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simple relay contact delay

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Zocco

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Hi,

I am working on a project that requires a relay to remain held on for a few milliseconds while a thermostat switches and picks up the power again. The relay coil is operating on 240v AC and there is no low voltage DC available to power a timer. I wish to use only passive components eg. resistors and or capacitors working at mains voltage.

You may discern from my question that I am pretty much a novice where electronics is concerned as I suspect there is probably a simple solution. Any help will be gratefully received. Thanks.
 
I'm out of my league here. But I have seen people use a Resistor and a Capacitor in circuit to create Voltages needed to power a circuit like what your describing.

The principle I think is to allow the time of the capacitor to discharge through the resistor over time. It's called and RC circuit. I don't know if you can do it with AC. How it is accomplished is when the switch powering your relay coil is open; the voltage from the Capacitor drains slowly through the resistor. This can be calculated to meet your requirements. Providing my answer is correct.

Edit: You might need to decouple the circuit and that is more complicated. This would keep the circuit from chattering the relay coil.

Edit:Edit:

RC Snubber circuits, wiki
 
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You can not use any passive circuit to delay an AC powered relay, only a DC relay. The simplest way to get an AC delay is to buy an AC delay relay built for that purpose.
 
Thanks, crutschow.


kv
 
Thanks for the information, much appreciated.
Unfortunately I had already bought my relay before I was aware of the problem and was hoping that I would not need to buy another. Am I correct in thinking that there is a type of relay with "make before break" contacts. This would do the job but at more expense and I would be left with a redundant relay.
 
A make-before-break relay will still require the purchase of another relay.

The only way I can think of to use your present relay is to rectify the line voltage and operate the relay from DC so you can use an RC delay circuit. The circuit would consist in order: mains, bridge rectifier, series resistor, capacitor to common, relay coil.

But an AC relay operating on DC requires significantly less voltage for the same coil current, since the relay inductance is no longer limiting the current. You would have to experiment by measuring (or knowing) the normal AC coil current and then adjusting the value of the series resistor (in the RC delay circuit) to get the same current with the rectified DC.
 
Thanks crutscow, I was already thinking along similar lines and thought that maybe one of my several redundant phone chargers would do the job of creating a suitable DC supply.
 
Thanks crutscow, I was already thinking along similar lines and thought that maybe one of my several redundant phone chargers would do the job of creating a suitable DC supply.
A 240VAC relay will likely require higher DC voltage to operate than a phone charger will output (which is typically less than 10VDC).
 
I have done this with a 120V AC relay. I hooked it up to a DC supply, and measured at what DC voltage it pulled in, and then added ~20%. I rectified the AC line with a simple half-wave rectifier into an electrolytic capacitor, something like 100uF@250VDC. The capacitor charges to pretty close to 160VDC , so I put a resistor in series with the relay coil to drop the coil voltage to what I measured in step one, above. The delay is determined by how long it takes the capacitor to discharge through the series resistance of the coil and the series resistor. Calculate the wattage of the resistor, as it will have to be like a 5W resistor... The delay is determined by the size of the capacitor.
 
I have done this with a 120V AC relay. I hooked it up to a DC supply, and measured at what DC voltage it pulled in, and then added ~20%. I rectified the AC line with a simple half-wave rectifier into an electrolytic capacitor, something like 100uF@250VDC. The capacitor charges to pretty close to 160VDC , so I put a resistor in series with the relay coil to drop the coil voltage to what I measured in step one, above. The delay is determined by how long it takes the capacitor to discharge through the series resistance of the coil and the series resistor. Calculate the wattage of the resistor, as it will have to be like a 5W resistor... The delay is determined by the size of the capacitor.
You can reduce the power in the resistor by eliminating the capacitor. That way the DC voltage is the average of the AC voltage, not the peak.

But if the half-wave voltage average is not sufficient to pull in the relay or the relay chatters from the half-wave AC, then you would need to go to a full-wave bridge.
 
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To avoid having to measure the required DC current to operate the relay, you can likely use the same current as the rated AC coil current. Just measure the DC coil resistance and then calculate the DC voltage required to give that AC coil current. Then calculate the series resistance required to give that coil voltage from the average DC voltage you are generating. For a full-wave rectified unfiltered sinewave the average voltage would be (0.636/0.707 ≈ 0.9) of the RMS voltage. For a half-wave rectified sinewave it would be ½ that.
 
Thanks Guys, Your efforts are much appreciated. It seems that things can't get much more simple, you have given me some good ideas to try.

Zocco.
 
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