generic snubber

The way I calculate an approximate value for a snubber is to choose a resistor which will limit the surge voltage to an acceptable value, and a capacitor which will store the energy of the inductive component. For example, a 120 VAC relay coil that draws 100 mA is about 3.2 H, and has a stored peak energy of 0.016 J. Choose the resistor to limit the voltage to

200 volts at 0.14 A (peak), or 1.4 k, and the capacitor for the energy would be 0.8 uF. The snubber in normal operation will draw about 36 mA, so the resistor will dissipate about 2 watts. You can use smaller values of resistance and capacitance. I have found about 100 ohms and 0.1 uF to work well.

Another way to look at it: For your 1000 VA loads, assuming 120 VAC at 8 amps, it could be a pure inductive load of 40 mH with a peak energy of 2.9 J at 12 amps. The 47 ohms in your example would limit the voltage to a reasonable 564 volts peak. This corresponds to 6.8 kW, so with 2.9 Watt-Seconds it would dissipate within 428 uSec. The capacitance corresponding to this TC is 9 uF. Based on stored energy, it would be about

18 uF at 600 V. The current draw for 18 uF would be 145 mA and the resistor would dissipate about 1 watt.

I'm not sure my calculations are correct, especially for the capacitance value. I have seen snubber calculations for SCRs, but they are used to limit dV/dT. It may be useful to do a simulation for this. There is also the question of placing the snubber across the switching device (as in the case for the SCR), or the inductive load (as for the relay coil). I am interested in more details on this because I deal with inductive load switching quite often in circuit breaker test sets, which may involve over

500 amperes at 480 VAC for the SCR and transformer primary, and 50,000 amperes at 5 VAC being interrupted by a breaker. We use an SCR snubber of 40 ohms and 0.0033 uF, and it works well enough. It is mostly to keep the SCRs from turning on when power is first applied to the circuit. We cannot tolerate much higher values of capacitance because the leakage current translates to unacceptably high output currents.

We are using a snubber of 1 ohm and 30 uF at 600 VAC across the output of a tap switch that selects 9 voltages from 0 to 540 VAC, and then controlled by an SCR. Before adding the snubber, the SCR would sometimes turn on from dV/dT at higher voltage taps. The snubber tends to hold the voltage at the input of the SCR to its value prior to tap changing, and also limits the dV/dT. It draws up to about 6 amperes, and the resistor dissipates 36 watts, but this is acceptable in a 50 kVA test set.

I will post this response as is, at the risk of criticism. I will investigate the details and also await any corrections and discussion you may be able to supply.

Thanks,

Paul E. Schoen

I did some simulation using Tina, and the snubber values I calculated seem to be fairly reasonable. On the 1000 VA inductive load, the 47 ohm 18 uF combination produces a quickly damped spike of less than 1 kV, while with

0.1 uF there can be 8 kV spikes. The worst condition is if the load is switched off at its peak current. For very pure inductances, this peak current can be 2.8 times the normal RMS current if the AC is initially applied at a voltage zero crossing. This illustrates why a zero-crossing triac is the worst way to switch an inductive load.

The snubber also works by placing it across the switch. However, this requires a very low impedance source, as it will see the inductive spike current.

A reasonable snubber would probably be 100 ohms and 470 nF, although the capacitor could see a 2.5 kV spike. Real world conditions are difficult to simulate. Actually, much of the energy stored in the inductance will be dissipated in the initial arcing of the opening contacts. A solid state device like a triac will continue to conduct until current reaches a low value, so snubbing will not need to be as robust. However, the devices are more fragile, so the energy must be absorbed without high voltage excursions.

Paul

about

resistor

Assuming you are referring to 18uF in series with a 47 ohm resistance, at

120V 60Hz, you might want to repunch those numbers...

Right. About 814 mA and power of 31 watts. A 3.3 uF capacitor would draw about 145 mA, and it also makes an effective snubber without excessive power, although a 1 kV spike could be generated compared to 500 V for the

18 uF.

A higher value of resistance can be used to reduce the voltage spike on the capacitor at the expense of a higher overall spike voltage on the load. 150 ohms with 3.3 uF limits the capacitor voltage to 600 volts, while the spike reaches 1.5 kV. The resistor will dissipate about 3.5 watts.

This is a useful exercise, and simulation clearly shows ringing and DC current offsets due to inductance.

Thanks,

Paul

Thanks Paul for your excellent explanation. I am afraid however that I am not able to supply much discussion nor corrections, sorry. All I know is that probable loads are 120VAC/240VAC/24VDC power supplies.

Thanks, Jenalee K