# Calculating the transresistance in a multistage voltage-shunt(shunt-shunt) feedback amplifier

I've been trying to analyze the following circuit using whatever books and resources I have to hand: simulate this circuit – Schematic created using CircuitLab

I've identified the topology as a voltage-shunt feedback amplifier (is this correct?) It looks as if the voltage is sampled at the output and the feedback current is subtracted (anti-phase) from the source, therefore voltage-shunt.

Any examples I've seen for this type of topology are either the collector feedback single stage amp or and classic inverting op-amp circuit. But, I haven't come across any multi-stage examples on which to base my analysis.

Applying the usual rules (short Vo to find input, short Vi to find output) to find the circuit to calculate the open loop gain for this topology, I get the following: simulate this circuit

Using AC analysis for each stage, I calculate the trans-resistance for stage 2 (Rm2) as 27k, given by:

Note: I've used certain sensible (I think) simplifications here.

$$Rm2 = \frac{Vo}{Ib2}=hFe\cdot R5=(100)(270)=27k$$

and the trans-resistance of stage 1 as:

$$Rm1=\frac{Vo1}{Is}=\frac{-Ic1\cdot RL}{Is}=\frac{-Ic1}{Ib1}\cdot\frac{Ib1}{Is}\cdot RL=hFe\cdot \frac{Ib1}{Is}\cdot RL = -73k\\\\ where RL = hFe\cdot R5 \parallel R1 = 27k \parallel 820 = 795\\\\ and Ib1 = \frac{Is\cdot Ri}{Ri + hFe\cdot re} where Ri=R2\parallel R3\parallel R5$$

I'm not sure I've gone about this the right way, but it seems that multiplying the two trans-resistance values is not the way to go (like you'd do for multi-stage voltage-series amps).

So, my question is: how do I proceed to get the open-loop trans-resistance in this (or any) multi-stage voltage-shunt configuration?

UPDATE: Thinking about it, I could use the same method as with a current-shunt feedback amplifier and express:

$$\frac{Vo}{Is}=\frac{Ib1}{Is}\cdot \frac{Ic1}{Ib1}\cdot \frac{Ib2}{Ic1}\cdot \frac{Vo}{Ib2}\\\\ =\frac{Ib1}{Is}\cdot hFe\cdot \frac{Ib2}{Ic1}\cdot \frac{Vo}{Ib2}$$

but I'm just not sure.

I think that the voltage gain cannot be larger than R2/R5 in this case.

If I do more "traditional" analysis (I'm not familiar with the TS method) simulate this circuit – Schematic created using CircuitLab

I get the voltage gain equal to :

$$A_V = \frac{R_1}{r_{e1}} * \frac{R_5||R_6}{R_5||R_6 + r_{e2}} \approx \frac{R_1}{r_{e1}} \approx 38\cdot I_C\cdot R_1 \approx 38\cdot 6.2mA \cdot 820\Omega \approx 193 V/V$$

And now using the Miller effect I can find the Voltage gain of the whole amplifier simulate this circuit

$$A_{V2} = \frac{R_3||\frac{R_2}{1+A_V}||r_\pi}{R_5 +R_3||\frac{R_2}{1+A_V}||r_\pi} *A_V \approx 5V/V$$

Additional I assumed $r_\pi =(\beta +1) r_e = (100 +1) \cdot \frac{26mV}{6.2mA} = 423\Omega$

• Not too familiar with the Miller effect method of finding the closed-loop gain. It seems there are multiple ways of working this out, but I'm surprised that the method I spoke of (see cc.ee.nchu.edu.tw/~aiclab/teaching/Electronics3/lect08.pdf page 8-57) is not as popular as I thought. I'll need to look up the miller effect method you mentioned. May 7, 2018 at 17:30
• Incidentally, this circuit appears as a buffer to an oscillator I'm interested in and I've tested it with a signal of ~50Mhz. It had an attenuating effect on the input and I wanted to analyze it to find its "theoretical" gain. I wanted to understand if the miller effect (collector to base of Q1) was having an effect. May 7, 2018 at 17:38
• – G36
May 7, 2018 at 17:49

My answer does not answer trans-resistance, rather closed loop gain of a negative feedback amp with Norton Equivalent feedback.

Feedback ratio=R2/(R3//R5)~12.3 = -gain

Input impedance is reduced as a consequence of negative feedback. The Rbe is reduced the excess gain removed by feedback which becomes essentially much smaller than the series input R5.

• Hi Tony, I was trying to follow the methodology given in a book I own, where the author gets the open-loop trans-resistance and then applies the de-sensitivity factor 1+bA. But calculating TR for cascaded stages isn't discussed (hard to find anywhere actually!) so I got a little stuck and I wondered what the standard approach was. I guess I'm looking for a methodology. May 7, 2018 at 13:59
• The methodology I used is the Emitter follower as a voltage source feedback, just like an Op Amp gain ratio but must be a fraction of open loop gain, which it is here. May 7, 2018 at 14:48

I finally figured out the answer to my own question and I thought I'd present it here alongside the fine answers already received. I was looking for a solution that followed the standard (textbook) type analysis of a CCVS/Voltage-shunt/Shunt-shunt, where the gain Rm is expressed as a trans-resistance. But extended for multiple stages.

The key to developing the open loop gain, is to treat the configuration as a current amplifier with the final stage output current multiplied by the load. So:

$$\frac{Vo}{Is}=\frac{Ic2\cdot Re2}{Is}=\frac{Ic1}{Ib1}\cdot \frac{Ib1}{Is}\cdot \frac{Ic2}{Ib2}\cdot \frac{Ib2}{Ic1}\cdot Re2=\\\\ hFe\cdot \frac{Ib1}{Is}\cdot hFe\cdot \frac{Ib2}{Ic1}\cdot Re2$$

So, that was really the answer to my question and plugging in the values I get: an Ai of -3 for Q2 and an Ai of 92 for Q1. Then...

$$Rm=\left ( 92 \right )\left ( -3 \right)\left ( \approx270 \right )=-74.5k$$ and $$\beta = \frac{If}{Vo}=-\frac{1}{47k}=-21.2uS$$ making $$D=1+\beta Rm=2.58$$ which makes the trans-resistance with feedback $$Rmf=\frac{-74.5k}{2.58}=-28.88k$$ To convert this to a voltage gain we say $$\frac{Vo}{Vs}=\frac{Vo}{Is\cdot 5k6}=\frac{Rmf}{5k6}\approx -5.1$$

which is in good accordance with the answer by G36. It's a bit (lot!) more involved, but I wanted a supplemental method for what's in my textbook so I'll know for again.

Now to find out why the built one is actually attenuating - but that's a different story.