72novaproject
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01-06-12 03:40 PM - Post#2176486
I don’t know much about LED light bulbs and I have a question.
Will an LED bulb work as a charge indicator light and provide the required resistance for the #1 terminal on a 12SI alternator?
I know LED stands for “Light Emitting Diode” and diodes only let current flow in one direction so I am a little confused.
Thanks for the help in advance,
Steve
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 Vaughn
"11th Year" Gold Supporting Member
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01-06-12 07:40 PM - Post#2176548
In response to 72novaproject
LEDs don't provide resistance as such, they are better described as "voltage drops", about 1.7 volts for most diodes (often rounded up to 2 volts). LEDs use almost no current - so there might not be sufficient current flow to energize the field inside the alternator.
If you remember, LEDs sometimes need to use special flashers, because there isn't enough current draw through the diode to make bimetallic springs inside standard flashers heat up and disconnect the switch contact points inside the metal flasher can.
You can try wiring up a simple led circuit from the battery to see if it will work, then measure the voltage output at the alternator output - if it is at 14 volts, the alternator is working.
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 acardon
"9th Year" Silver Supporting Member
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01-06-12 08:02 PM - Post#2176553
In response to 72novaproject
You can put a 50 ohm resistor in parallel with the LED to provide the current to excite the alternator. The LED may be too sensetive to give an accurate reading of the alternator output. You may also need a resistor in series with the LED to keep it from turning on when there is a small voltage drop across the 50 ohm resistor.
Don
66 Corvair (driving)
57 2dr HT (restoring)
56 2dr HT (waiting to be restored)
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72novaproject
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01-07-12 08:15 AM - Post#2176653
In response to acardon
Thanks guys,
Does the #1 terminal even need a resistor? I thought that it did. I guess when you see people loop back to the output terminal that is for the #2 “voltage sensing “ terminal.
Does this look correct?
Steve
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acardon
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01-07-12 08:29 AM - Post#2176655
In response to 72novaproject
I think it will work without the resistor but will draw excessive current when the key is on before the engine starts, plus the LED will never light up. (The old path of least resistance thing). #1 terminal is at ground until the alternator starts putting out voltage. When the voltage is equal at both ends of the resistor the LED will go out. Since you may have a little voltage drop between the ignition switch voltage source and the #1 terminal, the voltage may never be exactly equal on both ends of the resistor, thus the optional circuit. BUT I don't know the value of the resistor in series with the LED that may be required.  A slight difference in voltage on a incadesent bulb would not be noticeable or cause it to glow.
Don
66 Corvair (driving)
57 2dr HT (restoring)
56 2dr HT (waiting to be restored)
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72novaproject
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01-07-12 09:49 AM - Post#2176675
In response to acardon
Thanks Don but now I am confused.
QUOTE:
I think it will work without the resistor but will draw excessive current when the key is on before the engine starts, plus the LED will never light up. (The old path of least resistance thing).
Please help me understand this statement. In the non-optional circuit, is this the parallel circuit you described above? What do you mean by draw excessive current? As in it will blow fuses? Are you saying that current will flow through the resistor rather than the LED because the LED has a greater resistance value than the 50 Ohm resistor?
QUOTE:
#1 terminal is at ground until the alternator starts putting out voltage. When the voltage is equal at both ends of the resistor the LED will go out. (Understood)
Since you may have a little voltage drop between the ignition switch voltage source and the #1 terminal, the voltage may never be exactly equal on both ends of the resistor, thus the optional circuit. BUT I don't know the value of the resistor in series with the LED that may be required.
I think what you are saying is (optional circuit) that the resistance value of the LED + resistor in series must be equal to the resistance value of the resistor in parallel. Since the value of the voltage drop and the value of the LED are unknown, the required resistance value of the parallel resistor would need to be calculated to determine the one needed. Or, the resistor in series would need to be adjusted to equal 50 ohms with the LED included in the series circuit. Is this correct?
Steve
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65_Impala
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01-07-12 05:52 PM - Post#2176804
In response to 72novaproject
Yes, you are confused. I think the first statement meant that you can't connect the alternator directly to the ign switch without a resistor. The second statement was meaning that the voltage drop on the wiring between the alternator and the ignition switch may caouse the LED to glow while the engine is running. This won't happen because the alternator will produce a higher voltage and reverse bias the LED. There is no reason to expect the ignition switch voltage to ever be higher than the voltage at the alternator.
You need the first circuit and the resistor in series with the LED depends on the LED. If the LED is an assembly already made to operate on 12V then you don't need it. If the LED is a standard component LED then you need to divide 12V by the LED rated current. IF you don't know then use 10mA or 0.010A to give a 1200 ohm, 1/4 watt resistor. If you find the LED datasheet then use about 1/2 of the maximum rated current.
Also, the 50ohm resistor needs to be at least a 5 watt resistor. Make sure you mount this resistor so it won't burn other wires or parts.
I'd first hook the LED up with jumper wires and confirm it's the brightness you want if you are using your own resistor.
Peter
Edited by 65_Impala on 01-07-12 05:55 PM. Reason for edit: No reason given.
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72novaproject
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01-08-12 09:04 AM - Post#2176961
In response to 65_Impala
Thanks Peter!
I may have gotten deeper into this conversation than I am qualified to participate in. I am not all together sure this is worth the trouble. I would like to be certain of the results and only do this once. I may just keep an eye on the volt gauge but I haven’t given up on this altogether if you guys are willing to stay engaged.
The instrument cluster is not stock in my car so no charging indicator light circuit, bulb socket or lens currently exists. If it were a stock cluster I would just connect it the way the factory did it. The cluster comes with LED’s for turn signals and high beam indicator and I thought a charge indicator would be appropriate. I can’t really think of a way to use an incandescent bulb for the charge indicator.
This is the cluster that is going in the car. It is ABS plastic and if I mount something like a panel light the incandescent bulb might melt it.
A few more Q’s:
Whether or not I use a charge indicator I still need a resistor to the field terminal of the alternator. I have a very basic understanding of volts, amps, ohms and watts but I am not sure exactly how these values affect each other or what values are appropriate. Without an LED charge indicator, what value of resistor would I need for the field terminal and what should the voltage read at the field terminal with the wire disconnected? Would a typical ignition ballast resistor be appropriate?
When you said “Reverse Bias” the LED I am lost. Again, the “diode” part of "L.E.D." has me confused. I take it that it is important to connect the red wire on the LED in the proper orientation because there will always be a greater voltage on the alternator side of the circuit. Forgive me but I am trying to learn here. Which side of the circuit should the red wire go to? Alternator or ignition switch? And wouldn’t the resistor past the LED work the other way around to reduce the voltage that the LED see’s from the alternator once it begins to charge?
Let’s say I decide to bench test this set up which I can’t do with an actual running alternator as this car is under construction and I am about to wire the entire thing. How would I hook this up on a test bench?
Is this the circuit you suspect will work and how could I bench test it?
Thanks for your patience,
Steve
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Vaughn
"11th Year" Gold Supporting Member
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01-08-12 11:51 AM - Post#2177044
In response to 72novaproject
Your assumption that the incandescent bulb will melt the housing is wrong.
A charge light is only on for a limited amount of time - usually a matter of seconds. It will only be on for a few seconds before you start the car, and if your alternator goes out, you will probably shut the motor off almost immediately to see what the problem is. It won't be on long enough to overheat the plastic.
There are a number of press in bulb holders that will insulate the heat of the bulb from the dash. Radio Shack has a number of different ones, and you can find a few at your local parts house. This is an example:
http://order.waytekwire.com/productdetail2/M50/47 6...
If you want to go with LEDs for energy savings - this application isn't on long enough to see any appreciable energy savings - so the extra effort to go with LEDs over a standard lamp virtually isn't worth it.
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acardon
"9th Year" Silver Supporting Member
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01-08-12 12:24 PM - Post#2177061
In response to 72novaproject
Whether or not I use a charge indicator I still need a resistor to the field terminal of the alternator. I have a very basic understanding of volts, amps, ohms and watts but I am not sure exactly how these values affect each other or what values are appropriate.
I agree with Vaughn, a pilot light from Radio Shack would seem to be a viable solution. I think you would need an operational alternator to really bench test the LED hook up.
You will need a 35 to 50 ohm resistor wired like your first drawing without the LED. If you use a incadesent bulb, put it in place of the LED.
Don
66 Corvair (driving)
57 2dr HT (restoring)
56 2dr HT (waiting to be restored)
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65_Impala
Very Senior Member
Posts: 1607
Reg: 12-29-02
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01-08-12 05:26 PM - Post#2177163
In response to 72novaproject
The LED only works one way. The ign switch provides power and the alternator will ground the LED when the engine is not running. When the engine is running, the alternator voltage will likely be a little higher than the ign switch voltage which means the power is connected backwards and the LED won't light. The normal light doesn't light because the voltages on both ends are about the same. The LED won't light because the voltage at the wrong end will be a little higher.
The 3rd drawing you posted is wrong. You have to use the 1st drawing assuming the LED you use is a complete lamp assembly with a 12V rating. In this case, the lamp will already have the proper resistor inside it.
If you use a component LED which is not rated for 12V then you need to use a series resistor. Likely, a 1.2k, 1/4 watt resistor will work fine.
In all cases, you need a 50ohm, 5W resistor connected between the ignition switch and the alternator.
Personally, I would use a component LED and drill a small hole just large enough for the lens of the LED to poke through. I just have never found a light housings that I thought was particularily attractive and that would look right in the dash.
To bench test, just hook the LED to a 12V power source (battery) and see if you like the brightness.
I've been looking for custom indicator lights since I'm in the preliminary stages of planning a custom guage panel for a 65 Impala. No one makes a panel so I have to make my own. I'm thinking about milling a piece of aluminum using the CNC mill at work. I'm thinking a smaller hole at the front with a larger recess from the back. I would printing B&W negative indicator pictures onto a sheet of clear plastic and then glue them to pieces of acrylic rod before sticking the whole works into the back of the holes. Finally, drill a hole in the back of the acrylic rod and glue a LED into it. This would both get a nice looking light and I could also put things like turn arrows, engine symbol, temp symbol, oil press symbol and a batt symbol in for the indicators instead of having plain lights. But, I may go simpler too and just find bezel-less lights and do a stepped recess in the panel so they plug-in flush and then just put the lights in logical locations where you'd expect them to be, like the batt light near the volt meter. Or maybe just plain LED's poked through from the back. Haven't decided how hard I want to make it yet, lol.
Peter
Edited by 65_Impala on 01-08-12 05:27 PM. Reason for edit: No reason given.
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72novaproject
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01-08-12 10:46 PM - Post#2177265
In response to 65_Impala
In your last post you said I need to wire the led and resistor in series so I drew it that way and now you are saying parallel which is the first way I drew it. Which one is it?
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acardon
"9th Year" Silver Supporting Member
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01-09-12 08:56 AM - Post#2177337
In response to 72novaproject
You need the first circuit and the resistor (the 1200 ohm resistor) in series with the LED depends on the LED. If the LED is an assembly already made to operate on 12V then you don't need it. If the LED is a standard component LED then you need to divide 12V by the LED rated current. IF you don't know then use 10mA or 0.010A to give a 1200 ohm, 1/4 watt resistor. If you find the LED datasheet then use about 1/2 of the maximum rated current
The 3rd drawing you posted is wrong. You have to use the 1st drawing assuming the LED you use is a complete lamp assembly with a 12V rating. In this case, the lamp will already have the proper resistor inside it.
If you use a component LED which is not rated for 12V then you need to use a series resistor. Likely, a 1.2k, 1/4 watt resistor will work fine.
In all cases, you need a 50ohm, 5W resistor connected between the ignition switch and the alternator.
Steve, it depends on the diode, wheather you need a 1.2K, 1/4 watt resistor in series with the LED but you need the 50 ohm resistor in parallel in either case.
Don
66 Corvair (driving)
57 2dr HT (restoring)
56 2dr HT (waiting to be restored)
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72novaproject
Senior Member
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01-09-12 04:02 PM - Post#2177455
In response to acardon
Thanks Don! I think the light bulb is starting to come on (pun intended).
Does this look about right? I still need to know the red wire / black wire orientation for the reverse bias thing. Can you verify that on my diagram?
Sorry I am a little slow on this stuff.
Steve
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65_Impala
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01-09-12 04:08 PM - Post#2177458
In response to 72novaproject
That is correct. Just some confusion over which resistor since you may need 2 of them. Adjust the 1200ohm resistor to get the brightness you want.
I would assume the red wire would go to the ignition switch. You can test it on a battery if you want to know for sure.
Peter
Edited by 65_Impala on 01-09-12 04:10 PM. Reason for edit: No reason given.
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acardon
"9th Year" Silver Supporting Member
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01-09-12 04:21 PM - Post#2177462
In response to 72novaproject
Getting close.  Red and black are correct. IF the LED is a 12 volt LED assembly, you don't need the 1.2K ohm resistor. The purpose of the 1.2K ohm resistor is to drop the voltage across a 1.7 volt LED.
Don
66 Corvair (driving)
57 2dr HT (restoring)
56 2dr HT (waiting to be restored)
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72novaproject
Senior Member
Posts: 3186
Age: 57
Loc: D/FW Texas
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01-10-12 03:44 AM - Post#2177643
In response to acardon
Thanks Peter,
I ordered additional LED’s (beyond what is included) from the instrument cluster company. I assume these are 12 V rated as they wire directly in (according to their instructions) for turn signals and a high beam indicator.
Since the charge warning light will only be on in the event of a failure and when I start the car I am not too worried about the brightness of the bulb. If I did want to change the brightness of the 12V LED’s (or any of them for that matter) am I correct to assume the greater the resistance the dimmer the bulb?
Getting close. Red and black are correct. IF the LED is a 12 volt LED assembly, you don't need the 1.2K ohm resistor. The purpose of the 1.2K ohm resistor is to drop the voltage across a 1.7 volt LED.
Don, as always, thanks again. I am starting to detect a faint heartbeat. 
To get closer than close I would need to remove the 1.2K resistor in my latest schematic. “OR” I could install a standard component LED with the 1.2K resistor, BUT, I would need to move it to the ignition switch side of the LED to drop the voltage to it with the key on and engine off. The resistor shown in my schematic is on the wrong side of the LED. Correct?
Now, I suspect the LED’s I ordered are 12V but where do I get a 50 Ohm 5 Watt resistor (a part number or link would be nice) and how do I wire it in? Does it get hot? Do I need to separate it from my wire bundle? Should I wrap it in some type of high heat braded sheath?
This has been an interesting discussion and I hope others may benefit from it. I know I did. 
Steve
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65_Impala
Very Senior Member
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01-10-12 05:09 PM - Post#2177877
In response to 72novaproject
It doesn't matter which side the resistor goes.
Peter
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65_Impala
Very Senior Member
Posts: 1607
Reg: 12-29-02
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01-11-12 05:17 PM - Post#2178295
In response to 72novaproject
I didn't answer about the resistor. It will get hot if you have the key on and the engine off or the alternator not working. You will likely have to seperate it or wrap the other wires with a heat shield where it touches. I would try Ebay for the resistor.
http://www.ebay.com/itm/10-x-Axial-Leaded -Fixed-Ty...
You get 9 spares, but it was the first one I found.
Don't worry about it being exactly 50 ohms either, 47 ohms to around 75ohms give or take would work fine.
Try to buy a round body wire wound resistor though. Those things are really rugged. You can crank way over 5W of power to them until they even start to glow and they still don't fail.
Peter
Edited by 65_Impala on 01-11-12 05:21 PM. Reason for edit: No reason given.
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72novaproject
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01-12-12 07:36 AM - Post#2178506
In response to 65_Impala
Thank you for your help and patience Peter.
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