2006-12-28 18:56:44 · 5 answers · asked by BISWAJIT B 1 in Science & Mathematics ➔ Engineering
The heat is generated in the junction, and in watts is equal to the voltage drop (around 0.7 volts) times the current.
2006-12-28 19:03:07 · answer #1 · answered by Tech Dude 5 · 1⤊ 0⤋
Depends on the majority carriers. Actually, I would expect most of the heat would appear in the media between the anode and cathode, since when a current is flowing there is a voltage drop there.
2006-12-28 19:01:51 · answer #2 · answered by Anonymous · 1⤊ 0⤋
Thermionic (vacuum tube) diodes are still around in some audio amps such as Mesa Boogie guitar amps. In vacuum tube diodes the anode dissipates all the heat due to bombardment of electrons and ions.
2006-12-28 23:51:41 · answer #3 · answered by Del Piero 10 7 · 1⤊ 0⤋
Same as if you had faulty connections on your car battery the negative side. Gas being pressureized heats up. Gas moving from a high pressure to a low pressure creates cold. Electrons are negative "particles" and move away from the negative side of a battery into the postive side. My monies on the cathode. However the emitting crystal is mounted on one conductors with a thin wire or wires on the top of it. I would think that the base would conduct more than the wire? Which is which? I not got one to look at the moment...Have you?
2006-12-28 19:18:27 · answer #4 · answered by Al 3 · 0⤊ 2⤋
alan B may be a fine cook (looking at his avatar's headgear) but he aint no electronics engineer.
Tech Dude is spot on
2006-12-28 19:30:07 · answer #5 · answered by Vinni and beer 7 · 2⤊ 0⤋