For FETs they say P=I^2*Rds(on) and for BJTs they say P=voltage drop * I
For FETs Rds(on) is let's say 100mA and with current of 10A FET will generate 10W. TO220 housing has round 60C/W so with no heat sink it will quickly pass max operating temperature.
For BJT voltage drop should be 0.7V so at 10A is will generate 7W of heat (we neglect Ib that is some 100 times smaller). So still to hot for T0220 witn no heat sink.
so for feats I can use Rds(on) resistance when it is fully open and this resistance increases if it is not fully open but then also current decreases so P=I^2*Rds is sure enough equation to determine heat generation.
what about BJTs. Why 0.7 voltage drop? Does it have to do any with internal diodes? Does all BJTs have this voltage drop the same?
For TO220 either FET or BJT I found similar junction to case round 1.4 and junction to ambient round 60. Also max operating temp is 150-170C. No doubt that 10A current would need a heat sink. Also ambient temperature when all this is in a closed box can reach 60C easly. I just wanted to know how powerful heat sink do I need so that's why I am asking first how to calculate heat beiing generated internaly due to internal resistance.
For TO220 either FET or BJT I found similar junction to case round 1.4 and junction to ambient round 60. Also max operating temp is 150-170C. No doubt that 10A current would need a heat sink. Also ambient temperature when all this is in a closed box can reach 60C easly. I just wanted to know how powerful heat sink do I need so that's why I am asking first how to calculate heat beiing generated internaly due to internal resistance.
You should do the math to find the right heatsink.In order to find the right one, you have to carefully check out the heatsink manufacturers' specifications.They give these specifications being as Thermal Resistance ( as in case transistor) and you calculate the right one by inverse calculation from heatsink to junction.
The power dissipation is simply DC current ( or RMS Value of an AC current ) x DC Drop-Out Voltage ( or RMS value of an AC voltage) between C-E or D-S leads.You don't need to calculate the generated heat in the junction, all you need is to keep this temperature below the required absolute maximum value with a safety margin.
...Because you wrote it in post#1. There is no need to be 0.7V. It could be less.
But for darlingtons you have one PN-voltage drop added. Here you won´t go far below 0.7V
The current gain of a BJT is maybe 100 when it has plenty of collector voltage because it is a linear amplifier. But its current gain is very low (10 or less) when it is used as a saturated switch. If the base current is 1/10th the collector current then its saturation voltage drop will probably be a lot less than 0.7V and it will not be hot.
The graphs in a datasheet are for a "typical" device that you cannot buy. Some are worse and some are better. The text in the datasheet usually shows the worst one's spec's then you can design the circuit so that even the worst ones work properly."