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People keep only talking about assumptions where the open loop gain for a non-inverting op-amp is so high you don't have to consider it.

But suppose for a non-inverting amplifier that the open loop gain is something like 100. What would that mean for the resistances and input/output voltages?

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    \$\begingroup\$ "what's the actual formula for open loop gain in a non-inverting op-amp?" - is that what you really want, or do you want to know the effect it has on closed loop gain? \$\endgroup\$ – Bruce Abbott Dec 3 '17 at 5:07
  • \$\begingroup\$ Is either of these answers the correct one? \$\endgroup\$ – Harry Svensson Dec 14 '17 at 4:50
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\$A\$ = open loop gain

\$V_{out}=A(V^+-V^-)\$

Let's first assume \$A=\infty\$, we'll cover A=100 later, and we got an op-amp set up as a non-inverting amplifier.

This is the equation in an ideal case: \$V_{out}=(1+\frac{R_2}{R_1})V_{in}\$ where \$R_2\$ is the feedback resistor and \$R_1\$ goes to ground.


Let's see if we can get that same answer from our first expression.

\$V^+ = V_{in}\$

\$V^- = V_{out}\frac{R_1}{R_1+R_2}\$, I hope you can see that it is a voltage divider.

\$ \begin{align}\\ V_{out}=A(V^+-V^-) \rightarrow V_{out}&=A(V_{in}-V_{out}\frac{R_1}{R_1+R_2})\\ V_{out}&=AV_{in}-AV_{out}\frac{R_1}{R_1+R_2}\\ V_{out}+AV_{out}\frac{R_1}{R_1+R_2}&=AV_{in}\\ V_{out}(1+A\frac{R_1}{R_1+R_2})&=AV_{in}\\ \\ V_{out}&=\frac{A}{1+A\frac{R_1}{R_1+R_2}}V_{in}\\ \\ V_{out}&=\frac{A(R_1+R_2)}{AR_1+R_1+R_2}V_{in}\\ \end{align} \$


"Hmmmm that doesn't look like \$V_{out}=(1+\frac{R_2}{R_1})V_{in}\$ to me", well let's use limit and let \$A \rightarrow \infty\$. In other words, let's make this op-amp ideal.

\$\lim\limits_{A \to \infty}\frac{A(R_1+R_2)}{AR_1+R_1+R_2}=\frac{A(R_1+R_2)}{AR_1}=\frac{A}{A}\frac{R_1+R_2}{R_1}=\frac{R_1+R_2}{R_1}=1+\frac{R_2}{R_1}\$

"Ahhh! There it is!". This is more of a sanity check for me since I'm so damn rusty. Let's carry on and see what we get if A = 100. The answer to your question.


\$V_{out}=\frac{A(R_1+R_2)}{AR_1+R_1+R_2}V_{in} \rightarrow V_{out}=\frac{100(R_1+R_2)}{100R_1+R_1+R_2}V_{in}\$, hmmm doesn't look like I can make it look better than that to be honest.


But let's put some numbers to it to see what would happen. Let's say \$R_1 = R_2 = 1000 Ω\$ and that \$V_{in}=1 \$ V. In an ideal case \$V_{out}\$ should be 2 V.

\$V_{out}=\frac{100(1000+1000)}{100×1000+1000+1000}×1 ≃ 1.96\$ V, hmm not too bad for an open loop gain of 100.


If you want it to behave like an ideal amplifier, knowing that \$A=100\$, then you would have to set this equation:

\$\frac{100(R_1+R_2)}{100R_1+R_1+R_2} = 2\$, or whatever gain you wish to have.

Lock one of the resistors to some value, let's lock \$R_1\$ to 1000 Ω.

\$ \begin{align}\\ \frac{100(1000+R_2)}{100×1000+1000+R_2} &= 2\\ \\ 100(1000+R_2) &= 2(100×1000+1000+R_2)\\ \\ 100×1000+100×R_2 &= 2×100×1000+2×1000+2×R_2\\ \\ 100×R_2-2×R_2 &= 2×100×1000+2×1000-100×1000\\ \\ R_2(100-2) &= 2×100×1000+2×1000-100×1000\\ \\ R_2 &= \frac{2×100×1000+2×1000-100×1000}{100-2}\\ \\ R_2 & ≃ 1040.81 Ω \end{align}\\ \$

And voilla, now it behaves as if you have infinite open loop gain. If you plug in the numbers as I did before, then you will get 2.00 V, instead of 1.96 V.

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  • \$\begingroup\$ _At voilla... _ now it doesn't behave like an ideal op-amp. You denie yourself on next lines. Infact changing values to get the same gain has nothing to do with all the "benefits" like bandwidth, immettances and sensitivities improvements that higher loop gain could take in. \$\endgroup\$ – carloc Dec 3 '17 at 14:45
  • \$\begingroup\$ @carloc Hahaha, that's actually true. Thanks for pointing that out. \$\endgroup\$ – Harry Svensson Dec 3 '17 at 15:20
  • \$\begingroup\$ Hehehe :) I am afraid we have a little misurderstanding. You cancelled the right sentence (about bandwidth, immettances and so on) and left in the wrong one (changing values to get the same gain by no means makes it behave like an ideal op-amp) \$\endgroup\$ – carloc Dec 3 '17 at 17:25
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The result will be that there is some steady state error in the output. You can look at this as a closed loop P-only control system.

Say you have the following:

schematic

simulate this circuit – Schematic created using CircuitLab

OA1 is a bad op amp with an open loop gain of 100 at DC. It will always take the differential input voltage and multiply it by the open loop gain to determine the output voltage. See also a related answer of mine.

This logic holds in any op amp circuit, even if we add feedback resistors. It will always obey the closed loop transfer function $$\frac{G(s)}{1+G(s)H(s)}$$

Where G(s) is the open loop transfer function of the op amp and H(s) is the transfer function of the feedback circuit you put around the op amp.

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    \$\begingroup\$ Shouldn't the output voltage be 0.990099V? \$\endgroup\$ – Bruce Abbott Dec 3 '17 at 5:02
  • \$\begingroup\$ In this buffer configuration, \$V_{out}= \frac{A}{1+A}V_{in}\$, if we set \$A = 100\$ and \$V_{in}=1\$ V. Then we get \$V_{out}=\frac{100}{100+1}×1=\frac{100}{101}=0.990099\$ V. So @Bruce is correct. \$\endgroup\$ – Harry Svensson Dec 3 '17 at 5:17
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The op amp's open loop voltage gain is given by equation (1),

$$ A = \frac{V_o}{V_n-V_i} \;\;\;\;\;\;\;\;\;\;(1) $$

where

A  := The op amp's open-loop gain
Vo := The op amp's output voltage
Vn := The voltage at the op amp's non-inverting input
Vi := The voltage at the op amp's inverting input

Rearrange equation (1) to solve for \$V_i\$, the voltage at the op amp's non-inverting input, as shown in equation (2):

$$ V_i = V_n - \frac{V_o}{A} \;\;\;\;\;\;\;\;\;\;(2) $$

But suppose for a non-inverting amplifier that the open loop gain is something like 100. What would that mean for the resistances and input/output voltages?

Consider the non-inverting op amp circuit shown in Figure 1:

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Op amp non-inverting voltage amplifier circuit.

The voltage \$V_i\$ at the op amp's inverting input can now be expressed two ways, as shown in equation (3):

$$ V_i = V_n - \frac{V_o}{A} = \frac{V_o\,R2}{R1+R2} \;\;\;\;\;\;\;\;\;\;(3) $$

Solving equation (3) for \$V_o\$ yields equation (4),

$$ V_o = \frac{A\,V_n(R1+R2)}{R1+R2+A\,R2} \;\;\;\;\;\;\;\;\;\;(4) $$

Equation (4) holds the mathematical answers to the questions you posed.

Note that taking the limit of equation (4) as the op amp's open loop voltage gain \$A\$ tends to zero, or toward +infinity yields the results shown in equations (5) and (6), respectively:

$$ \lim_{A\rightarrow 0} V_o(A) = 0 \;\;\;\;\;\;\;\;\;\;(5) \\[0.2in] \lim_{A\rightarrow +\infty} V_o(A) = \frac{V_n\,R2}{R1+R2} \;\;\;\;\;\;\;\;\;\;(6) $$

From equation (6) it is evident that we want the op amp's open loop voltage gain to be very high because we can then leverage the simplified voltage gain model shown on the right-hand side (RHS) of equation (6), instead of using the more complicated model shown on the RHS of equation (4).

One can perform parametric sensitivity analysis on equation (4) to determine the sensitivity of the op amp's output voltage \$V_o\$ with respect to (for example) the op amp's open loop voltage gain \$A\$, i.e., \$S_{V_o,A}\$. There are various methods for performing sensitivity analysis—e.g., absolute, relative, semi-relative, etc.—none of which can be explained here (by me) in "100 words or less".

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    \$\begingroup\$ Jim, an SE tip: if you wish to avoid the "schematics for the deaf" appearance when you have only a simple schematic you can add 's', 'm' or 'l' (small, medium or large) before the '.png' in the image URL. It works with .jpg too. \$\endgroup\$ – Transistor Dec 3 '17 at 8:09

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