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Relay delay circuit - please help.

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Odysseas

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Hi everyone,

I need some help with a fairly simple circuit, that is supposed to activate a relay a few seconds after power is applied to the entire circuit. The delay should be controllable through the potentiometer.

Now my questions: Are the values for both the potentiometer and the capacitor appropriate for a darlington pair? I was thinking of using a pair of 2N3904 - would these work?

Am I right in thinking that the relay should open again, immediately when power is removed from the circuit?

Thanks for any help!! Odysseas
 

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Any general purpose transistor should be fine. Like BC548, 2N2222, 2N3904...

But you shoul swap the Capacitor and Potentiometer positons.
The capacitor must be in place of the potentiometer and vice-versa.
 
A 4.7k pot and the 100uF capacitor will result in a delay of only milli-seconds. Use a 100k pot for a few seconds.

If the pot is turned too low then the current will be too high and the pot and transistors will blow up. Connect a 4.7k current-limiting resistor in series with the pot.

If the 2nd transistor gets warm and the pcb is slightly conductive due to humidity then it will not turn off. Connect a 100k resitor from its base to its emitter to turn it off when it should be off.
 

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Okay, I have constructed it, and it works. However, when I disconnect external power, and reconnect it immediately, the relay-delay is very short (capacitor is still charged.)

Would it help if I placed a 1k resistor across the capacitor, to discharge it at least after a short while?
 
That won't help, it will stop the capacitor charging!.

The classic method is a diode, connect it anode to +ve of the capacitor, and cathode to the +12V supply (across the pot and series resistor). When the supply is removed the capacitor will discharge through the relay and transistor - until the transistor switches off. In any case, it will discharge a LOT better than now.

But I'm with everyone else, this is a really crude and not very effective circuit - but it's an interesting learning experience for you!.
 
As Nigel implied, the diode only works if there is a return path to the other end of the capacitor (GND) after you remove power. If you disconnect power with a switch, you can use the circuit below. It requires a DPST switch. The 100 ohm resistor limits the peak current in the diode and the switch.
Alternately, if you can afford to waste some power, you can eliminate the 100 ohm resistor and the switch pole to ground. Add a resistor of a few kohms from +12V to ground instead.
If a recovery time of a couple of seconds is OK, you can omit the 100 ohm resistor, and just connect the diode across the pot. The 4.7k resistor becomes your discharge current limiter.
 

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