Differential amplifier voltage controlled gain miscalculation

[setq]

First post, so hello and I'll make this a good one

I'm working through a relatively ancient 1st edition copy of The Art of Electronics and the old Lab manual, not homework. Too old for homework now. I've just completed the section on transistors and am doing the additional exercises at the end and am stuck on their differential amplifier voltage gain determination.

The question (6) describes the following:

In this problem you will ultimately design an amplifier whose gain is controlled by an externally applied voltage. (a) Begin by designing a long tail pair differential amplifier with emitter current source and no emitter resistors (undergenerated). Use +/-15 volt supplies. Set Ic for each transistor at 1mA and use Rc=1k. Calculate the voltage gain from a single ended input (other input grounded) to a single-ended output.

Did that and it works in LTspice and on paper.

Now modify the circuit so that an externally applied voltage controls the emitter current source. Give an approximate formula for the gain as a function of controlling voltage.

This is described in the following LTspice screencap. As Gm is related to total current of the long tailed pair I thought we can just adjust the current source:



I used the following calculations:

Re = .025 / Ic  -> This is the emitter resistance at the collector current specified by Ebers Moll.

Ic = (Vc - 0.6) / R1 -> This is the collector current controlled by the current source

Gm = R2 / 2rE -> This is the gain ignoring any emitter degeneration resistors as there aren't any.

Substituting these in for the above gives me:

Gm = (0.00125 * R2 * R1 ) / (Vc - 0.6)

When I throw this in LTspice with a voltage of 1.6v and resistors of 1k, 500 ohms as calculated I'm getting a gain of 625

If I put a 10mV p-p sine into the simulation however, I'm getting a p-p of 350mV which is a gain of 35.

Can anyone see where I've gone wrong here - I'm completely stumped? Any help appreciated!

[PChi]

Isn't Gm = R2 x (Vc - 0.6) / ( 2 x 0.025 x R1 ) ?
I make that 40 and LTSpice is adding in the effect of the finite collector resistance of Q4 dropping the voltage gain to 35.

[setq]

Interesting and much closer! How did you derive that from the schematic?

[PChi]

I just rearranged the equations that you listed.
Gm = R2 / 2Re
Re = .025 / Ic so 1 / Re = Ic / 0.025
Substituting for Re in the equation for Gm = R2 x Ic / (2 x 0.025)
Ic = (Vc - 0.6) / R1
Substituting for Ic in the equation for Gm = R2 x (Vc - 0.6) / ( 2 x 0.025 x R1 ).

[setq]

Gotcha that makes sense now. I fudged my algebra completely.

Thanks for your help - much appreciated!