# Inverse feedback with amps question

Discussion in 'Homework Help' started by Paulo540, Dec 7, 2009.

1. ### Paulo540 Thread Starter Member

Nov 23, 2009
188
0
Hi,

thanks in advance for your help. I have been racking my brain forever with this lesson and have one question left to answer. I don't just want the right answer but moreso help with realizing how to get there. I've read the net and my book a hundred times and would really appreciate some input.
I drew the schematic from the book since my scanner isn't hooked up.

The question is as follows :

In this figure, inverse feedback could be eliminated without in any way interfering with d-c base biasing and temperature compensation by...

A. removing Re and Ce, then grounding emitter
B. removing Ce
C. removing R1
D. shunting a cap across R3
E. replacing R3 with two resistors, their sum = to R3 and connecting a cap from their junction to ground.

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2. ### JDT Well-Known Member

Feb 12, 2009
658
87
Inverse feedback is usually known a negative feedback. In your circuit, negative feedback is happening in two ways:- DC negative feedback - stabilising the DC operating point; and signal (and DC) negative feedback - feeding back some of the signal thus reducing gain.

To eliminate all negative feedback two of the answer options will have to be selected. Can you answer with more than 1 option?

Does that help?

3. ### steveb Senior Member

Jul 3, 2008
2,432
469
But, he needs to choose the answer that does not interfere with DC bias and temperature compensation. I only see one answer that does this.

4. ### steveb Senior Member

Jul 3, 2008
2,432
469

A. removing Re will definitely affect the DC biasing, and removing Ce will actually increase the negative feedback on the AC signals.

B. Again, removing Ce will increase the negative feedback for AC signals

C. R1 is a critical biasing resistor. Removal of it will drastically affect DC bias.

D. Shunting a cap across R3 will increase negative feedback for AC signals.

E. This is the only one that could possibly work. AC signals will be prevented from traveling from output back to the input; hence no negative feedback is possible. The DC conditions are the same because the two new resistors add to equal R3.

The key thing here is that there are two negative feedback mechanisms in this circuit. The emitter impedance forms a feedback mechanism, and the collector to base resistor forms a feedback mechanism. The DC biasing uses both of these mechanisms; however, the AC amplification only uses the collector feedback mechanism because the emitter impedance is zero for AC signals. (note there is a crossover range, but it's clear the question is ignoring that)

5. ### Paulo540 Thread Starter Member

Nov 23, 2009
188
0
Very cool, I appreciate the quick responses.

It makes sense now that you explained it that way. I definitely was leaning towards E, but it was mainly because the other 4 seemed to harm the parameters in some way and not because I really knew, haha.

I didn't really grasp why E would work but now I do,

Thanks again