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I'm working through this handout Feedback In Amplifiers and learning how to visualize simple transistor circuits and their negative feedback. I'm perplexed by the example given for current-shunt feedback. The example circuit simply includes the symbol \$\beta\$ in a box, labelled as the "feedback network." Below is the schematic in question (last page of PDF).

enter image description here

With where the feedback network is connected, it looks sort of like it might be the portion of \$R_E\$ that was not bypassed by \$C_E\$. But this doesn't seem right because \$R_E\$ usually provides feedback in series, right? And it wouldn't be connected as it appears before \$R_1||R_2\$ either.

What components exist within this "black box" of feedback? And, if no one knows, is anyone able to point me towards a single-BJT example of current-shunt feedback?

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The problem of the shown system is that it constitutes a mixture between circuit diagram and block diagram. This should be avoided. Where did you get this circuit from?

Nevertheless - certainly, the shown beta-network is not a simple resistor (part of RE) because the output of the beta-block is indicated as a feedback CURRENT If. And this is a problem - why?

For an increase of the emitter curent Ie - and NEGATIVE feedback - the ouput of the beta-block (current If) must decrease (as indicated also in the figure). Hence, the current If cannot be simply a part of Ie. Instead, a kind of signal inversion is required within the beta network.

I think, this is a - more or less - academic example only without much practical meaning. Normally, the emitter is used for negative current-controlled voltage feedback (resistor RE) or we use the collector node for negative voltage-controlled current feedback (RB between C and B).

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  • \$\begingroup\$ I agree that the diagram I included is faulty. It was suggested on SE in a previous question. electronics.stackexchange.com/questions/528064/… I just can't find anything better. Does there exist a bjt circuit that uses this form of feedback? If so, please point me to one that I can study. \$\endgroup\$
    – nuggethead
    Dec 2, 2020 at 2:32
  • \$\begingroup\$ ...."this form of feedback" ? When the feedback signal goes to the base it must be derived from the collector (for negative feedback). As another alternative, a second transistor in common-collector configuratiuon can be used. In this case, the feedback signal (over the two stages) can be directed to the base of the 1st transistor. \$\endgroup\$
    – LvW
    Dec 2, 2020 at 7:49
  • \$\begingroup\$ Can you post an example? \$\endgroup\$
    – nuggethead
    Dec 2, 2020 at 10:37
  • \$\begingroup\$ eeeguide.com/dc-feedback-pair-with-two-amplification-stages \$\endgroup\$
    – LvW
    Dec 2, 2020 at 12:47
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Negative feedback uses a comparison ( a subtraction) of actual output to desired output.

The subtraction is done across the base_emitter junction.

The feedback is the emitter voltage, generated across that parallel RC network.

The OUTPUT of this amplifier is the collector current.

That collector current depends almost solely upon the Vbe.

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  • \$\begingroup\$ I don't understand. When you write "the substraction is done across the base_mitter junction," do you mean that some of the current \$i_i\$ is being taken away by feedback? And what is meant by "The feedback is the emitter voltage, generated across that parallel RC network?" I'm confused because the "bottom" of the feedback box is connected to ground. Can you explain any further? \$\endgroup\$
    – nuggethead
    Dec 1, 2020 at 3:02
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Imagine the circuit before the feedback network was installed: that is, as just a common-emitter amp, with the lower junction of emitter resistor RE and bypass capacitor CE connected to ground instead of to the feedback network.

Next consider the feedback network in isolation. Imagine looking into its input: that is, from right to left, into the two terminals on the network's right-hand side. That input will present some small impedance.

Now reinstall the feedback network: break the imaginary connection between RE//CE and ground, and insert the input terminals of the network into the gap. You've now restored the original circuit: the transistor's emitter current iE is flowing in series through the small input impedance of the feedback network. That is, the feedback network is sampling the emitter current.

That explains the "current" part of what your handout described as "current shunt feedback". (But wait: the amp's output is a voltage at its collector, not a current through its emitter. How does sampling the emitter current equate to sampling the output? A common-emitter amp delivers a voltage output by sinking collector current through load resistor Rc. The transistor's high gain means its emitter current iE is almost identical to its collector current. By sampling the emitter current therefore, effectively you're sampling the amp's output.)

Now look at the output of the feedback network - the terminal on top. A common-emitter amp embiggens a small base current to deliver a larger collector current. Before source current iS can reach the base and be amplified, the feedback network steals a small amount "if" of it - "if" is shunted away through the network's output terminal. That explains the "shunt" part of the "current shunt feedback" description.

The feedback is negative: greater emitter current - i.e. greater amplifier output - causes more of the input signal to be stolen by the network, with less remaining to reach the base and be amplified.

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