[ExI] Electronic Circuit Question - Resolved

[Lee Corbin]

The question (See Below) is now entirely resolved. The answer
was much along the lines that Jef had suggested. The load
(showed as disconnected in a diagram) plays the decisive
role. 

The ground holds one end of the 1K load resistor at 0V, and the
emitter resistor's low end is held at -10V, and that creates a
voltage divider.  So the emitter can never go more negative
than half way between (since each is a 1K resistor).  Thus the
emitter can go no lower than -5V, though when the base swings
above -5V, more current is drawn from that "-5V point" and
it rises too.

Thanks to Bryan, Jef, Stathis, and Ed McHale, and also to
some fine folks at sci.electronics.basic, who got it through
my thick head that the ground symbol always stands for 0V
in such a circuit, regardless of any higher or lower "rails".

Sorry for the off-topic post, and apologies for having not
quite correctly conveyed all the particulars.

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."