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I am trying to figure out the gain of the following circuit: enter image description here

What I have done so far is looking at each amplifier independently: $$ U_3=-R_9/R_8=-4 $$ $$ U_4=R_{10}+R_{11}/R_{10}=3 $$ $$ V_o=-R_9/R_8 \times (R_{10}+R_{11}/R_{10})=-12 $$ And have gotten -12 as the final answer. However, the questions hints at "cascading the transfer functions". Does my method look correct? What am I doing wrong here?

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    \$\begingroup\$ I'm baffled as to why you think U3 inverts but U4 doesn't. \$\endgroup\$
    – Finbarr
    Aug 31, 2021 at 14:46
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    \$\begingroup\$ This looks like two inverting amps to me. So your -12 gain is obviously incorrect. \$\endgroup\$
    – jwh20
    Aug 31, 2021 at 14:49
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    \$\begingroup\$ Gain = -4 * -2 , Offset = 2V * (2+1) \$\endgroup\$ Aug 31, 2021 at 15:02
  • \$\begingroup\$ @Finbarr I have been staring at the circuit and the only conclusion I can come to is I am neglecting the V1, so does that mean I will have to multiply it by -1 and V1 to get my corresponding gain? The V1 is throwing me off... \$\endgroup\$
    – thediyer
    Aug 31, 2021 at 15:02
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    \$\begingroup\$ Use KCL at U4's inverting input. \$\endgroup\$
    – Null
    Aug 31, 2021 at 15:10

2 Answers 2

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It's a bit of a tricky case. You actually have two inputs: \$V_{in} \$ and \$V_p \$, the voltage source connected to + in your second op-amp. Both these sources affect \$V_o \$ so to find the effect from each source you need to use superposition.

schematic

simulate this circuit – Schematic created using CircuitLab

Case 1: \$V_p =0\$

The node equations become $$\begin{cases} \frac{V_1-V_{in}}{R_8}+\frac{V_1-U_3}{R9}=0\\\\ \frac{V_2-U_3}{R_{10}}+\frac{V_2-V_o}{R_{11}}=0 \end{cases} $$ With \$V_1=0 \$ and \$V_p=0 \$ and the principle of a virtual short you get \$V_o=V_{in} \frac{R_{11}R_9}{R_{10}R_8} \$.

Case 2: \$V_{in}=0 \$

The node equations are the same $$\begin{cases} \frac{V_1-V_{in}}{R_8}+\frac{V_1-U_3}{R9}=0\\\\ \frac{V_2-U_3}{R_{10}}+\frac{V_2-V_o}{R_{11}}=0 \end{cases} $$ With \$V_1=0 \$, \$V_2=V_p\$ and \$V_{in}=0 \$ and the principle of a virtual short you get \$V_o = V_p \frac{R_{10}+R_{11}}{R_{10}} \$

The total response: $$V_{o,total}=V_{in} \frac{R_{11}R_9}{R_{10}R_8}+V_p \frac{R_{10}+R_{11}}{R_{10}} $$

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  • \$\begingroup\$ Thanks for the detailed explanation! Your a lifesaver \$\endgroup\$
    – thediyer
    Aug 31, 2021 at 19:00
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    \$\begingroup\$ For me, the gain is only the factor multiplying \$V_{in}\$, in this case \$\frac{R_{11}R_9}{R_{10}R_8}\$ everything else is just an offset. @thediyer the key technique for the analysis (already mentioned by Carl) is "Superposition" worth reading about it. \$\endgroup\$
    – Raul M.
    Aug 31, 2021 at 19:49
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The "aggregate", "net", "combined", or "compound" transfer function of cascaded amplifier stages is sometimes quoted to be the simple product of the transfer functions of the individual stages, but this is only true if every stage implements a simple direct proportionality, like \$ V_{OUT} = -3 \times V_{IN} \$.

Your second stage introduces an offset \$ V_1 \$ which invalidates that approach.

Here you must use function composition to obtain the ouput \$ V_{OUT} \$ as a function of some input \$ V_{IN} \$:

$$ V_{OUT} = h( V_{IN} ) $$

To illustrate, I'll redraw your circuit with some extra annotations, to visually segregate the stages, and give us some variables to work with:

schematic

simulate this circuit – Schematic created using CircuitLab

The transfer functions of stages 1 and 2 are \$ f \$ and \$ g \$ respectively, so that:

$$ V_F = f(V_{IN}) $$ $$ V_{OUT} = g(V_F) $$

The two functions must be composed to obtain the overall transfer function:

$$ V_{OUT} = h(V_{IN}) = g(V_F) = g(f(V_{IN})) $$

Sometimes you see this written as \$ h = g \circ f \$, which I read as "g of f" or "g about f".

The first stage transfer function \$f\$ is very simple. I assume you are familiar with the opamp configured as a simple inverting amplifier:

$$ f(V_{IN}) = -\frac{R_9}{R_8} \times V_{IN} $$

The second stage function \$g\$ is less trivial, because of the offset \$V_A\$ due to \$V_1\$. To simplify the algebra we can employ what we know about opamps with negative feedback, which is that the opamp adjusts its output to equalise the potentials at its inverting and non-inverting inputs:

$$ V_A = V_B = 2V $$

Then construct the usual KCL and Ohm's law equations, and solve for \$ V_{OUT} \$ as a function of \$ V_F \$ and \$ V_A \$:

$$ I = \frac{V_{OUT} - V_B}{R_{11}} $$

$$ I = \frac{V_B - V_F}{R_{10}} $$

$$ V_{OUT} = \frac{R_{11}}{R_{10}}(V_A - V_F) + V_A $$

Intuitively, you can interpret this to mean that the stage amplifies the difference between \$V_A\$ and the input, with the usual gain of \$ -\frac{R_{11}}{R_{10}} \$, but also offsets the output by an additional amount \$V_A\$.

The penultimate step is to compose the functions:

$$ \begin{aligned} h(V_{IN}) &= g(f(V_{IN})) \newline \newline &= \frac{R_{11}}{R_{10}}(V_A - f(V_{IN})) + V_A \newline \newline &= \frac{R_{11}}{R_{10}}(V_A + \frac{R_9}{R_8}V_{IN}) + V_A \end{aligned} $$

Lastly we plug in the resistances to get:

$$ \begin{aligned} V_{OUT} &= 2(2V + 4\cdot V_{IN}) + 2V \newline \newline &= 6V + 8 \cdot V_{IN} \end{aligned} $$

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