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?
Thanks a lot.