# Related to the opamp question solution (a point I don't understand in the difference amplifier)

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

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

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.

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:

With that we get the circuit parametrization:

k = R13 / R14 = R16 / R15


and

Vo = k * (Vy - Vx)


simulate this circuit – Schematic created using CircuitLab

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