I am trying to activate a 12V Relay by simulating a switch. The relay must trigger but only once but the 12v will remain live in the circuit. How can I activate the relay as if it was energised for a split second

Cheers

MrPad

 

How can I activate the relay as if it was energised for a split second

What is the duration of a "split second"?

The relay must trigger but only once but the 12v will remain live in the circuit.

Does this mean the relay is to be self-holding?

 

I have got a feed of 12v that stays on for 15 seconds and I want that feed then to feed another part of the circuit that will emulate a switch, I had thought of using a relay, but the relay will stay on for the period the voltage is present and then go off, I only want the "switch" to activate when the voltage is present

So as you can see from above the voltage comes in but I also want to emulate the pushbutton so the voltage/current flows only for that split second to the part of the circuit where a 12v trigger is required.

Hope you can help

MrPad

 

Use a capacitor from the 12V to the relay coil. If the capacitor is large enough, it will provide one pulse of current large enough to activate the relay for "a split second".

 

what do you recommend using does it need to electrolytic..?

 

Is there any other way I can do this without using large capacitors



MrPad

 

A transformer perhaps?

The initial change of 0 volts to 12 volts is seen by a transformer primary, and the relay is connected to the secondary. The secondary will only have a voltage present for a very short period in response to the changing flux introduced by the initial voltage change. After the voltage becomes steady the secondary will cease creating an output and the relay will release.

 

That'll work. I didn't think of it because I have large electrolytic capacitors laying around all over the place. Easier than thinking about what kind of transformer will do the job, and if I have one. Hopefully, MrPad will have a pile of transformers laying around.

This could also be done with a small capacitor and a transistor to amplify the pulse, but that solution is more complicated than the original circuit!

 

That'll work. I didn't think of it because I have large electrolytic capacitors laying around all over the place. Easier than thinking about what kind of transformer will do the job, and if I have one. Hopefully, MrPad will have a pile of transformers laying around.

This could also be done with a small capacitor and a transistor to amplify the pulse, but that solution is more complicated than the original circuit!

I really don't think this is such a good idea. A transformer used in this way is likely to draw a large current from the supply, because the current in the primary would build up to a value limited only by the winding resistance. This could lead to a dangerous situation, as the transformer and whatever was feeding it could burn out. Even if the current is limited by a suitable external resistor, the consumption is likely to be heavy.

A bipolar transistor or MOSFET driver would indeed be the normal way to do this, but might be more difficult for the OP: the fact that he has to ask would suggest he has not had much experience with this sort of thing.

 

I agree with Aduster. Even if the primary current doesn't burn the transformer out it's wasting current and will probably be as large #12's capacitor. Actually, since this is only a 12VDC circuit and a "split-second" contact closure, the electrolytic cap can be physically, relatively small.

Like this, for a few seconds of on-time. The resistor provides a discharge path for the cap when the switch opens. And the diode prevents the relay from momentarily closing a second time when the switch opens. The on-time depends on the relay's coil resistance, the relay's drop-out voltage and the size of the capacitor.

Ken

 

That's where I drop the ball. I couldn't quantify the capacitor without knowing about the normal current the relay requires. You're pretty good at this!