Opamp Inverting Amplifier: The case of non-inverting input not being connected to ground

Question

What is the relation between \$V_i\$ and \$V_o\$ when the non-inverting input is supplied with a non-zero voltage level?

I have shared my formula derivation below, but I can't validate it since I can't find this special case anywhere on the internet.

Note: My interest comes from the circuit in this document (page 32, figure 25).

^{simulate this circuit – Schematic created using CircuitLab}

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My work:

Let \$A\$ be the gain of the opamp at linear region, and \$\pm V_{cc}\$ is large enough not to saturate the opamp.

$$ V_o = A(V_p - V_n) $$

From node voltages method:

$$ V_n = \dfrac{\dfrac{V_i}{R_i} + \dfrac{V_o}{R_f}}{\dfrac{1}{R_i} + \dfrac{1}{R_f}} = \dfrac{R_iV_o + R_fV_i}{R_i + R_f}$$

Then we have:

$$ V_o = A\left(V_p - V_n\right) = A\left(\dfrac{(R_i + R_f)V_p}{R_i + R_f} - \dfrac{R_iV_o + R_fV_i}{R_i + R_f} \right) $$

Rearranging the terms:

$$ V_o + \dfrac{AR_iV_o}{R_i + R_f} = A\left(\dfrac{(R_i + R_f)V_p}{R_i + R_f} - \dfrac{R_fV_i}{R_i + R_f} \right) \\ \left[ \dfrac{AR_i}{R_i + R_f} + 1 \right] V_o = A\left(\dfrac{(R_i + R_f)V_p}{R_i + R_f} - \dfrac{R_fV_i}{R_i + R_f} \right) \\ \dfrac{(A+1)R_i + R_f}{R_i + R_f} V_o = \dfrac{A(R_i + R_f)V_p}{R_i + R_f} - \dfrac{AR_fV_i}{R_i + R_f} \\ \left[(A+1)R_i + R_f\right] V_o = \left[A(R_i + R_f)V_p\right] - \left[AR_fV_i\right] \\ V_o = \dfrac{A(R_i + R_f)V_p}{(A+1)R_i + R_f} - \dfrac{AR_fV_i}{(A+1)R_i + R_f} \\ $$

If the gain \$A\$ is large enough, we can write:

$$ \lim\limits_{A \to \infty} V_o = \lim\limits_{A \to \infty} \left[\dfrac{A(R_i + R_f)V_p}{(A+1)R_i + R_f} - \dfrac{AR_fV_i}{(A+1)R_i + R_f}\right] = \dfrac{R_i + R_f}{R_i} V_p - \dfrac{R_f}{R_i} V_i $$

Then the formula is:

$$ \boxed{V_o = -\dfrac{R_f}{R_i} V_i + \dfrac{R_i + R_f}{R_i} V_p} $$

If \$V_p=0\$ we get the inverting amplifier equation:

$$ V_o = -\dfrac{R_f}{R_i} V_i $$

And, if \$V_i=0\$ we get the non-inverting amplifier equation:

$$ V_o = \dfrac{R_i + R_f}{R_i} V_p = \left(\dfrac{R_f}{R_i}+1\right) V_p $$

This looks like to be a mixed case in which it both works as inverting and non-inverting amplifier.

Answer 1

You can verify it using the common techniques for ideal opamps:
1) Having negative feedback makes the voltages on both terminals equal
2) There is no current flowing into or out of the terminals.

So, if \$I\$ is the current flowing through \$R_i\$ and \$R_f\$, $$V_n=V_p$$ $$I=(V_n-V_i)/R_i=(V_o-V_n)/R_f$$ Rearranging: $$V_o=\frac{R_f}{R_i}(V_n-V_i)+V_n=\frac{R_f}{R_i}(V_p-V_i)+V_p = -\frac{R_f}{R_i}V_i+\frac{R_f+R_i}{R_i}V_p$$ Which is exactly your formula.

Answer 2

The first thing you need to do when trying to understand an unfamiliar circuit is not to analyze it blindly but to try to see some familiar (sub)circuits in it.

First, the superposition principle can help as to see an inverting amplifier (assuming Vp = 0) and non-inverting amplifier (Vi = 0).

Then, combining them, we can see an unfinished op-amp differential amplifier... in the sense that it has slightly unequal input gains for the inverting input (Rf/Ri) and non-inverting input (Rf/Ri + 1). It is interesting why the gain of the non-inverting input is one unit greater than that of the inverting input... but this is another topic.

There are two options to align the two input gains - by reducing the gain with 1 of the non-inverting input or by increasing the gain with 1 of the inverting input. The first requires an attenuator and the second - an amplifier. Of course, we choose the simpler solution and connect a correcting voltage divider with gain of Rf/(Ri + Rf) before the non-inverting input of the op-amp. Thus the non-inverting gain becomes Rf/(Ri + Rf) x (Ri + Rf)/Ri = Rf/Ri and becomes equal to the inverting gain.