Delay on relay - Page 1
Forum section: electronics / beginners

author: Leuven
author_rank: Regular Contributor
timestamp: February 14, 2012, 12:31:25 pm
content: Hi, can anyone take a few moments to scribble down a schematic for a 12V delayed-on relay?From the moment it's triggered with a 12V source, the relay must still stay open for 1-10 sec (pot adjustable) and only close after the set time. If trigger voltage last lesser than the set time, cycle resets.It's for some car audio components which turn on at different times and I want to get the amp to turn on after everything else.Many thanks!

author: Jimmy
author_rank: Regular Contributor
timestamp: February 14, 2012, 12:35:28 pm
content: Try asking mr googleR1 1 1 Meg PotR2 1 10 K 1/4 Watt ResistorC1 1 10uf 25V Electrolytic CapacitorC2 1 0.01uf Ceramic Disc CapacitorD1,D2 2 1N914 DiodesU1 1 555 Timer ICRELAY 1 9V RelayS1 1 1A 120V SPST SwitchMISC 1 Board, Wire, Socket For U1R1 adjusts the on time.You can use a different capacitor for C1 to change the maximum on time.S1 is used to activate the timing cycle. S1 can be replaced by a NPN transistor so that the circuit may be triggered by a computer, other circuit, etc

author: Leuven
author_rank: Regular Contributor
timestamp: February 14, 2012, 01:48:52 pm
content: Thanks Jimmy, can you confirm this is a delayed-on circuit and NOT a timed-on (meaning it stays on for a set time)?

author: sacherjj
author_rank: Frequent Contributor
timestamp: February 14, 2012, 06:03:09 pm
content: I believe that is a timed on. Shorting trigger to ground will activate output. Discharge C1 until below the threshold and turns off the output.The Industrial Term you are looking for is Delay-on-Make.Here is a circuit that will turn on the relay after a certain delay of holding the button down. It is discharging the cap through the 1M resistor, and turns on the output when it reaches the trigger threshold (I think that is 2/3 Vcc if I'm remembering correctly.) Look at the LM555 Data sheet.The problem with something for this in your solution is when the switch is released, the cap will charge through the diode and 10K, turning off the relay pretty quickly. You would need to either use a latching relay or modify the trigger portion to stay on. But it sounded like you are using voltage to drive this anyway, so if you trigger with a MOSFET instead of the switch, with a signal that stays on, I think you will be good.Note: This doesn't "reset" the delay until a little bit of time. It is kind of accumulating on time for the button press. I think this will still work in your situation.

author: Jimmy
author_rank: Regular Contributor
timestamp: February 14, 2012, 11:35:47 pm
content: Sorry Leuven after reading your post again it seams I gave you the wrong circuittry

author: vk6zgo
author_rank: Super Contributor
timestamp: February 15, 2012, 01:42:02 am
content: All we did in the old days,was to parallel a big cap with the relay coil.The cap has to charge before the relay pulls in.The downside is that there is a delay on switchoff as well,as the cap sources current to keep the relay made.It is usually shorter than the on time delay.If you have a few spare contacts on the relay,you can probably get around this.

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 15, 2012, 01:57:34 am
content: Quote from: Leuven on February 14, 2012, 01:48:52 pmbuy a relay with 3 pin.gndcon, normally opened, and normally closed pins. just switch the NO and NC pin if the circuit is the other way round. with one circuit you can make both timed-on and timed-off function.

author: DrGeoff
author_rank: Frequent Contributor
timestamp: February 15, 2012, 01:59:00 am
content: Try using an NPN transistor with the pot in series with a suitable base resistor and the cap across BE. E to GND, C in series with relay to +V. Add a diode from base to the other end of the pot to discharge the cap rapidly when the input goes low.The actual component values can be calculated by the OP to suit the application.p.s. Should have added a diode across the relay coil. Use your imagination and pretend it is there

author: Jimmy
author_rank: Regular Contributor
timestamp: February 15, 2012, 05:55:07 am
content: Quote from: Mechatrommer on February 15, 2012, 01:57:34 amwhat happens when the radio/ circuit is off the amp will be on

author: Leuven
author_rank: Regular Contributor
timestamp: February 15, 2012, 02:40:16 pm
content: Thanks a lot guys. I took the one easiest for me to understand and added some notes on it, am I on the right track? Any suggestions for the Mosfet and the R2 resistor?Also quick question about the mosfet, I read somewhere I have to put a 100k resistor between the gate and source to have it off (non-conducting) by default, but I thought that thing is off by default unless I apply voltage to the gate, right??Alternatively, looking at diagram 2, would this help to reset the capacitor quicker at power off?

author: Leuven
author_rank: Regular Contributor
timestamp: February 15, 2012, 03:55:24 pm
content: Quote from: DrGeoff on February 15, 2012, 01:59:00 amIf I understand correctly, in this diagram the relay will take place of the LM555 in that when I apply voltage it will be transferred through the mosfet immediately. But because of the slow build-up due to the capacitor/resistor, relay won't turn on until a certain critical voltage is reached.

author: sacherjj
author_rank: Frequent Contributor
timestamp: February 15, 2012, 04:33:00 pm
content: Quote from: Leuven on February 15, 2012, 03:55:24 pmWell, it is a BJT, not a MOSFET. However, you could also do this with a MOSFET.What is happening is voltage of the input would normally turn on the transistor to activate the relay. However, the capacitor is "stealing" the current that makes it through the resistors to charge. Once the capacitor gets full, the current it requires goes down and is used to activate the transistor. The time to reach this point is determined by the values of R and C.The transistor is just acting as the switch when we reach a threshold. For a MOSFET, it wouldn't be the current flowing through the transistor, but the gate voltage of the transistor that does the actual switching, but essentially the same thing occurring.The relay isn't acting as the LM555 in other schematics, the transistor is. It is acting similar to what exists in the LM555 at the threshold pin, just much simpler.You MAY required protection on the input side with a small resistor in front of the diode, if the discharging capacitor through the diode is too high of a current surge.

author: Rudolfo
author_rank: Contributor
timestamp: February 17, 2012, 07:46:34 am
content: To avoid overvoltage for the BC639 a freewheeling diode is recommended across the relay coil. You may also take a look at this article:http://www.powermanagementdesignline.com/showArticle.jhtml;jsessionid=PKP0IC5ROGR3ZQE1GHOSKH4ATMY32JVN?articleID=170101102&queryText=iC-Haus

author: Zero999
author_rank: Super Contributor
timestamp: February 18, 2012, 03:28:34 pm
content: The disadvantage of the transistor circuit is it's not very accurate or stable - the delay will depend on the supply voltage and gain which can vary widely.How much power can the trigger signal provide?Another option is to connect the relay from the positive to the 555's output, as per the circuit posted by Jimmy.https://www.eevblog.com/forum/beginners/delay-on-relay/msg90954/#msg90954If the trigger signal can't provide enough current to drive a relay, the 555 timer circuit can be combined with a transistor or MOSFET.