Which circuit is better for a basic time delay - Page 1
Forum section: electronics / beginners

author: Grayfox
author_rank: Regular Contributor
timestamp: April 05, 2016, 12:51:02 pm
content: Which circuit is more suitable for a time delay and are there any issues to using one over the other?The power source is 12v, the relay is a NO type with a coil resistance of 1.2KOhm and the capacitor will range from either 1800µF to 5600µF(5seconds to 20seconds)

author: alexanderbrevig
author_rank: Frequent Contributor
timestamp: April 05, 2016, 01:05:49 pm
content: Typically, this is something that comes to mind for the best time delay circuit that uses relays:

author: Seekonk
author_rank: Super Contributor
timestamp: April 05, 2016, 02:46:43 pm
content: They are not the same type of delay and neither of them actually work as shown. You should do more reading before you ask a question.

author: MattHollands
author_rank: Frequent Contributor
timestamp: April 05, 2016, 03:27:08 pm
content: As Seekonk points out, I'm not sure if this is what you're trying to do.In the case of the first circuit, initially, there is no charge across the cap so the voltage across the "resistor" (the relay coil), will be your battery voltage. Therefore current will flow, and the relay will close. This current will begin charging up the capacitor and so the voltage across capacitor will increase, the voltage across the "resistor" will drop and eventually the current will become too low to turn on the relay, and the relay will turn off.In the case of the second circuit, you can pretty much ignore the capacitor because it is in parallel with the "resistor" and so the voltage across the "resistor" is always your battery voltage so current will always flow and your relay will always be closed. The only time the capacitor is interesting is when you remove the battery after some time. In this case, the capacitor will contain enough charge to keep current flowing in the "resistor" for a while, so the relay will stay closed for some time before opening..

author: vk6zgo
author_rank: Super Contributor
timestamp: April 06, 2016, 04:18:31 am
content: Quote from: MattHollands on April 05, 2016, 03:27:08 pmTherewill bea very short delay on turn on with the second circuit,as real batteries/supplies have some internal resistance.I have found,to my chagrin, in the past that this can catch you out in situations where you have groups of cascaded relays.

author: Grayfox
author_rank: Regular Contributor
timestamp: April 09, 2016, 01:44:38 pm
content: I found this schematic on youtube for a delay off and made it up using some spare parts.Only issue is the relay stays energised even when the cap is 470uF and resistance on the trim pot is 600ohm.An online cap discharge calculator says the charge should last less than 2 seconds or does this calculator not apply in this situation.

author: Seekonk
author_rank: Super Contributor
timestamp: April 09, 2016, 01:56:51 pm
content: I think you destroyed he transistor with high base current. I would use a 10K resistor to the base. Then Place your variable resistor across the cap with at least several hundred ohms in series to limit current.

author: Grayfox
author_rank: Regular Contributor
timestamp: April 09, 2016, 02:04:40 pm
content: Quote from: Seekonk on April 09, 2016, 01:56:51 pmThing is if I disconnect power from the constant side the relay switches to the open position and if I reconnect power it stays in the open until the cap gets recharged.Unless I am using the wrong type of transistor.The original design just said 3K-10K resistor, but I want this time delay to be some what adjustable so I swapped it for a trimpot.I first tried 9.5K the circuit stayed powered for a 3min so I kept reducing the resistance on the trimpot till I got to 600ohm