Can a Resistor Simulate a Light Bulb Load for LED Bulb Installation?

AI Thread Summary
A shunt type regulator in a snowmobile requires a constant load, typically provided by stock 55-watt bulbs, to function properly. When replacing these bulbs with LED alternatives, a resistor can be used to simulate the load. The power drawn by the LED bulbs must be calculated, and the difference from the system's designed load (110 watts) determines the power to dissipate in the resistor. A sample calculation shows that a resistor rated for at least 100 watts is necessary, with options available from suppliers like Digikey. It is crucial to study the resistor's datasheet for proper heat dissipation and consider higher wattage options for safety.
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Snowmobile, shunt regulator needs load to work. Replacing 2- 55 watt bulbs with LEDs

Need to replace draw on system to keep regulator working correct.
Hello All,

I have a snowmobile that uses a shunt type regulator to control voltage. They require a constant load (headlights) to work proper.

The 2 stock bulbs are 55 watts on low with the system running 12v.

Racers install another set of hand warmers, wired to high, and tape them to the tunnel to keep the system loaded when they remove the bulbs on mod sleds.

Is there a resistor I can wire into lighting circuit when I install LED bulbs that will simulate one 55watt light bulb?
 
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You need the actual current drawn by the LED bulbs, then the actual voltage with the engine running at cruise speed. Calculate the power drawn by the LED bulbs, subtract from the 110 watts the system is designed for, and the result is the power to dissipate in the resistor. Then calculate the resistance in the resistor.

Sample calculation with made up numbers:
LED bulbs draw 2 amps at 14 volts. That's 2 * 14 = 28 watts.
110 watts total minus 28 watts = 82 watts.
82 watts divided by 14 volts = 5.86 amps.
5.86 amps divided by 14 volts = 0.42 ohms.

These made up numbers need a 0.4 ohm resistor rated for at least 100 watts.
Looking in the Digikey catalog, I find a 0.39 ohm resistor rated for 100 watts: https://www.digikey.com/en/products/detail/te-connectivity-passive-product/HSC100R39J/5587123?s=N4IgjCBcoExaBjKAzAhgGwM4FMA0IB7KAbRADYBOAdhgFYQBdfABwBcoQBlVgJwEsAdgHMQAX3wwADAA560EEkhoseQiRABmCpIAEAWwDyACz2YQ%2BACwbdhk2aYg2HAKoC%2BrA8gCy2VJgCuPNhi%2BAC0cPKKvP6qRJCk9A5OkCBi4iARpAAKBADu2Dw6AErYmHyYrAQ89qJAA. This is not the only power resistor in that catalog, it is just the one that I found in a quick search. This is not the only supplier, there are others just as good that may have a better selection. You will need to do your own calculation with real numbers.

Study the datasheet VERY carefully. That resistor will only dissipate 100 watts with an adequate heat sink. You might be better off to look for a 200 watt resistor, or a pair of 0.84 ohm 100 watt resistors in parallel.
 
Thank you, I will look at power draw for my LEDs and follow your formula above to find the correct one for me.
 
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