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9827304 No.9827304 [Reply] [Original]

Hi just looking for a bit of quick electrical engineering advice before I make this little circuit.

I want to combine a small switch with a large heavy duty PNP transistor to make a switch that will control a 1.8 Amp Delta fan that runs at 12V.

So I have a choice of two circuits.
Pretty simple.
Which should I make, or does it not really matter.

>> No.9827311

La corriente es mayor en el emisor, a fin de cuentas no importa mucho.

>> No.9827325

Para corrientes altas Ic=~Ib

>> No.9827380

The current is higher at the emitter, in the end it doesn't matter much.

High current Ic≈Ib.

Hmmm..

Was just playing with the circuit and I destroyed my 500Ω potentiometer, dammit.... will be using the first option so current isn't wasted since they seem to behave pretty much the same.

>> No.9827972

>>9827380
Ie ~= Ic, not Ib.

Base current is virtually negligible, since the ratio of base current to collector current is usually 100-200, you'll really only be losing out on like 1% efficiency when using the collector vs the emitter.

It'll be easier to bias the circuit using the second configuration. You don't know what voltage is going to be generated across the fan, so figuring out how to bias the base will be difficult with the fan attached to the emitter.

With the second configuration, you can assume a Veb drop of ~0.8V & beta of 100, then calculate your bias resistor as (12V - 0.8V) / R = <desired_fan_current> / 100. Obviously, actual transistor parameters vary so try it out with a potentiometer first and then place a fixed resistor.

>> No.9827983

>>9827380
Also circuit 2 will give your fan more voltage to work with, since circuit 1 will need at least 0.8V at the PNP emitter, you'll only really have 11.2V across the fan, whereas circuit 2 can have 11.8V across the fan before the PNP begins to enter saturation.

>> No.9827988

>>9827380
Maybe the power was too much for your potentiometer? at 500ohms with the second circuit configuration you'd be putting 1/4Watt through the pot. Assuming it was set at less than 500ohms to begin with that number can only go up.

>> No.9829132

>>9827983
Ok thx, I'm just gonna leave the circuit the other way since the fan is a little bit more powerful than I need it to be.

I'm using a 1 ohm resistor, and Ib is about 200mA. It seems to refuse to go above 250mA even with no resistor. Not sure quite why the circuit is behaving like that, but maximum current for Ib is 1A according to the spec sheet for my transistor.
>>9827988
That's exactly what happened, a spark flew out of it plus some burning smell. The potentiometers are only rated to 1/10th of a watt or something, I gotta make some resistor ladders to replace them because I'm always frying them by mistake.

>> No.9829650

>>9829132
Interesting, idk what effect would be limiting the base current to 250mA. Theoretically the base-emitter junction should function like a diode so with 12V across it the current should be exponentially high! Did you try a larger resistor?

>> No.9829663

>>9827304
PNP transistors are high side switches, only option 2 will work. Also is that switch an actual switch or is it a stand in for later pulse width modulation?

>> No.9829665

>>9829663
Sorry I got that back to front, PNP transistors are low side switches, NPN are high side switches.

>> No.9829695

>>9829665
You got it right the first time. Although it could kind of work as a low side switch, it would just have a greater voltage across it, and wouldn't need a base resistor.

>> No.9829725

>>9829132
You should be fine keeping the potentiometer at the full 500 ohms, it will give you an Ib of 24mA and, if your transistor is any good, that should be plenty to run 2A at the collector (You should need no less than 100 ohms though).

Your pot probably fried because you turned it down too much. Remember, between the emitter and the base, the transistor is essentially a diode. You are hooking the pot straight across 12V (autists beware), so lowering its resistance will gove you large heat dissipation fast (>1W at 25%).

Circuit 2 is the way to go if you want efficiency. It lets you keep Ube at operating voltage independently from the fan's power curve. In curcuit 1, you will always sacrifice about 1V to keep the base at a lower potential than the emitter.

Hope this helps, happy soldering