For a project I'm currently working on I want (need) to relocate the charge indicator light. I'd like to do away with the bulb and run an LED.

As you may or may not know, the small amount of voltage through the light is what excites the normal alternator. I don't believe, from what I've read, that there needs to be a specific resistance, but I don't know that for sure. If there does need to be some, that could be substituted by simply wiring in an inline resistor, so that's really not an issue.

What MAY be an issue is that since the LED is a DIODE, the current can only flow one way. I don't see how this could affect charging since it's a DC circuit and the polarity must always be the same, but electronics is not my strong suit so I'm asking those of you who do have a deeper understanding than I do.

John Stricker

Both the LED and the alternator need resistance in that circuit. Problem is the LED needs at least several hundred ohms or it will fry. The alt wants about 30 if I remember right. If that circuit is not wired right the alt will not turn on.

The only way around it may be if you use an I regulated CS alt AND you wire the I circuit. Then the dash lamp is mostly just an indicator. The I circuit will still turn on the CS alt. (See the CS article in my cave.)

Note that you have to make certain the regulator is an I config. You can't go by the markings on the alternator plug. (See the Rabid Wombat CS conversion info in my cave.) I don't know of any way to tell what one is what for sure. You'd have to contact someone that rebuilds the CS units. You didn't say what engine this is on.

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The only thing George Orwell got wrong was the year...

The Ogre's Fiero Cave (It's also at the top of every forum page...)

[This message has been edited by theogre (edited 01-21-2005).]

Ogre,

The alternator is an SI. On almost all of my farm equipment, and most of that uses an SI alternator from the factory, there is no charge light and no resistance between the switched 12V and the alternator terminal. There is a diode in line to eliminate battery discharge with the unit off and, on some pieces of equipment and how it's wired, to let it even turn off when you turn off the key.

I've read before that the excitation circuit needs some resistance but have never heard why and, knowing how the alternator works at a basic level, don't really understand why it would need any resistance. It was common several years ago when converting from an old generator on old tractors and stuff to run the sense wire right back to the charge pole on the alternator and put a jumper in the form of a diode across the sense and excite circuits, put some tape over it, and run it for many hundreds of hours with no effect.

One of the reasons I asked the question was to find out WHY it was recommended to have some resistance in the circuit, if, in fact, it is necessary.

As far as the LED is concerned, if I have to make up an entire circuit with several components to have and LED indicator, I'm not going to bother and I'll just rely on the digital voltmeter I'm using. On the Fiero, the diode for the excite circuit is in the harness somewhere back by the firewall and I'm using the stock engine wiring to the alternator so it seemed logical to keep the indicator light, but I'd prefer something that has a longer life than the lightbulb. If an LED isn't practical, I'll probably just measure the resistance across a stock bulb and solder in a resistor of like amount and capacity and forget the light.

Thanks for the input, I've read your alternator article before but I'll check it out again to remember what I forgot.

John Stricker

quote Originally posted by theogre: Both the LED and the alternator need resistance in that circuit. Problem is the LED needs at least several hundred ohms or it will fry. The alt wants about 30 if I remember right. If that circuit is not wired right the alt will not turn on. The only way around it may be if you use an I regulated CS alt AND you wire the I circuit. Then the dash lamp is mostly just an indicator. The I circuit will still turn on the CS alt. (See the CS article in my cave.) Note that you have to make certain the regulator is an I config. You can't go by the markings on the alternator plug. (See the Rabid Wombat CS conversion info in my cave.) I don't know of any way to tell what one is what for sure. You'd have to contact someone that rebuilds the CS units. You didn't say what engine this is on.

Yay! A practical use for my in-depth electronics knowledge!

Ok, it needs 30 ohms? No prob.

Six 2.2V LED's in series creates a 13.2V rated load... perfect for a car.

Each one rated at 70mA means the amps stay constant across the whole circuit (in series circuits, voltage is additive, current is constant).

So, 13.2V divided by 70mA equals about 190 Ohms... too much for what you need.

So, how do we drop the resistance to 30 Ohms? Simple.

Take this example: Two 4 ohm resistors in series equals 8 ohms, BUT two 4 ohm resistors in series equals 2 ohms (HINT HINT: This is useful for car audio, as most car speakers are 4 ohm impedance).

Now, if we take a 35 ohm resistor and put it in parallel with the 6 LED cluster (190 ohms), that creates a 30 ohm load right where you need it.

How? Simple. 1/190 + 1/35 = 1/x = 30 Ohms
---
Digikey.com

Part number: 160-1506-ND ...for the LED's
Part number: RHM35.7FCT-ND ...for the 35.7 ohm resistors

Order each in qty of 10. Each thing is about 50 cents each. Really cheap.
---
Hope this helped!!!
----------------------------------------
Petty Officer Michael C Casaceli
Aircraft Electronics Technician Third Class
Patrol Squadron Ten
United States Navy

I've got a better solution. Place a 200 to 2000 ohm resistor in series with any LED, then put a 30 ohm(5w minimum 10W suggested) resistor in paralell with those.

The 30 ohm resistor will effectively simulate the lamp for the alternator while the led and resistor will work as expected. Just try a few different values from 200 to 2000 ohms to adjust the brightness to taste.

Yep, I was going to say exactly what 'Ka' did. Put the variable resister in series with LED, all in parrallel with the 30ohm R. Just like the above should work great, and simple.

The idea behind the multiple leds was so he could make a cluster (sized for a Fiero warning light, hint hint) like what you see on the led shift lights for tachometers.

And the idea behind NOT having the brightness adjustable is so that it stays perfectly at 30 ohms... the point to all of this. LED's arent that bright (at least the ones I picked aren't), so no worries there.

With such a small resistance amount, it probabbly needs to be pretty precise. Usually, the higher the resistance, the less precise it needs to be.

[This message has been edited by fieroturbo (edited 01-22-2005).]

Well, first thing is all you guys got +'s from me for trying to help.

I did some homework last night and it would seem that what the alternator wants to see is 35 ohms, but I'm going to measure that this morning and find out how much resistance there really is. Personally, I doubt that's real critical, if important at all, since in my other examples where we've used these alternators earlier we didn't see any ill effects with no load in the line and also, just how close to 35 ohms will the circuit stay with the length of wire, the really most excellent connections there are on the printed circuit of the instrument panel and where the lamp holder screws in, and all the other factors. I'll be that resistance is anywhere from less then 30 ohms to 3-400 ohms and the thing would charge. But if a job's worth doing it's worth doing right.

I also learned just a tiny bit about LED's last night not the least was finding about 6 online calculators. With a FV of 3.6V and a draw of around 30 ma, it shows resistance needed of about 350-400 ohms at 15V and about 250 ohms (going off memory here) at 12V. It seems like I should probably set the resistance for about 13V since the only time the LED should be lit is when the alternator isn't charging. Does that make sense to you guys as well? The circuit drawn above with one resistor in line to the excitation pole and the circuit with the LED in parallel is a very simple, and easy thing to do so it's worth pursuing, in my opinion, but the reason I want to do it is so I don't have to ever fool with chainging bulbs or having bad mechanical connections and I don't want to push the LED's so hard by selecting the wrong resistor value that I burn them out in short order.

BTW, I still haven't found ANY reason for there to be a necessary resistance of 35 ohms in the excitation circuit, but since a resistor is only a few cents and the load through it is in milliamps, I might as well just put it in.

John Stricker

sounds dumb but try triggering the led with a relay and should be ok it should make up for the resistance value needed for the bulb. Only bad thing will be when the light flashes on as you start it etc you will hear the relay click. Learned that a couple years ago on the 03 -96 camaros when replecing the factory wing with an SS wing wiht LEDs instead of bulbs it will trigger the ABS light adding a relay is enough to keep the ABS light off . you will need to use a resistor if using a small single LED to reduce the voltage on most. I beleieve most small ones can handle around 3 volts or you will just pop the LED.

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'87 GT
93 3800sc series I / 4t60e
7.597 1/8mile ET
1.579 60'
88.53 mph

[This message has been edited by jb1 (edited 01-22-2005).]

For SI or CS... The sense wire can go anywhere on the output cable. It is very common to see that one one directly to the output lug. Some aftermarket SI and CS conversion/repair plugs include a lug on the sense wire to run it right to the output. Even GM usually just splices it to the output wire near the main lug or even into the main lug when they crimp it. You don't want any resistance at all in this wire or you will make the alternator bump up the output voltage. This wire is how the regulator reads the output voltage. (Regulators in "One Wire" alternators make this connection internally.)

On the lamp circuit... I've never had occasion to test it. I've always been told that if you don't have enough resistance in the circuit the regulator can be damaged. Too much and it might not turn on. I don't know that it has to be an exact value, just in a certain range. I've seen the diode drawn/built into most of these circuits but the thing can be burried anywhere in the harness. Some are really hard to find.

quote Originally posted by theogre: On the lamp circuit... I've never had occasion to test it. I've always been told that if you don't have enough resistance in the circuit the regulator can be damaged. Too much and it might not turn on. I don't know that it has to be an exact value, just in a certain range.

I too have heard this... I think it either directly, or indirectly controls the field coils. When connected directly to +12v, it draws about 0.25 amp. I've read that connecting it straight to +12V is "okay", but it will reduce your alt's life expectancy.

OK, in your schematic, isn't the diode backwards? Shouldn't it be like this:

Won't light the way you have it, will it? (I know, if it doesn't light, reverse the leads, but it does make a difference where the resistor is at).

John Stricker

quote Originally posted by KA: I've got a better solution. Place a 200 to 2000 ohm resistor in series with any LED, then put a 30 ohm(5w minimum 10W suggested) resistor in paralell with those. The 30 ohm resistor will effectively simulate the lamp for the alternator while the led and resistor will work as expected. Just try a few different values from 200 to 2000 ohms to adjust the brightness to taste.

Forgive my ignorance but I have no idea what you guys are talking about. Are you talking about the light on the dash that indicates a failure of the charging system? And you are telling me that light is needed to turn on the alternator?

I don't think so, I have been driving my 88 for 2 years with no such light. As far as I know its just an indicator.

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1988 GT (Firebird Interior),1988 Fastback (4.9 Chop-top #1 of 1), 1984 Indy...Firebird Interior Installation Website

quote Originally posted by jstricker: OK, in your schematic, isn't the diode backwards? Shouldn't it be like this:

Yes/no/maybe. In it's normal operation, the LED should not be on.

In normal operation, the alternator draws a small current through the lamp. I do not know what this current is, as I only checked the draw from the battery, and that was 1/4 amp. If I had to guess, I would think it was around 50 mA. The lamp, in order to light, requires at least 250mA, give or take a lot. Since only 50mA flows through it, it is not enough to light. BUT, in the event of a fault, or engine-off condition, it draws more than that 250mA, effectively shorting the alternator end to ground, and lighting the lamp.

So...... it is like an over current detector of sorts......

At least that is how I think it operates. I could be completely off base with this. But if I am not, then you will need to change the schematic. If you give me a little bit of time, I could probably figure something out that will work for you.....

On the 88 V6 there is no charge indicator light. On page 8A-30-0 of the Helms is says L4 VIN R only.

As for whether it's needed on the 88 L4 I couldn't tell you since I don't have an L4 anymore.

So if the charging (actually "no" charging) indicator bulb burns out, the alternator stops charging. Then how would you know that the alternator is not charging, because the bulb would be burned out unless you has aux. gauges. This all sounds a little wierd ....but then again, it WAS designed, approved and built by GM ....

The charge lamp only provides the initial excitation to the field coil in the alternator - the field is also fed directly from the output of the alternator (internal connection) so as soon as it starts producing output the field terminal goes to +12 volts (yes, I know it's not really 12 volts, but for this discussion it'll do).

The charge lamp is connected from ignition 12 volts to the field terminal. When you turn the key on, there's 12 volts on the ignition side of the lamp but the field is at 0, so the bulb lights. This provides bulb test, and the current flowing through the bulb provides the startup "kick" to the field. When the alternator spins up and starts producing output, there's 12 volts on both sides of the bulb so it goes out.

That's normal operation. If the alternator quits producing output (failure) the field terminal is no longer fed from the alternator output so the charge lamp now has 12 volts on the ignition side and 0 volts on the field side. Now the bulb lights up to inform you that something's wrong.

Some Fieros don't have charge lamps - these cars have a 180 ohm resistor connected in place of the bulb to provide that startup field excitation.

If the charge light bulb is burned out or missing, the alternator will still work but it'll act strangely. What you'll notice is that when you start the car your volt gauge stays low until you rev the engine a time or two. There's enough stray magnetism in the rotor that the alternator can self-excite if you spin it fast enough. Once it's up and running, the voltage will pop up to normal and stay there; the alternator will work normally now. At least it will until you shut the car off and restart it...

Wanna replace the charge light with a LED? I'd suggest replacing the bulb with a 180 ohm resistor, then connect the LED and appropriate dropping resistor in parallel with the 180 ohm resistor. That'll work OK.

Paul,

This is true, IF the car was equipped with a charge indicator light to begin with. The current to excite the alternator has to pass through the bulb and that's where the 35 ohms of resistance comes from. If the bulb burns out, no current is supplied to excite and turn on the alternator.

Now this is where I've had my question on the resistance in the wiring. I went through about 50 different charge system schematics for mid/late '80s gm cars today (on Mitchell on demand). On cars that don't have a charge indicator, there is NO notation on the wiring diagram that there is a resistor in line. That really doesn't mean much, though, because the wiring diagrams also don't show the diode that must be in that circuit either, and if it's not there then you won't be able to shut off the engine with the key because the alternator will back feed the electrical system.

So far, all of my factory shop manuals that I have from GM all show some type of charge light in the circuit. The factory manuals are a lot better about telling you details like this than Mitchell or Alldata.

John Stricker

Edited to answer your direct questions that I overlooked.

As was pointed out in another post when I was typing mine, the alternator will not stop charging when you're on the road if it's already charging because your bulb element breaks, but it won't start charging (normally) again after you shut it off. How do you know if it's the alternator or light? When you turn the key on, the light should come on. This is the bulb test mode and is just a by-product of it doing what it normally does. If you turn the key on and the charge light doesn't come on, you either have a bad alternator (won't go to ground), bad wiring, or burned out bulb. Bulb is the first thing to check.

quote Originally posted by PaulJK: So if the charging (actually "no" charging) indicator bulb burns out, the alternator stops charging. Then how would you know that the alternator is not charging, because the bulb would be burned out unless you has aux. gauges. This all sounds a little wierd ....but then again, it WAS designed, approved and built by GM ....

[This message has been edited by jstricker (edited 01-23-2005).]

Whuffo,

Just curious where you found the 180 ohm figure. I'm not saying you're wrong because I've found figures all over the place looking this stuff up on the net, everything from 30 ohms to 35 ohms to 380 ohms, to 140 ohms and now your figure. Did you find the resistor and measure it? (Personally, that seems to me it might be the only way to know for sure what the factory wants) Or did you find it documented somewhere?

BTW, the method you described is what I intend to do, because it's simple and has such a low parts count, but it would be nice to know for sure what resistance the factory wants there. I was going to measure the resistance in a spare dash panel today from the circuit connectors, but didn't get around to it. (Fought trying to get my foam hot wire cutter to cut, damn transformer).

John Stricker

quote Originally posted by Whuffo: The charge lamp only provides the initial excitation to the field coil in the alternator - the field is also fed directly from the output of the alternator (internal connection) so as soon as it starts producing output the field terminal goes to +12 volts (yes, I know it's not really 12 volts, but for this discussion it'll do). The charge lamp is connected from ignition 12 volts to the field terminal. When you turn the key on, there's 12 volts on the ignition side of the lamp but the field is at 0, so the bulb lights. This provides bulb test, and the current flowing through the bulb provides the startup "kick" to the field. When the alternator spins up and starts producing output, there's 12 volts on both sides of the bulb so it goes out. That's normal operation. If the alternator quits producing output (failure) the field terminal is no longer fed from the alternator output so the charge lamp now has 12 volts on the ignition side and 0 volts on the field side. Now the bulb lights up to inform you that something's wrong. Some Fieros don't have charge lamps - these cars have a 180 ohm resistor connected in place of the bulb to provide that startup field excitation. If the charge light bulb is burned out or missing, the alternator will still work but it'll act strangely. What you'll notice is that when you start the car your volt gauge stays low until you rev the engine a time or two. There's enough stray magnetism in the rotor that the alternator can self-excite if you spin it fast enough. Once it's up and running, the voltage will pop up to normal and stay there; the alternator will work normally now. At least it will until you shut the car off and restart it... Wanna replace the charge light with a LED? I'd suggest replacing the bulb with a 180 ohm resistor, then connect the LED and appropriate dropping resistor in parallel with the 180 ohm resistor. That'll work OK.

Okay I just read up on this in Ogre's cave and he wasn't too sure about 88V6...

I can 100% guarantee that the 88 V6 does not use such a charge indicator light. And even if it did, mine doesn't. I replaced my Fiero gauges 2 years ago and I didn't hook up any resistor to that line (Connector C3 pin 6). According to my Helms I didn't need to. I have over 5,000 miles on my dash I think I would have noticed if I needed that bulb or not.

Maybe on earlier cars the charge indicator light is needed. I never even knew the function existed until 30 minutes ago.

[This message has been edited by jscott1 (edited 01-23-2005).]

Whoops..... The circuit is correct if you flip the LED. I'd peg the largest resistor in series with the LED at about 500 ohms. Depending on what you want to put in there, and it's voltage/current requirements, you might need something much lower than that. I don't know what kind of space requirements you have, but 5 watt 30 ohm resistors are going to be fairly large. 1" long x .25" or there abouts. This should do the trick for you:

http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=345794&Row=46070&Site=US

That is, if you decide to go with the ~30ohm......

oww my brain... It's too late for this kind of math

[This message has been edited by ryan.hess (edited 01-23-2005).]

I found a calculator on line to help you size the resistor.

http://ourworld.compuserve.com/homepages/Bill_Bowden/led.htm

That should be more than enough current to turn on the alternator. Assuming you need such a thing.

I will look (and even measure if I think about it) the current draw through that wire but I'm pretty sure it's in milliamps, like maybe 200 ma or some such number?? That's fairly easy to check but I'm sure it's nowhere near 5 watts. How many watts is that light bulb??? Yeah, the resistor for the LED will depend on the LED (size, color, and FV), but there are pleny of online calculators for that so I don't even have to use arithmetic.

I stopped at Radio Shack today and the brightest LED they showed was a yellow one, which would be fine for an annunciator like this, with 1900 mcd output, almost twice the next brightest 5mm LED they normally keep. It has a max FV of 3.0V with a 2.1V typical FV, 40 ma current draw, then if we took a worst case over voltage of 16 volts to the LED and ran the numbers, it would need a 1 watt, 10%, 330 ohm resistor. For a more normal input voltage of 13.6V and the normal FV of 2.1, it would still need almost a 300 ohm resistor so it should stay very bright. Only problem was, the Radio Shack didn't have any in the bin so I couldn't get one to play with. They were pricey compared to digikey at $2.69, but still nothing that would break the bank. Doesn't matter, though, if they don't have any of them. I'm going to have to go to another town tomorrow with another Radio Shack and I'll see if they have them there.

I appreciate all the comments and help, this has been a fun learning experience for me.

John Stricker

quote Originally posted by ryan.hess: Whoops..... The circuit is correct if you flip the LED. I'd peg the largest resistor in series with the LED at about 500 ohms. Depending on what you want to put in there, and it's voltage/current requirements, you might need something much lower than that. I don't know what kind of space requirements you have, but 5 watt 30 ohm resistors are going to be fairly large. 1" long x .25" or there abouts. This should do the trick for you: [URL=http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=345794&Row=46070&Site=US]http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=345794&Row=46070&Site=US[/UR L] That is, if you decide to go with the ~30ohm...... oww my brain... It's too late for this kind of math

Radio Shack should have some LEDs with the appropriate dropping resistors built-in to make them 12V. They would be drop in replacement for the incandescent bulb. So you wouldn't have to worry about the math.

quote Originally posted by Whuffo: If the charge light bulb is burned out or missing, the alternator will still work but it'll act strangely. What you'll notice is that when you start the car your volt gauge stays low until you rev the engine a time or two. There's enough stray magnetism in the rotor that the alternator can self-excite if you spin it fast enough. Once it's up and running, the voltage will pop up to normal and stay there; the alternator will work normally now. At least it will until you shut the car off and restart it...

Thank you for stating this! I have said it so many times and it always seems that no one will believe it. It is actually common in the Fiero to have the excitement line failed. The alternator will self excite at about 1800 rpm, there is no need for the excitement wire, but it is better if it is there. This is only on the early style alternators.

quote Originally posted by jstricker: Whuffo, Just curious where you found the 180 ohm figure. I'm not saying you're wrong because I've found figures all over the place looking this stuff up on the net, everything from 30 ohms to 35 ohms to 380 ohms, to 140 ohms and now your figure. Did you find the resistor and measure it? (Personally, that seems to me it might be the only way to know for sure what the factory wants) Or did you find it documented somewhere?

I think 180 ohm is high too. We used to have trouble on the early S10's having too much resistance and the alternators had to self excite, If memory serves me correctly, anything much over 70 ohms and they would not work. This was a long time ago though.

[This message has been edited by Electrathon (edited 01-23-2005).]

Jonathan,

Actually, I don't think it would be enough if you just ran it through the LED. The LED's don't use 12V, that's the function of the resistor, to cut the voltage down before it gets to the LED to whatever the max Front Voltage (FV) of the LED is. For instance, the max FV of the Yellow LED I was looking at today had a MAXIMUM FV of 3.0V so whatever is coming OUT of the LED is going to be something less than 3.0 V or you're going to pop the LED. That's what the resistor ahead of the LED is for, to cut down the voltage seen by the LED. I'm pretty sure that 3.0 V isn't enough to excite the alternator, I think I read somewhere it takes over 11 or 11.5 volts to do that. That's the reason the LED will have to be in parallel and not in series like the light bulb was, but we still need to figure out what the resistance needs to be (if any) in the line between the 12v and the excitement terminal.

John Stricker

quote Originally posted by jscott1: I found a calculator on line to help you size the resistor. http://ourworld.compuserve.com/homepages/Bill_Bowden/led.htm That should be more than enough current to turn on the alternator. Assuming you need such a thing.

quote Originally posted by jstricker: Jonathan, Actually, I don't think it would be enough if you just ran it through the LED. The LED's don't use 12V, that's the function of the resistor, to cut the voltage down before it gets to the LED to whatever the max Front Voltage (FV) of the LED is. ...

This isn't quite right. The whole point of the resistor is to drop the voltage down to the point of the LED. Which means you have about 9 volts across the resistor by itself...however because the resistor is in series with the LED the load as seen by the alternator is 12V not the 3V front side voltage of the LED by itself. Remember when you have two components in series the voltage drop is additive.

This whole thing is starting to make my head hurt... I think I would just wire a 12Volt line with a 30 ohm resistor in parallel to the light and call it a day.

EDIT - I just checked my 84 HELMs and it shows a 10 ohm resistor wired into the Alternator from the ignition switch. It does not show the Charge Indicator to be an integral part of the circuit.

I seldom disagree with Theogre, but I have a real hard time believing that even GM would design a car that depends on an indicator bulb on the dash to activate the charge circuit. If you feel the need you can replace the bulb with the appropriate resistors/LEDs but I don't believe it makes a whole lot of difference to the alternator.

[This message has been edited by jscott1 (edited 01-23-2005).]

Jonathan,

I don't have an '88 factory service manual, but I did pull up the '88 charging circuit on my Mitchell on Demand and here's what they say:

Delco-Remy SI series alternators are equipped with a solid state nonadjustable internal regulator. The Brown Field wire to the alternator is used to turn on the alternator. A 10-ohm resistance is provided by the alternator warning light, or choke heater relay when vehicle is equipped with a voltmeter, to protect the diode trio. (bold emphasis mine)

I do have to note that when one particular section covers several models of GM, Mitchell is pretty bad about just cutting and pasting as evidenced by the reference to a "choke heater" in a car that never had choke.

I went out to the shop and found an old cluster and checked some resistance. The bulb itself has about 4.4 ohms but when checked in the holder and through the printed circuit board it showed about 13.4 ohms so that's pretty close to what the Mitchell manual claims should be there.

What I'm going to do is put about a 20 ohm 1 watt resistor in my charge circuit and stick the LED in parallel. We'll see what happens, but I don't see why it would cause any problems.

Whether you agree with ogre or not just look at your schematics for any GM car with a charge indicator and read the troubleshooting and you'll find that they did indeed wire the excite wire through the charge light. If the bulb burns out, it won't charge (normally). Test it out some time. Pull the bulb on a car equipped with the charge indicator light and start it up, you won't get a charge unless you rev the engine up to around 2-3000 rpm and there is some residual magnetism left in the alternator (which there will be if you've just had it running).

John Stricker

quote Originally posted by jscott1: This isn't quite right. The whole point of the resistor is to drop the voltage down to the point of the LED. Which means you have about 9 volts across the resistor by itself...however because the resistor is in series with the LED the load as seen by the alternator is 12V not the 3V front side voltage of the LED by itself. Remember when you have two components in series the voltage drop is additive. This whole thing is starting to make my head hurt... I think I would just wire a 12Volt line with a 30 ohm resistor in parallel to the light and call it a day. EDIT - I just checked my 84 HELMs and it shows a 10 ohm resistor wired into the Alternator from the ignition switch. It does not show the Charge Indicator to be an integral part of the circuit. I seldom disagree with Theogre, but I have a real hard time believing that even GM would design a car that depends on an indicator bulb on the dash to activate the charge circuit. If you feel the need you can replace the bulb with the appropriate resistors/LEDs but I don't believe it makes a whole lot of difference to the alternator.

Just a heads up -

The reason for the 5W resistor is for when the engine is off, or the charging circuit failed..... You guys keep telling me that the charge indicator lamp side of the alternator is then grounded. If that's the case, then you have 12 volts across 30 ohms, or 12^2/30=4.8 watts.

If you go with the 20 ohms, it's 7 watts. I think the indicator lamps are somewhere between 1-5 watts (sorry I can't be more helpful, I'm unfamiliar with them) which probably means that there is some internal resistance in the alternator. I'd rather err on the cautious side than starting a cabin fire. When you do your testing, leave the engine off for a while (key on), and check the resistor's temp.

As I said earlier my 84 shows a 10 ohm resistor wired to the ignition switch. That is a much more reliable way than to depend on a bulb in the dash.

If you have seen my car you will know there is no such bulb, as I have a firebird cluster with no charge indicator. My alt has been operating normally ever since.

I'm just saying that the only two data points I have for Fieros, (84 and 88) and neither have a change indicator light.

[This message has been edited by jscott1 (edited 01-23-2005).]

As stupid as it seems more than a few GM vehicle owners have found out the hard way that the SI and some configurations of CS will not work properly if at all when the dash lamp has failed. There are at least a few threads on this forum about it.

Problems with C500 can also keep the alt from working right since the Lamp circuit passes thru there.

The spec for standard 194 dash lamp is 3.78W at 14V (.27A) At specified voltage and current that gives it a hot resistance of about 52 ohms. (V/A=R) Tungsten filaments go up in resistance when hot. Depending on the amount of current flowing at any given time the resistance will vary between the cold 4.4 ohms measued above and the hot resistance. Even if the current isn't enough to make the lamp glow the heat will increase the resistance.

The SI wiring is shown here for those that haven't seen it. (Bottom diagram.) The lamp wire is connected directly to the Field coil and diode trio according to the diagram. Alldata GM and every other SI diagram I've seen is the same.

The diode in the lamp circuit may have two purposes. One could be to keep battery voltage from feeding back thru the lamp when the car is off. It could also be keeping the alt from trying to use the lamp as an output path when the car is running.

IF there is any current flowing into the alternator on the lamp wire once the is car running, an LED may never turn off. The 194 lamp can flow several volts before it is visibly glowing. My guess, and this is only a guess, is that once the alternator lights off the lamp carries no current as long as the blocking diode in the harness is working. In that case... The question is only if the LED and required ballast resistor pass enough current to get the alternator to start. Answering that is a fairly simple matter of connecting the LED and resistor across the existing lamp hole in the dash. You can use a volt meter to determin polarity of the terminals. I have an I wired CS configuration so I can't test it.

quote Originally posted by theogre: The spec for standard 194 dash lamp is 3.78W at 14V (.27A) At specified voltage and current that gives it a hot resistance of about 52 ohms. (V/A=R) Tungsten filaments go up in resistance when hot. Depending on the amount of current flowing at any given time the resistance will vary between the cold 4.4 ohms measued above and the hot resistance.... The diode in the lamp circuit may have two purposes. One could be to keep battery voltage from feeding back thru the lamp when the car is off. It could also be keeping the alt from trying to use the lamp as an output path when the car is running.

OK, guys. TheOgre is correct; he just beat me to it.

1) Replace the lamp with a 5 watt resistor. The purpose of this resistor is to limit the excitation current provided to the alternator at startup. Any resistance value between 33 ohms and 47 ohms (both standard values) should be fine; you only need to approximate the resistance of the original lamp. (As noted elsewhere, a 10 ohm resistor is appropriate as a safety device in an alternator test rig, but not in normal operation.) When the ignition switch is on and the alternator is not generating output, full battery voltage will appear across this resistor; when the alternator is generating output, approximately zero volts will appear across it.

2) Parallel the 5 watt resistor with an LED in series with another resistor. The purpose of the second resistor is to a) limit the current through the LED to a safe value, and b) to adjust the LED's brightness. Something in the range of 270 to 470 ohms and a dissipation rating of 1/2 watt should be about right. The cathode of the LED (the end marked by a band) must connect to the alternator end of the circuit. The current through this second path will be much smaller than the current through the 5 watt resistor, so its contribution to the alternator excitation current can be disregarded.

Refer to the original drawing provided way back when by 'KA.' It is also correct, provided that the + end of his circuit should connect to the alternator, while the - end of his circuit should connect to switched 12 volts (i.e. HIRS ... "Hot in Run and Start" ... in GM-speak).

quote Originally posted by KA:

Finally, you might want to add your own reverse-blocking diode to the circuit ... especially if you're replacing any of the original wiring harness. Unlike an incandescent lamp or a resistor, sufficient reverse voltage can zap an LED in milliseconds. Any common diode like a 1N4001, 1N4002, etc. should be fine. A blocking diode can be connected to either end of KA's circuit, but its cathode must always point (electrically speaking) toward the alternator.

[This message has been edited by Marvin McInnis (edited 01-25-2005).]

If I could swear here it would accentuate my term.

$*$#& GREAT THREAD!

This one is in my favorites. Thx guys.

[This message has been edited by Soelasca (edited 01-23-2005).]

I'd really like to thank all of you guys. Marvin, thanks for reinforcing some things. Yes, I see what you're sayin that KA's drawing would be correct if the + on his drawing is the alternator. That's counter-intuitive to me because of the way the circuits are drawn and how the polarity would have to be to make the LED light, but I see what you mean.

This thread has taught me a lot and made me do some research that I needed to do anyway, but you guys shamed me into it. I'll probably use the circuit drawn and a 33 ohm 5 watt resistor in parallel with the LED. I'm going to test the circuit and make sure the diode that's in the factory harness (wherever it physically is) is working correctly and if it is, I won't use a second one.

Thanks again everyone, it's appreciated.

John Stricker

PS: anyone that didn't already have +'s got one for their help or even just questions