I was trying to derive the non-inverting amplifier equation and I ran into some trouble deriving it.

non-inverting amplifier

By making the voltage between R1 and R2 VRef, I set my equation to

Vin - VRef = Vout.


VRef = Vout(R1/(R1+R2)), this means

Vin - Vout(R1/(R1+R2)) = Vout.

By simplifying the equation and moving Vout to the other side of the equation, I am getting

Vin = Vout (1 + R1/(R1+R2)) which simplifies to

Vin = Vout (2*R1 + R2)/(R1+R2).

When I then move it over to the other side of the equation, I get

Vin (R1+R2)/(2*R1 +R2) = Vout

and I am stuck with an ugly fraction I can't seem to simplify or reduce. Does anyone have any advice on how to simplify this fraction to (1 + R2/R1)?

  • \$\begingroup\$ Just a note, since you received pretty good answers: don't take equations as good without questioning yourself. Why should the first be true? What does the OPAMP do? \$\endgroup\$
    – clabacchio
    Oct 24, 2012 at 18:35

4 Answers 4


To derive this properly from basic principles, you have to work with the amplifier's open loop gain. You seem to be confusing open loop and closed loop gain, or perhaps misunderstanding how op-amps work. The output voltage is not the difference between + and -. It is the difference between + and - multiplied by the open loop gain.

In amplifying configurations of op-amps, the difference between + and - is always very tiny! Fractions of a millivolt. It is so small, that we can understand many circuits with the helpful simplifying assumption that the voltage is the same at + and -. This very tiny voltage difference between + and - is multiplied by the huge open-loop gain to produce the output. It is feedback which calibrates this tiny differential voltage so that a reasonable output voltage is obtained in spite of the huge gain.

So, let us define some variables:

$$A_0 = open\ loop\ gain$$ $$V_+ = voltage\ at\ +\ terminal$$ $$V_- = voltage\ at\ -\ terminal$$ $$V_{out} = output\ voltage$$

Now, we have:

$$V_{out} = A_o (V_+ - V_-)$$

But, since we have a feedback path, the voltage at the - terminal is established by the output voltage by the voltage divider:

$$V_- = {R1\over{R1 + R2}} V_{out}$$

For simplicity, let us reduce this fraction formed by the resistances to a single variable that we call f, for feedback:

$$f = {R1\over{R1 + R2}}$$ $$V_- = f V_{out}$$

So now we can substitute this V- into the first formula:

$$V_{out} = A_o (V_+ - f V_{out})$$

Output voltage is the V+ voltage, minus the feedback voltage, scaled by the open loop gain. We factor in the Ao and then get Vout terms together:

$$V_{out} = A_oV_+ - fA_oV_{out}$$

$$V_{out} + fA_oV_{out} = A_oV_+$$

$$V_{out} (1 + fA_o) = A_oV_+$$

$$V_{out} = {A_oV_+\over 1 + fA_o}$$

Now the next step requires us to make an assumption: the amplifier's open-loop gain Ao is very large, like 100,000 or more. What this means is that the 1 + on the bottom makes no difference, because fAo is a large number (unless f is a very small number, but we are only interested in using significant feedback, rather than negligible feedback). So what we do is simply remove the one:

$$V_{out} = {A_oV_+\over fA_o}$$

Now the open loop gain on top and bottom cancels out, leaving us with:

$$V_{out} = {V_+\over f}$$

The output voltage is the input voltage divided by the feedback. If the feedback is 1/5, the output voltage is five times the input voltage, et cetera. Now one more step: substitute the resistor fraction for f:

$$V_{out} = {V_+\over {R1\over{R1 + R2}}}$$

$$V_{out} = {{R1 + R2\over R1}}V_+$$

And of course

$${{R1 + R2\over R1}} = 1 + {R2\over R1}$$

which is what you're looking for.

Do not ever forget that this simple formula relating input and output voltage only works because both the feedback and the open loop gain are so large that we're able to ignore the 1 in 1 + fA0. This assumption can break. For instance, at higher and higher frequencies, op-amps have less and less open loop gain. At some frequency, the open loop gain drops all the way down to 1, and then drops some more at even higher frequencies.

  • 1
    \$\begingroup\$ ++ Nicely done. \$\endgroup\$
    – markrages
    Oct 24, 2012 at 20:29
  • \$\begingroup\$ Thank you for clarifying this. I see where I went wrong was I forgot to multiply (V+ - V-) by the open loop gain A which arrives to the right answer. Since a lot of the websites I looked at solve the equation with the current through R1 and R2, I wanted to verify if it could be solved with a voltage divider approach. \$\endgroup\$ Oct 25, 2012 at 15:25
  • \$\begingroup\$ Looking at R1 and R2 current here is completely counter-intuitive and unhelpful here. For one thing, the current through R1 and R2 depends on their absolute values, but the feedback doesn't. If the op-amp is ideal, it has infinite input resistance, and so R1 and R2 can be arbitrarily large without being disturbed. If R1 and R2 are arbitrarily large, then almost no current flows through them. On the other hand, currents are very useful for understanding the inverting stage, because the input sinks current. \$\endgroup\$
    – Kaz
    Oct 25, 2012 at 15:35

You have negative feedback, so the op-amp is going to try to drive the output to where the negative input matches the positive input: $$ V_{in} = V_{ref} $$

The current through R1 can be found using ohm's law. $$ I_1=\frac{V_{ref}}{R_1} $$

The current through R1 has to also be the current through R2, since no current can flow in or out of the input of an ideal op amp. $$ I_2=I_1\\ $$

The relationship between the output voltage and Vref can be found with ohm's law: $$ V_{out}=V_{ref}+I_{2}R_{2} $$

Substitute the first three into the fourth, and you get $$ V_{out}=V_{in}+\frac{V_{in}}{R_1}R_{2} $$

Factor out Vin: $$ V_{out}=V_{in}(1+\frac{R_{2}}{R_1}) $$

Real-world caveat: an op-amp also has power rails, not shown in your schematic. The op-amp output will always stay between these rails. If the amp rails out, the first equation above ceases to be true.


Seems like Vref in the O.P. is the voltage on the inverting input. (Not the best choice of a variable name, by the way. ref implies a known fixed preset voltage. V(-) would look better.)

The first equation in the O.P. is incorrect.

The output attempts to do whatever is necessary to make the voltage difference between the inputs zero. This is on of the ideal OpAmp rules. Therefore,

Vref = Vin or, in a different naming convention V(-) = V(+)

The second equation in the O.P. is correct. It comes from the voltage divider formula. Combine the two equations into a system

\$\begin{cases} V_{ref}=V_{in} \\ V_{ref} = V_{out}(R_1/(R_1+R_2)) \end{cases}\$

and solve for Vout/Vin


Your very first equation is wrong; it should be Vout = A(Vin - Vref)

This can be rewritten as Vout/A = Vin - Vref

With opamps, A is assumed to be very large (infinite), so Vout/A is essentially 0. This means that the equation simplifies to

0 = Vin - Vref

or simply, Vin = Vref

Now, if you substitute Vref = Vout × R1/(R1 + R2), you should get the result you're looking for.


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