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I greet everyone,

I am in the process of learning opamps, I got stuck here while solving a question about the difference amplifier. When finding the Vo value, we do the following operation;

\$V_O = \left(\frac{-R_{13}}{R_{14}} \times V_X\right) + \frac{R_{16}}{R_{15} + R_{16}} \times \left(1 + \frac{R_{13}}{R_{14}}\right)\times V_Y = 4V + 4.5V = 8.5V\$

This is how it is solved. But I don't understand the logic of how we remove this part here;

\$ \frac{R_{16}}{R_{15} + R_{16}} \times \left(1 + \frac{R_{13}}{R_{14}}\right)\times V_Y\$

How can we remove this part? I would be grateful if someone from the community could explain

enter image description here

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  • \$\begingroup\$ Please use mathml/TeX; it's hard to read the equations as written. \$\endgroup\$
    – Hearth
    Commented Jul 6 at 19:55
  • \$\begingroup\$ When you say this: R16 / R15 + R16 don't you actually mean this: \$\dfrac{R_{16}}{R_{15} + R_{16}}\$? ----> \$\dfrac{R_{16}}{R_{15} + R_{16}}\$ \$\endgroup\$
    – Andy aka
    Commented Jul 6 at 19:56
  • \$\begingroup\$ I don't see it removed. You still calculate something related to Vx and something related to Vy and sum them together. Please explain what you mean. \$\endgroup\$
    – Justme
    Commented Jul 6 at 19:58
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    \$\begingroup\$ Remove? We do not remove anything? What you see is a superposition principle in action. \$\endgroup\$
    – G36
    Commented Jul 6 at 19:58
  • \$\begingroup\$ @Andy , yes that's exactly what I mean, I added it as latex but it didn't convert. \$\endgroup\$ Commented Jul 6 at 20:02

2 Answers 2

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Trying to read between the lines, I think you are wondering how \$\left(1+\frac{R_{13}}{R_{14}}\right)\$ gets removed from the \$V_x\$ term in the equation. Perhaps the confusion so far is that you cited the \$V_y\$ term that shows this factor, but you are wondering why you don't also see the same factor with the \$V_x\$ term.

I may be totally wrong about that. But that's my guess.

By superposition, you know that:

$$V_-=V_x\cdot\frac{R_{13}}{R_{13}+R_{14}}+V_o\cdot\frac{R_{14}}{R_{13}+R_{14}}$$

You also know that:

$$V_+=V_y\cdot\frac{R_{16}}{R_{15}+R_{16}}$$

And those are equal:

$$V_x\cdot\frac{R_{13}}{R_{13}+R_{14}}+V_o\cdot\frac{R_{14}}{R_{13}+R_{14}}=V_y\cdot\frac{R_{16}}{R_{15}+R_{16}}$$

Multiplying through by \$\frac{R_{13}+R_{14}}{R_{14}}=\left(1+\frac{R_{13}}{R_{14}}\right)\$, another way to look at this is:

$$\begin{align*}V_o&=\left[-V_x\cdot\frac{R_{13}}{R_{13}+R_{14}}+V_y\cdot\frac{R_{16}}{R_{15}+R_{16}}\right]\cdot\left(1+\frac{R_{13}}{R_{14}}\right)\\\\&=-V_x\cdot\underbrace{\frac{R_{13}}{R_{13}+R_{14}}\cdot\left(1+\frac{R_{13}}{R_{14}}\right)}_\frac{R_{13}}{R_{14}}+V_y\cdot\frac{R_{16}}{R_{15}+R_{16}}\cdot\left(1+\frac{R_{13}}{R_{14}}\right)\\\\&=-V_x\cdot\frac{R_{13}}{R_{14}}+V_y\cdot\frac{R_{16}}{R_{15}+R_{16}}\cdot\left(1+\frac{R_{13}}{R_{14}}\right)\end{align*}$$

Not sure if that addresses your question. But it's the best guess I have at this time.

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Your question is not very precise.

I can only guess, what you want. The opamp circuit you have drawn delivers:

Vo = ky * Vy - kx * Vx

Would you like to know how to calculate ky and kx from R13 - R16 - or even, when kx = ky?

The special case kx = ky happens always when:

R13 / R14 = R16 / R15

then even

kx = R13 / R14
ky = R16 / R15

You'd like to know how to convert your complex start formula until you get this reduction? I did it here:

enter image description here

With that we get the circuit parametrization:

k = R13 / R14 = R16 / R15

and

Vo = k * (Vy - Vx)

schematic

simulate this circuit – Schematic created using CircuitLab

PS: if you choose kx not the same as ky then the formula stays complex!

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