6 ms·
>The workaround of a resistor seems to defeat the purpose. P=VI. I=V/R. V=12 => P=24/R. I don't quite understand how a resistor defeats the purpose of LEDs,
by swiley 8y ago
>The workaround of a resistor seems to defeat the purpose.
P=VI.
I=V/R.
V=12 => P=24/R.
I don't quite understand how a resistor defeats the purpose of LEDs, they're more efficient because you get more light per watt not because resistance is low.
- tyingq 8y agoLikely talking about load resistors that are installed in parallel with the LEDs, converting energy to heat solely for the purpose of making the "incandescent bulb is bad" sensor happy.
- ridgeguy 8y agoIf as GP said the bulb fault is sensed as resistance is too low, putting a resistor in parallel with the LED will reduce the resistance even further. I think it's the I^2R loss in a (series) resistor that makes using a resistor problematic from the POV of energy savings.
- tyingq 8y agoThe sensor is looking at current draw. If it's too low, it assumes a burned out incandescent. Basically, the LEDs are too efficient and look like a burned bulb. The resistor in parallel draws enough current to make it happy.
- solarkraft 8y agoThen resistance is not too low, but too high.
- taneq 8y agoThe reason you need the resistor is that the LEDs pull too little current, not that the voltage drop across them is too low.