> If you have a diode passing a large current it will get hot.
> Wouldn't this thermal energy be from the "in" current, so
> the "out" will be lower than the "in" ? I understand some
> junction electro-chemistry, but not that much. Surely a
> diode is not a perfect conductor

The heat does not involve any electrons, only photons, thus any electrons
entering on one leg also exit on the other, therefore the same current
flows in both legs. The heat is caused by 'collisions' in the junction
material (and some of it probably by IR quanta emission from electrons
falling between stable states). The energy loss comes from the potential
difference across the device times the current. If you want to picture it
graphically you throw balls into a long horizontal pipe. All the balls
come out the other end but they are somewhat slowed down. The electrical
equivalent of slowing down is potential loss (voltage). The balls are
equivalent to unit charges (electrons, current = charge * time). The pipe
heats up if you throw enough balls into it.

Peter

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