help for snubber design

Discussion in 'General Electronics Chat' started by dileepchacko, Aug 30, 2008.

  1. dileepchacko

    Thread Starter Active Member

    May 13, 2008
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    1
    Hi All

    I need help for designing a RC snubber for non potential contact of a simple switch with at least ( 4KV AC/DC isolation. The contact rating of the switch is 250V/5A. consider an inductance load is connecting through the switch and 220V DC is applying for the load. Consider the switch is closed for 50ms, suddenly it is opening. The problem is , a high voltage is around 600kV is developing across the switch for an instant of time. I need to suppress that transient voltage to at least 280V. The inductance details are L = 1.09H and Resistance of the inductor coil is R = 350 ohm. I have simulated a rough RC snubber in PSPICE. From the simulation it is clear that, 1.487KV is developing across the switch. I have attached the details of the simulation.
     
  2. bertus

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    Apr 5, 2008
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  3. SgtWookie

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  4. sunsetb

    New Member

    Apr 11, 2017
    2
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    Hi,
    It it probably too late for an answer, but still for future member this coulb be useful

    First, what is happening: Basically you connect a resistor and an inductor in serie with a voltage source (220V)
    This is a basic RL circuit which is exponentially rising to its maximum current (0.656 A, Vsource/R = 220/335 = 0.656)

    Then, you suddently disconnect the voltage source to the inductor. Physics tells you that the voltage over an inductor is equal to L*dI/dt, which means that the more the current quickly change in the inductor, the more voltage you have accross the inductor. In you case you go from 0.656A to 0 in a very short time, inducing a reeeeaally high voltage accross the inductor, thus making you switch voltage increase (remember, your switch is directly connected to your inductor, thus if your inductor voltage rise, so will your switch voltage)

    What to do:
    Think about what the energy does: when your inductor current is flowing at its value of 0.656 A (before opening the switch), you have an energy of 0.5*L*I² stored in your inductor (magnetically stored). When your current is suddently stopped, this energy is converted in voltage, and try to escape by whatever means necessary (here, by blowing your switch and taking electron wherever it can. Physic always win).


    So why not offer to this energy a new path?
    The snubber capacitor is in the correct place, but have the wrong value.

    Keep thinking about the energy: your inductor has stored 0.224 Joules (=0.5*L*i²). So your capacitor must be able to store that much energy (actually, when closing the switch, the inductor will take its needed energy from the capacitor. If nothing is done, the capacitor will take this energy back. Fortunately, the resistor will progressively dissipate the energy, so the whole system will go down to 0V)
    So what about this capacitor? If the inductor has stored E= 0.224Joule, the capacitor must be able to give 0.224Joule.
    The energy inside à capacitor is E= 0.5*C*u²

    So here, the desired capacitor is 2*E/u²=5.715F if you want to have only a 280V rise accross your inductor.
    The resistor in the snubber is here to dissipate energy, so the whole system will go to 0V more quickly

    Hope that it helped =D
     
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