[ExI] Electronic Circuit Question

[Lee Corbin]

Yeah, this is pretty off-topic, and I do apologize to the list.
But it was a quiet Saturday afternoon, and I thought it might
provide some interest---besides, I was getting damn sick
staring at the circuit diagram.  :-)

Jef writes

> You didn't mention the value of the load impedance,

It didn't say. The circuit is supposed to have high output impedance,
so I thought it wouldn't matter---and I'm pretty sure that that's what
the authors are trying to communicate.

> but if it were equal to the emitter resistor value, then you would in effect have a
> voltage divider putting the emitter at approximately -5V, and (when
> the transistor is forward-biased) the base at approximately -4.4.

Hopefully I have "drawn" the circuit intelligibly for you. A load
impedance (load resistor) that you are describing would be
parallel to the 1K resistor I described, right? If so, I don't see
any voltage divider. 

Besides, in this example, the base is at the whim of the investigator,
the emitter is therefore .6 volts below that.  The resulting voltage
draws amps from the 1K resistor (and definitely not from anything
else). Did I communicate that properly?

> Wow, this takes me back more than a few years.

Thanks for your kindly assistance,

Lee

>>  The bottom of the diagram is at -10 volts and the top is at
>>  +10 volts (i.e. a 20volt supply somewhere). Just above the
>>  -10 volts is a 1K resistor, and above that the emitter of an
>>  NPN transistor.  There is no resistor between the collector
>>  and the +10 volts.  The experiment is to let the base voltage
>>  (input) vary between +10 and -10.  The output is taken
>>  (hence "emitter-follower") at the top of the 1K resistor.
>>
>>  Because the base-emitter voltage is always around .6 volts,
>>  the output naturally follows the input, but at .6 volts less.
>>
>>  But the book says that when the input voltage drops down
>>  to -4.4 volts, the base-emitter junction gets back-biased,
>>  (and the transistor turns off?).  I don't understand why the
>>  voltage on the base cannot keep going down, say to -6V,
>>  with the output voltage continuing to keep in step, say at
>>  -6.6.  Even at -6 volts, there seems to me to be plenty
>>  of leeway between that and the -10V source below it.
>>
>>  Here is their explanation:
>>
>>    "The output can swing to within a transistor saturation
>>    voltage drop of VCC (about +9.9v) but it cannot go
>>    more negative than -5 volts. That is because on the
>>    extreme negative swing the transistor can do no more
>>    than turn off, which it does at -4.4 volts input (-5V
>>    output).  Further netgative swing at the input results in
>>    back-biasing of the base-emitter juntion, but no further
>>    change in output."
>>
>>  I still don't see why the base could not be at, say, -6v
>>  and the output .6 lower.  Why should the base-emitter
>>  junction be back-biased when there is still a big voltage
>>  difference between the base and the -10 volts at bottom?