# Operational amplifier output voltage calculation

This problem asks for $$\ v_O \$$ in terms of circuit problems.

I solved the problem. But my solution has

$$\ \large v_O = -R_4 (I_1 + \frac {V_1}{R_2} + \frac {V_2}{R_3}) \$$

$$\ I_1 \$$ term as you'd see.

But according to solution manual it is like that.

I wonder whether my solution is correct.

• Try to use a superposition principle and you should be able to see why I1 don't influence the output voltage. – G36 Jun 26 '19 at 18:39
• Thanks. If we apply superposition V1 would be short circuit. If I am not mistaken. – Erdem Jun 26 '19 at 18:59
• Yes, exactly. V1 will "sink" all I1 current. – G36 Jun 26 '19 at 19:01
• It doesn't matter what you put across V1, it will not affect the output voltage. It's an ideal source so the voltage across it is independent of what it's attached to. The exception is another ideal voltage source of a different voltage. In that case the result is undefined or maybe the universe collapses into a tennis ball. – John D Jun 26 '19 at 19:44

Lets do the superposition method from Helmholtz:

For only $$\I_1\$$, the circuit is:

simulate this circuit – Schematic created using CircuitLab

The short-circuit, which replaces source $$\V_1\$$, short-circuits $$\I_1\$$. If you also assume that the OpAmp is ideal, the output voltage is zero.

For only $$\V_1\$$, the circuit is:

simulate this circuit

Again, as we have an ideal OpAmp, $$\V_{R3}\$$ and \\$V_{R1} is zero.

Therefore:

$$\ V_1=V_{R2} \$$

$$\I_{R2} = \frac{V_1}{R_2} = I_{R4}\$$

$$\V_{out}=-V_{R4}=-R_4 \cdot I_{R4} = -V_1 \cdot \frac{R_4}{R_2}\$$

For only $$\V_2\$$, the circuit is:

simulate this circuit

As we have an ideal OpAmp again, the voltage between "-" and "+" is zero and therefore:

$$\ V_2 + V_{R3} = 0\$$

$$\V_{R2} = 0 \$$ and thus $$\I_{R2} = 0\$$

$$\I_{R3}=I_{R4}=\frac{V_2}{R_3}\$$

$$\V_{out}=-V_{R4}=-R_4\cdot I_{R4}=-V_2 \cdot \frac{R_4}{R_3}\$$

Adding the results for $$\V_1\$$ and $$\V_2\$$ together, we get the final result:

$$\ V_{out}=-V_1 \cdot \frac{R_4}{R_2} -V_2 \frac{R_4}{R_3}\$$

I hope I could help you. If you still have questions, feel free to write a comment.

Regards,

Daniel

$$\I_1\$$ should have nothing to do with it. Ohms law directly tells you the current through $$\R_2\$$ and $$\R_3\$$. Therefore, you know the current through $$\R_4\$$ and thus, the output voltage.