2Amp CW Current Source, BJT or MOSFET?

This is one of the things you were supposed to have already learned from your textbook or in class.

Hint: Ohm's law is involved.

Good Luck! Rich

BJT might be Darlington. So back to the question, in the linear region, which one has less resistance? BJT or MOSFET? CW means continuous, i.e., charging MOSFET gate capacitance is not a problem here.

I built several driver about 1A without forced air cooling. For 2A I am doubted. I heard some company make devices called linear mosfet, anybody used it before?

To summarize several of the clarifications:

You're making a linear constant-current source, and you want to choose the pass transistor that's going to burn up the least heat because you want to avoid using heat sinks.

So you're going to drive 2A into some load, and that load is going to respond with whatever voltage it needs across it for 2A to flow. That voltage is going to be fixed by the characteristics of the load and the fact of the 2 amps.

Then you're going to power the transistor from some supply, whose voltage is going to be determined by the 2A that the transistor is drawing.

So the voltage across the transistor will be fixed by the load and the supply, and the current through the transistor will be fixed at 2 amps. Power = voltage * current.

Now, You tell Us how changing the transistor is going to changed the power dissipation with a fixed voltage and a fixed current.

BJT heat feels like a different kind of heat, just like a wood stove :-)

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Yea, this is right. My mistake.

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maybe I need to lower the supply rail or change the components working in wider temperature range. Irlr3105pbf works from -55 to + 175C, but high temperature will degrade the performance.

I've answered this in another sub-thread. Resistance is irrelevant. The device will adjust its own "resistance" to whatever it needs to be to pass the needed (regulated) current.

You _did_ say a current source/sink, right?

Hope This Helps! Rich

And don't forget the clean, bright crispness of that BJT sound! ;-)

Cheers! Rich

But only if it's thems Ge-trainsistahs :-)

Depending on the voltage source you use, the transistor will have to dissipate 2A * V watts, and the usual "performance" requirement is going to have a lot to do with heatsinking and price, not much to do with the other transistor properties. If you start with 20V unregulated power (use a regulated 12V for the op amp and such), the transistor will shed up to 40W.

TIP120 darlington transistor will do it: $0.49 each. IRF520 NMOSFET will do it, too: $0.56 each.

The heatsink will be the expensive part.

The main deal here is drive current. With a bjt, you'll need 200mA. So, use a darlington or a mosfet, whichever is cheaper. Either will work. Use the one with the best heat transfer characteristics.

I would reference the output ground at the transistor's drain or collector. Use a small shunt (maybe .1 ohm) and a single supply opamp. An LM324 works well for low frequency loads, and you can't beat the price.

Sizing the heat sink will mean predicting how much voltage will be across the device at max power. If your load is a short to 100V, then the mosfet + heatsink will need to dissipate nearly 200W at 2A. If the load is a short to 10V, then they will only need to dissipate 20W at

Regards, Bob Monsen