Y' canna break the laws of Physics :-) / DO try this at home:
_____________________________________________

Sorry - memory mis-served me.
Cct is similar but not as I said.
I've re-written this below

This is well worth doing whether you are a beginner OR if you think
you have a good grasp of basic electronics (or better) and that you
know how good-old-transistors function.This takes under 5 minutes to
try and should greatly widen your electronic mental filters.

If you have not met this before and are not surprised I will be
surprised.

Take any small signal NPN transistor.
Connect BASE to ground.
Connect EMITTER to about +12 volts DC through about a 1K resistor
The base-emitter junction is biased opposite to normal.
There has to be enough voltage to break down the reverse biased
base-emitter junction - somewhat like a zener diode - 12 volts should
suffice in most cases.

1. Predict the voltage expected on the (open circuit) COLLECTOR
Do this 1st before you measure.
No cheating y' hear.

2. Take an oscilloscope or standard electronic multimeter (typically
10M ohm plus input resistance)
Measure voltage between COLLECTOR (floating) and ground.

Explain your result.

Clue -  you cannot break the laws of Physics, even though it appears
that's what's happening here :-) - but you do need to KNOW them.

have fun



Russell McMahon

PS
1.    I just tried this again - it behaves as I remembered and (now)
expect.
2.    It fooled me when I first tried it.
       I have an ME(electrical), which didn't save me here :-).



-----Original Message-----
From: Steve Thackery <steve.thackery@BTINTERNET.COM>
To: PICLIST@MITVMA.MIT.EDU <PICLIST@MITVMA.MIT.EDU>
Date: Saturday, July 03, 1999 11:43 PM
Subject: Re: Circuit Challenge - was Re: [OT] A little help? or.... s


>As I see it the base-collector junction is forward biased, so you
would
>expect to see the base sitting at about 0.7V positive wrt the
collector
>(ground).
>
>There is also a diode equivalent between the base and emitter, which
would
>normally drop 0.7V if conducting.  The base is at 0.7V wrt ground,
so I
>would expect to see 0V wrt ground at the emitter.
>
>I set up the circuit as described, and got something very similar:
the base
>sits at 781mV, and the emitter (which is dangling in the air) has
2.5mV on
>it, as measured by my digital multimeter.  The transistor I used was
a
>2N2222A.
>
>The only bit I'm puzzled about is why Russell said it appears to
break the
>laws of physics.  In my view, it is doing just what you'd expect.
Is my set
>up behaving differently to his?
>
>Regards,
>
>Steve Thackery
>Suffolk, England.
>Web Site: http://www.btinternet.com/~stevethack/
>