Let's says I have a gate driver and a MOSFET. I have made the decision to drive my FET using 5V. The gate driver peak current is 2A. Do I size my 5 volt power supply at 2A plus my other stuff? Do I need more or can I get away with less but at the price of, say, max permissible switching frequency?


2 Answers 2


The PEAK driver current is 2A, not the continuous current. Your power supply needs to supply the continuous current to recharge the decoupling caps and the caps are what actually supplies the peak currents to the gate-source capacitance in the MOSFET since the power supply is too slow (or unavailable if it's a high-side bootstrap)

Also, it's not actually your gate driver peak current you need to be concerned about. As long as its high enough to turn your MOSFET on and off fast enough then what you really need to be worried about is your gate charge \$Q_{gs}\$ and switching \$f\$ frequency which determines the actual current your power supply needs to be capable of.

\$I_{average} = \frac{Q_{gs}}{T_{switch}} = Q_{gs}\times f\$

  • 1
    \$\begingroup\$ Wow, this makes a lot of sense. I don't normally think about a chip pulling over 2A from a capacitor so it didn't occur to me. \$\endgroup\$
    – foreverska
    Apr 19, 2020 at 3:35

If use of a gate-driver requires peak-performance (minimum switching times, quickest possible charging of the FET gate, minimum of ringing), then you need to design a LOW INDUCTANCE power supply.

That means

(1) you use a Ground Plane under the gate-driver and under the FET;

(2) you have 0.1uF SMT from the gate-driver VDD lead to the underlying Ground plane

(3) you bring a clean rail-rail logic-drive signal from the pulse-source with sharp edges and low over/undershoot; you may want to source terminate this pulse trace on the PCB.

(4) you include Miller Effect Plateau in the FET gate drive requirement


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