latching relay circuit

Hi,

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! :)

cheers, Jamie

Reply to
Jamie Morken
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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.

Reply to
Homer J Simpson

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:

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Reply to
mkaras

"Homer J Simpson" schreef in bericht news:qCF_g.26798$P7.25514@edtnps90...

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

Reply to
petrus bitbyter

Try this!

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Reply to
Paul Taylor

"Paul Taylor" schreef in bericht news:ehi4kt$au9$ snipped-for-privacy@south.jnrs.ja.net...

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

Reply to
petrus bitbyter

On a sunny day (Mon, 23 Oct 2006 17:41:46 +0200) it happened "petrus bitbyter" wrote in :

C1 R1 ---------------||----------===-------------------------------0 | | | ------------------- | | | push button | Thyristor LOAD 16A | | \\ R3 --- | | D2 |- \\ ----===--- \\ / Th1 | 230V AC |--|>|--| | \\ --- -------

12V | | + | |------------/ \\ \\ / zener --- \\ === | | ------- 20A triac / \\ --- [ ] [ ] | | D1 | C2 | R2 | R3 | | | | | | -------------------------------------------------------------0

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.

Reply to
Johnny Chipmelter

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.

Reply to
Homer J Simpson

Hi,

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

cheers, Jamie

Reply to
Jamie Morken

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

Reply to
Homer J Simpson

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