I want to dim a commercial LED strobe module and can't seem to figure out what wattage the resistor needs to be. Most posts are about single or multiple LEDs with their known forward Voltage etc. like e.g. this post here demonstrates.

My module however is a self contained unit with, as it seems, Voltage regulation and micro controller integrated.

The LED strobe module is an emergency vehicle strobe mini light bar type:

Voltage: 12-24 / 10-30 VDC, 3 x 3 watt LED elements, Average current draw: [email protected] VDC, Max current draw: [email protected] VDC, Self-contained 22 flash patterns

I will run it on approx. 13.9V on a motorcycle (both battery and running engine) and found that a 400 to 500 Ohm resistor gives me the desired brightness.

I arrived at 400 to 500Ohm by trial and error using 3W rated resistors, which I believe are overkill for the final circuit.

My test rig is an 18V Li Rechargeable for a power drill, actually giving 19.2V and using the 500ohm the voltage drops to about 9.35V for the light module.

I tried to get the Wattage using this power source thinking it'll be save enough on the lower Voltage on the Moto:

W = 0.7A * (19V-9V) = 7W ?

It's huge for a LED light plus the 3W resistor barely gets warm. I suspected something in the range of about 1 to 2 W hence testing with 3W resistors.

Can somebody shove me into the right direction? What have I missed here?


1 Answer 1


I take it you measured the voltages. The current was probably and assumption. You can solve this just using voltages. Formula is V^2/R. So you have 100/500 Watts.

Something seems wrong with your numbers however. This would also imply that 20mA. Are you sure the value of your resistor isn't 50 Ohms?

  • \$\begingroup\$ 100% sure, the resistor is big enough to have it written on and measuring through is dead on 500 Ohm. You are correct as well thinking that I assumed the current. I took the led module spec max value for I. \$\endgroup\$
    – reeen
    Commented Nov 3, 2016 at 2:26
  • \$\begingroup\$ Looking at your equation, to be on the save side with a 1W resistor is about right then? \$\endgroup\$
    – reeen
    Commented Nov 3, 2016 at 2:32
  • \$\begingroup\$ Ohms law says current = voltage / resistance. 10/500 = 20mA. That doesn't seem right for your module Resistors are marked with the last digit as a multiplier of 10 to that digit. So 500 Ohms should be marked 501. 50 Ohms would be marked 500 because it means 50x10^0. When you say measure through, you mean you measured it with an Ohmmeter? \$\endgroup\$
    – owg60
    Commented Nov 3, 2016 at 2:35
  • \$\begingroup\$ Thanks so much for looking into this. You're correct, measured with Multimeter on Ohm setting. I meant the resistor is big enough so it actually has 500Ohm written on it, numbers and Ohm sign not the ring codes. the other resistors used have 560, 400, 300 ohm written on. I too felt something doesn't line up hence my post. I look at the LED integrated module not like a single LED and it surpassed my basic understanding of LED dimming. \$\endgroup\$
    – reeen
    Commented Nov 3, 2016 at 4:07

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