The Cellar - View Single Post - Big, Fat Electronics question for all y'all.

tw · 03-29-2013, 08:49 PM

I ignored numbers for that audio tapered pot. 10k and 0.25 watts to drive a 12 volt 0.3 amp fan? Let's say the pot reduces 12 volts to 11. Then the pot must be turned 0.1 degrees. Or 0.0003 of a turn. That cannot happen.

Suppose the pot is turned a tiny bit more. At 0.1% of a full circle turn, it is now heating to well over its .25 watt limit (almost 1 watt).

An adjustable pot must be in the tens of ohms range (not xK ohm range) to control a 0.3 amp fan. And somewhere at or above 4 watts so that it does not burn.

Relevant equations are V=IR (R=pot resistance, I=fan current, V=number of volts below 12).
And W=VI (those same V and I. And W= watts heating the pot).

03-29-2013, 08:49 PM	#15
tw Read? I only know how to write. Join Date: Jan 2001 Posts: 11,933	I ignored numbers for that audio tapered pot. 10k and 0.25 watts to drive a 12 volt 0.3 amp fan? Let's say the pot reduces 12 volts to 11. Then the pot must be turned 0.1 degrees. Or 0.0003 of a turn. That cannot happen. Suppose the pot is turned a tiny bit more. At 0.1% of a full circle turn, it is now heating to well over its .25 watt limit (almost 1 watt). An adjustable pot must be in the tens of ohms range (not xK ohm range) to control a 0.3 amp fan. And somewhere at or above 4 watts so that it does not burn. Relevant equations are V=IR (R=pot resistance, I=fan current, V=number of volts below 12). And W=VI (those same V and I. And W= watts heating the pot).