Quote:
Originally Posted by Xringer
Using diodes to drop voltage is smart, since there is hardly any power loss to heat.
Using a "smaller fuse" won't work, since a fuse is a safety device, not a voltage dropper. The fuse element melts in half to protect from over-current. That causes an open circuit. Lights out..
A long thin "feed wire" would work. If it were long and thin enough to drop 3.5 volts,
the remainder of the voltage would be applied to the headlamps.
BUT, the wire would get warm. 3.5V x ~10 Amps =35 watts of heat. Wasted heat.
~10A is assuming 2 lamps at 5A each. 13V x 5=65W lamp..
I have no idea what wattage headlamps are used these days.  . |
High school physics teaches us that P = V * I. This is true even if you drop the voltage across a diode instead of a resistor. You will need huge heat sinks to dissipate that kind of power.
A better way is to use a PWM circuit with a power MOSFET. You can look for DIY kits/circuits or DC motor speed controllers. You might still want a bypass switch for 100% brightness since the controller is not 100% efficient and it probably won't give you 100% duty.