Calculation the power of resistor in a snubber

Hi
For rc snubber I have used a 100nf@600v cap and I want to consider the 100 ohm resistor as a SMD (1/4 w).
For max surge voltage (600v) max current that flows to the resistor is [Xc=1/(2*pi*60*100nf)-->Imax=600/Xc] 0.023 so the power that the resistor should resist is [100*(Imax)^2] 0.06W.
It seems to me that I can use the SMD resistor?
Does my calculation correct or I should take sth into account ?
Thanks

The voltage rating is important, because high voltage transients will be diverted through the capacitor and will appear across the resistor. AS these transients might only last for microseconds the average energy in them is low, but if the voltage is too high for the resistors packaging, then it will flash across in microseconds . Look up the voltage ratings of high value resistors in the same packaging.
Frank

It could be SMD 100mW, but 600V. Or You can put 2x50Ohm 300V in series, or 3x33Ohm 200V.

The maximum energy in Joules (not power) that is dissipated in the resistor when the capacitor is charged is 1/2 CV^2. This same energy is dissipated when the capacitor is discharged. The resistor power (Joules/sec or Watts) dissipated is the energy times the number of transients per second that occur.

Thanks Guys For your reply
@chuckey I have check The SMD resistors type, there is conventional and fine line technology that the fine line has higher withstand voltage.
@crutschow for instance in this case I have (0.5*100nf*600^2) 0.18*60(freq) so the power is almost 1 watt, am I write?
if so should I consider both constraint into account (voltage and power)?

Actually for both charge and discharge of the capacitor for each cycle the power is fCV^2 = 60*100nF*600^2 = 2.16W so you need at least a 3W resistor. The value of the resistor is no effect on the power dissipated unless it gets large enough that the RC time-constant is approaching the charge time due to the applied frequency.

I must disagree with this calculation. 1/2 CV^2 is energy stored in the capacitor when charged. So this energy was not dissapated on resistor, but transfered from signal source to capacitor. Same applies for discharge: Only part of this energy will be dissapated on resistor, remaining energy will be returned to signal source. Power dissipated on resistor depends on signal shape and resistor value. 0.06W which was calculated at first, applies only to 60Hz sine signal.
Maybe we are talking about different things as there are different snubber configurations. It would be good to see the circuit diagram to see if snubber is RC in parallel or in series and what is the signal source.

GUMY said:

I must disagree with this calculation. 1/2 CV^2 is energy stored in the capacitor when charged. So this energy was not dissapated on resistor, but transfered from signal source to capacitor. Same applies for discharge: Only part of this energy will be dissapated on resistor, remaining energy will be returned to signal source.................

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Yes, that would be true for a 60Hz sine wave. For fast spikes that have a rise-time and fall-time significantly faster than the snubber RC time-constant, most of the energy would be dissipated in the resistor. So the question is: What is the shape of the transient he is trying to snub?

GUMY said:

Only part of this energy will be dissapated on resistor

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I think so.

GUMY said:

It would be good to see the circuit diagram to see if snubber is RC in parallel or in series and what is the signal source.

Click to expand...

Here you are
The AC line has 220v However I note it as a 600v to be in a safe margin.


@crutschow The shape is sine wave. but I'm worried about spikes

Ok, I assume You have inductive load and situation is as follows:
1. Steady state: -When triac ic off, RC is forming conducting path from AC line to load and there will be about 8mA current flowing throug RC and into the load, dissipating around 6mW on the 100 ohm resistor.
- When triac is on there is negligible current flow trough RC
2. Transient: When triac switces from on to off, voltage spike will force the current to flow through RC, dissipating some energy on R and charging C. When the spike will dissapear C will discharhe through R and load and situation will return to steady state. Power dissipated on R depends on spike voltage and duration (load inductance).

The main question is, do You use this circuit for dimming (60 on-off cycles per second) or only to switch the load on and off in longer time period?

@GUMY Exactly as you said and I want to use it as a dimming circuit However I don't know about the absolute value of the Load.

I did some more thoughts and calculations... 2.16 W is worst case situation assuming that You have pure inductive load (R=0) and spike voltage and duration is big enough to charge capacitor to 600V.
It is important to know also operating load current as voltage of the spyke is proportional to that. If You have maximum current 1A at the peak of the period, and You disconnect load at that moment, 100V spyke will be formed, to assure continous current of 1A through 100Ohm resistor at the first moment.

You can find some useful hints here: