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Right, done it. Now that's very interesting !!!!!!
Exactly as you specified, even down to the same
2N2222 device. The B-E junction breaks down at 7.56V, ie. Vbe =
(minus) -7.56V With a DVM, 10M input resistance, Vcb = (plus)
+0.35V almost as if it is acting as a voltage inverter, no wonder
you ripped the can apart. That shouldn't be, BUT...
The tunnelling (or whatever the current carriers do
in a zener) are probably wizzing across the B-E junction and hitting the
C-B junction, then overshooting and slowing down or something, so
generating a voltage. Just inspired guessing. Any solid state
physicists out there ?....
How can anyone want puzles like this removed from
the reflector John ?
Andy G4JNT
----- Original Message -----
Sent: Monday, November 11, 2002 7:58
PM
Subject: LF: Tuesday's riddle
Hi group,
this little experiment was pointed out to
me by Ralph, DL2NDO a few years ago:
You need an ordinary NPN
transistor (eg. 2N2222), a 9V battery, a 1k resistor and a voltmeter. Connect
the negative battery terminal to the base and the positive to the emitter, via
the resistor. Now predict the voltage drop between collector and base. Can't
be too hard...
Then measure it - you'll be surprised. Any explanations?
We finally ended up opening the poor transistor's case.
Have
fun
Markus, DF6NM
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