Re: Clipping in Simple Emitter Follower Circuit


"John Popelish" <jpopelish@xxxxxxxx> wrote in message
news:TOCdnZ1hT8_HY1fanZ2dnUVZ_rKtnZ2d@xxxxxxxxxxxxxx
lcorbin@xxxxxxxxx wrote:
I have a question. On pages 54-55 of Horowitz and Hill's classic "The
Art of Electronics" (first edition), 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.
That I understand.

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?

Could it be that if an NPN transistor,in an emitter follower
configuration say, does not get enough current, will that alone will
cause it to turn off? In other words, perhaps the emitter voltage
can go so low that not enough electron current is drawn through
the emitter resistor to keep it on, even though the base/emitter
voltage is still well above .6V?

Perhaps someone with a copy of the first addition can double check your
reading, but if it is exactly as you say, I agree that there is no good
reason the output voltage cannot swing almost all the way to the negative
supply rail (all the way down to about zero load current).

Are you sure there is nothing else shown connected to the emitter? A
capacitively coupled additional load, perhaps?

In addition to the emitter resistor which has already been mentioned, there
is a 1k load resistor returned to ground and the explanation quoted above
actually begins:

"For instance, in the loaded circuit shown in Figure 2.7 the output can
swing to within a transistor saturation ..."


.

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