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I'm working on understanding the four negative feedback topologies. My process has been to first analyze the circuit's dc and ac characteristics for quiescent current, gain, impedance, etc. Then, I like to compare the circuit to the common "block diagrams" that show the internal amplifier and the feedback network. It was very helpful to me to be able to see that (for instance) the emitter resistor provides the feedback for the common collector and common emitter. One such example is here.

I'm struggling with the following circuit. enter image description here I know a little about it already, but just enough to confuse me. I know that I must use a current source for the input signal, as the feedback provides a current. I know that the internal amplifier will see a current and provide an amplified voltage, so that makes this a transimpedance amplifier.

But here's what I can't figure out. For the transconductance and voltage amplifier topologies, I was able to remove \$R_E\$ in order to view the circuit without feedback for comparison. With that I was able to verify that the gain of the internal amplifier and the feedback network proved out. But in this circuit, I can't seem to figure out how to remove the feedback.

How do I analyze the circuit to get values for the gain of the internal amplifier without feedback (open-loop) in ohms and the feedback factor, \$K\$ in seimens?

Here is the block diagram I am referring to (top-right picture).enter image description here

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  • \$\begingroup\$ But if this circuit is driven from an ideal voltage source no AC feedback. You only have feedback for DC. Thus the AC gain is \$A_V = g_m \times R_C||R_F - \frac{R_C}{R_C + R_F}\$ \$\endgroup\$ – G36 Jan 17 at 13:04
  • \$\begingroup\$ If your input is a voltage source (as shown) then that is a given. If you want to connect this to a current source using Cc in parallel then that's an option. \$\endgroup\$ – Andy aka Jan 17 at 13:30
  • \$\begingroup\$ @Andy aka are you able to edit the original post to show current source (connected correctly) instead of my voltage source? I wasn't sure how to do that but I think it should be a current source. \$\endgroup\$ – nuggethead Jan 17 at 13:34
  • \$\begingroup\$ I just wouldn't analyse it as a current source; I would treat the base as a virtual ground instead. OK it isn't totally accurate to do it that way but, then again, who can totally accurately analyse a single BJT stage given all the inconvenient nuances it has. My way would be good enough for me. \$\endgroup\$ – Andy aka Jan 17 at 13:36
  • \$\begingroup\$ Oh boy, @Andy aka I must be in over my head! What is a virtual ground??? \$\endgroup\$ – nuggethead Jan 17 at 13:50
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The problem you are facing stems from the fact that the ideal feedback relation

$$A_f=\frac{A_{ol}}{1+A_{ol} B} $$

is derived from block diagrams, and block diagrams have the peculiarity of unilaterally transferring signals. Block diagrams do not model load effects or the inherent bidirectionality of power transfer enacted by two-ports. And two-ports - which relate pairs of variables (voltage, current) at the input and output are what we naturally use to solve circuits. When we solve a feedback circuit using two-ports we implicitly take into account bidirectionality and loading effects and that might introduce spurious terms in the feedback relation.

Since this feedback configuration samples the output voltage and compares the input current, it is best explained by using an input current generator (as you ask in the comments above). So, here is your circuit with an input current source

BJT feedback circuit

Once we have solved biasing and found the values of the small signal parameters, we can solve it with a reasonably detailed model for the BJT (I assume that the input capacitor is there for decoupling purposes, and since it is in series with an ideal current source I will neglect it in the analysis - simulation with an ideal current source confirms it won't affect the output)

small signal circuit

KCL at node 1 says

kcl1

KCL at node 2 says

kcl2

by eliminating unneeded variables and after a bit of algebraic massaging we get

vout/iin take 1

an expression that we can recast as

vout/iin similar to ideal feedback

Note what happens if we remove the feedback by making \$R_f\$ go to infinity: \$R_c/R_f\$ goes to zero and \$r_\pi //R_f\$ becomes \$r_\pi\$, while in the denominator the whole middle term is turned into nothingness. Hence the open loop gain becomes

open loop gain

Now, let's get back to the complicated feedback relation we have found above and let's see what happens when we choose the feedback network in such a way that it is the least disturbing as possible (while still performing its function).
If \$R_f\$ is much bigger than \$R_c\$ we can neglect the \$R_c/R_f\$ term and approximate the parallel of \$r_\pi\$ and \$R_f\$ with just \$r_\pi\$. Being \$R_f\$ still finite, the middle term in the denominator won't go to zero, though. We get an approximate feedback relation that can be cast in the form we have derived with block diagrams:

when two-ports behave like block diagrams

where

approximate block transfer functions

Note that I could have spared a bit of algebraic mess, had I chosen to use a BJT model with current control (\$i_c = \beta i_b\$ would have avoided bringing \$v_\pi\$ and \$r_\pi\$ along), and chosen the opposite conventional sign for \$i_f\$ (in that case the block diagram would have has a + summing node and the ideal formula would have been

$$A_f=\frac{A_{ol}}{1-A_{ol} B} $$

and we would have obtained a positive B.

Moreover, had I realized from the start that I wanted to avoid loading effects, I could have used a simplified and idealized version of the two port representing the amplifier stage with a zero input resistance (what we ideally want in an amplifier that accepts an input current) and a zero output resistance (what we really want in amplifier that produces a voltage output - note the by not including \$r_o\$ we already had that simplification). The analysis would give directly the simplified formula.

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  • \$\begingroup\$ Why didn't you include f and Xc? It is just as significant as r_e in your example \$\endgroup\$ – Tony Stewart EE75 Jan 21 at 20:16
  • \$\begingroup\$ @Tony From the comments it is clear that this is an exercise in feedback configurations and that the input gererator should have been a current generator. Makes sense because we have a current comparison at the input. 10 uF at 1 kHz is what, 16 ohms? I consider it ininflient with a current drive. The other answers were considering a voltage signal generator and in that case you need a series resistance. \$\endgroup\$ – Sredni Vashtar Jan 21 at 20:37
  • \$\begingroup\$ And, shame on me, that was not re but rpi. I had all the solution computes with re instead of rpi and now there's only a trace in the small signal diagram. I'll change it later. \$\endgroup\$ – Sredni Vashtar Jan 21 at 20:42
  • \$\begingroup\$ Normally lab experiments are from a 50 ohm voltage generator so the ratio of Rfb/Rs and open loop gain is most important. Confirm your results with this tinyurl.com/yxosgqnf \$\endgroup\$ – Tony Stewart EE75 Jan 21 at 20:56
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 that's a different circuit. I agree that if you use a voltage source and a divider at the input the new circuit becomes sensitive to the series resistance, but this is not the circuit I used. I just simulated the circuit in my answer with Ci changed from 0.1 to 100 uF and it does not budge a iota. What changes is the voltage 'vin' before the input cap. This is a circuit best explained with current control. :-) It's good to be agnostic, in this sense. (Edit: still using an ideal current source, to be clear) \$\endgroup\$ – Sredni Vashtar Jan 21 at 21:16
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I don`t know if the following can answer your question:

  • The circuit as shown (with a "ideal" signal voltage source Vs) will provide DC feedback for stabilizing the DC operational point. But (above the high-pass cut-off frequency) there will be no signal feedback - unless the signal source has a finite source resistance.

  • Both feedback loops can be removed when you connect (after redesign) the resistor Rf to the DC supply voltage.

  • As an alternative (for removing signal feedback only) you can split the existing Rf into two (app. equal) resistors and connect a large capacitor between the midpoint and ground.

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    \$\begingroup\$ I don't think I get it. If it doesn't provide ac feedback, how does it "count" as a negative feedback topology? \$\endgroup\$ – nuggethead Jan 17 at 13:26
  • \$\begingroup\$ if I edit my original schematic to show a current source with finite source resistance would I be more likely to be able to get the answer I'm seeking? \$\endgroup\$ – nuggethead Jan 17 at 13:39
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    \$\begingroup\$ I do not know which answer you are "seeking". What I did is the following: I have decribed the shown circuit which has DC feedback and no ac feedback. Now -when you add a source resistance or if you replace the voltage source with a current source, you have also signal feedback. \$\endgroup\$ – LvW Jan 17 at 13:56
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All resistors affect the gain and missing is your source Rb resistor for feedback ratio. I prefer to pull down the base to bias the collector better as well as attenuate feedback.

The Rc collector affects the open loop gain ratio by current thru the emitter and Rbe. Since that gain is not enough , Rc affects the closed loop gain.

Here by changing the base resistance only, the gain changes from 8 to 350 using Rin = 100 to 100k

The negative feedback improves linearity greatly and the base pull down improves Vc stability to centre the swing since the base voltage is lower than the collector.

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I would analyse the DC case first. Here's a simulation that helps you to understand that analysis: -

enter image description here

So, the analysis goes like this, in stages: -

  • I expect the collector voltage to be close to half the supply voltage i.e. 6 volts
  • Given the forward diode characteristics of the base-emitter junction (0.7 volts), I would expect to see 5.3 volts across the 390 k&ohm resistor.
  • That's a DC base current of 13.5897 μA.
  • To drop 6 volts across the collector resistor (1.76 kΩ) requires a current of 3.4091 mA.
  • 3.4091 mA less 13.5897 μA needs to drawn by the collector (3.3955 mA)
  • Therefore, a BJT would have a current gain (beta) of: -

$$\dfrac{3.3955\text{ mA}}{13.5897\mu A} = 249.86$$

That's how I would analyse the DC and the simulation ties in nicely with it (as expected!). It yields a current gain of 249.86 that exactly produces mid-rail DC on the output (the collector).

  • If the current gain was in fact only 200, I could reverse the process described above and calculate the output voltage to be 6.625 volts (not far from mid-rail).
  • If the current gain were 100, the mid-rail voltage would be 8.462 volts (a little high).
  • If current gain were 400, "mid-rail" (collector) would be 4.722 volts (a little low).

For the AC analysis I would assume a virtual ground at the base i.e. the AC voltage seen at the base is virtually zero. We can then use the well-trusted op-amp formula for AC gain magnitude: -

$$V_{GAIN} = \dfrac{R_F}{X_C}$$

And, if that was deemed too imprecise, I would dig into negative feedback theory to find the formula when the active device has finite voltage gain. Of course a BJT does have finite voltage gain because there is always an internal emitter resistor. The well-trodden formula for \$r_E\$ is this: -

$$r_E = \dfrac{26\text{ mV}}{I_C} = ~7.65\Omega$$

And, that would make the BJT open loop voltage gain = 1760/7.65 = 230. But, for a reasonable input frequency that isn't too high, the gain magnitude would be very close to \$R_F/X_C\$.

But, it's all rather hypothetical because a real BJT has many more nuances about it that make a few scrawls on paper a bit pointless in reality. The answers/numbers above are basically approximations so please do remember that.

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