Hi .
I'm looking for a simple time delay "like with a capacitor and a resister" for a 12v DC relay to energize a few seconds after power has ben applied.
Can someone help?
Thanks.

 

Due to the current requirements of the relay, you will need active components, such as a 555 timer.

 

Due to the current requirements of the relay, you will need active components, such as a 555 timer.

The relay is a printed circuit board relay with only milliamp on the coil to energize it.

 

Even at milliamp levels, you'd still need a large fractional farad capacitor to provide a multi-second delay.

If the relay is already mounted on a board, what is the purpose of wanting a delay? What is the project? There may be an easier solution to hack the circuit.

 

You already have a resistor in the relay winding,all you need is the capacitor.

Measure the resistance of the coil,& if it is around 1kΩ or more you are probably good to go with a parallel capacitor,-----try a 2000uf electrolytic.

The downside of this circuit is that it will not be a delay followed by a fast switch,as the actual switching may be a bit slow once the "operate" point is reached.
You will also have a delay when the relay releases.

If you want to avoid these things,it gets a bit more complicated.

I would try the simple method first,as it has been used successfully for many years.

 

You can't use the relay coil's resistance as part of the RC circuit. The only way to do that would be to put the capacitor in series with the relay coil, and that would cause the relay to close immediately when power is applied, then open.

I don't think there's any practical way to avoid an active component of some kind, even though the relay coil current is very low, but the active component doesn't need to be an IC. A transistor or MOSFET can be used to control the current to the coil, driven from an RC circuit that uses a reasonable capacitor value.

There are various ways to do it. A MOSFET is easier to drive, but also easier to damage. Probably my recommendation would be a Darlington.

When power is applied, current through R charges C until its voltage reaches about 1.2V, when Q1 starts to conduct, and remains conducting until power is removed.

Q1 is a Darlington which has an extremely high gain (10k or more), so the base current needed is tiny, so you can use a high value for R (e.g. 1 megohm or more), so you can use a relatively small value for C.

The time taken for a capacitor to charge through a resistor to about 1/10th of the supply voltage is about 0.1 RC. So if you want a two second delay, RC must be about 20. (R in ohms, C in farads.) So if R is 1 megohm, C should be about 20 uF (22 uF is a standard value).

D1 is there to discharge C when the supply is removed, but won't be very effective in this circuit - it can only discharge C half way because of its 0.6V forward voltage. I don't see any simple way to avoid this problem. If C isn't discharged fully when power is removed, then when power is next applied, the delay before the relay activates will be shorter than it should be. It could take minutes or even hours to fully discharge by itself.

Edit: You could add a resistor across C, and it could be quite a bit lower than R. For example, 220 kilohms. This would affect the delay time slightly; C would have to be recalculated. D1 could then be omitted.

D2 is there to protect Q1 against back EMF from the relay coil when Q1 turns off. But in this application, Q1 never turns off. So arguably it isn't needed. But I recommend keeping it.

This might be a better option:

Here the transistor is used as an emitter follower. The voltage across the relay coil will rise gradually as C charges through R, until it reaches the relay's pull-in voltage, where the relay will close.

With this circuit the time delay is affected by the actual pull-in voltage of the relay. Assuming that it's around 9V (typical for a 12V relay), the delay will be about 1.6 RC. So for a two second delay, RC must be about 1.25. Assuming R is 100 kilohms, C must be about 12.5 uF.

Edit: Or you could use R=120k and C=10 uF. It's not that important, since electrolytics usually have a tolerance of +/- 20% or worse anyway.

D1 will be more effective at discharging C in this circuit. Again D2 is arguably not needed but I would keep it.