Those look tiny for the power they are rated for.
I had some idea that production errors or something made larger transistors impossible or very difficult/expensive. I guess not. Just add silicon.
Whilst it's true manufactures often exaggerate when it comes to power ratings, a transistor doesn't have to be that big, to switch a relatively large motor. Remember most of the power goes to the motor. The amount of power lost in the transistor should be as little as possible.
Take the STP45N40DM2AG I linked to in my previous post.
https://www.st.com/content/ccc/resource/technical/document/datasheet/ff/da/51/30/5f/70/4e/77/DM00213649.pdf/files/DM00213649.pdf/jcr:content/translations/en.DM00213649.pdfSuppose you use it to switch a 2400W motor, running off 240V at 10A. Calculate the power lost in the transistor.
According to the data sheet, it has an on resistance of 0.072Ω.
P = I
2R = 10
2*0.072 = 100*0.072 = 7.2W
Oh, that's a little optimistic, reading down the data sheet, it also says the on resistance increases with temperature and it's specified at 25°C, but that's no good, since it's going to be near hot motor. Assuming the transistor might get very hot, say 125°C, the on resistance and thus the power loss will double. Refer to the graph Figure 10 Normalized on-resistance vs temperature on page 7 of the data sheet.
P = 14.4W
A heat sink is used to dissipate the power lost in the transistor, which is normally much larger than the transistor.
According to the transistor's data sheet, it has a thermal resistance of 0.5°C/W. This means for every W of power dissipated the temperature of the transistor itself, inside the case will increase by half a degree, when on a perfect heat sink. The maximum operating temperature of 150°C, but it's never a good idea to work at the maximum temperature and the power losses increase, so lets limit it to 125°C. Assuming it needs to work on a hot day, next to a hot motor, say an ambient temperature of 65°C.
Maximum temperature rise
T
r = T
max - T
ambient = 125-65 = 60°C
Maximum thermal resistance to ambient
TR = T
r/P
d = 60/14.4W = 4.17°C/W
The thermal resistance of the transistor is 0.5°C/W so that needs to be subtracted.
4.17-0.5 = 3.67°C/W.
So your heat sink needs to have a lower thermal resistance than that.
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So the gate voltage on one of these, Vgs is +/- 25v, and the threhsold is from 3-5v? So you would need up to 25 volts to drive this transistor?
I think it's finally clicked with me how a transistor works... it breaks the rules of electrical circuits. This was never taught in schools b/c nobody actually knows anything about them.