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Firstly my electronics knowledge is limited.

I am trying to work out a way to use the existing Rheostat (10 ohm) that is used to dim the instrument lights in my 1978 car. I have changed the globes to 12V dimmable LEDs, but the voltage drop on the dimmer is not sufficient to dim the LEDs. I have tested the LEDs and they are turned off at about 4.5V and are almost at full brightness at 10V.

I could use a PWM device with a different trimpot, but I would like to use the existing dimmer switch. I thought maybe an Op amp with a 800mA transistor on the output might work, but I am struggling with that concept. Any ideas or available modification circuitry available?

Thanks.

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  • \$\begingroup\$ Your car in older than me. \$\endgroup\$ – Gregory Kornblum Aug 21 '18 at 15:18
  • \$\begingroup\$ I have made up Henry Crun's Op Amp circuit on a breadboard and have done some testing with some interesting results. With 1 LED across the output Vout doesn't change with a change in R1. With no load Vout does change. Any thoughts? \$\endgroup\$ – Geoff H Aug 23 '18 at 12:39
  • \$\begingroup\$ With a 10 ohm resistor as R1 and no load on the output, Vout is 7.5v and with R1 = 2.5 ohms Vout = 6.97V. With a 10 ohm resistor as R1 and 1 LED on the output, Vout is 5.26V and with R1 = 2.5 ohms Vout = 5.26V. Note the same Vout with the LED connected at Vout \$\endgroup\$ – Geoff H Aug 23 '18 at 12:43
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    \$\begingroup\$ But still younger than me \$\endgroup\$ – Geoff H Aug 23 '18 at 12:47
  • \$\begingroup\$ @GeoffH: Is the rheostat (small 'r') connected to +12 or chassis? \$\endgroup\$ – Transistor Aug 23 '18 at 21:25
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The 10 ohm Rheostat will be in series with the light, that would work with light bulbs.

Suppose the light bulbs are 10 Ohms in total (that's 10 ohms for all of them in parallel) then with Rheostat = 0 ohm you get full power and with Rheostat = 10 ohms you would get about 1/4 of the full power. (1/4th since the voltage is halved but then the current halves as well).

The LEDs consume much less power so less current flows so less voltage is dropped. Suppose the LED us 1/10th of the original light bulb current then the voltage drop over the Rheostat is also 1/10th. So 6 V (12 V / 2 for bulbs) becomes 0.6 V. That's useless for dimming !

So if we could increase the current back to it's old value then dimming would be much better. What you could do is connect light bulbs in parallel with the LEDs. Those give light and might break. Instead of bulbs you could use resistors, these behave similarly to bulbs but give no light and should not blow either (assuming you're using them correctly).

So what value resistor should you use?

That depends on how much current was flowing originally, we do not want to exceed that current as that will damage the Rheostat. Suppose there used to be 4 light bulbs of 2 W each that is 8 Watt in total

8 Watt / 12 V = 0.67 A then an equivalent resistor (to "fake" the load of the 4 2 W bulbs) is 12 V / 0.67 A = 18 ohms

Since the power is 8 w you need a 10 W resistor of 18 ohms.

If you do not have one of those lying around, try a 12 V 8 W (or 10 W) bulb, connect it in parallel with the LEDs and see how that works.

Note that this isn't a energy efficient solution ! Normally we use LEDs to save power. My proposal is an "easy fix", it is not intended as an energy efficient solution. Then a car isn't energy efficient anyway. Also the 8 W for dashboard illumination pales to the 2 x 50 W of your car's headlights (assuming they're still bulbs).

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    \$\begingroup\$ +1 This is probably the best solution for this OP, as ugly as it is. \$\endgroup\$ – Spehro Pefhany Aug 21 '18 at 14:47
  • \$\begingroup\$ This would work depending on how the 12V dimmable LED is dimmed. It appears the LED's intensity is voltage controlled. But I do not see how you connect the LED to the resistors that replace the light bulb(s). \$\endgroup\$ – Misunderstood Aug 21 '18 at 15:03
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    \$\begingroup\$ @Misunderstood The LED's aren't (designed to be) dimmable. Yet I think you would agree that I can dim any LED if I just lower the current through it. My guess is that the LEDs used here are a couple of SMD LEDs in series with a series resistor and a bridge rectifier in a "bulb" housing (for easy upgrade). You can find this cheap on Ebay. My solution basically suggests to connect the LEDs in parallel with the original bulbs but replace the bulbs with a resistor. The resistor increases the current so the series variable resistor has more influence on the voltage. \$\endgroup\$ – Bimpelrekkie Aug 21 '18 at 16:49
  • \$\begingroup\$ I agree with what you are saying as that makes sense. The OP said "12V dimmable". When an LED light bulb replacement says "dimmable" that means there is additional circuitry to do the dimming. Unless there was some significant advantage, I'd opt for a light bulb rather than LED. \$\endgroup\$ – Misunderstood Aug 21 '18 at 17:52
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Here's a circuit that should work but you may need the help of someone with a little electronics experience to put it together.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A variable current source.

How it works:

  • Q1 is turned on by R4 drawing current from its base.
  • Current now flows through Rmin + R1 (your dimmer), Q1 and out to the LED lamps.
  • As the current increases through Rmin + R1 the voltage across them increases. This voltage is applied to Q2's base.
  • When Q2's base voltage drops about 0.7 V below its emitter voltage it starts to turn on, reducing the emitter-base voltage on Q1 and preventing Q1 turning on any further.
  • The circuit settles with about 0.7 V across R1.

You mention 800 mA in your question so let's say we want to adjust between 80 and 800 mA using as much of the range of the dimmer as we can.

  • At 10 Ω we will get 700 mV at \$ I = \frac {V}{R} = \frac {0.7}{10} = 70 \ \text {mA} \$ which is very close to what we want.
  • To get 800 mA we need a resistance of \$ R = {V}{I} = \frac {0.7}{0.8} = 0.875 \ \Omega \$. So setting Rmin to something between 0.5 to 1 Ω should work.

If you want to be picky then recalculate the 10 Ω with 10 + Rmin but I'd be surprised if you could see the difference.

A couple of notes:

  • Calculate the power rating of Rmin from \$ P = I^2R \$ and buy one with twice the rating so that it runs cool.
  • Q1 should be rated at twice the max current you expect to draw.
  • Q1 should be mounted on an isolated heatsink.
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  • \$\begingroup\$ +1 I like this solution. This may not work with the "12V dimmable" the OP purchased. I assume the OP's ight bulb replacement LED has additional circuity (maybe similar to your circuit) rather than just an LED and resistor. Nice explanation too. \$\endgroup\$ – Misunderstood Aug 21 '18 at 17:59
  • \$\begingroup\$ I like that R1 can be left connected to +12 the way it is in the car. (if you swap it with Rmin) \$\endgroup\$ – Henry Crun Aug 21 '18 at 20:29
  • \$\begingroup\$ Good point. I'll swap Rmin and R1 to leave R1 connected to the battery +. \$\endgroup\$ – Transistor Aug 21 '18 at 20:31
  • \$\begingroup\$ One minor change to improve reliability: add a 1k or so resistor in series with the base of Q2. This protects Q2 against damage if the output is accidentally shorted to ground. The added resistor has no other effect on the circuit. \$\endgroup\$ – Dwayne Reid Jun 13 at 21:51
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You could use a cheap microcontroller (e.g. ATTiny85 or similar) to measure the voltage output of the existing circuit using its analog to digital converter (using a voltage divider to reduce the input voltage to a range the microcontroller will tolerate) and use that to set the duty cycle of a PWM (which the microcontroller also has). Then just use a transistor or mosfet to drive your LEDs from the PWM output.

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    \$\begingroup\$ He said his electronic knowledge is limited. I don’t think this is a viable solution in this case. \$\endgroup\$ – PDuarte Aug 21 '18 at 12:55
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    \$\begingroup\$ Microcontrollers can be handled by children with elementary language skills, this is 100% a viable solution. +1 \$\endgroup\$ – Passerby Aug 21 '18 at 16:11
  • \$\begingroup\$ Use a PICAXE. It has an inexpensive and simplified programming hardware and has a simplified language. If that's too much, it has a visual flowchart method you can use intsead. \$\endgroup\$ – DKNguyen Jun 12 at 18:20
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Well this is what you asked for: an opamp circuit.

R7 is setting what the voltage will be when R1 is a max R. R4/6 set the gain

U1 is acting as a voltage follower somewhat like using an NPN transistor here. I use an LM317 as it has current limiting and thermal shutdown, and is kind of an easy way to get these.

schematic

simulate this circuit – Schematic created using CircuitLab

This is a constant current circuit. It is a bit inflexible since R1 has a fixed value and Vref is fixed, so you don't have too much range to adjust. However it might be in the correct range for you. Note that this is really the same circuit as Transistors, but using LM317 with a 1.2V reference instead of Q2 with a 0,6V reference. So current will be 2x. Again, using lm317 gives thermal protection. If you add D1, then current will approximately halve.

schematic

simulate this circuit

You can run the simulator ("DC solve") to look at the voltages, and see what changing R does.

You need to adjust R4 and R6 to set the dim and bright points. When R7=R1, Vout is ~250mV.

Op amp output should be 1.2V below VOUT. Opamp outputs can't go to V+, nor can LM317, so max Vout will be ~10V.

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  • \$\begingroup\$ I will try one of these circuits over the weekend, probably the opamp circuit first and see if it does the job. I will post my results after the test. Thanks \$\endgroup\$ – Geoff H Aug 22 '18 at 23:34
  • \$\begingroup\$ I have made up Henry Crun's Op Amp circuit on a breadboard and have done some testing with some interesting results. With 1 LED across the output Vout doesn't change with a change in R1. With no load Vout does change. Any thoughts? \$\endgroup\$ – Geoff H Aug 23 '18 at 12:48
  • \$\begingroup\$ With a 10 ohm resistor as R1 and no load on the output, Vout is 7.5v and with R1 = 2.5 ohms Vout = 6.97V. With a 10 ohm resistor as R1 and 1 LED on the output, Vout is 5.26V and with R1 = 2.5 ohms Vout = 5.26V. Note the same Vout with the LED connected at Vout \$\endgroup\$ – Geoff H Aug 23 '18 at 12:49
  • \$\begingroup\$ @GeoffH Not wired right I expect. You can run the simulator ("DC solve") to look at the voltages. You need to adjust R4 and R6 to set the dim and bright points. Op amp output should be 1.2V below VOUT \$\endgroup\$ – Henry Crun Aug 23 '18 at 20:57
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    \$\begingroup\$ I wish this website was around in the mid 1980s when I did a subject on analogue electronics & the simulator would have been fantastic. Actally I wish the internet was available in the 80s! \$\endgroup\$ – Geoff H Sep 10 '18 at 9:59

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