Am 18.05.2016 um 16:51 schrieb scott:
>>> Yes, of course -- the resistance does indeed vary with gate voltage.
>>> However, given a constant gate voltage (which is kind of a prerequisite
>>> for the "digital on" state), the resistance across the drain-source
>>> channel remains pretty much constant.
>>
>> I thought Scott was talking about the linear amplification state.
>>
>> Which is where this part of the thread started. Transistors have three
>> states. On, off and in between. The in between state is always there
>> even is the transition between on and off is very small. You even get
>> ringing in FET circuits.
> 
> I *think* I understand clipka's point now though. In digital circuits,
> because the output of a FET is usually connected to something with
> extremely high impedance (eg the gate of another FET, or through another
> FET that is off), even in the "in between" state still almost no current
> will flow and hence no power will be dissipated.

Um... no. I was contesting the following statement of yours:

"[...] a transistor used in digital circuits is either in a state where
the current flow is zero ("off") OR the voltage drop is zero ("on"), so
heat output is usually zero."

More specifically, I contested the claim that the normally-zero heat
output in the "on" state was due to the magnitude of the voltage drop. I
made no claims about the "in between" state of the FET.

I may also like to point out that I did not address the case when
/another/ FET, with the drain-source channel wired in series with the
drain-source channel of the FET in question, is switched on. In that
case, a transient current does indeed flow through the FET in question,
and yes, if the FET in question is in fully "on" state, as long as the
other FET is still in the in-between state, that one will have a
sufficiently high resistance to bear the majority of power dissipation.

In such a transitory state of /other/ parts of the circuitry, a totally
switched-on FET will indeed not consume any noteworthy power thanks to a
much lower voltage drop /compared to its neighbor/ -- but power
consumption will still happen, namely in that neighbor.

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