HalfSpec Posted May 15, 2005 Report Share Posted May 15, 2005 Hi, I’m very new here, and I basically found this forum while searching for answers concerning my latest project. I’m trying to build an automotive “turbo timer.” For those of you who don’t know Commercial turbo timers consist of 3 wires; a 12v wire, a ground wire, and an ignition monitor wire. While the key is turned the ignition wire supplies 12v. When the key turned back the ignition wire goes to 0v and the turbo timer immediately routes 12v back into the ignition wire for the set amount of time.This is what I’m trying to build. I am only a sophomore electrical engineer, so all I’ve been taught so far is theory. I need someone who knows electronics (specifically the LM555 IC) to help me out.I’ve attached a schematic of my proposed circuit. In theory it should work as follows:Key is turned in vehicle and 12v goes through ignition wire, diode, relay, resistor, and then to pin 2 on LM555. When key is turned back ignition wire goes to zero making pin 2 go low and causing pin 3 to fire and power the relay, which in turn directs 12v back to the ignition wire for the amount of time dictated by R2 and C2. The diodes and C1 are there to help prevent voltage spikes. http://www.halfspec.com/sc2.GIF (Full Size)That’s it! I was hoping it would work, but right now it’s not. I’m pretty sure the timing setup dealing with pins 4,6,7, and 8 is correct, because this is pretty standard in other LM555 circuits. I appreciate any input on this, because I’ve been trying to diagnose the problem all day and haven’t been able to make any progress. Lane Simmons[email protected] Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 15, 2005 Report Share Posted May 15, 2005 The diode should be in reverse paralell with the relay.Pin 5 should be by passed to 0V with 10nf.An AC coupling capacitor, configuration like the one in the circuit I've posted can help if pin 2 is held low for too long. Quote Link to comment Share on other sites More sharing options...
HalfSpec Posted May 15, 2005 Author Report Share Posted May 15, 2005 Alun, Thanks for the fast response. Here is my updated schematic:http://www.halfspec.com/sc3.GIF.I'm assuming you mean the diode on the output pin of the LM555. Is this what you mean by reverse parallel? Also, did I correct my mistake by connecting pin 5 to the ground with the 10nf capacitor?I'm not really sure what you mean by holding pin 2 low for too long. Is there a set amount of time that trigger pin should be held low? Thanks for your help, and sorry for my ignorance. I really should have been further a long in my electrical knowledge before I started this project, but I couldn't wait to get started.Lane Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 15, 2005 Report Share Posted May 15, 2005 The diode needs to be connected in reverse paralell with the relay as I've shown, your modified circuit had the diode shorted so it wouldn't do anything.If you hold pin 2 low the timer will remain on untill pin 2 is released, the 1nf AC coupling capacitor and 10k resistors solve this problem with my circuit. Quote Link to comment Share on other sites More sharing options...
HalfSpec Posted May 15, 2005 Author Report Share Posted May 15, 2005 Alun, Thanks for being patient with me. I think I've got it straight now, although it seems like this would just short the circuit out at the diode:http://www.halfspec.com/sc4.GIFOk, what you're telling me seems to go against what I've based my design on. This is how I understood the trigger pin to work. When pin 2 is pulled low pin 3 fires for the present time. I thought that if pin 2 was pushed high while pin 3 was still firing the process would start over. Is this not true? I designed this circuit so pin 2 would NOT be brought high again until pin 3 was done and the whole process shut down. Is this bad logic? If it is I just need to scrap this circuit and start over.Thank you for your timeLane Quote Link to comment Share on other sites More sharing options...
Guest Alun Posted May 15, 2005 Report Share Posted May 15, 2005 A 555 timer is triggered when the voltage on pin 2 goes negitive for a short period, if it remains negative for longer than the timing cycle then the timing cycle will carry on untill it goes high. If the cycle is not finished when pin 2 goes high then the cycle will finish as it should.When pin 2 briefly goes low the output of the 555 will remain high for 1.1RC seconds, if pin 2 is low for longer than that it will remain high untill pin 2 goes high.So yes I'm sorry your logic is flawed you'll have to start over I'm afraid. Quote Link to comment Share on other sites More sharing options...
HalfSpec Posted May 15, 2005 Author Report Share Posted May 15, 2005 Thanks for clearing up everything for me Alun. I'd rather be wrong and have direction than think I'm right and looking for solder splashes :)I truly appreciate your insight.Lane Quote Link to comment Share on other sites More sharing options...
HalfSpec Posted May 16, 2005 Author Report Share Posted May 16, 2005 How about now? :)http://www.halfspec.com/sc5.GIFI removed the diodes for simplicity. I'll figure them out once I get a working circuit. ThanksLane Quote Link to comment Share on other sites More sharing options...
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