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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 Feb 23 '15 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\$ – hkBattousai Feb 23 '15 at 22:41
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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\$ – George Herold Feb 24 '15 at 0:58

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