Current capacity of RC snubber

I need to design an RC snubber circuit for a multilevel inverter operating at 28V input, 400Hz frequency. Current flowing through the switch is 7A max. .. How do i determine the current flowing in the RC circuit. So how do i choose the wattage of the resistor.

When the transistor is off (not conducting), the RC circuit will be charging. We know that the R limits the amount of current that can can flow through the path. From what I see you can have 4 paths of conduction.
1.) Both transistors off - Resistors and capacitors are in series
2.) top transistor on - Only the bottom R and C are charging
3.) bottom transistor on - Only the top R and C are charging
4.) both on - Neither are charging

You also have switching transients from when the capacitor is discharged that needs to be considered. The path of conduction is still limited to the resistor though so it is covered already.

So sizing for the worse case would be the lowest resistance and highest current, so when one branch is on. When the branch is on, the current will only flow for a short amount of time until the capacitor is charged. The maximum current that the resistor could see is 2.8A assuming a 28V input. From here you can do one of two things.
1.) Size the resistor to handle that at steady state
2.) Calculate the amount of energy that the resistor would really see based on the charging of the capacitor

Number 1 is easier and on the safe size. W = 2.8^2*10 = 78.4W of dissipation. That is a beastly resistor.

Number 2 is harder where you would analyze the capacitor current up to 5 time constants. You can determine the voltage and current draw and from there you can use the on time to determine the amount of energy the resistor would see. You can then look for energy ratings on power resistors to determine what pulses they can handle. This may give you a more friendly sized resistor because of the short duration of the energy. But also doing it this way has some issues because your resistor is going to slowly heat up from the pulses over time (no time to cool).

I wonder what's meaned with "switching time constant" in the quoted paper. I don't expect that the thesis will be a big help in choosing snubber values.

Resistor rating can be easily calculated referring to stored capacitor energy. If the capacitor is fully charged/discharged (the usual case), you get P = f*CU².

The lower the ohm value, the more effective it is at reducing voltage spikes.

The RC snubber dissipates the wattage that flows in the inductor.
At 7A and 28V it will dissipate 196 W, in a few mSec.

Suppose you cannot allow greater than 100V spikes. The resistor might need to be 18 ohms in that case, to carry upwards of 7A (because that is the amount going through the inductor). The duty cycle is very short, but the dissipation is several hundred watts while it lasts.

It is not a science. Try values in the 10, 20, 50W range. Use the power rating that stays within your temperature requirements.

The lower the ohm value, the more effective it is at reducing voltage spikes.

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If so, you would get best snubber operation with zero ohms, but that's not the case. Instead the snubber is most effective when the R value is equal to the characteristic impedance of the parasitic LC circuit.

The RC snubber dissipates the wattage that flows in the inductor.

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What's "the inductor" in the present circuit?

There are many configurations of snubbers for many reasons to limit current, voltage and di/dt,dv/dt and discharge fully before the next switch to minimize the Switch Safe Operating Area (SOA) V*I product impulse power.

Therefore there is an optimal RC time constant which depletes the charge to say 2% before switching.
I think, this time constant, T = RC <= 43% *1/f

There is stored energy in the capacitance of the snubber diode which affects reverse recovery time and is the main cause of oscillation with low values of R. So choose the fastest recovery time you can afford. ($


Even more important than snubber is management of the critical dead-time between commutation so that the impedance of the driver does not stay shorted across the supply too long or the bridge stay open circuit too long as this will only increase the losses in the drivers and snubber. or increase the ringing. So if you can figure out how to control prop delay of switching asymmetrically or use a chip that has it built in, aim for 1 us or 2% of dead time and allow for thermal drift. But watch out for SHoot-thru failures !! Snubbers wont help here. Generally ESR and ESL are critical in megawatt inverters, so they use Litz wire, but lets take advantage of them in this small inverter.

Thus ½LI² = ½CV² and RC =43% 1/f is my best guess

I haven't compare losses to FVM's Q=1 solution which is critical damped, with R-C-D loop parasitic inductance in each part summed, as L, impedance at resonance (LC)²=1/(2πf) matches R for low Q dampening.

FvM said:

What's "the inductor" in the present circuit?

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Since the OP spoke of making an inverter, and since the diagram shows only a few components that might go toward making an inverter, it seemed natural to suppose there is an inductor (or transformer) somewhere, and that it generates spikes which need snubbing.

If I had seen more articles telling how mosfets need a snubber network...
But it doesn't seem to be common. I guess I had the impression the body diode reduces spikes, to the extent that the orientation permits.

If so, you would get best snubber operation with zero ohms, but that's not the case. Instead the snubber is most effective when the R value is equal to the characteristic impedance of the parasitic LC circuit.

Click to expand...

Yes, this applies to the damping resistance. Its value will depend on L and C (if those can be determined).

For me, again I was thinking of an inductor. A very high impedance encourages high voltage spikes. A low impedance reduces the amplitude of the spikes.

It is true, zero ohms is unrealistic.
It should probably be several times the resistance which is suggested by the amounts 7A and 28V. In other words it should be several times 4 ohms.