Hello,
I used a BJT (BC337) and a 100 K potentiometer to control the speed of a fan (12 V, 0.45 Amp). I used a 12 V, 2 A power supply and connected everything as below.
It controls the speed from zero (fan is off) to the highest speed when the potentiometer is half way. My problem is that when I change the potentiometer knob to the other end, it starts getting very hot and starts to burn (some smoke coming out). Is there anybody explaining why it is happening?
Thanks
If you connect the transistor base lead directly to +12 Volts, there is no current limiting resistance and the base-emitter junction draws all the current it can. That leads to smoke!
Thanks. Sounds completely correct. I was thinking that there is a current limit between transistor's base and emitter pins. What range of resistance do you think might be a good fit to be between the middle pin of potentiometer and base pin?
Thanks a lot
You could put a 470 ohm but that will not solve the transistor getting hot problem.
You need a different transistor that can dissipate 2W or more.
A TIP41C will work
Place the fan in the emitter lead to create an emitter follower circuit. Connect the collector to +V. The output voltage will be approximately 0.7V below the base voltage, depending on the fan's load.
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