It's always a pleasure to read this forum and this time I'm making the step to ask a question :) ! I'm currently working on a pre-amplifier for microphones based on a schematic seen in a famous sound console!

I'm currently having troubles to realize a small signal study of the stage. The main schematic is this one:

Main schematic

Where V1 is the input and voltage across R8 and R9 are the outputs. Inputs are the same but -V1 is 180° shifted.

I give you now the schematics in AC and with the equivalent models of the BJT (where output impedance and capacitances are negliged for now)

AC analysis

Small signals

The values for hfe and hie are from datasheet. And current sources are inversed in order to work with positives currents. For information when RV1 is shorted I found 68 dB of gain from simulation and measure.

So my questions are:

Is my equivalent model right or have I forgotten something? How could I express Vs from V1 and RV1?

  • I tried expressing the potential in RV1 when shorted with Millman to be able to express Veb but got stuck
  • I tried replacing the current sources by voltages sources but where not able to find the equivalent Norton model...
  • I tried applying a superposition theorem by shorting either V1 or -V1 but I don't know how to deal with current sources in this case...

If someone would gently give me just a clue on how to solve this question I would be very pleased.

  • \$\begingroup\$ That's an interesting design--I almost thought it was a long-tailed pair at first, but then I saw that there's no shared emitter resistance, so it's more like just two common-emitter amplifiers with some coupling between their emitters. What famous sound console is this from, out of curiosity? \$\endgroup\$
    – Hearth
    Oct 27, 2023 at 15:21
  • \$\begingroup\$ That's this emitter coupling wich is driving me crazy. It's from a Midas XL3 ;) \$\endgroup\$
    – TooT
    Oct 27, 2023 at 16:09
  • \$\begingroup\$ Structure is reminiscent of the mc1496. \$\endgroup\$ Oct 27, 2023 at 20:04
  • \$\begingroup\$ I should have taken a moment to say more. Here's why I'm so reminded. \$\endgroup\$ Oct 27, 2023 at 21:13
  • \$\begingroup\$ @Hearth With the gain resistor at zero, it's a long-tailed pair for AC. The gain resistor adds to the (nonlinear) differential emitter resistance, reducing gain and improving linearity for large signals. \$\endgroup\$
    – John Doty
    Oct 28, 2023 at 13:15

2 Answers 2


I've finally worked this out ! Turning properly -V1 off and working on a half branch made the trick.

I find a voltage Gain expressed by :

$$ A_V = - \frac{R_7//Z_e \cdot \beta}{h_{ie} + \frac{R{V1} \cdot R_3}{R_{V1} + R_3} \cdot (\beta +1)} $$

With \$Z_e\$ being the input impedance of the next stage.


Due to circuit symmetry, you coud simplify the small signal model further. We can split RV1 in half and connect the middle point to GND. And now finding the voltage gain is easy.


simulate this circuit – Schematic created using CircuitLab

$$\frac{V_{out1}}{V_{in1}} = \frac{R_7||R_9}{r_{e1} + 0.5R_{V1}||R_3} * \frac{\beta}{\beta +1}$$


$$r_{e1} \approx \frac{26mV}{I_{E1}}$$

The \$Q_1\$ emitter DC current will be equal to:

$$I_{E1} \approx \frac{V_{CC} - V_{BE1}}{R_3 + \frac{R_6}{\beta +1}} \approx \frac{18V - 0.7V}{7.5k\Omega} \approx 2.3mA$$

  • \$\begingroup\$ Thanks I arrived at the same conclusion :) Just a question about your calculation of $\R_{ie}\$, I keep seeing these 25~26mV for the delta Vbe where does it come from ? I found my answer here ;) electronics.stackexchange.com/questions/194831/… \$\endgroup\$
    – TooT
    Oct 28, 2023 at 14:40
  • \$\begingroup\$ After closer looking, we don't have the same result. You have Ie going through Re1 wich is I guess the input resistance of the BJT as you calculate it with Vt/Ie1, should'nt it be Ib instead of ie ? \$\endgroup\$
    – TooT
    Oct 28, 2023 at 14:43
  • \$\begingroup\$ You have used \$h_{ie}\$ also knows as a \$r_{\pi}\$ or \$h_{11}\$ And \$ \Large h_{ie} = (\beta +1)*r_e\$ Take a look here electronics.stackexchange.com/questions/617038/… \$\endgroup\$
    – G36
    Oct 28, 2023 at 14:56
  • \$\begingroup\$ I prefer using the T-model (small-signal model) instead of Hybrid-pi. electronics.stackexchange.com/questions/518032/… \$\endgroup\$
    – G36
    Oct 28, 2023 at 15:00
  • \$\begingroup\$ Many thanks ! I'll give it a look when my brain will have rested ;) \$\endgroup\$
    – TooT
    Oct 28, 2023 at 15:16

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