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Clipping in Simple Emitter Follower Circuit

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.

Reply to
lcorbin
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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?

--
Regards,

John Popelish
Reply to
John Popelish 

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LTspice agrees with you:

Version 4
SHEET 1 880 680
WIRE -80 48 -208 48
WIRE 144 48 16 48
WIRE 144 224 144 48
WIRE -208 352 -208 48
WIRE -32 352 -32 112
WIRE 144 352 144 304
WIRE -208 464 -208 432
WIRE -32 464 -32 432
WIRE -32 464 -208 464
WIRE 144 464 144 432
WIRE 144 464 -32 464
WIRE -208 512 -208 464
FLAG -208 512 0
SYMBOL res 128 208 R0
SYMATTR InstName R1
SYMATTR Value 10k
SYMBOL npn -80 112 R270
SYMATTR InstName Q1
SYMATTR Value 2N3904
SYMBOL voltage 144 448 R180
WINDOW 0 24 104 Left 0
WINDOW 3 24 16 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 10
SYMBOL voltage -208 336 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 10
SYMBOL voltage -32 336 R0
WINDOW 3 24 104 Invisible 0
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SYMATTR InstName V3
SYMATTR Value PULSE(-10 10 0 1)
TEXT -242 536 Left 0 !.tran 2
Reply to
John Fields

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 ..."

Reply to
Andrew Holme

Maybe I should have paid more attention to that, because you are right! There is a "dotted line" leading to a 1K resistor that also goes to ground, and it's labeled "R-load". But I supposed that the next stage to be at high impedance, and that besides, a 1K resistor in parallel to a 1K resistor would just be a .5K resistor, and wouldn't the problem still remain?"

Thanks again.

Reply to
lcorbin

So the emitter voltage that corresponds to zero emitter current (when the two emitter resistors connected in series from ground to -10 and form a voltage divider) is -5 volts, not -10. The emitter can only pull positive from that voltage.

--
Regards,

John Popelish
Reply to
John Popelish

If they both connected to the same DC voltage, they would be equivalent to a 0.5k resistor to that voltage. But if they connect to different voltages, they are equivalent to a 0.5 k resistor connected to a voltage that is the average of the two actual voltages the two separate resistors connect to. Make that replacement and go over your analysis, again.

--
Regards,

John Popelish
Reply to
John Popelish

Well that explains it ! The furthest it'll swing negative is set by potential divider effect.

Graham

Reply to
Eeyore

Exactly. The OP should read up on Thevenin and Norton analysis.

Graham

Reply to
Eeyore

A point I would like to make is that this example is exactly the lesson the original poster, M. Hamed complained that he had never seen in any tutorial material.

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Regards,

John Popelish
Reply to
John Popelish

Well, despite the best efforts of you gentlemen, the situation still strikes me as murky. The load indicated in the diagram (which it was very remiss of me to fail to mention) is a simple

1K resistor connected downwards to the ground symbol. But as I said (and thanks to the poster with the book handy), there is a "rail" (I guess you say) of +10V at the top of the diagram directly connect to the collector and a -10V connected to the bottom of the 1K emitter resistor.

Is "ground" supposed to be -10V or 0V here? If it's -10V, then the load 1K and emitter 1K are merely in parallel and the problem remains. If it is 0V, then yeah, there is a voltage divider effect I can calculate (after all, the emitter is swinging to negative voltages). Is this latter the solution to the problem?

Thanks, Lee

Reply to
lcorbin

It is the zero volt reference that each of the two power supplies is measured with respect to. you have to have some place to connect that other volt meter lead.

Yes.

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Regards,

John Popelish
Reply to
John Popelish

Many thanks to all, especially John Popelish. Yes, the 1K between the

-10V and the emitter, together with the 1K between the emitter and ground, holds the emitter at -5 volts. So the base cannot cause the emitter (which follows it) to go below -5. Very nice.

Thanks again, yours, Lee Corbin

Reply to
lcorbin

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