Car stopped charging while out on a run, I managed to get home but when I checked voltage across battery without car running and with the car running was the same, 12.35 volts.

To lazy to check wiring I ordered a new alternator, fast forward 3 days and in the garage with the new alternator fitted and checking voltage across battery to find exactly the same readings.

I walked back into the kitchen because everything looks better with a cup of tea, 15 minutes later when I started checking wiring. I found a connector at the alternator end burned out as the wire had come loose. Great I thought easy but being the person I am and 3 years ago I completely rewired the car with a gbs wiring harness but that harness does not cover the alternator wiring, so I thought lets rewire the alternator wiring and be done with it.

Now I'm running a pinto engine with a bosch alternator. On the bosch are 3 connectors two large and 1 small. One large connector went straight to the battery live the other large connector went to my kill switch, so when the key is connected it goes straight to the battery. The little connector goes to my warning battery charging light but I haven't connected this yet as I'm waiting for a 86 ohm resistor to put in line before it goes to the warning lights as l have smith flight dials and this is what they say, this light wasn't connected before but as I am rewiring I wanted the light to work.

All excited turn the key on stared the car checked the battery and went to make a cuppa as things always look better after a cuppa.

Still 12.56 volts with car running and the voltage drops to 12.34 volts with lights on so not charging. After a lot of swearing I check the two large connectors when unplugged from alternator, I have continually across both  to the battery and also 12 volts with battery connected, the alternator has a good earth back to the battery as I  checked continually. Last chance I put the old alternator back on with the same reading the voltage is the same with car running as not.

Sorry for the long description but my question is have I made a school boy error or are both alternators  faulty. Any help gratefully received as I'm all tea out. 

Cheers andy

I am no great expert but I'm pretty sure the alternator relies on the warning light circuit to excite it to start charging. You probably need to connect yours up. Worth a Google or confirmation from another source. Good luck.

Yes, that’s correct, the light is essential.

Thank you gentlemen, I shall wait until I have wired that ignition light and let you know how I get on. Very much appreciate your input.

Cheers andy

Any idea why my light stays on for ages after the car has started? It doesn't seem to be causing any problems.

Jonty this could be the brushes not making good contact due to either sticking in their guides or just being worn to end of life. As they warm up they get better contact.

You are saying you are using a resistor in the circuit, have you tried just using a 5W bulb, it could be that your circuit is not drawing enough current until the revs increase or the voltage drops.

Caveat I am not an expert or even claim to know much about it, But as I understand it the alternator dumbly measures the voltage on the direct from battery wire, it compares that with the voltage on the wire that goes via the Exciter bulb it then uses these values as to whether it needs to charge or not. If the Exciter wire is not connected, the bulb blown, or it's not drawing enough current it can't do that, so doesn't charge.

When we used to use Acewell dash's we used to either hide a 5W bulb behind the dash painted black, or add a resistor the value of which now escapes me, but I did think it bigger than 86ohm.

I have recently been investigating this subject due to having destroyed an alternator trying to change the brushes,.... we don't talk about this. However, here is what I think I believe is relevent to this thread. The connections on the alternator are a group of three spade terminals in a line, two large and one smaller. The two large terminals are connected together, that is they are essentially one terminal, check this out with your meter. On my car they are both connected to the same heavy duty cable going directly to the positive terminal on the battery. The small terminal is connected to the alternator regulator circuit, the circuit that controls the current feed to the field coils which regulates the output from the alternator (on the big terminals). When the alternator is generating power, it uses the generated power to feed the field coils under control of the regulator. So it's a chicken and egg situation. If the alternator is not generating power it can't engergize the field coils. This is the situation when the engine is first starting up. A small current is needed by the regulator in order to slightly energize the field coils to generate a tiny current that is then fed back into the coils etc. until the process builds up to the large current output produced by the alternator. This small current of course comes from the battery. It is convenient to use the resistance of the filament in an incandesant light bulb to allow only a small current to the regulator because as soon as the alternator starts to generate a greater current at 12 volts, no current will flow to/from the battery to the regulator and the light bulb will go out. We call this bulb the "ignition light". On the Sierra donor, this bulb is rated at 2.2 watt. At 12 volts this means that the current flowing to the regulator at startup is 183 milliamps, and implies that the bulb provides a resistance of 66 ohms. It's important not to feed too much current to the regulator, it might cause harm, so any lower resistance in this circuit should be avoided. Higher resistance is safe as long as enough current flows to get enough initial field in the field coils. Replacing the Ignition Light with an LED or any high resistance device/circuit usually prevents enough current reaching the regulator, accordingly a resistance can be introduced IN PARALLEL, remember the formula 1/R = 1/r1 + 1/r2, such that the total resistance is reduced towards the nominal above.

17 minutes ago, Sparepart said:

However, here is what I think I believe is relevent to this thread.

Why is there no 'Like' option for this post?

Needs 12 v via batt light as others say to excite the alternator try reving it to 5-6 k mine used to kick in at that , ive now wired in batt light 

BTW. Any old type of resistor won't do. It must be able to operate without overheating and burning out. The 2.2 watt ignition light gets quite warm, eventually can burn out and stop the alternator charging. It is not expensive to buy a resistor that will more than cope with the power. I attach photo of an example below, only costs around 3 quids and should last until the end of the universe. You would need one at whatever ohms you are after of course not necessarily the one shown.

Edited May 14, 20251 yr by Sparepart
add

On 5/14/2025 at 9:40 AM, Sparepart said:

BTW. Any old type of resistor won't do. It must be able to operate without overheating and burning out. The 2.2 watt ignition light gets quite warm, eventually can burn out and stop the alternator charging. It is not expensive to buy a resistor that will more than cope with the power. I attach photo of an example below, only costs around 3 quids and should last until the end of the universe. You would need one at whatever ohms you are after of course not necessarily the one shown.

I have a 4W bulb currently which I want to swap for a LED, is my logic correct?

4w / 12v = 0.333A

12v / 0.333A = 36 Ohms resistor required?

They seem to come in 10W, 25W, 50W  variant's, does it matter which one, I can get 36Ohms in 10W but only 33Ohms in 25W or 50W?

Edited September 28, 2025Sep 28 by phaeton
spilling mustacks

The calculation of the current flowing through a 4 watt 12 volt bulb =0.333 amp looks okay, and that it has a resistance of 36 ohms is good. I am wondering why you are looking for 360 ohm resistors. It is important to understand that this replacement is not just pull out a bulb and plug in a LED with a resistance in series. If you search the web for circuit diagrams showing how to replace the ignition warning bulb with a LED then you will find the following sort of circuit.

 

As I mentioned previously the resistance (here R1) is in parallel with the LED. The current flowing through the LED will be limited by a relatively large resistance (here 500 ohm) so typically only a small current (here .024 amp) would flow, not enough for the field coil. The effective resistance R is calculated by 1/R = 1/R1+1/R2, here 1/R = 1/50+1/500, i.e 1/R = 0.022, i.e R = 45.46 ohms. If you want the LED replacement to give the same current to the field coil on your car as the current bulb does, then you are aiming at R = 36 ohms.  You can see by the example that to do this the R1 (the resistor that you will buy) will need to be a bit more than 36, and will depend on the resistance of the LED circuit (R2) which you probably don't know, but can assume is high like above. Bottom line is a resistance of something like 45-50 ohms would do. This is then going to use about 4W just like before, but just as the bulb gets hot you need the resistance to be rated as being able to take at least 5W ... but the more the better higher wattage capability will lengthen the life of the resistor. Say aim for 50 ohm resistor with a 10W capacity. By the way the diode shown in the circuit was added by the designer (not me) as protection against any reverse polarity voltage spikes coming from the regulator in the alternator. I don't know how likely that is. I realise that this is a VERY long winded answer, but I can't stop once I get started, unlike my car.

46 minutes ago, Sparepart said:

The calculation of the current flowing through a 4 watt 12 volt bulb =0.333 amp looks okay, and that it has a resistance of 36 ohms is good. I am wondering why you are looking for 360 ohm resistors. It is important to understand that this replacement is not just pull out a bulb and plug in a LED with a resistance in series. If you search the web for circuit diagrams showing how to replace the ignition warning bulb with a LED then you will find the following sort of circuit.

 

As I mentioned previously the resistance (here R1) is in parallel with the LED. The current flowing through the LED will be limited by a relatively large resistance (here 500 ohm) so typically only a small current (here .024 amp) would flow, not enough for the field coil. The effective resistance R is calculated by 1/R = 1/R1+1/R2, here 1/R = 1/50+1/500, i.e 1/R = 0.022, i.e R = 45.46 ohms. If you want the LED replacement to give the same current to the field coil on your car as the current bulb does, then you are aiming at R = 36 ohms.  You can see by the example that to do this the R1 (the resistor that you will buy) will need to be a bit more than 36, and will depend on the resistance of the LED circuit (R2) which you probably don't know, but can assume is high like above. Bottom line is a resistance of something like 45-50 ohms would do. This is then going to use about 4W just like before, but just as the bulb gets hot you need the resistance to be rated as being able to take at least 5W ... but the more the better higher wattage capability will lengthen the life of the resistor. Say aim for 50 ohm resistor with a 10W capacity. By the way the diode shown in the circuit was added by the designer (not me) as protection against any reverse polarity voltage spikes coming from the regulator in the alternator. I don't know how likely that is. I realise that this is a VERY long winded answer, but I can't stop once I get started, unlike my car.

Do not worry about the long winded answer, although I didn't mention 360 Ohms? Maybe 36Ohms come over as 360 Ohms instead of the 36 Ohms is what I meant.

But the upshot is I want a 50 Ohm, 10W resistor which is the answer I was looking for Thank you very much.

The bootstrap current does not need to be as high as 0.3A my car has a 1.5W bulb an still starts charging almost as the engine starts.

The alternator is a 3 phase AC generator feeding into a 9 diode bridge rectifier. in effect 2 bridge rectifiers sharing the negative legs.

1 Rectifier goes direct to the battery. The second feeds the control circuit which then drives the rotor with a controlled current to generate the correct voltage to charge and keep charged the battery..

The difference between your 0.3A start current and my 0.1A start is at the most 0.1 second in the alternator becoming self sufficient.