how to calculate resistor needed

I have a vintage car with 6v Lucas electrical system. The specified bulb for the ignition warning lamp (charging indicator) is 2.5v 0.5w. Originally the lamp holder had a fine resistor wire wound around it to accommodate the lower voltage.

The resistor wire has long since been damaged, and I'd like to replace it with a modern resistor. What's the correct way to calculate the value of the resistor required?

In case it makes any difference, the charging system consists of a 3-brush generator and cut-out unit (not a voltage regulator).

Any help with this knotty problem greatly appreciated.

Richard

I would start by calculating the current needed for the bulb -

P = V(A)
0.5 = 2.5A
A = 0.2

So, now you know the bulb needs 200mA to shine it's brightest at 2.5V.  That's not saying it won't work with less current, it just won't burn as bright.

Now that you know the voltage and current the resistance of the bulb can be determined using Ohm's Law -

R=V/I
R=6/0.2
R=30 ohms

You could then pull up a voltage divider calculator like this one: https://ourworld.compuserve.com/homepages/Bill_Bowden/r2.htm

Plug in your battery voltage (6), your R2 (30) and your voltage out (2.5) and calculate.

The end result is a 1/2 watt resistor of 42 ohms.  If you look at all the values given, though, you'll see a slight problem - the current through the bulb will only be 0.083A - under half of what the rated current is.  The bulb may or may not work at this current - you'll have to experament with it.  The problem with voltage dividers is their limited current output.

To incrase the output you could go with an adjustable voltage regulator and set it to 2.5vdc.  At that point, your current wouldn't be limited and you would give the bulb max power at all times.

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Kevin Pierson

Kevin,

Many thanks indeed. That explains why, when I tried a 20 ohm resistor (got that info. from a car restoration book) the resistor got very hot very quickly and the bulb barely glowed. I had no idea it was so complicated - I'm afraid my understanding of electricity is limited to comparisons with plumbing.

I'm not familiar with the voltage regulator you describe (only those used in vehicles). Will make enquiries at the electronics shop. Alternatively it might be a simpler scheme just to fit a 6v 1.2w bulb without the resistor.

Thanks again for your help,

Richard

If a 20 ohm resistor made it glow dimly then a 40 ohm is going to make it glow even less.  The heat associated with the resistor would be from the fact that the current going through the resistor would be .120mA at 3.6vdc which would be just under 1/2 watt.  You would need a 1 watt resistor to prevent it from getting hot.

An adjustable voltage regulator would be something like the LM317 - it uses a few components to select the desired voltage output.

Another solution may be retrofitting it with an LED instead of an incandescent bulb.  LEDs are bit easier to work with when it comes to weird voltages as they are discreet components (instead of resistive).  There are a few tricks you can use to make them work.

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Kevin Pierson

Hullo Kevin,

Hmm... had a feeling a 40 ohm resistor might get a bit hot. I've been using a resistor of 8 ohms 5w, though I've no idea how I arrived at those figures (it will be obvious that I'm well out of my depth with all this). It appeared to be working well, but now the bulb has burned out after a relatively short time so the logical conclusion is that it couldn't cope.

That type of bulb is hard to come by now (looks like an old-fashioned screw-base torch bulb), and I got that one from a vintage parts specialist in England. I've replaced it with a 6v bulb which seems to work well, even with the resistor still in place! However, this bulb, like the 2.5v, glows faintly when the dynamo is charging - theoretically, everything should balance out and the light should be completely extinguished unless the points of the cut-out unit are open, thus preventing the battery from discharging through the dynamo.

The thought of using an LED had crossed my mind, but it would need to be in the shape of the bulb to fit into the unit. And perhaps this 1930s system needs a bit of resistance?

Probably my next move will be to remove the resistor and just use the 6v bulb I have. I ran it like that for years and covered a lot of mileage, without any apparent trouble.

Thanks again for your advice. I really appreciate the time you've given me.

Richard