I didn't get anything like that, but the negative voltage generator has a
pronounced threshold as the device starts to avalanche.
I wonder if you are getting an effect caused by the capacitive loading of the
scope probe coupled with, perhaps, a negative resistance region over the few mV
close to avalanche start. Again a guess. Steve's memory of Bob Pease's
explanation is a bit unexpected, I wouldn't have thought a reverse biassed
junction would generate light. After all, LEDs are forward biassed in
operation.
Another explanation that someone came up with is that the avalanche generates
noise (it certainly does, noise diodes are just RF diodes with reverse
breakdown) and that this noise is then being rectified in the C-B junction to
give the -ve voltage. I still wonder if it could be due to energetic
electrons hitting the base junction and passing though the thin layer, then
building up a charge on the reverse biassed C-B junction.
Or perhaps a combination of all three !
Andy G4JNT
-----Original Message-----
From: Rik Strobbe [SMTP:[email protected]]
Sent: 2002/11/13 12:35
To: [email protected]
Subject: LF: Tuesday's riddle (part 2)
Hello Marcus,
very interesting.
I did the experiment with a BC547c and as Andy I got a negative voltage.
While playing with it I reduced the voltage to +/- 5V and saw an even more
interesting effect, over a range of a few 10mV I got a nice sawtooth
signal. I wonder if someone can reproduce it ...
73, Rik ON7YD
At 14:58 11/11/2002 -0500, you wrote:
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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