How can I calculate capacitance value for DIY impulse/oneshot relay?

Discussion in 'General Electronics Chat' started by strantor, Jul 7, 2012.

  1. strantor

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    We occasionally get posters who ask for something along the lines of "I have put a DC actuator on my door, and I want to be able to press a button and have it close, and when I press that same button again, I want it to open." In these I usually recommend an impulse relay, like this one. It fits the job perfectly; it changes state on energize,and remembers its state with no power. Only problem, it's prohibitively expensive and uncommon, so not available in many voltages.

    I was thinking, if you were to put a capacitor in series with coil of a DC relay, the capacitor should pass a brief pulse through the coil. hopefully enough to briefly energize the relay. If my theory is right, then we just made a oneshot relay without a 555, pretty handy.

    Now, take that idea and apply it to a dual coil latching relay. send pulse from the capacitor to the common terminal of one of the sets of contacts; from the SET contact, go to the RESET coil, and from the RESET contact go to the SET coil. Now you have an impulse relay.

    Only problem I can see, is that the capacitor would have to be sized properly or else the latching relay might switch more than once, maybe even oscillate several times if the capacitor was too big. This wouldn't be too big of deal for the simple oneshot, but for the impulse relay it is.

    So how would you calculate the size of the capacitor so that the latching relay will switch once and only once? I assume you would need to know the DC resistance and maybe the inductance of the coil?
     
  2. Externet

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    Applying a pulse to a solenoid is not repetitive trough a series capacitor fed by DC.

    The formula you are looking for is

    C=Q/V ---->Capacitance = Charge/Voltage

    And

    Q=It ---->Charge = Intensity x time

    Replacing the second Q into the first formula, you obtain

    C=It/V

    So the time needed to kick your coil is

    t = VC/I ----->time = Voltage x capacitance / intensity

    but Voltage / Intensity = resistance

    then, -----> time = resistance x capacitance

    or, the well known t=RC

    ----------------------------------------------

    If your coil needs 0.1 seconds pulse to engage and behaves properly,

    Make 0.1 = RC

    The resistance of the coil is known, -say 100 Ω- then

    C = 0.1 / 100

    C = 0.001 Farads = 1000μF

    Hope it is clear and helps.
     
  3. gerty

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    Also don't forget you have to allow for the difference between the "pull in" voltage and the "drop out" voltage .
     
  4. THE_RB

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    I've done it in practice using a charged cap to pulse a mechanical counter relay, the cap was charged with a very small current as operation was infrequent, then contacts closed between cap and relay coil and the cap held enough charge to operate the counter relay. The benefit was that a small battery could be used, and also was that the contacts could be held closed for seconds but the relay coil only got 100mS or so of current, so those seconds didn't drain the battery.

    Testing was very easy I just charged a test cap using a variable PSU, then touched the charged cap on the relay terminals. This found the size cap that was needed to reliably work the relay at minimum battery voltage, and from memory the cap was 2200uF. If you have one of those little latching relays it might work from 220uF or 470uF. Because they don't have a normal relay's armature spring they don't need as much energy.
     
  5. strantor

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    Externet, Thanks! the math makes sense, but I don't understand this statement:
    Are you saying that it would never oscillate or that it wouldn't work or what?
     
  6. strantor

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    RB, I'm not sure, but it sounds like you're referring to having charge stored in a cap and then discharging it across the coil using a switch of some sort. I'm talking about putting that cap in series with the relay coil. No stored charge, no switch. As soon as DC hits the cap, it should permit a brief pulse and then block the DC.
     
  7. strantor

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    Can you expound on that? I'm not sure what you mean.
     
  8. Externet

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    Correct, will not oscillate. Feeding DC trough a capacitor in series will yield only one burst of current to the coil. When the capacitor reaches charge, there is no more current flowing.

    It is convenient to leave a 'large' resistor in parallel to the capacitor to discharge it for the next time to operate. Such 'large' resistor value not to affect the desired performance.

    (+)------------||-------------coil------------(-)
     
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  9. THE_RB

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    Yeah its six of one, half dozen of the other. :)

    If you leave the cap in series you need the resistor across the cap (as Externet said) you have the same function but the 4 parts arranged differently.

    The exception would be if you are using a push-pull driver like a CMOS output pin, that would apply + or - current through the coil, I have some tiny DIP toggle style relays here like that that work with bidirectional current.

    The larger toggle relays tend to have two separate coils or a centre-tapped coil.
     
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  10. gerty

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    I was going under the assumption (bad move, I know) that you wanted to put the cap in parallel with coil. In doing so, when removing power, you would have the stored charge in the cap to keep the relay pulled in for a short time .
    The point I was trying to make is the pull in voltage for a relay is higher than the drop out voltage, and the title of this post is about calculating capacitance values.Therefore you would be using 2 different voltage values when calculating capacitance for the relay coil.
    Now tell me you're not confused :D
     
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