Making a lamp dimmer - resistor?
Discussion
I don't know if I'm in the right place, but I couldn't see a section for electrics.
It's not for a proper car, but I'm making a toy car with working lights. I have some reversing lamps from a caravan, like this .
They are being connected to the wiring used by an electric wheelchair, which has LED lights. When I wired up one of the Jokon lamps, it was very bright, and I don't want to be blinding anyone with these things - they're just for effect. The LEDs seem to run on the full 24V from the batteries, but are 0.7W according to the back of the housing. The lamp I have is for 12V, and is a P21W.
I can't really get involved with reducing the supply to 12V because the switching is done through a control box which also controls the motors. Can I reduce the brightness by wiring something into the circuit with the bulbs? I'm a bit sketchy on my electronics with stuff like this, but was wondering if I just add enough resistance whether that will achieve what I want. My next thought is that I'll be making that resistor very hot presumably, which might just create a different problem.
Any ideas or thoughts welcome. Cheers.
It's not for a proper car, but I'm making a toy car with working lights. I have some reversing lamps from a caravan, like this .
They are being connected to the wiring used by an electric wheelchair, which has LED lights. When I wired up one of the Jokon lamps, it was very bright, and I don't want to be blinding anyone with these things - they're just for effect. The LEDs seem to run on the full 24V from the batteries, but are 0.7W according to the back of the housing. The lamp I have is for 12V, and is a P21W.
I can't really get involved with reducing the supply to 12V because the switching is done through a control box which also controls the motors. Can I reduce the brightness by wiring something into the circuit with the bulbs? I'm a bit sketchy on my electronics with stuff like this, but was wondering if I just add enough resistance whether that will achieve what I want. My next thought is that I'll be making that resistor very hot presumably, which might just create a different problem.
Any ideas or thoughts welcome. Cheers.
Thanks for the replies. The bulbs I have are designed for 12V. Any idea what sort of resistors I would need to look at if I wanted to go down that route? I'm guessing the tiny ones I have to put in breadboards aren't the thing to use 
Wiring in series is an idea, but the same problem applies to the indicators, which are also far too bright, and they need to be wired up individually. Looking on parts websites, it occurred to me that you can get bulbs designed for 24V. I'm trying to work out if I bought those instead, with the same wattage rating, whether they would be dimmer or exactly the same
I'll have a look at the LED conversion route, thanks. As it's a Toylander (classic Land Rover model), incandescent bulbs would suit the look better, but not if they're belting out a million candle power.
Wiring in series is an idea, but the same problem applies to the indicators, which are also far too bright, and they need to be wired up individually. Looking on parts websites, it occurred to me that you can get bulbs designed for 24V. I'm trying to work out if I bought those instead, with the same wattage rating, whether they would be dimmer or exactly the same
I'll have a look at the LED conversion route, thanks. As it's a Toylander (classic Land Rover model), incandescent bulbs would suit the look better, but not if they're belting out a million candle power.
Edited by Prawo Jazdy on Monday 29th November 12:16
When you have a 24V power source, then you should use 24V bulbs. They're used in trucks, so good chance you can pick some up near you. Then they should be a lot dimmer than 12V bulbs, and last considerably longer. If you want have dimmer 24V lights, you'd choose a lower Wattage.
Your P21W fits into a BA15s holders, and there are bulbs with 24V and less than 21W that fit, such as https://www.truckjunkie.eu/HELLA-GLOEILAMP-5W-BA15...
Your P21W fits into a BA15s holders, and there are bulbs with 24V and less than 21W that fit, such as https://www.truckjunkie.eu/HELLA-GLOEILAMP-5W-BA15...
What he said ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
If you want to use a resistor to dim them, you will need a minimum 12W resistor, which is quite large, and will suffice for a single bulb at a time.
The other alternative, is to get a 'buck' regulator and use it to step down to 6v for everything on the car - however you will need quite a hefty device.
EDIT: Something like this might suffice, depending on total load.
If you want to use a resistor to dim them, you will need a minimum 12W resistor, which is quite large, and will suffice for a single bulb at a time.
The other alternative, is to get a 'buck' regulator and use it to step down to 6v for everything on the car - however you will need quite a hefty device.
EDIT: Something like this might suffice, depending on total load.
Edited by TonyRPH on Monday 29th November 17:24
The simplest thing to do is to place a resistor in series with each bulb. The battery voltage will be split across the bulb and the resistor, such that bukb voltage + resistor votage = battery voltage.
We want 12V across the bulb. That means the voltage across the resistor will be :
Resistor voltage =Battery voltage - bulb voltage
Resistor voltage = 24V - 12V
Resistor voltage = 12V
Next we need to work out the current that flows out of the battery, through the resistor, then through the bulb and goes back into the battery. We can get that from the bulb power.
A P21W bulb is 21 Watts at 12 V.
P=I*V
Where P is power in watts , I is current in Amps and V is voltage.
So at 12V, the current through the bulb is P/V = 21/12 = 1.75A, since we have a series circuit, this is also the current flowing through the resistor.
To work out the value of the resistor, we use Ohms law, V= I * R (where R is resistance in ohms.)
If V = I* R then R = V / I
R = 12V / 1.75A = 6.85 Ohms.
Since the voltage across the resistor is the same as the voltage across the bulb & the current through the resistor is the same as the current through the bulb, then we can immediately say that the power dissipated by the resistor will be the same as the power dissipated by the bulb, 21W.
If the resistor is dissipating 21W, the resistor fitted should be capable of handling at least twice this. In practice,use this:
https://cpc.farnell.com/welwyn/wh50-6r8ji/resistor...
And mount it to a metal plate to dissipate the heat.
That's the simplest solution, but it is a poor one.
If you have two headlights on, the draw from the battery is 3.5A, or 84W. Half of this is wasted in the resistors. That alone will drain the battery quickly.
Going one step better, you can purchase a P21W (also known as a 382 bulb) that was designed for 24V. Doing so will halve the power drain from 84W (resistor solution) to 42W. It's still a lot of power.
The best solution is to buy a 382 LED bulb for 24V operation. Power drain should be ~3W per bulb, so that's down from 84W to about 6W.
You could buy 382 LED bulbs for 12V operation and add a resistor. Not as good as buying 24V ones as you'll still be dissipating about 6W in the bulbs, & you'll also be wasting 6W in the resistors. If you go down this route, we will need to recalculate the resistor, which means we'll need to know how much power the LED bulb actually uses. 3W is just my guess.
Addressing other proposals in this thread:
1/ Use a regulator to drop the voltage from 24V to 12V for the lighting system
Yeah, this would work. It's overkill though and given the OP's level of expertise, I'd advise against it. It would likely be just as expensive as fitting 24V LED bulbs, but it would be more complicated. The OP would need to correctly determine the load he's placing on the regulator, source an appropriate part & wire it up. 24V LED bulbs are far simpler.
2/ Tap into the battery to get a 12V feed.
Bad idea.
It will work, for a bit. Then you'll find the battery is rapidly losing capacity.
What will happen is the bulbs will discharge one battery faster than the other. The controller is looking at the pack voltage to determine when to stop discharging and when to stop charging. When both 12V batteries int he 24V pack have the same voltage, this is fine. But if one has a lower voltage thant he other then the lower voltage battery will be overdischarged in use, and the higher battery will be overcharged when charging.
It is unlikely that a wheelchair battery pack has automatic cell balancing circuitry to correct this issue.
We want 12V across the bulb. That means the voltage across the resistor will be :
Resistor voltage =Battery voltage - bulb voltage
Resistor voltage = 24V - 12V
Resistor voltage = 12V
Next we need to work out the current that flows out of the battery, through the resistor, then through the bulb and goes back into the battery. We can get that from the bulb power.
A P21W bulb is 21 Watts at 12 V.
P=I*V
Where P is power in watts , I is current in Amps and V is voltage.
So at 12V, the current through the bulb is P/V = 21/12 = 1.75A, since we have a series circuit, this is also the current flowing through the resistor.
To work out the value of the resistor, we use Ohms law, V= I * R (where R is resistance in ohms.)
If V = I* R then R = V / I
R = 12V / 1.75A = 6.85 Ohms.
Since the voltage across the resistor is the same as the voltage across the bulb & the current through the resistor is the same as the current through the bulb, then we can immediately say that the power dissipated by the resistor will be the same as the power dissipated by the bulb, 21W.
If the resistor is dissipating 21W, the resistor fitted should be capable of handling at least twice this. In practice,use this:
https://cpc.farnell.com/welwyn/wh50-6r8ji/resistor...
And mount it to a metal plate to dissipate the heat.
That's the simplest solution, but it is a poor one.
If you have two headlights on, the draw from the battery is 3.5A, or 84W. Half of this is wasted in the resistors. That alone will drain the battery quickly.
Going one step better, you can purchase a P21W (also known as a 382 bulb) that was designed for 24V. Doing so will halve the power drain from 84W (resistor solution) to 42W. It's still a lot of power.
The best solution is to buy a 382 LED bulb for 24V operation. Power drain should be ~3W per bulb, so that's down from 84W to about 6W.
You could buy 382 LED bulbs for 12V operation and add a resistor. Not as good as buying 24V ones as you'll still be dissipating about 6W in the bulbs, & you'll also be wasting 6W in the resistors. If you go down this route, we will need to recalculate the resistor, which means we'll need to know how much power the LED bulb actually uses. 3W is just my guess.
Addressing other proposals in this thread:
1/ Use a regulator to drop the voltage from 24V to 12V for the lighting system
Yeah, this would work. It's overkill though and given the OP's level of expertise, I'd advise against it. It would likely be just as expensive as fitting 24V LED bulbs, but it would be more complicated. The OP would need to correctly determine the load he's placing on the regulator, source an appropriate part & wire it up. 24V LED bulbs are far simpler.
2/ Tap into the battery to get a 12V feed.
Bad idea.
It will work, for a bit. Then you'll find the battery is rapidly losing capacity.
What will happen is the bulbs will discharge one battery faster than the other. The controller is looking at the pack voltage to determine when to stop discharging and when to stop charging. When both 12V batteries int he 24V pack have the same voltage, this is fine. But if one has a lower voltage thant he other then the lower voltage battery will be overdischarged in use, and the higher battery will be overcharged when charging.
It is unlikely that a wheelchair battery pack has automatic cell balancing circuitry to correct this issue.
Edited by BrokenSkunk on Thursday 9th December 11:16
Prawo Jazdy said:
This is really helpful stuff, thanks very much. Re: rewiring onto a single battery - I don’t think I can because the lighting circuit comes out of a control box, and I don’t want to mess about with the input voltage to the control box, because a lot of other stuff runs off it.
You can use the lighting out of the control box to pull in a mini relay which connects to the 12v as described above. That way the control box is left untouched and the light on line pulls in the mini relay. You could even have a tiny standalone lipo of the required voltage of your choice and that would last for years.Choice of bulbs here.
R149 is 24v 5w single contact R150 is the same with dual contact.
https://www.ringautomotive.com/en/product/r209
Edited by netherfield on Friday 10th December 16:48
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