Electronic – Transistor power dissipation rating

powerpower-dissipationtransistors

I'm trying to build the following circuit

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Except mine is a 24V ~20A (500W) motor, so I need to swap the components. I'm currently looking into transistors and I've stopped on the 2N6284 (NPN) transistor, which is rated at 100V 20A. This basically means it's good to up to 2000W (right?), but it says the maximum power dissipation is 160W. Now this makes no sense so I'm guessing it's how much of the 2000W going through it will be lost as heat? I've looked all over the data sheet and there are a lot of things I don't understand on top of which is this diagram that suggests that temperature increases as power dissipation decreases? Isn't the relationship supposed to go the other way around, the more power lost as heat the higher the temperature?

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If any of my assumptions is correct then how can I calculate how much power will be lost running my 500W motor at 24 volts?

Best Answer

First, the power burnt up in a component is equal to the voltage drop of that component times the current going through it. The transistor has a \$V_{CE}\$ of 3V when the collector current is 20A, so that works out to 60W.

Second, the power derating curve works by telling you the dissipation you can allow for a given case temperature. You read across until you find your design case temperature, then up to find the allowable dissipation (or, usually, the other way -- in your case you'd start at 60W, then read across to find that you should design for a case temperature of around 135 degrees C).

Third, you'll need heat sinking to keep the case temperature down -- that thing won't dissipate 60W by itself.

Fourth (and beyond what you were asking), you need a base current of 200mA to maintain that 3V \$V_{CE}\$, and that's pushing into a base at a voltage of 4V. You are not going to do that with a microprocessor GPIO pin.