RC snubber calculation ??

Discussion in 'General Electronics Chat' started by capacitor1, Jun 3, 2013.

  1. capacitor1

    Thread Starter New Member

    Jan 12, 2013
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    Hi everyone

    I found on the internet this information about snubber RC circuit :
    there's 2 rules here:

    [​IMG]
    for a inrush current of 7A and coil 220V/50HZ

    we found that C=5uF
    and R=2.5;

    no problem here
    but the Power of resistor should be 220^2/2.5=88 watts !!!!?

    I don't know where's the error in my calculation and can i use this rule for AC coil ??

    help me to understand please

    thanx
     
    Last edited: Jun 3, 2013
  2. capacitor1

    Thread Starter New Member

    Jan 12, 2013
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    No answer ??!

    or I put my thread in the wrong forum (general electronics chat)??
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    No it shouldn't!!!?
     
  4. capacitor1

    Thread Starter New Member

    Jan 12, 2013
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    ok but how , how i can calculate that ?? i found other answer that it's enough to put 1W resistor !??why?

    please help

    thank u
     
  5. crutschow

    Expert

    Mar 14, 2008
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    The peak power may be 88W but it's only for a very short time. The maximum energy dissipated in the resistor is no more than the energy stored in the inductor. Since this is typically small, the resistor dissipation will also be small.

    If you want to get technical you would need to know the surge energy (or peak transient power) that the resistor can tolerate. Otherwise you just make an estimate and go from there.
     
  6. capacitor1

    Thread Starter New Member

    Jan 12, 2013
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    Mr Crutschow

    thanx for your reply

    but the energy stored in coil isn't big and equal to L*I^2/2??

    for me , Inrush current 7A , and L= 0.55H

    E=13.475 J

    what about this number ??


    please help me to know
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hello Capacitor1,

    You need to explain your numbers carefully - they make no sense & possibly for several reasons.

    Your main mistake seems to be in failing to understand or consider the transient nature of the response of this circuit.

    Present your calculations in a more logical sequence and it might be possible to offer you some guidance.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    13.475 J equals 13.475 watt-seconds. So the resistor would have to dissipate, at most, the equivalent of 13.475 watts for 1 second. Likely a 1 watt resistor could handle that if it didn't have to do it very often. What's the rep rate on the transient?
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I think the OP's problem is that they want to design a snubber for an AC [220V / 50 Hz] solenoid and they are using design information for a DC situation. As they indicated in the first post this is their uncertainty.

    With 220V 50 Hz the notional steady state [zero offset] current in a 0.55H solenoid would be ~1.3A RMS or 1.8A peak. So the peak energy storage in the inductor would notionally be ~0.9 Joules.

    The situation would be somewhat worse when the inductor is energized from the AC supply such that there is a steady state offset in the AC current - albeit an impractical situation. In any event the energy would be at most ~1.8 Joules were the switch opened at the highest energy condition.
     
  10. capacitor1

    Thread Starter New Member

    Jan 12, 2013
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    Hi Eng T_n_K

    I can understand from your reply that:

    1- when the switch is closed the current is AC , and all rules will be on AC current.
    2-when the switch is opened the kickback current is DC ,and all rules will be on DC current.

    Please help me to understand

    thanks
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You are basically correct.

    The AC analysis will tell you the circuit conditions from the time when the switch is closed until the switch is opened.

    Once the switch is opened you take the "initials conditions" to be ...
    * Inductor Current
    * Capacitor Voltage

    You then perform a transient analysis to determine what happens as the stored energy in the RLC loop is dissipated in the snubber resistor.

    The chief concern of the design process would be to protect the switch from excessive over-voltage. Using your given circuit parameters the peak switch voltage could potentially exceed 850V.

    Strictly speaking both the AC and DC analyses might need to be a transient solution since there is no indication from you as to the time frame in which the switch is closed and then opened. One would need to know the particular application you have in mind. It may well be that the switching device is semiconductor in type and in addition to the over-voltage stress issue one would have to ensure the snubber design allowed for the device peak current ratings.
     
    Last edited: Jun 5, 2013
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