Smokin' POTentiometer

Repost, I believe the message got lost in the SPAM...

Hello everybody. About a month ago I made a post titled "How do I hook up my potentiometer?"

Now, I ended up hooking it up John Fields' way with the diode in parallel with the motor [12V]:

.V+>---+---------------+ . | | . [220] | . | | . P /c . O--+---[Diode>]----+

Now, I have been using a 12V power supply and a NPN with a high power rating and I got a hold of a 500ohm POT (unknown power rating). I hooked it up and all and it runs fine when the POT resistance it

Thanks guys!

Several ways, but the original answer was general and described a way to control "a lot more current" with a pot. It was not a specific solution for your individual application.

Knowing the specs would be helpful: how much current does the motor draw and what transistor are you using? With the specs in hand, we can determine what should be used.

Given the symptom (the pot smoking when turned about 1/2 way) and assuming things are wired correctly, it means the transistor draws more base current than the pot can handle. A transistor with higher gain, or a darlington transistor, or a circuit with

Ed

Add a 100 ohm resistor to the pot as shown. Measure the voltage drop to see how much current is being drawn. A 500 ohm 1 watt pot is rated at I = sqrt(P/R) = 44 mA. If the motor draws 1 amp and the transistor has a beta of 20 you will just start to overload the pot.

It would be much better to use a MOSFET and get rid of the 220 ohm resistor so you can get closer to the supply voltage out to the motor.

Paul

"However, using a more powerful transistor to run a more powerful fan is not normally a viable option. The gain will typically be much lower so the needed base current very soon gets to a level which will damage the potentiometer track."

The above three methods covered

A swap to a MOSFET voltage follower isn't really a good option, the voltage drop is several volts rather than 0.7 or so.