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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).

schematic

simulate this circuit – Schematic created using CircuitLab


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.

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  • \$\begingroup\$ Figure 28 on page 32 maybe? \$\endgroup\$
    – Andy aka
    Commented Feb 23, 2015 at 22:27
  • \$\begingroup\$ @Andyaka I meant the page of PDF, not the page numbers seen on the page footers. F28/P32 contains the same circuit though. \$\endgroup\$ Commented Feb 23, 2015 at 22:41
  • \$\begingroup\$ "This looks like to be a mixed case in which it both works as inverting and non-inverting amplifier." The term you're looking for is "differential amplifier". And yes, you can consider the configuration to be just that. It's not the most generally useful example, since the common-mode and differential gains are different, but it is sometimes seen, especially in level shifters. \$\endgroup\$ Commented Nov 21, 2019 at 21:16

2 Answers 2

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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.

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  • \$\begingroup\$ To OP the first equation is key! With negative feedback the opamp is trying to keep it's inputs the same. \$\endgroup\$ Commented Feb 24, 2015 at 0:58
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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.

Op-amp differential amplifier - idea

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.

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