[ExI] Electronic Circuit Question (Emitter Follower)

[Jef Allbright]

On Sat, Mar 1, 2008 at 5:25 PM, Lee Corbin <lcorbin at rawbw.com> wrote:
> I have a question for people who know about transistor
>  circuits. On page 56 of Horowitz and Hill's classic "The Art
>  of Electronics", there is a nice description of the emitter
>  follower circuit.  The following is used as an example:
>
>  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?
>

Lee -

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

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

Does this answer your question?  Does this topic belong on the Extropy list?

- Jef