Why do MOSFET drivers need so much current?

I'm using a MOSFET driver to drive a 6 MOSFET H-bridge, with a 40V supply. At 25 KHz, the driver gets very hot, almost too hot to touch. The MOSFETs have a gate charge of 350 nC. Why does it take so much current to switch the MOSFETs? If the average switching current can be calculated with QxF (gate charge times frequency, which yields 9 mA), why does the driver become so warm? Does the power dissipation in the driver depend on the peak current or the average current? It seems that the average current should be the same regardless of the gate resistor, because the same charge needs to be delivered at the same frequency.

I'm using an Allegro A4935 MOSFET driver and IRFS7530TRL7PP MOSFETs. Here is the schematic:

• please define "very warm". 60°C? 100°C? 40°C? Could you provide a schematic? Do you have some copper for heat dissipation on your PCB? Which mosfets and which mosfet drivers do you use? EDIT: how much current are you using, actually? At which voltage? – jwsc Jan 15 '16 at 7:34
• Do you have resistors on the gate to control inrush current? Heating is $I^2R$ and you might have quite a lot of I for a short amount of time... – Daniel Jan 15 '16 at 7:39
• Which driver, what MOSFETs? A circuit diagram would help. Note that calculating the power dissipation needs to add the driver characteristics as well. It is not uncommon for gate drivers to supply amps when turning FETs on (and sinking that when turning them off). – Peter Smith Jan 15 '16 at 7:45
• It's not unusual for the gate drivers to run warmer than MOSFETs whom they are driving. – Nick Alexeev Jan 15 '16 at 8:01
• Schematic required for decent answer. – Andy aka Jan 15 '16 at 10:38

A current of the gate charge *switching frequency is consumed from the supply. With 6 FETs, this is then 6*350nC*25kHz = 53 mA. This generates a power of 53mA*40V=2.1 W. This power is dissipated between the IC and the gate driver resistors. That total power can be calculated without knowing the exact waveform shape.

To know how much is in the gate resistors vs. the IC, you would have to know the driver on-resistance of the IC, and then the power is shared in proportion to the resistances. The A4935 has gate driver resistance of about 10 ohm, and you have about 5 ohm gate resistors, so 67 % of the power is in the IC, and 33 % is in the 6 gate resistors, so the IC dissipates 1.4 W, and each resistor dissipates about 0.1 W.

'Too hot to touch' is about 70-80 C for a plastic package. Depending on your PCB heat sinking it is possible that a temperature rise from ambient (25 C) is caused by 1.4 W.

When a capacitor is charged via a resister you will find that 1/2 the energy ends up in the capacitor and 1/2 is burnt in the resister. When you discharge the capacitor all the capacitors energy is burnt in the resister.

In your setup most of the resistance is in the driver chip so it gets hot.

Remember that Miller effect is bad for power MOSFETs. This means that your driver will run cooler when no drain voltage is present.

Gate resisters will help run the driver chip cooler by transferring some of the heat to them. You could start with gate resisters of roughly the resistance of the MOSFET driver chip. At 25Khz you can probably get away with this .

• On top of that, if you look at the block diagram of a gate driver, it's basically a push-pull amplifier that's designed to run at 100s of kHz -- driving MOSFETs isn't cheap energetically, especially coupling in the Miller effect which can 'magnify' published gate capacitances. – Krunal Desai Jan 16 '16 at 1:38