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latching relay circuit

Discussion in 'Electronic Design' started by Jamie Morken, Oct 22, 2006.

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  1. Jamie Morken

    Jamie Morken Guest


    I am trying to make a circuit that will function so that when the power
    is disconnected from the running device, and then the power is
    reconnected, the device will stay turned off, until it is manually
    turned back on with a push button switch. I was told that this could be
    done with a latching relay and a DPDT switch, what would the circuit
    look like for this? The device runs on 120VAC and draws 15Amps. I have
    seen some latching relays on ebay for $1, but I can't figure out how
    they would be used to build this circuit! :)

  2. Nope. Just a relay and a push button.

    Wire the push button and the NO relay contacts in parallel, and that pair in
    series with the relay coil. Wire the load across the relay coil.
  3. mkaras

    mkaras Guest

    If you need to isolate the load from the push button circuit, which is
    often the case when the relay coil and push button curcuit may operate
    on an low voltage DC and the load may be an AC motor on mains power
    then it is best to use a separate set of NO contacts on the relay to
    switch the motor power.

    Look here for some ideas on the relay latch circuit with a push button:
  4. Will work fine as long as the pushbutton is rated for the same current (15A
    or more) as the relay contacts. Otherwise you better take a double pole
    relay and power the load and the relaycoil separately. Keep in mind that the
    inrush current may be much more then 15A and that the contacts must be able
    to handle it.

    petrus bitbyter
  5. Paul Taylor

    Paul Taylor Guest

    Try this!
  6. Nice circuit but the relay used will be too light to switch 15A. To switch
    15A (or more, inrush current you know) you need thick contacts and a high
    contact pressure. If you can find a relay that can switch the required
    current and has a 9V coil, it will need quite some current from the 9V
    battery to be activated.

    petrus bitbyter
  7. C1 R1
    | |
    | ------------------- |
    | | push button | Thyristor LOAD 16A
    | | \ R3 --- |
    | D2 |- \ ----===--- \ / Th1 | 230V AC
    |--|>|--| | \ --- -------
    12V | | + | |------------/ \ \ /
    zener --- \ === | | ------- 20A triac
    / \ --- [ ] [ ] |
    | D1 | C2 | R2 | R3 |
    | | | | |

    Solid state solution:
    C1 must provide enough current to keep thyristor on and drive triac, should be HV cap.
    Resistors R2 R3 only against leakage, some kOhm.
    R1 fusible carbon for current limit in case something shorts.
    D1 creates stable 12.7V AC wave -0.7 to +12V.
    D2 rectifies this wave.
    C2 smoothes the rectified wave.
    Th1 will latch if button pressed.
    R3 limits peak gate current.

    IIRC triac can be driven positive on gate in all quadrants?
    This solution can sometimes be cheaper then a relay.

    Circuit not tested, use at own risk.
  8. Rube Goldberg would be proud.

    Use a regular contactor, wire a set of NO contacts in parallel with the
    pushbutton - normal motor starter set-up.
  9. Jamie Morken

    Jamie Morken Guest


    On the relay it says 150V/15Amps or 300V/10Amps. Why does the current
    rating go down with increasing working voltage?

  10. It's more the ability to open the circuit that is the key.
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