## Easy delay ON using SPDT relay for DRL setup with 12V DC

 I will cut to the chase. I do not know a lot about electronics. I got some basic knowledge and I am capable of learning, but I am too busy with other stuff. Hence, I am simply looking for an easy solution. Attached, I hope I can do that... is a diagram of my current setup. In short - DRL LED rings on my car (I did bi-xenon projector retrofit and added DRL LED rings). I want to have them ON only when engine is RUNNING. Initially I used a fuel pump feed, but it was poor design - I was simply taking current "after" the relay by wire attached to "exit/fuel pump supply" post on the relay. I does not look good. Add a circuit - well, no fuse is HOT when running. They are energized when ignition is ON. I checked all of them. So, the idea is to use a SPDT relay, +12 feed from a fuse HOT when ignition is ON, and ground from oil pressure switch. How it should work: 1. Ignition is OFF - no 12V power, hence LED is OFF. 2. Ignition ON, engine OFF. Both 12V and ground energize the relay that would switch to "empty post". LED remains OFF. 3. Ignition ON, engine RUNNING. 12V remains ON, but ground is gone (oil pressure switch receives pressure when engine is running what will switch it off). However, I am afraid that they will be a quick blink when I turn the ignition ON. LED will react faster than the relay. Hence, I want to add something on the 87 post to delay the activation of LED. I imagine it would work like that - turn ignition ON and for a fraction of a second +12V will reach the 87 post, but a capacitor with resistor will not pass it through to the LED, in the meantime relay will energize. In result the capacitor did not charge. When I start the engine, relay will go back to resting position allowing +12V to flow and charge capacitor and activate LED. Time? 1-2 seconds will be enough... by PLP April 24, 2013 This'll do what you want. :) by signality April 25, 2013 simple... I guess it is. Simple for you to understand. Just like chelating ligands for me. OK, please walk me through. I get the LEDs. They have some driver built in so no need of extra resistors. However, although I do not think it matters, I put a 50Ohm, 5W resistors to bring the current down when in position light operation mode. It is switched by a SPDT relay. Let's go back here. Engine off - I get ground through oil pressure switch, but no 12V (call the positive 12V for short). Ignition ON - both ground and 12V present. LED remain OFF - am I correct? Once the engine is running, ground is gone, 12V remains and transistor will switch allowing the current to pass through. Is that correct? I am a chemist, well, chemical engineer so as physics is OK, electronics is limited... Thanks for helping. by PLP April 25, 2013 Sorry .... I've been in too much of a hurry. I think I got the logic sense of the oil pressure switch the wrong way round. Let me redraw and explain a bit more. by signality April 25, 2013 thank you in advance. Maybe you have some other idea as what to use as a trigger "engine is running"? by PLP April 25, 2013 OK. Have another look: I've modified the circuit and added a description. You should be able to build this on 0.1" stripboard. I don't know what current your LEDs draw but the MOSFETs have << 10mR on resistance so their I^2R power dissipation will be negligible for I < 10Amps, so the one you choose should not need a heatsink. :) by signality April 25, 2013 thanks for the new design. Unfortunately I must add new information to it. When cranking engine the +12V turns off. Once I release the starter and engine begins to run the +12V is back. Would it affect the circuit? Also, can I add anything to make the "turn on" smooth without making the circuit too complicated. Finally, to answer your question. The LED pulls about 2A@13.5V, hence no heatsink as I understand. by PLP April 29, 2013 "When cranking engine the +12V turns off. Once I release the starter and engine begins to run the +12V is back. Would it affect the circuit?" The simulation starts the 12V feed coming up from 0V so the circuit will work just as the simulation shows. "... can I add anything to make the "turn on" smooth without making the circuit too complicated." Not easily. One way is to add a capacitor across D3 to slow down the gate voltage pulling down to 0V at LED turn-on. That slows down the turn-on of M1. At present M1 is either fully with a few mR (say 10mR) between source and drain pins => (2*2A)^2* 0.01R = 160mW i.e. negligible power dissipation; or fully off with a drain current of <<1uA => 13.5A*1uA = 13.5uW i.e. negligible power dissipation. However if M1 turns on slowly it dissipates power because at the mid point of the transition from on to off it will have a significant voltage across it and a significant current through it => gets hot => needs heatsink &/or bigger device. The "proper" way to do it is to use pulse width modulation to switch the LEDs on and off. Start with LEDs off all the time then switch them on for say 100us every 1ms then 200us every 1ms then 300us every 1ms all the way up till they are on for 1ms every 1ms. i.e. 100% duty cycle = fully on. That way the MOSFET is still only ever fully off or fully on => low dissipation. But that is a more complex circuit. It at least adds a 555 timer, maybe another device too. There may be a way to use the two transistors that are already there to make a self oscillating PWM circuit but I haven't time to play with that now. Ramping the LEDs off again would be much more complex because your 12V feed is either on or off and off = no power to drive the down ramp. :) by signality April 30, 2013 It has been a while, but I wanted to thank to @signality for the diagram. It has been over two years and the circuit works just fine. by PLP December 22, 2015 @PLP, That's great news! Many thanks for letting me know. Simple little circuit but it very neatly showcases the power of simulation. :) by signality December 22, 2015