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I would like to know how to reduce the brightness of the running lamp in the tail lights of my tow car. It is a MINI Cooper and each tail light contains one single filament running lamp. These lamps also act as the brake lights in normal(not towed) driving. I believe the car's electronics change the voltage from say 5.5v as a running light to 12v as a brake light although I haven't confirmed this. Since I will tow this car with my RV, the running light feed from the RV will range 12v-14.6v. This high of a voltage will appear that the brake lights are illuminated and also create unnecessary heat within the tail light. I don't know how to calculate the proper size resister to safely reduce the brightness to say maybe half. There are two lamps, 21w each that are fed from one line. I have diodes in the circuit so there won't be any type of back feed into the car's electronics. Thank you for your help. Howard

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  • \$\begingroup\$ get a lower power lightbulb ... or use external lights that attach to the tow car \$\endgroup\$
    – jsotola
    Apr 21, 2020 at 5:02
  • \$\begingroup\$ The Mini Cooper's rear lights I've seen are a normal rear light cluster. The brake light is brighter than the running light because it is a separate element. on a different wire. Which MINI Cooper do you have? And a normal single filament light will need at least half voltage -- if you are correct, it's more likely to be 7 than 5.5V. \$\endgroup\$
    – david
    Apr 21, 2020 at 10:04
  • \$\begingroup\$ It is a Countryman. Each tail light has a reverse lamp, turn lamp and two additional single filament lamps which illuminate at the same time as running and brake lights. The tail light wiring harness has separate running and brake light wires that operate the same lamps. This is known as separate wiring whereas the RV is combination wiring, the turn and brake are on the same wire. It's just the running light power from the RV will be full voltage which are too bright and hot for short or long periods of driving. This is why I need to reduce the voltage. Thanks. \$\endgroup\$
    – Howard G
    Apr 21, 2020 at 12:05

3 Answers 3

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I don't know how to calculate the proper size resister to safely reduce the brightness to say maybe half. There are two lamps, 21w each that are fed from one line.

The relevant formulas are \$R = V / I\$ and \$P = V * I\$ (where V = voltage in volts, I = current in amps, R = resistance in ohms, and P = power in watts).

Calculating the required resistance for an incandescent lamp is a little tricky though, because the filament resistance has a positive temperature coefficient so it reduces at lower voltage (causing higher than expected power draw), while the light output spectrum shifts more towards infrared which is less visible. So you might have to experiment with different resistor values to get the exact effect you want.

Another option might be to simply wire an identical lamp in series. Each lamp should then drop half the total voltage, and you could use the 'ballast' lamp as an indicator that the tail light is on.

So how much current and power would each lamp draw, and what would be the equivalent resistance to do the same job? I tested a 12 V 15 W festoon lamp on a range of voltages from 0 to 12 V, and this is what I got:-

enter image description here

At 12 V the lamp drew 1.22 A, so its resistance was 12/1.22 = 9.84 Ω. At 6V it drew 0.86 A, ~70% of the current at 12 V (a fixed resistor would draw 50%). Its resistance at 6 V was 6/0.86 = 6.98 Ω, ~70% of the value at 12 V.

We can expect that a 12 V 21 W lamp would act similarly. At 12 V it should draw 21/12 = 1.75 A, and the resistance should be 12/1.75 = 6.86 Ω. 70% of that is 1.23 A and 4.8 Ω. So a 4.7 Ω (standard value) power resistor should drop the voltage from 12 V to ~6 V. It would dissipate ~6*1.23 = ~7.4 watts, so should be rated for 15 W or higher (derated 50% to keep surface temperature down), and kept away from anything that doesn't like high temperatures.

So you could put either a 4.7 Ω 15 W resistor or a 12 V 21 W lamp in series to reduce the tail light voltage to ~6 V.

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  • \$\begingroup\$ Wow, thanks Bruce. That took me back to my years in electrical school learning Ohms Law and series/parallel circuits. Your answer was clear and concise and exactly what I was looking for. Would a 10 ohm 30w resistor do the same job if I use one resistor prior to both lamps? Thanks \$\endgroup\$
    – Howard G
    Apr 21, 2020 at 12:12
  • \$\begingroup\$ If you feed both lamps from a single resistor then it will have to pass twice as much current, so the resistance needs to be halved. Power loss will be doubled, so the calculated values are 2.4 ohms and ~15 watts. A 2.2 ohm 30W resistor would be close enough. \$\endgroup\$ Apr 21, 2020 at 20:00
  • \$\begingroup\$ Great, thank you Bruce. I would have responded earlier but since I'm new to this site, I didn't know where to look for your responce if you had one. Thanks again. \$\endgroup\$
    – Howard G
    Apr 22, 2020 at 19:12
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Would it be safer to wire across the Mini's brake light switch, using a relay controlled by your RV's brake light? Then you don't have to break into the Mini's wiring loom (always a bad idea when dealing with a BMW product.) The brake light circuit should be working regardless of the car's key switch position.

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  • \$\begingroup\$ Thank you for your help. Apparently I cannot use the brake light switch since it is not active without the key on. It's too bad because part of the tow braking system involves an air cylinder attached to the brake pedal that is actuated by the RV's air brakes. That would have illuminated the brake lights if it wasn't so computerized. \$\endgroup\$
    – Howard G
    Apr 21, 2020 at 3:19
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Diodes might be more effective in reducing the voltage and current (and therefore power) through your bulbs.
Each diode would drop about 0.7-0.9V, so you would need about 5-8 pieces.
You might need to use 3A diodes, especially if using the same diodes for both bulbs.
A better solution would be to get lower wattage bulbs or replace the bulbs with LEDs, which would use a lot less energy (about 10 times less) and would be easier and less wasteful to control.

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