I'm slowly working my way through understanding negative feedback in simple BJT amplifiers. I'm studying this circuit here: enter image description here

I read that one of the advantages to negative feedback in this topology is that it raises the input impedance by a factor of \$1+KA_V\$ (where \$K\$ is the feedback factor and \$A_V\$ is the open-loop gain of the internal amplifier). I did a simple analysis of the input resistance \$R_{in}\$ and found that



\$R_{in}\approx 45k\Omega\$

I was able to verify that this is the correct input impedance. By simulating a test signal \$v_s=20mV\$, I measured the current entering the circuit at \$i_{in}=444nA\$, which checks out.

Here's the part that doesn't make so much sense to me. If the negative feedback increases the input impedance by a factor of one plus the loop gain:


What was the internal amplifier's input resistance before the negative feedback was added? From my previous work with this circuit, I know that ...

\$K=(\frac{1}{A_F})-(\frac{1}{A_V})\approx 1\$

\$A_V\approx g_m(R_E) = 0.12 * (4000) = 480\$

\$A_F\approx 1\$

Plugging all this in, I should be able to find the internal amplifier's nominal input resistance \$R{in'}\$before the feedback network was added.



\$R_{in'}\approx 93.6\Omega\$

For the life of me, I can't see how the internal amplifier could be analyzed to have an input resistance of 93.6. I'm fairly certain my loop gain is correct as all those calculations check out in a previous question (here).

As an aside, I did try comparing this circuit to one with \$R_E\$ removed to see if that helped any (it didn't, I don't think). I reduced \$V_{cc}\$ to a low enough level that my \$I_{CQ}\$ remained at \$3mA\$. My input resistance fell as expected to \$R_{in}\approx \beta*r_e \approx 2080\Omega\$. So, again, my analysis of input resistance proved out OK and checked out in simulation. But adding the feedback resistor \$R_E\$ only increased the input resistance by a factor of around 21, not the expected 1 + loop gain factor.
Can anyone explain what expression for an internal amplifier's input resistance will work with the \$R_{IN}=R_{}in'(1+KA_V)\$ equation?

  • \$\begingroup\$ Just curious. What do you think \$A_v\$ means (I think that you think you know, but if you read your writing and check the schematic again you might realize a mistake? -- Or you will point out something that tells me I'm in error.) And what exactly are you considering as counting as negative feedback? \$\endgroup\$
    – jonk
    Commented Dec 30, 2020 at 4:03
  • \$\begingroup\$ I wouldn't call this a voltage amplifier; it's an emitter follower, which is much more of a current amplifier. \$\endgroup\$
    – Hearth
    Commented Dec 30, 2020 at 5:11
  • \$\begingroup\$ It is the input resistance at the base that is increased due to negative feedback. (not the total input resistance of the stage). The influence of the base divider is to be considered (in parallel) at the end of the calculation only! \$\endgroup\$
    – LvW
    Commented Dec 30, 2020 at 9:25
  • \$\begingroup\$ @jonk I think \$A_v\$ is the amplification of the tiny portion of \$v_s\$ that appears at the input to the internal amplifier. It's much smaller than the resolution of my simulator will show accurately, on the order of perhaps 50 to 100uV. This voltage, multiplied by Av, gives the expected output near 20mV. \$\endgroup\$
    – nuggethead
    Commented Dec 30, 2020 at 12:37
  • \$\begingroup\$ @Hearth yes, it amplifies current in its use as an emitter follower, as voltage gain is effectively unity. I only call it a voltage amplifier because of the topology, where the internal amplifier receives a voltage signal and outputs a voltage. This is different than other amplifier topologies like transconductance, transimpedance, etc. \$\endgroup\$
    – nuggethead
    Commented Dec 30, 2020 at 12:46

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

Closed-loop gain Acl=gRE/(1+gRe)

Loop gain: Aloop=-gRE

Input resistance (without external base divider) rin=h11(1+gRE)

Comment: The "open-loop gain" (Aol=gRE) is a fictitious expression because the circuit cannot be operated without feedback. This is due to the fact that there is no "outer" feedback loop that could be cut. Output and inverting input nodes are identical (internally connected).

  • \$\begingroup\$ What is h11 in your last step? Is it beta? \$\endgroup\$
    – nuggethead
    Commented Dec 30, 2020 at 13:02
  • \$\begingroup\$ Sorry....I should mention that h11=hie=rbe (the dynamic input resistance between base and emitter). We have hie=beta/g with g=Ic/Vt. \$\endgroup\$
    – LvW
    Commented Dec 30, 2020 at 13:15

enter image description here If we start out with the definition of closed-loop input impedance as \$R_{IN}=\frac{v_s}{i_{in}}\$, then we can see that the input impedance \$R_{in'}\$ of the internal amplifier would give us the voltage \$v_{in}\$ as \$v_{in}=i_{in}R_{in'}\$. With this, we can use the open-loop gain to find the voltage output. \$v_{out}=A_vR_{in'}i_{in}\$. The voltage presented as feedback \$v_f\$, then, is \$KA_vi_{in}R_{in'}\$.

Performing KVL on the input side, \$v_s = i_{in}R_{in'}+KA_vi_{in}R_{in'}\$.

Rearranging slightly, \$v_s = i_{in}R_{in'}(1+KA_v)\$.

And dividing both sides by \$i_{in}\$, we arrive at this expression for the overall closed-loop impedance, or the impedance seen at the input.


To verify all of this, I solved for the overall closed-loop impedance \$R_{IN}\$ using the formula \$R_{IN}=R_{in'}(1+KA_v)\$ using \$R_{in'}=\beta(r_{e})\$. Note that in computing the internal amplifier's input impedance, I treated it as if there was no feedback -- no degeneration resistor \$R_E\$.

\$R_{in'}=\beta(r_{e}) = 250*8.33 = 2083\Omega\$

Then, using \$R_{IN}=R_{in'}(1+KA_v)\$

\$R_{IN}=2083(1+(1*482))\$ \$R_{IN}\approx1M\Omega\$

Note that I used values for \$K\approx1\$ and \$Av\approx482\$ above from a previous question here. This overall impedance checks out in simulation. (To test it, I modified the original circuit to have the feedback component removed. I replaced \$R_E\$ with a short to to ground and adjusted \$Vcc\$ down to approx. 2.37v. This kept the remaining resistor values unchanged and kept the quiescent current steady at 3mA.)

Finally, if I return the supply voltage and \$R_E\$ to the circuit, I get this value for \$R_{IN}\$


\$R_{in'}=\beta(R_E+r_{e}) = 250*4008.33 \approx 1M\Omega\$


So, the internal amplifier without feedback is roughly equivalent to the circuit without the emitter resistor. Yes, this would be a useless circuit! But now the effect of feedback can be seen more clearly. And to answer the original question, the input impedance "before" feedback was added is lower by a factor of \$A_v\$, which would be equivalent to the circuit running without the emitter resistor in place.

The reason I had trouble simulating this in my original posting above is that I had erroneously included the effect of the biasing resistors \$R_1||R_2\$. Taking these into account, we see that the circuit without feedback is hardly changed at all by the divider:


but in the circuit with feedback, the divider network significantly reduces the effective input impedance:


  • \$\begingroup\$ @jonk does my thinking make sense in this answer? \$\endgroup\$
    – nuggethead
    Commented Jan 11, 2021 at 21:13
  • \$\begingroup\$ @LvW this is the way I have been thinking about feedback. Am I on the right track? \$\endgroup\$
    – nuggethead
    Commented Jan 11, 2021 at 21:14

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