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I have two questions about the following JFET-BJT amplifier circuit. This circuit is presented and analyzed in paragraph 3.2.3D of the Art of Electronics.

figure 3.31 AoE

The book comes up with the following design equations:

Design equations from the book

My first question is how do they arrive at the open loop gain equation \$G_\text{OL}\$? I do not understand how the transconductance of \$Q_3\$ (\$g_{m3}\$) and \$R_4 \parallel R_6\$ come in. Isn't \$Q_3\$ just a follower and therefore not contributing any voltage gain?

My second question is about the last equation for \$V_\text{out}\$. If I put in the values from the figure I arrive at \$V_\text{out} = 1.1\text{ V}\$. Also, if I use KCL at the source of \$Q_1\$ (\$I_{R1} = I_D + I_{R6}\$), I arrive at

$$V_\text{out} = V_S\left(1 + \frac{R_6}{R_1}\right) - \frac{R_6}{R_2}V_\text{BE2}$$

assuming \$V_\text{EE}=0\text{ V}\$.

Am I missing something here?

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    \$\begingroup\$ I think that the formula for a Gol is a rough approximation only. The proper way of doing this was shown in the 2.5.5 chapter (AoE). Also, notice that R3 is a bootstrapping resistor. Thus, R3 will be seen as a larger resistor R3/(1 - Av) where Av is an emitter follower voltage gain RE3/(re3 + RE3). Therefore we have this R3 * (re3 + RE3)/re3 ≈ R3 * RE3/re3 ≈ R3 * gm*RE3. Of course, this is just a rough first approximation (ballpark calculations). \$\endgroup\$ Commented Dec 11, 2023 at 16:43
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    \$\begingroup\$ I forgot to add that your equation for Vout is correct. $$V_{OUT} = V_S \left(1+ \frac{R_6}{R_1}\right) - \frac{R_6}{R_2}V_{BE2}$$ Good job for catching this small error. \$\endgroup\$ Commented Dec 14, 2023 at 15:43
  • \$\begingroup\$ To determine the center-band voltage gain, you should first plot the equivalent circuit for center-band incremental signals. Having done this, it is understood that the amplifier (with negative feedback) has as its reaction network H, the divider made up of R6, R5||R1 so that the transfer function of H is: H=(R5||R1)/[(R5||R1)+R6]. If the voltage gain of the open loop amplifier is very high, then the feedback voltage gain is approximated to 1/H =1+R6/R5||R1. \$\endgroup\$ Commented Dec 15, 2023 at 14:41
  • \$\begingroup\$ @Franc I am afraid you have forgotten the (pretty low) input impedance at the source node of the FET. \$\endgroup\$ Commented Dec 15, 2023 at 15:04
  • \$\begingroup\$ @LvW Are you referring to R1 ? In the formula of the approximated feedback gain, I considered the small resistance R1 forming the parallel (R5||R1). Otherwise, I've used the transistors' center band approximated dynamic models . \$\endgroup\$ Commented Dec 15, 2023 at 15:46

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Q3 is not being driven by a voltage - it is being driven by the current from Q2. It is not just an emitter follower.

The gain of Q3 affects the impedance that Q2 sees and so its voltage gain.

The gain of Q3 is affected by the load at its emitter, ie R4 in parallel with R6.

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  • \$\begingroup\$ I see that the collector current of Q2 sets V_BE of Q3 through R3, so Q3 turns on and additional (Q3 emitter) current is flowing through (R4||R6). This is still very qualitative though, I still cannot derive the nice multiplications in the equation for G_OL. \$\endgroup\$ Commented Dec 10, 2023 at 19:07
  • \$\begingroup\$ @Kevin White The emitter current of Q3 is controlled by the VOLTAGE across R3 (as each BJT is controlled by a voltage). When you think it would not be an emitter follower - what is it in your opinion? \$\endgroup\$ Commented Dec 11, 2023 at 9:48
  • \$\begingroup\$ @LvW - The bottom end of R3 is connected to the emitter of Q3, not ground. The overall effect is that the output from Q3 emitter is basically dependent on the current from Q2 collector. You can treat it as an emitter follower but the voltage at the base needs to be calculated based on the gain of Q3 and the load of R4||R6, not just R3 * Q2 current. \$\endgroup\$ Commented Dec 11, 2023 at 18:45
  • \$\begingroup\$ @Kevin White I completely agree - and the first two sentences repeat what I have stated already in my former comment. So I cannot consider your comment as a kind of "counter argument". There is no doubt that the Q3 works in common-collector configuration. And the voltage at the emitter of Q3 - as the only signal output privided by Q3 - is the voltage that acts as a feedback signal. Therefore, it is clear to me that Q3 (in common-collector configuration) works as an emitter follower. \$\endgroup\$ Commented Dec 12, 2023 at 8:20
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In response to LvW's previous message. I based myself on the simple approximation of the voltage gain of a feedback system with the gain of the direct action chain much greater than unity (see figure):

enter image description here

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  • \$\begingroup\$ The above feedback model for voltages is valid only under the assumption that all blocks - including the summing block - do not load each other (zero output and/or infinite input impedance). However, this is not the case for the circuit under discussion. At the feedback node (S of Q3) we have a parallel connection of R1 and the low input resistance (1/gm) at the source of Q3. \$\endgroup\$ Commented Dec 16, 2023 at 11:58
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The equivalent circuit I used is the following:

enter image description here

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  • \$\begingroup\$ I suppose that this contribution is an answer to our discussion (comments) above. The shown equivalent circuit is valid for closed-loop applications only (input signal Vg).However, when we speak about the feedback factor (you call,it "reaction network H") the signal source Vg must be set to zero and the relevant input is at the source node of Q1 (similar to common base configuration). \$\endgroup\$ Commented Dec 16, 2023 at 9:13

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