viven44 Posted March 29 Share Posted March 29 As I was thinking about it more I realized that I had to be making a bad assumption. I was thinking that an LED had the same 0.7V drop as a regular diode. But that clearly can't be right or it would act about like a dead short if it was put between 12V and ground. Having an internal resistor makes a lot of sense! Yes it does make sense to sell it with a built-in resistor. A stand-alone LED is practically a transistor. The ON resistance aka. RDSon (drain to source resistance) when ON would be negligible. Maybe 10s of milli-ohms. That is why I was struggling to comprehend why the relay wouldn't be pulling-in everytime as it was basically a dead-short everytime the clutch was pressed. With the built-in resistor (which makes that LED a "module") it makes total sense now... of course, none of this matters with the diode in the unarmed-clutch-press scenario. Link to comment Share on other sites More sharing options...
Nothing Special Posted March 29 Share Posted March 29 In that schematic, the LED would definitely come on when the system is armed. The LED is connected to ground on one side and 12V would be supplied via the bosch relay on the other side. But agreed that your current schematic would work as well with the diode! The LED would come on when the system is armed. But it would also come on when you were starting the truck normally (ignition switch and clutch switch both closed). Of course you could prevent that by adding a diode in the wire going from the module out to the "bottom" of the clutch switch (pointing the opposite way of the one that's now shown in the wire above that). Lots of ways to skin the cat (which is a rather morbid saying...). Link to comment Share on other sites More sharing options...
Gary Lewis Posted March 29 Author Share Posted March 29 In that schematic, the LED would definitely come on when the system is armed. The LED is connected to ground on one side and 12V would be supplied via the bosch relay on the other side. But agreed that your current schematic would work as well with the diode! Yes, the LED would definitely come on when the system is armed. But so would the starter. Anyway, I plan to add Bob's blocking diode, exactly as you've shown it, today. And then test the system and button everything up 'cause I'm sure it is going to work. Link to comment Share on other sites More sharing options...
viven44 Posted March 29 Share Posted March 29 In that schematic, the LED would definitely come on when the system is armed. The LED is connected to ground on one side and 12V would be supplied via the bosch relay on the other side. But agreed that your current schematic would work as well with the diode! Yes, the LED would definitely come on when the system is armed. But so would the starter. Anyway, I plan to add Bob's blocking diode, exactly as you've shown it, today. And then test the system and button everything up 'cause I'm sure it is going to work. Oops I put the resistor in the wrong spot, it was supposed to go where the LED used to be. Yes what you have is going to work perfect Link to comment Share on other sites More sharing options...
Nothing Special Posted March 29 Share Posted March 29 .... And speaking of simple vs. elegant, rather than capacitors is there a "cleaner" source of power you could use? Something that doesn't drop out when switching to Start? I'm thinking your three general options are Ignition power, Aux power or unswitched power.... Only because this microanalysis isn't up to 8 pages yet... There is a fourth power option. Or more accurately, another permutation of option 3. Rather than use a capacitor you could use unswitched power but put a switch on it. That would give you a clean power signal without the "complexity" of a capacitor but still no power draw when the truck is shut off. This does require you to remember to turn it on when you want it and off when you don't. But the LED not coming on will remind you to turn it on, and using a lighted switch (lights up when turned on) or a separate indicator light can remind you to turn it off. The capacitor is still the more elegant solution, and I'm not trying to talk you out of it. Just trying to help get to 8 pages! Link to comment Share on other sites More sharing options...
Nothing Special Posted March 29 Share Posted March 29 Oops I put the resistor in the wrong spot, it was supposed to go where the LED used to be.... That would put the full 12V drop across the LED when the module was armed and the key was turned to start, which would blow the LED. Link to comment Share on other sites More sharing options...
Gary Lewis Posted March 29 Author Share Posted March 29 .... And speaking of simple vs. elegant, rather than capacitors is there a "cleaner" source of power you could use? Something that doesn't drop out when switching to Start? I'm thinking your three general options are Ignition power, Aux power or unswitched power.... Only because this microanalysis isn't up to 8 pages yet... There is a fourth power option. Or more accurately, another permutation of option 3. Rather than use a capacitor you could use unswitched power but put a switch on it. That would give you a clean power signal without the "complexity" of a capacitor but still no power draw when the truck is shut off. This does require you to remember to turn it on when you want it and off when you don't. But the LED not coming on will remind you to turn it on, and using a lighted switch (lights up when turned on) or a separate indicator light can remind you to turn it off. The capacitor is still the more elegant solution, and I'm not trying to talk you out of it. Just trying to help get to 8 pages! You may have missed that I was sadly mistaken - I was not using a circuit for power that is hot in both Run & Start. I am now, and it doesn't drop out between Run & Start. I tried it yesterday with and w/o the diode and/or capacitor and it worked perfectly in all cases. I cannot explain how the circuit was working at all since the module had no power in Start. But it does now and all it needs is your diode. Link to comment Share on other sites More sharing options...
viven44 Posted March 29 Share Posted March 29 Oops I put the resistor in the wrong spot, it was supposed to go where the LED used to be.... That would put the full 12V drop across the LED when the module was armed and the key was turned to start, which would blow the LED. The resistor would be behind the LED in that current path, thereby limiting the current through the LED. If I’m not mistaken it isn’t the voltage that kills them but the current. I made that schematic saying it would take both elements out of the current path but then I had concocted that in my mind a couple days ago where I did predict that scenario where the current has to be limited when armed or when the clutch was pressed sans diode anyways I’ve contradicted myself here Link to comment Share on other sites More sharing options...
Nothing Special Posted March 29 Share Posted March 29 The resistor would be behind the LED in that current path, thereby limiting the current through the LED. If I’m not mistaken it isn’t the voltage that kills them but the current.... If I'm understanding your adjustment to your schematic correctly, you are moving the resistor from the wire on the bottom back up to the wire on top. Yes, that will protect the LED when the system is armed, but when the key is turned to start there will be a different source of power that will kill the LED: And voltage / current, to-may-to / to-mah-to. Yes it's current that kills it, but it's voltage that makes the current. With no resistor there's the full 12V (or 14.2V or whatever the alternator gives) across the LED, and with almost no resistance in the LED there's a ton of current and the LED blows. With the resistor the current through the resistor causes a voltage drop so there's much less voltage across the LED. It's the resistor that's limiting the current, but that also limits the voltage across the LED. So whichever way you want to look at it works. Link to comment Share on other sites More sharing options...
viven44 Posted March 29 Share Posted March 29 The resistor would be behind the LED in that current path, thereby limiting the current through the LED. If I’m not mistaken it isn’t the voltage that kills them but the current.... If I'm understanding your adjustment to your schematic correctly, you are moving the resistor from the wire on the bottom back up to the wire on top. Yes, that will protect the LED when the system is armed, but when the key is turned to start there will be a different source of power that will kill the LED: And voltage / current, to-may-to / to-mah-to. Yes it's current that kills it, but it's voltage that makes the current. With no resistor there's the full 12V (or 14.2V or whatever the alternator gives) across the LED, and with almost no resistance in the LED there's a ton of current and the LED blows. With the resistor the current through the resistor causes a voltage drop so there's much less voltage across the LED. It's the resistor that's limiting the current, but that also limits the voltage across the LED. So whichever way you want to look at it works. You got me. I didn't think of that other route. I was too hung up on that clutch press scenario also in that route you pointed out, it would need its own resistor as well for about 2 scenarios. Else about half of the current would go there when cranked (until the LED is killed!).. that schematic isn't ideal at all That LED better stay where it is currently Link to comment Share on other sites More sharing options...
Recommended Posts