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I am using a 3-phase MOSFET driver to drive power MOSFETs for a motor controller. When I connect the circuit to a 24V supply (H-bridge supply) and enable the switching at 25 kHz, the driver gets warm. However, when I connect it to a 40V supply (for the final application) the MOSFET driver gets much warmer. Is the driving current supposed to increase with increased supply voltage? I am not sure because the gate-source voltage on the high-side MOSFETs is the same; the driver just needs to supply 40+Vgs to the gate instead of 24+Vgs to the gate. If the gate charge is the same, why would it take more current to drive it at a higher drain voltage?

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  • \$\begingroup\$ To clarify, you're not asking if changing the gate driver supply will cause a current increase, right? \$\endgroup\$ – W5VO Jan 15 '16 at 7:09
  • \$\begingroup\$ Even if the current stays same, the dissipated power (in the driver) does increase, because while the gate is charging, the driver is half-open and dissipates some power; the higher voltage the more power is dissipated while switching. \$\endgroup\$ – ilkhd Jan 15 '16 at 7:41
  • \$\begingroup\$ The MOSFET driver usualy has supply voltage of +15V or +12V or double voltage supply +15V/-10V as the MOSFET ratings for Vgs max are almost identical for all MOSFETs. \$\endgroup\$ – Marko Buršič Jan 15 '16 at 9:21
  • \$\begingroup\$ Q=CV. I = dQ/dT = CV * switching frequency. And C includes the drain-gate capacitance (Miller capacitance) which sees the supply voltage. So, yes. \$\endgroup\$ – Brian Drummond Jan 15 '16 at 13:05
  • \$\begingroup\$ So is the Q used in current calculation not the "gate charge" from the data sheet, but some other function of C and V? \$\endgroup\$ – Raphael Chang Jan 15 '16 at 20:47
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There is a small change in total gate charge with the FET's supply voltage. It is not particularly large, because the difference in Gate-Drain capacitance when going from 24 V to 40 V is relatively small. However, the gate drive current is generated in your IC from a linear regulator from the 40 V input. Therefore with a constant gate drive, the IC will still dissipate more power at higher VIN.

There is an NXP Three Phase Driver IC -- MC34937 which allows the gate drive power to be distinct from the FET bridge supply -- you can use that to either 'move' some of the dissipation outside the IC, or supply it from a DC/DC converter which reduces the power dissipated considerably.

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The miller effect depends on the gain. For a switch, the gain is proportional to Vds, which is approximately the "supply voltage". (I interpreted your "supply voltage" as the voltage that is switched by the power MOSFETs to the load.) Therefore, the "gate charge" increases with "supply voltage".

The "gate charge" from datasheets is specified for a specific Vds and Vgs. It changes with actual operating Vds and Vgs. (By the way, it is easy to see that the "gate charge" also depends on Vgs given Q = C V for a given capacitance, so the "gate charge" also increases with "supply voltage of the gate driver").

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You are hard switching and there is nothing wrong with that .It is the miller effect that makes the effective gate capacitance much greater than what you thought.The modern mosfets have much more capacitance than the old ones .This is the price of the lower on resistances that are available.

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  • \$\begingroup\$ Generally, modern FETs have less capacitance for a given on resistance; however, now there are devices available with lower on resistances than earlier. \$\endgroup\$ – jp314 Jan 16 '16 at 16:03

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