Sizing the gate resistor for power

Question

There are many questions and answers regarding the value of the gate resistor, ranging from "anything will do" to "put something and see what happens". But even to put something and do experiments I need a footprint on the board, and I could not find anything on power rating for it.

Let's say the driver is powered by 32 V and can supply 2.5 A to the gate. That's a whole 80 W. Often seen on the schematics 100 Ω gate resistor will limit that to 320 mA, which is still 10 W. I've never seen 10 W gate resistors though, so these primitive calculations must be wrong.

One idea is to use gate charge to get energy, then calculate slope and get time, then from energy and time get power and finally multiply it by frequency to get required resistor power rating. Sounds way too complicated, considering that most gate resistors I've seen were no larger than basic 0603 package.

So, how do I size the gate resistor?

Answer 1

One idea is to use gate charge to get energy, then calculate slope and get time, then from energy and time get power and finally multiply it by frequency to get required resistor power rating. Sounds way too complicated ...

Sometimes, life is complicated.

Each time the gate is charged or discharged, an amount of energy equal to the gate energy is dissipated in the gate resistor. From this energy, and the switching frequency, you get a power.

Let's take a crude approximation to get into a reasonable ballpark. If you had an equivalent gate capacity of 2 nF, charging to 15 V, that's a gate energy of 225 nJ. With 100 kHz switching, that is being dissipated 200k times per second, or an average power of 45 mW.

The word 'equivalent' is doing a lot of heavy lifting there. If you want an exact gate energy, then you'll need to integrate the gate charge/voltage curve for a particular drain load. But you'll often find that a back of the envelope approximation will show you're well within dissipation limits for standard components.

The power calculated here is an upper limit, assuming the gate driver has zero output resistance, and the FET has zero real part to its gate impedance. Finite values for both of those will result in less dissipation in the actual gate resistor. This will be a more significant saving at lower gate resistor values like 10 ohms or so.

Answer 2

You're concentrating on peak power, but average is what matters.

It's simply Icc = Qg(tot) * Fsw, and further, * (Vgs(on) - Vgs(off)) or * Vcc to get P.

If the driver has an internal regulator or other means of limiting output, its power dissipation will be proportionally higher (as expressed with the latter case), when Vcc exceeds Vgs(on) - Vgs(off). I've used several SMPS controllers that work this way, where Vin might be rated 30V or higher, while an internal 8-15V regulator furnishes internal logic and gate drive supplies. (Note that driver power can be much higher than gate resistor power, but you asked about gate resistor, so this is all technically extraneous information.)

The gate resistor must also be rated for the peak power and current, or energy, under relevant pulse conditions (i.e. a pulse duration of ~Rg * Qg(tot) / Vgs(on)). This can be hard to find, but it's such a small duration, I don't think it matters, at least not for any size resistor you'd be able to get away with in the first place (say 0402 up).

Finally, note that Rg here is total gate circuit resistance, which includes driver output and transistor internal resistances. Only the external resistor dissipates its fraction of this total. Obviously in the limiting case of zero external resistance, all is dissipated by the driver and transistor.

There is also the edge case where less power can be dissipated overall, when a slowly-varying gate voltage is used, with reactive drive (such as from a series inductor with diode clamping). When Ciss * Rg^2 < Lg, dissipation will be lower. This is almost always impractical to use (drain switching speed is slowed, greatly increasing switching losses for a tiny savings in control losses), and I'm not aware of any applications offhand (but, there might be some oddball cases where it was used, and, literally, I'm just not aware of them), but it's at least a curiosity that it can be done.

Answer 3

I've never seen 10 W gate resistors though, so these primitive calculations must be wrong.

That's because the resistor's peak and average power dissipation are different things. As you might already be aware, a 1206 resistor, for example, has generally a 0.25W power rating, but they allow a peak power of even 10 times but for some limited time because the "average" power should be kept within limits.

It also depends on the application. For example, in DC-DC converters that benefit from ZVS (zero voltage switching), the MOSFET's turn-on speed is generally not relevant, so the gate resistor for turn-on can be kept high which allows you to select even the smallest package, and then you can use transistors that can handle peak currents for turn-off.

For high-speed switching applications, the driver's output impedance, the MOSFET's input capacitance (Ciss), as well as the trace inductance (or complex impedance), should be kept in mind as the resistor is there not just to limit the peak current, but also to prevent potential oscillations (damping) caused by parasitic reactive components.

Answer 4

I think what you're missing is that the gate in a MOSFET is pretty much insulated from the other two terminals. So it will have a very high resistance, and a modest capacitance (see the answer by Neil_UK).

So it doesn't really matter how much current your gate driver can deliver. Unless you're switching at a really high frequency, the average current through the resistor will be tiny.

Answer 5

The 2.5A current rating for the driver is all about how fast it can put charge on that gate and how fast it can pull it off again. The purpose of a low impedance driver is to slew the power FET through the transition region as rapidly as possible so that the drain-source channel isn’t a resistance that’s dissipating power. Adding a series gate resistor will increase the dissipation in the power FET. If the design needed that driver, (similar specs to the antique LH0060) it will be because it needs to slew the power transistor.

If the OP is mistaken about the circuit and the transistor is bipolar then it’s a more complicated problem and the transition problem isn’t solved just by increasing the driver output current without controlling the Miller effect. (The base region charge increases dramatically when Ice is in saturation. It all has to be pulled out before Ice changes. It’s a can you shut it off at this rate problem but it doesn’t so much effect on dissipation in the transition region. It’s more of a minimum pulse width/ minimum regulatable output current problem for a power supply.)