into it, then there would be no path for current to flow. The inductance of the motor and its cabling would respond with a dramatic increase in voltage. The result will often be the destruction of one or more mosfets, and/or other devices connected to the power supply.

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  • \$\begingroup\$ Isn't that what the flyback diodes are for? \$\endgroup\$
    – php_nub_qq
    Feb 19, 2021 at 14:35
  • \$\begingroup\$ The flyback diodes allow current to flow in the reverse direction of the mosfet if there is a path for that current to flow. Take the power supply, C1 and the mosfets out of the circuit, and where is current going to flow? \$\endgroup\$ Feb 19, 2021 at 14:42
  • \$\begingroup\$ In your circuit, you only have current flowing in one direction. In that particular case you can use a flyback diode in parallel to your motor (in the reverse direction) to absorb current when the mosfets are shut off. \$\endgroup\$ Feb 19, 2021 at 14:48

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