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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.

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  • \$\begingroup\$ How often are you switching the MOSFET and how long does it take to switch? \$\endgroup\$ – user253751 Sep 26 '18 at 3:01
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    \$\begingroup\$ Average power should be fine. Average power is energy stored in the gate capacitance * switching frequency * 2. What you are going to find is that any SMT resistor will do. \$\endgroup\$ – mkeith Sep 26 '18 at 3:02
  • \$\begingroup\$ The resistor value does not affect the power calculation as long as the resistor is not so big that the gate does not fully charge and discharge during the switching cycle. \$\endgroup\$ – mkeith Sep 26 '18 at 3:03
  • \$\begingroup\$ I agree that the resistor should not be large enough to cause timing issues, also it should be able to damp out the inductive ringing at the gate. But at the same time it can't be too small or else the peak current through the gate driver can cause the gate driver to burn out, right? So when you say any SMT resistor, this still needs to be accounted for? \$\endgroup\$ – Alex Sep 26 '18 at 3:53
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    \$\begingroup\$ I am not saying that the resistor value doesn't matter. It definitely does. I am saying that over the range of useful values, the dissipation in the resistor does not depend on the resistor value. Whether it is 10 Ohms or 200 Ohms, the power dissipation IN THE RESISTOR will be the same. But of course, there will be lots of effects elsewhere in the circuit. \$\endgroup\$ – mkeith Sep 26 '18 at 6:11
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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.

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  • \$\begingroup\$ Simulations don't seem to suggest that the expression you have mentioned is correct, or is there something overlooked? I have posted it as a solution below \$\endgroup\$ – Alex Sep 27 '18 at 2:26
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@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: enter image description here

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: enter image description here

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...

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  • \$\begingroup\$ The way I look at it is power = energy / time. So consider the power dissipated in a resistor to discharge a capacitor. Basically all the energy in the capacitor is dissipated as heat energy in the resistor. That is 0.5CV^2 dissipated in the capacitor during discharge. How much is dissipated when the capacitor is charged up through the gate resistor? By symmetry, it has to be the exact same amount. So 0.5CV^2 + 0.5CV^2 = CV^2. Then divide by period, T, or multiply by F (same thing). \$\endgroup\$ – mkeith Sep 27 '18 at 2:35
  • \$\begingroup\$ Thanks for pointing that out @mkeith, the simulation is only charging not a complete charge-discharge cycle. \$\endgroup\$ – Alex Sep 27 '18 at 2:41
  • \$\begingroup\$ @mkeith, I guess there shouldn't be any problem with the solution now... \$\endgroup\$ – Alex Sep 27 '18 at 14:17
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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}

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