Let's suppose the following circuit:
I am trying to find a relation between Vs, V1 and V2 in this circuit. Please note that Vs is measured in relation to ground. I have declared some extra points (A,B,C) on the circuit and using voltage divider laws and the properties of an ideal operational amplifier I came up with the following solution:
Is this correct?
Another solution with a different result is the following:
I am really confused.
Voltage division isn't a great approach to hang this on. Just build the output from the bottom up. You just need to know that current doesn't enter the input terminals of an op amp.
You know Vc, so you know the current through R3. That has to be the same as the current through R4, so now you know Vb. You also know Va, so now you can calculate current through R2, which has to be the same as the current through R1, which would give you Vs.
Looks a little like the input stage of an instrumentation amplifier,but there's an extra resistor. CORRECTION This is a two op-amp instrumentation amplifier. Full discussion of the circuit at http://www.analog.com/static/imported-files/design_handbooks/5812756674312778737Complete_In_Amp.pdf on page 2-4.
-- though the resistor numbers and inputs are not quite the same as what you use
Taking a shot at your derivation, starting from the line i2=i1=V2/R3, lets have a go at it
$$ V_B = V_2 + i_2R_4 = V_2 + \frac{V_2R_4}{R_3} $$ $$ = V_2 \left ( 1+ \frac {R_4}{R_3} \right ) $$
$$ i_3 = \frac{V_A-V_B}{R_2} = \frac{V_1 - V_2 \left ( 1 + \frac{R_4}{R_3} \right )}{R_2}$$
$$V_S= V_1 + i_4 R_1 =V_1 + \frac{V_1R_1}{R_2} - V_2 \frac{R_1}{R_2} \left ( 1+ \frac{R_4}{R_3}\right )$$ or just shifting to make this look a bit more differential: $$ V_S = V_1 \left ( 1 + \frac{R_1}{R_2} \right ) - V_2 \frac{R_1}{R_2} \left ( 1 + \frac{R_4}{R_3} \right )$$
That's a quick pass, and something feels wrong. Feel free to correct.
A clever circuit... beautiful and symmetrical (usually, R1 = R2 = R3 = R4 = R) like all the circuits of instrumentation amplifiers... It also gives a good opportunity to show how to make the unfamiliar circuits familiar - by dividing them into more well-known functional blocks instead of blindly analyzing them...
We can discern in this circuit of a perfect instrumentation amplifier two sub-circuits - an imperfect (unbalanced) differential amplifier (the top part consisting of the upper op-amp and the resistors R1, R2), and an ordinary non-inverting amplifier (the bottom part consisting of the lower op-amp and the resistors R3, R4).
Let's first consider the upper circuit part. With respect to V1, it is a non-inverting amplifier with gain of 2, and with regard to the lower input (from the side of the lower non-inverting amplifier) - an inverting amplifier with gain of -1. As the lower non-inverting amplifier has a gain of 2, the two gains (inverting and non-inverting) of the imperfect differential amplifier are equalized... the two partial voltages superimpose and mutually neutralize at the op-amp output... and it becomes a perfect balanced differential (instrumentation) amplifier. From this perspective, the analysis is very simple:
Vs = V1*(R1 + R2)/R2 - V2*(R3 + R4)/R3*R1/R2 = 2V1 - 2V2 = 2(V1 - V2); Happy New Year!
It is interesting to reveal the evolution of the op-amp differential amplifier to see where this circuit solution stays. I will use figurative (not generally accepted) names of the particular circuit solutions that are more meaningful. Also, to simplify this qualitative explanation, I suppose equal resistances (R).
1. Unbalanced "differential amplifier". To make a differential amplifier, we need simply to subtract two input voltages. First we introduce a negative feedback by two resistors to obtain a fixed stable gain and then apply the two voltages to the inverting and non-inverting inputs of this "differential amplifier". Here the low resistance of the inverting input is a problem... but the big problem is that the two gains are not equal - the inverting gain (1) is less than the non-inverting gain (2). So, we have two choices to equalize them - to decrease (two times) the non-inverting gain or to increase (two times) the inverting gain. Let's consider them below...
2. Differential amplifier with non-inverting attenuation. To decrease (two times) the non-inverting gain, we can connect a voltage divider (with two equal resistors) before the non-inverting input thus obtaining the classical 1-op-amp differential amplifier. The two gains are now equalized... but the high resistance of the non-inverting input is decreased...
3. Differential amplifier with inverting amplification. With the same success we can increase (two times) the inverting gain if we connect a non-inverting amplifier (with a gain of 2) before the inverting input (the discused here solution). The two gains are equalized again... and the both inputs have high resistance... It is a real 2-op-amp instrumentation amplifier.
4. Buffered differential amplifier with non-inverting attenuation. Finally, we can modify the classic 1-op-amp differential amplifier (case 2) by including non-inverting amplifiers before its inputs; this will solve the problems of the low input resistances. But another problem is that they will ampliify both differential and common-mode input voltages.
5. Buffered differential amplifier with interconnected non-inverting amplifiers. If we are inventive enough, we will combain the two lower resistors of the voltage dividers (inside the input non-inverting amplifiers) into one resistor (Rgain) that can regulate simultaneosly both the input gains. Thus we will obtain the classic 3-op-amp instrumentation amplifier.
It is interesting that (only) in differential mode, there is a virtual ground in the middle point of Rgain; it has replaced the real ground. As a result, the common-mode input voltages are not amplified (k = 1); only the differential input voltage is amplified.
The answer is obvious:
It is a "differential amplifier with inverting amplification" (number 3 above).
Finally, let's generalize these techniques in a rule:
If a differential circuit is unbalanced, you have two options - to connect
What is the idea behind the op-amp instrumentation amplifier? (my Codidact paper)
i3=i2at node B, but you neglect the current from the output of the lower op-amp, which is not negligible in general. \$\endgroup\$