Quote:
Originally Posted by dhutton
Pretty much all sounds right.
Also, Watts = amps squared x ohm
Or power = I^2 x R
Your load on the resistor is very short term so you will be able to get away with a power rating smaller than the calculation would tell you.
You can make 1 ohm putting two .5 ohm in series and .25 ohm placing two .5 ohm in parallel...
Those up down currents foreshadow the problem I was predicting, the voltage drop will be higher going up than it is going down and the window won’t raise properly. You can try the Schottky diode if you find that to be the case.
Don
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Great advice, again, Thanks Don