Determine MOSFET switching speed

Question

How can I determine the switching speed of a MOSFET, with given Qg (gate charge), Vgs (G-S junction voltage) and gate resistor ? How should I determine the required value of the gate resistor?

Answer 1

From the point of view of the driving circuit, the gate looks like a capacitor to the source. It actually has some capacitance to the drain too, but that has been taken into account in the total gate charge figure. You know the voltage the gate must change and the charge that must be transferred to accomplish that. From there it's simple to compute the equivalent capacitance: Farads = Couloumbs / Volt. Once you have the capacitance, The R*C time constant gives you some idea how fast the gate will slew given a step input on the other side of the gate resistor. To reach 90% of the final gate voltage, for example, takes 2.3 time constants.

When the FET actually "switches" is more tricky. The FET won't suddenly go from full off to full on at a particular gate voltage, but there is a gate voltage at which a small incremental change will make the most difference in the FET output characteristic. You have to decide how full on and how full off "switching" really means, and then decide what gate voltage range that represents. Then you can use the equivalent R-C model to decide how quickly a step input will cause it to slew accross this region. For example, if you decide most of the switching happens between 20% and 80% of the gate voltage, then that would be 1.4 time constants.

Answer 2

The bulk of the switching action happens when the gate voltage plateaus at the threshold voltage Vgsth, at which point the drain voltage drops quickly and the so-called Miller effect keeps the threshold there until the drain reaches its minimum:

(from https://web.archive.org/web/20120324165247/http://www.ti.com/lit/ml/slup097/slup097.pdf)

Just for a practical example, let's say you have an IRL540N that you are driving with a 5V source with 100 ohms series resistance.

Gate threshold is spec'd at between 1 and 2 V. This means that the gate charging current would be 30-40mA. Total gate charge is spec'd at <74nC, so the time to for the drain-to-source voltage to drop during the Miller plateau is approximately t = Qmax/Imin = 74nC/30mA = 2.47usec, and I would expect the turn-on time to be a little bit larger than that because it includes the other time intervals shown in the figure above.

Why wouldn't one use a gate resistor with zero resistance?

Several reasons:

• parasitic source inductance in the MOSFET can cause high-frequency oscillations, or at least a highly underdamped turnon

• You generally want to tune the turn-on time appropriately for EMI reasons.

• And in a half-bridge gate drive, you usually use a diode in parallel with the turn-on resistor so that the turn-off is quick but turn-on is slow. Otherwise you can get shoot-through, for reasons beyond the scope of this post. (When I originally posted this answer, I stated that if I had time I'd write a blog covering this topic; I finally did in 2024: See Gate Drives for Dummies.)