MOSFET gate resistance's power rating

Question

I wanted to verify the following or get some helpful pointers.

So, I have a gate driver which can supply a maximum current of 0.5 A , and I am driving a MOSFET Vgs of 15 V, based on these two specifications I should be selecting a gate resistance of at least 30 ohm (preferably even higher). This way the gate driver doesn't get damaged when the MOSFET is switched ON and its Cgs is charged up.

The peak power that the resistance should be able to handle is

\$ 15 W \cdot 0.5 = 7.5 W \$.

So while selecting the SMT resistance, I was looking for something like a R~50 ohm, P=10 W and guess what, its hard to come by something like that (in a 1210 package size or similar), and not to mention they are pretty expensive (highest I found was 3.5 W for $ 3!)

So, instead of considering peak power, I assume we should consider avg. power?

\$P_{\text{avg}} = \frac{V_{\text{rms}}^2}{R} \$

While considering Pavg in LTSpice, it came to around 0.5 W (when Vgs was charged to around 80%)

So, is it fine to use a resistor of 1 W rating in this scenario?

Any other pointers would be much appreciated.

Answer 1

Normally you can size the resistor for the average power rather than the instantaneous power. And given certain assumptions, there is an easy way to calculate the average power dissipated in the resistor:

\$P = CV^2F\$

Where P is power, C is gate capacitance, V is the gate voltage, and F is the switching frequency. Note that the resistor value is not part of the formula. That is because given certain assumptions, the resistor value doesn't change the average power dissipation in the resistor.

Of course, the resistor has a strong effect on the overall power dissipation because it affects the turn-on and turn-off time of the transistor. As the gate resistor gets larger, the transistor power dissipation increases (because it switches more slowly). But if the resistor is too small, then there can be other undesireable effects such as ringing or miller capacitance coupling into the driver IC through the output, etc. But that is not what you asked about.

Answer 2

@mkeith mentioned the following expression for power: \$ P=CV^2f \$

More information on power can be found at: Power consumed by a CPU

In the case of an RC circuit, if we first look at energy, i.e. \$ CV^2 \$ is taken from the supply and half of that is stored across the cap, while the other half is lost in the resistor.

So can one say that the power burnt by the resistor is based on that \$ 0.5CV^2f \$ of energy? The following simulation suggests the same:

The green curve is the voltage across the resistor, if you consider its rms value, it is \$ 2.37V_{rms} \$. The average power burnt across the resistor would be:

\$ P_{avg}=\displaystyle{\frac{V^2_{rms}}{R}} \$ ; \$ (R=1 \Omega) \$

\$ P_{avg}=\displaystyle{\frac{2.37^2}{1}} = 5.61W \$

But, if one was to use the expression directly, i.e \$ P=\displaystyle{CV^2f} \$ OR \$ \displaystyle{\frac{(CV^2)}{t}} \$

\$ P = \displaystyle{\frac{1\mu * 15^2}{20\mu}} = 11.25W\$

This is twice of what the simulation suggests...

So whats going on here??

Well as suggested by @mkeith, the expression for power is actually the expression for a complete charge and discharge cycle. The \$ 0.5CV^2 \$ of energy stored on the capacitor when discharged, does so through the resistor. Based on symmetry, that would mean that the same amount of power would be once again burnt across the resistor during the discharge phase.

This can be verified through simulation as well:

The RMS value of the resistor voltage is \$ 3.354V_{rms} \$

Therefore, \$ P_{avg}=\displaystyle{\frac{V_{rms}^2}{R} = \frac{3.354^2}{1}} = 11.25W \$

I guess that should clear up things...

Answer 3

Power into a capacitor (that MOSFET gate) is

Frequency * Voltage * Capacitance

[ error; its F * V^2 * C]

The gate resistor will also dissipate that exact same power.

Thus 1MHz * 10 volts * 10,000 picoFarad [* 10] will dissipate

1e+6 * 10 * 1e-8 = 0.1 watt. {* 10, = 1 watt}