Need help with simple problem . . .

Hi, new here. My working electrical knowledge is minimal at best (hence my user name) and I am hoping some here take pity on me. I am wanting to replace a few 12v incandescent bulbs with LEDs. The circuit contains a flasher unit that senses load and when there is a burnt bulb, a circuit in the flasher causes the unit to speed up the flashing cycle as a warning of a bulb failure. My problem is I am at the moment stumped as what to do to create a load in the circuit to replace that of the incandescents since the LEDs draw far less.

Please have pity on me!

Use a resistor.

LEDs generally draw 10-20 mA. If the voltage drop across the LED is 1.5V, then you would want a 1Khm: resistor. How many LEDs to replace one 12v incandescent bulb? Are they in series or parallel?

Mike

If you need an additional load just add a resistor in parallel with the LED-resistor combination. Since an incandescent bulb takes a lot of power, the resistors will need to be power resistors.

For example to replace a 3W bulb would require a 56 ohm, 3W resistor. But it may be possible to get by with a higher value, lower wattage resistor and still make the flasher work properly. You'd have to experiment with different resistor values to determine that.

Thanks. I figured resistors but didn't think power resistors. Now to find where I stashed Ohm's Law in my brain and dust it off! Haha!!

Here are a few conversion formulas you might find useful...
https://www.aaroncake.net/electronics/math.htm

You could use larger LEDs, or parallel-string LEDs to increase current draw without re-wiring the circuit with smaller power transistors.