Clutch Switch Bypass For Big Blue

[Gary Lewis]

Time for a break for water.  You can see below how far I've gotten.

No, I'm not a fan of fuse taps.  I want things done in a it-can-never-fail way, and that takes solder, shrink tubing, etc.  This thing goes everywhere and I don't want things to quit in the back of beyond.

As for the relay's pull-in current, they are mechanical devices, not solid state.  The coil moves contacts within the relay, which takes power: wattage = ExI.  Small micro relays take a few ma, a Bosch relay might take 250 ma, and the fender-mounted starter relay I have on the bench pulls 3.5 amps @ 14v when pulled in.  That's 49 watts.  The LED only pulls 1/100th of that so the LED lights but the relay doesn't pull in.

[Gary Lewis]

Got the relay module tapped into Ckt 640 and tested it - perfect.  So now it is going back together, as shown below.

And at the bottom of the dash you can see the module's box is open, ready for the diode - tomorrow.  Gonna call it quits for the day as Janey and I are going out tonight.  

[viven44]

I am the same way. I have used fuse taps, and do it very much so on modern vehicles where wires aren't accessible... but for anything mission critical or easily accessible, I either solder and heat shrink or crimp with marine grade water-proof heat shrink tubing that completely 'glassifies' and seals...

And have a great evening!

[Nothing Special]

Gary Lewis wrote

....  As for the relay's pull-in current, they are mechanical devices, not solid state.  The coil moves contacts within the relay, which takes power: wattage = ExI.  Small micro relays take a few ma, a Bosch relay might take 250 ma, and the fender-mounted starter relay I have on the bench pulls 3.5 amps @ 14v when pulled in.  That's 49 watts.  The LED only pulls 1/100th of that so the LED lights but the relay doesn't pull in.

I understand Vivek's question, and I don't really understand Gary's answer.  Not that I really need to, it seems to work now, and it definitely will with the addition of the new diode.

But after reading Vivek's question I see that there seems to be a circuit from VCC, through the LED, through the (closed) clutch switch to the starter relay.  Yes the LED only draws a small amount, but it sort of seems like every time the clutch pedal is pressed (whether the timer is in use or not) the starter relay will draw its 3.5 amps through the LED (which would likely be pretty unhappy about it).

But as Vivek and I said, the addition of the diode will shut off that path, so there's no long-term concern.

[Gary Lewis]

Well, the schematic is now incorrect.  It used to have a resistor in series with the LED.  It is actually there but it is built into the LED's housing so I eliminated it in the schematic.  My bad.  I'll put it back.

But it limits the current through the LED to ~20 ma if you put the LED across 12v.  I don't have any way of knowing for sure on this one, but most LED's are designed to run at 2v.  Assuming that is the case then the internal resistor will be 500 ohms: R = E/I = 10/.02 = 500 ohms.

As for the primary coil of the starter relay, if it conducts 3.5 amps at 12v then it has the equivalent to 3.4 ohms.

I've drawn the circuit w/o the switch, or anything else, to keep it simple.  But with the total of 503 ohms there would be roughly 20 ma, or .02 A, flowing through the circuit.  Far less than the 3.5 amps needed to pull in the starter relay.

Does that help?

[Nothing Special]

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!

[Gary Lewis]

LEDs blow quite easily.  One designed for 30 ma fails before you get to 100 ma.  DAHIK.  My memory is going, but that was only last week, so...  But they are BRIGHT for a very brief period.  

As for the resistor, there are LEDs with an internal one and LEDs w/o.  I started this project with one that didn't have a resistor and quickly rued that decision as I blew one up and broke the leads off of another one.

So I quickly moved on to the one I have now which not only has the resistor built in but also has strain relief on the leads.

[viven44]

Addition of the resistor totally makes it clear for me..

But... Would this work ??? Takes out the LED out of the current path so you don't have to worry about voltage drop through a resistor or LED.
 
Edit: I don't think the voltage drop matters as that voltage drop won't affect anything.. anyways. Just add the diode Bob mentioned to your original schematic, should be good!

[Gary Lewis]

No, that won't work the way I planned it.  That plan is to have the LED come on when the system is armed so people understand that danger is at hand.

The system works that way now with the exception of the LED coming on each time the clutch is depressed, which is annoying - especially at night.  But Bob's diode should fix that.  I'll add it tomorrow.

[viven44]

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!

[viven44]

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.

[Nothing Special]

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...).

[Gary Lewis]

viven44 wrote

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.  

[viven44]

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

[Nothing Special]

Nothing Special wrote

....  
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!

[Nothing Special]

viven44 wrote

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.

[Gary Lewis]

Nothing Special wrote

Nothing Special wrote

....  
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.

[viven44]

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

[Nothing Special]

viven44 wrote

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.

[viven44]

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