For a non-ideal op-amp, I'm trying to troubleshoot the signs of my derivation for this op-amp equation by using superposition.

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According to this, the output is:

$$ V_o = -I_{B2}R_3+R_2\left(I_{B1}-\cfrac{I_{B2}R_3}{R_1}\right) $$

But through my derivation with the superposition theorem, I got:

By opening \$I_{B2}\$, no current passes through the + input hence the voltage is zero there. The voltage for both inputs, \$U_x = 0\$, is therefore 0, and no current passes through R1 (since is 0V left and right of R1). Since - input of the op-amp is 0V, current flows from \$V_o\$ to the input and is imposed to be IB1. Therefore:

$$u_x = 0, V_o = R_2I_{B1}$$

Now by opening the circuit in IB1, no current passes there. The input at the positive input is IB2*R3, which is \$U_x\$: the voltage also at the negative input of the op-amp. So \$U_x = IB2*R3\$.

Now it comes to the part I'm probably confused about: \$U_x\$ is at the negative input of the op-amp. Writing the node equation:

$$ \cfrac{U_x-V_o}{R_2} + \cfrac{U_x}{R_2} = 0, U_x = R_3I_{B2} $$

I get the final result

$$ V_o = R_2I_{B1}+R_3I_{B2}\left(1+\cfrac{R_2}{R_1}\right) = I_{B2}R_3 + R_2\left(I_{B1}+\cfrac{R_3}{R_1}I_{B2}\right) $$

I'm clearly mistaken by a minus sign - maybe at this last step of considering \$U_x\$ and having to do with its current direction, but I'm not able to make it make sense to me. IB1 is draining the current in that direction, which is coherent with IB2 current.


1 Answer 1


Is your problem to do with the fact that the current through R2 can flow in either direction and which direction it flows in depends upon whether the output voltage is at a higher voltage than the input voltages or at a lower voltage and this depends on the value of R3. At low and zero values of R3 the output will be above the input voltages and as R3 increases, the output will move downwards to be below the input voltages which will change the sign of the R2 current equation. I would think that you need to cater for both cases with separate derivations.


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