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Would just like to know, why on the first picture RE1 is "part" of the feedback and is RE3 not? Then on the second picture is RE part of the feedback as well? Is there any way generally how to say?

Link since where it is from is important, please watch: First picture https://youtu.be/5cnFBZrlYfM?si=E2evQ1elZbQ0ZwI7

Second picture is book material.

enter image description here

enter image description here

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  • \$\begingroup\$ The emitter is pretty close to a voltage source. That explains both cases. \$\endgroup\$ Commented May 25 at 6:07
  • \$\begingroup\$ @periblepsis thank you, thought there is more to that. \$\endgroup\$
    – user354575
    Commented May 25 at 6:11
  • \$\begingroup\$ Well, there is always more. Reality is a mess. But close enough for this use since details are absent. \$\endgroup\$ Commented May 25 at 6:21
  • \$\begingroup\$ What details? I mean, I have asked pretty straightforward question, only including what I do not know since I guess it is expected I do not write whole paragraphs here. \$\endgroup\$
    – user354575
    Commented May 25 at 6:43
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    \$\begingroup\$ Where does fig.1 come from? The circuit itself seems really bogus. As shown it has no DC point so it's not even an amplifier. Plus, FB configuration looks very odd to me. \$\endgroup\$
    – edmz
    Commented May 25 at 12:12

1 Answer 1

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In the first picture, RE3 is, in fact, part of a feedback loop, but not part of the overall (or global) feedback loop. It is part of a local feedback loop associated with Q3 (the transistor it is connected to), which may do nothing more than stabilise the bias point of Q3. But we can only speculate since this circuit does not show any details; I suspect this picture is being used to teach the fundamentals of feedback amplifiers.

In the second picture, Re2 (I will call it this because Re1 is reserved for any emitter resistor of the first transistor) samples the output, and then subtracts a current from the input via Rf. The action of the current in Rf is to reduce, not increase, the amount of signal (current) being applied to the input of the amplifier.

With feedback loops, they can be characterised by two main features:

  1. what variable is being sampled at the output (is it voltage, or current), and
  2. how that sample is mixed with the input signal to create the error signal.

The mixing is either in series with the input signal, or in parallel (aka "shunt") with the input signal. This error signal is then applied to the "bare amplifier without feedback".

By combining these terms, we get four (4) different types of feedback amplifiers:

  1. voltage-series
  2. voltage-shunt
  3. current-series
  4. current-shunt

Sometimes the first term is changed to shunt (for voltage) and series (for current), but I prefer using the terms voltage and current.

Sometimes it can be confusing to determine what is being sampled at the output, if it is a voltage or a current.

For example, in the first picture, the voltage across Re3 is the sensed variable, but since the load is shown as a resistor, then that sampled variable (voltage on Re3) is directly proportional to both load current and load voltage. One trick is to imagine a small change (wiggle) in the output variable, and then seeing if the circuit tries to reduce or increase that change by the way the input is affected. If the current in RL increases, then Q3 collector voltage goes down, but Q3 emitter voltage goes up (goes more positive away from gnd); this causes voltage at Q1 emitter to go up also by resistor divider action between Rf and Re1, which reduces Vbe applied to Q1. This is negative feedback, of type either voltage-series or current-series.

As for second picture, the load is not shown, but is marked as Iout which is collector current of Q2. This tells me that the current is being sampled. This current is being sensed as a voltage across Re2, and which modifies the current through Rf, which changes the current flowing into the base of Q1. If Iin increases in the positive direction, Q1 turns on harder, which reduces the voltage at Q2 base, which reduces the current in Re2, which reduces the voltage on Re2, which causes Rf to pull more current from the base node of Q1, which reduces the current into Q1 base, which is negative feedback.

Some other links that can help you understand feedback amplifiers:-

How do you apply the negative feedback laws when there are two feedback networks?

Art of Electronics Ex.2.20 Series negative feedback amplifier

Trouble visualizing voltage-shunt mixing

Why is it desirable in an amplifier to have high input impedance and low output impedance?

https://www.electronics-tutorials.ws/systems/feedback-systems.html

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    \$\begingroup\$ Thank you so much! You have really explained it so well! \$\endgroup\$
    – user354575
    Commented May 25 at 8:01
  • \$\begingroup\$ @Victor You're welcome. Please read the links, the topic of feedback amplifiers has been answered many times before. Cheers. \$\endgroup\$ Commented May 25 at 8:10
  • \$\begingroup\$ I must admit not to agree with the statements concerning the 1st circuit ("..not part of the overall feedback loop".). This would be true only if the resistor Rf would be iseally isolated from the output of Q3 (using a buffer). Only in this case, RE3 would have no influence on the feedback factor. However, as the Rf-path is in parallel to RE3 the current through Rf will act back to the output voltage at the emitter of Q3. With other words: We have no "ideal" voltage sampling. \$\endgroup\$
    – LvW
    Commented May 25 at 8:50
  • \$\begingroup\$ @LvW Agreed, this circuit does not ideally sample (or "measure") the output voltage or current; the feedback path could influence the variable being measured. Also, the diagram leaves out bias details which could also impact either or both of (a) the measurement of the variable being controlled, and (b) the mixing of the feedback signal with the input signal. However, I suspect the original source of these figures was not looking into that level of detail at this step; I figured that what the OP sought was an overview of what feedback is. Cheers. \$\endgroup\$ Commented May 25 at 9:01

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