This is probably more of an algebra question.

I do not understand where the "1"s come from in equation 1.10 given the much simpler voltage divider equation in the line just above it. And why do they convert a simple resistor division problem, into a division over a division problem?

The second part that loses me is the jump from equation 1.11 to 1.12. I have been trying and trying and I just cant get 1.11 to look like 1.12.

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2 Answers 2


For formula 1.10 they have taken this part of the equation: -


And divided top and bottom by \$R_1\$ to get this: -


Equation 1.11 evolves like this: -

$$V_o = a\left(V_i- \dfrac{V_o}{1+ \frac{R_2}{R_1}}\right)$$

$$V_o\left(1 + \dfrac{a}{1+\frac{R_2}{R_1}}\right) = aV_i$$

$$\dfrac{V_o}{V_i} = \dfrac{a}{1 + \frac{a}{1 + \frac{R_2}{R_1}}}$$

$$\dfrac{V_o}{V_i} = \dfrac{a(1 + \frac{R_2}{R_1})}{1 + \frac{R_2}{R_1} + a}$$

$$\dfrac{V_o}{V_i} = \left(1+\dfrac{R_2}{R_1}\right)\cdot\dfrac{a}{1+\frac{R_2}{R_1}+a}$$

$$\dfrac{V_o}{V_i} = \left(1+\dfrac{R_2}{R_1}\right)\cdot\dfrac{1}{\dfrac{1+\frac{R_2}{R_1}}{a}+1}$$

And that is equation 1.12. So, when a is very large we can approximate 1.12 to this: -

$$\dfrac{V_o}{V_i} = \left(1+\dfrac{R_2}{R_1}\right)$$

  • \$\begingroup\$ thanks for the break down. \$\endgroup\$ Feb 27, 2021 at 18:31

There is a great idea behind this famous circuit solution... and it is not just electrical but universal... and can be seen all around us. It is based on the main property of negative feedback systems (including human beings) to compensate for all kinds of harmful disturbances. In their quest to do this, they turn from followers into amplifiers.

So, from this system viewpoint, the non-inverting amplifier is a disturved follower. Let's see how the op-amp does this "magic" in your specific circuit - Fig. 1.6 (a).

If you were connected the op-amp output directly to its inverting input (without a voltage divider), it will easily make its output voltage equal to the input voltage Vi... and will act as an op-amp voltage follower.

When you insert a voltage divider between the op-amp output and the inverting input, it will "disturb" the op-amp by decreasing its output voltage (R1 + R2)/R1 times. The op-amp reacts to this "intervention" by increasing its output voltage so many times. As a result, the equality will be restored... and the voltage after the disturbance (at the inverting input) will follow the input voltage as before... but we take the voltage before the disturbance (at the op-amp output) as an output voltage.

In this way, the negative feedback has "reversed" the voltage divider with transfer ratio of R1/(R1 + R2) converting it into an amplifier with a transfer ratio of (R1 + R2)/R1... or 1 + R2/R1.

The advantage of this intuitive approach over the blind derivation of the formula is that it gives understanding and not just knowledge... and this is something very necessary in circuitry...

You may be interested in attending a lab exercise that I did with my students in 2008. During the lab, we built and investigated this circuit step by step. Then, we described it in a Wikibooks story to share it with curious people on the web.

Another Wikibooks story where I have described a part of correspondence between me and Gordon Deboo (the inventor of Deboo integrator), is dedicated to this great circuit phenomenon. Here is a text that is probably an excerpt from his book - Fig. 1.

Deboo's picture.jpg

Fig. 1. Scanned text that is probably an excerpt from a Deboo's book (Wikibooks).

And finally, here is an ResearchGate question dedicated to this topic: Can we "reverse" a voltage divider by applying the input voltage to its output and taking the output voltage from its input?

  • \$\begingroup\$ @LvW, I have moved my comment here... I think it is valid for any negative feedback circuit (system). Examples: an op-amp integrator is a reversed differentiator and v.v., an op-amp logarithmic converter is a reversed antilogarithmic converter and v.v., an analog-to-digital converter is a reversed digital-to-analog converter, the so-called trans-diode is a reversed transistor... also the input part of the BJT current mirror... But really, I need some time to answer your question what is the case of the op-amp inverting amplifier because it consists of only resistors... \$\endgroup\$ Feb 27, 2021 at 20:29
  • \$\begingroup\$ @LvW, Your comment appeared to me as a "disturbance" and I had to compensate for it like an op-amp... but with a hot bath, a sip of Jim Beam and hard thinking in the spirit of brainstorming:) Here is the result: Note that in all these examples, an imperfect converter (driven from the op-amp output) is transformed to a perfect inverse converter (driven by the input voltage source)... \$\endgroup\$ Feb 27, 2021 at 22:02
  • \$\begingroup\$ ... So, in the inverting amplifier, an "imperfect voltage divider" driven by the op-amp output from the side of R2, with a transfer ratio of R1/(R1 + R2), is transformed to a "perfect inverse voltage divider" driven by the input voltage source with a transfer ratio of only -R2/R1. It is a "perfect" circuit because the voltage drop across R2 is removed and, as a result, the input current depends only on R1... and this annoying "1" in the formula disappears. \$\endgroup\$ Feb 27, 2021 at 22:09
  • \$\begingroup\$ I must admit that I am not entirely convinced by the formalism of this model with inversion. With the non-inverting circuit it works quite well - but with the inverter? Why is R1/(R1+R2) an "imperfect" voltage divider? And why do you call R2/R1 a "perfect inverse voltage divider"? How does such a view help to better understand the principle of feedback? More than that, I think the "1" in case of a non-inv. circuit is not "annoying" but necessary as you have shown.... \$\endgroup\$
    – LvW
    Feb 28, 2021 at 9:55
  • \$\begingroup\$ @LvW, There is something symbolic and very exciting in the fact that we meet many times in different places on the web to discuss the same great circuit phenomena... I think the inverting amplifier exploits the same "reversal" idea as the non-inverting amplifier. In both circuits, the op-amp adjusts the input voltage of the R1-R2 voltage divider (across the whole resistor network) so that to keep its output voltage VR1 (across the resistor R1) equal to the input voltage. Only, there are two differences: \$\endgroup\$ Feb 28, 2021 at 14:41

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