The closed-loop gain of op-amp is calculated by the ratio of Vout/Vin. What about the open-loop gain? How does the value of open-loop gain and closed-loop gain affect the performance of op-amp ? What is the relationship between open-loop and closed-loop gain of op-amp?


5 Answers 5


Closed loop gain is the gain that results when we apply negative feedback to "tame" the open loop gain. The closed loop gain can be calculated if we know the open loop gain and the amount of feedback (what fraction of the output voltage is negatively fed back to the input).

The formula is this:

$$ A_{closed} = \frac{A_{open}}{1 + A_{open} \cdot Feedback} $$

The open-loop gain affects the performance generally like this. Firstly, look at the above formula. If the open loop is huge, like 100,000, then the 1 + does not matter. \$A_{open} \cdot Feedback\$ is a large number, and it doesn't matter whether or not we add 1 to this large number: it is like a drop in a bucket. Thus the formula reduces to:

$$ \begin{align} A_{closed} &= \frac{A_{open}}{A_{open} \cdot Feedback} \\ &= \frac{1}{Feedback}\\ \end{align} $$ So, with a huge open-loop gain, we can easily get the closed loop gain if all we know is the negative feedback: if it just the reciprocal. If the feedback is 100% (i.e. 1) then the gain is 1, or unity gain. If the negative feedback is 10%, then the gain is 10. With a huge open-loop gain, we can precisely set up gains: as precisely as we care to design and build our feedback circuit. With open-loop gain which is not that large, we may not be able to ignore that 1 +. All the more so if \$Feedback\$ is small.

Okay, so far that's more of an issue of clean math and design convenience. Big open loop gain: closed loop gain is simple. But, practically speaking, small open-loop gains means that you must use less negative feedback to achieve a given gain. If the open loop-gain is a hundred thousand, then we can use 10% feedback to get a gain of 10. If the open loop gain is only 50, then we must use much less negative feedback to get a gain of 10. (You can work that out with the formula.)

We generally want to be able to use as much negative feedback as possible, because this stabilizes the amplifier: it makes the amplifier more linear, gives it a higher input impedance and lower output impedance and so on. From this perspective, amplifiers with huge open loop gains are good. It is usually better to achieve some necessary closed loop gain with an amplifier that has huge open loop gain, and lots of negative feedback, than to use a lower gain amplifier and less negative feedback (or even just an amplifier with no negative feedback which happens to have that gain open loop). The amp with the most negative feedback will be stable, more linear, and so on.

Also note that we don't even have to care how huge the open loop gain is. Is it 100,000 or is it 200,000? It doesn't matter: after a certain gain, the simplified approximate formula applies. Amplifiers based on high gain and negative feedback are therefore very gain-stable. The gain depends only on the feedback, not on the specific open-loop gain of the amplifier. The open loop gain can vary wildly (as long as it stays huge). For instance, suppose that the open loop gain is different at different temperatures. That does not matter. As long as the feedback circuit is not affected by temperature, the closed-loop gain will be the same.

  • \$\begingroup\$ the open loop gain is determined by the IC inside the op-amp, is it? we just can determine the closed-loop gain by using Vout/Vin where it is determined by the input resistor and feedback resistor, is it? \$\endgroup\$
    – nee
    Oct 10, 2012 at 6:34
  • 2
    \$\begingroup\$ The open loop is indeed determined by the IC inside an op-amp. Although op-amps do not have many stages, they use active loads instead of passive load resistors to achieve great gains, which multiply to large numbers over just a few stages. The simple resistor calculations only work accurately because op-amps have such high gains. They are linked to the 1/feedback formula. The terminology you're using for the resistors suggests that you're visualizing an inverting op-amp configuration. This is a little tricky because the input and feedback mix to the same virtual ground at the - terminal. \$\endgroup\$
    – Kaz
    Oct 10, 2012 at 6:43
  • 1
    \$\begingroup\$ Take a look at noninverting stages first. They are simple. Input and feedback are totally separate. Input goes to +, feedback goes to -. The feedback fraction is given trivially by the voltage divider: the output is dropped over two resistors, R1 and R2. The feedback is the ratio between R2 / (R1 + R2). Since gain is 1/feedback, the gain must be (R1 + R2) / R2, or 1 + R1/R2. \$\endgroup\$
    – Kaz
    Oct 10, 2012 at 6:47
  • \$\begingroup\$ what the inverting stages? \$\endgroup\$
    – nee
    Oct 10, 2012 at 6:53
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    \$\begingroup\$ You might want to start new questions for these instead of going through comments. The inverting configuration has a different input impedance also. Common-mode gain is a non-ideal behavior of real op-amps. If we send the same input to both + and -, there is some amplification, although smaller than the differential gain. In an ideal op-amp, there would be no common mode gain. This is what the CMRR (common-mode rejection ratio) is all about. en.wikipedia.org/wiki/Common-mode_rejection_ratio \$\endgroup\$
    – Kaz
    Oct 10, 2012 at 16:33

My answer covers the non-inverting as well as the inverting opamp-based amplifier.


  • \$A_{OL}\$ (ope-loop gain of the opamp)
  • \$A_{CL}\$ (closed-loop gain with feedback)
  • \$H_{IN}\$ (input damping factor)-
  • \$H_{FB}\$ (feedback factor).

For resistive feedback: \$H_{FB} = \dfrac{R1}{R1+R2}\$

A) Non-inverting

Because the input voltage is directly applied to the summing junction (differential input) the classical feedback formula from H. Black applies:

\$A_{CL} = \dfrac{A_{OL}}{1+H_{FB} \cdot A_{OL}} = \dfrac{1}{\dfrac{1}{A_{OL}} +H_{FB}} \$

For \$ A_{OL} >> H_{FB}\$ we have

\$ A_{CL} = \dfrac{1}{H_{FB}} = 1+ \dfrac{R2}{R1} \$

B) Inverting

Because now the input voltage is NOT applied directly to the summing junction (diff. input pair) but through a resistive voltage divider to the inverting terminal the input voltage is correspondingly reduced before the formula for Acl may be applied. Because of the superposition rule we set (assuming \$V_{OUT} =0\$)

\$H_{IN} = \dfrac{-R2}{R1+R2}\$

Hence we have:

\$A_{CL} = \dfrac{H_{IN} \cdot A_{OL}}{1+H_{FB} \cdot A_{OL}} = \dfrac{H_{IN}}{\dfrac{1}{A_{OL}} +H_{FB}}\$

For \$A_{OL} >> H_{FB} \$ we have

\$A_{CL} = \dfrac{H_{IN}}{H_{FB}} = - \dfrac{\dfrac{R2}{R1+R2}}{\dfrac{R1}{R1+R2}} =- \dfrac{R2}{R1} \$

C) Final remark: Taking into account that the feedback factor acts back to the negative (inverting) opamp input the product \$ -H_{FB} \cdot A_{OL}\$ is defined as the loop gain.

EDIT: "How does the value of open-loop gain and closed-loop gain affect the performance of op-amp ? "

D) The following answer concerns the availabel bandwidth for the non-inverting amplifier as a function of the open-loop bandwidth Aol (real opamp):

In most cases, we can use a first order lowpass function for the real frequency dependence of the open-loop gain:


Thus, based on the expression for Acl (given under A) we can write


With 1/Ao<< Hfb and 1/Hfb=(1+R2/R1) we arrive (after suitable re-arranging) at


The expression in brackets is a first order lowpass function having the corner frequency


Hence, the due to negative feedback the bandwidth wo (open-loop gain) is enlarged by the factor AoHfb.

More than that, we can write


This is the classical constant "Gain-Bandwidth" product (GBW) which can be written also as

w1/wo=Ao/Acl(ideal) .


It can be helpful to think of this in terms of excess gain, that being the difference between open loop and closed loop gains. For example, if the open-loop gain is 100,000 and the closed-loop gain is 10, the difference is 99,990 or nearly 100 dB. (Read this essay if it is not clear how I converted gain to dB.) If the closed-loop gain is 1,000 instead, that barely reduces excess gain, because the difference is still very large. You have to get within a factor of 10 difference in this case to reduce the difference to below 99 dB.

The open-loop gain of this example amplifier is so high that we can just call the excess gain 100 dB for all practical purposes.

This excess gain contributes to an improvement in performance parameters. As an example, if the offset voltage of the amplifier is 30 mV and you have an excess gain of 60 dB, the offset voltage of the closed loop system would be improved by a factor of 1000 to 30 µV. But one must take into account the frequency of operation, as the open loop gain has difference dominant poles and zeros, so if you are operating significantly close to those the explanation becomes less simple.

Also, the concept of open loop gain only applies to voltage feedback, voltage mode amplifiers. Norton amplifiers, current feedback amplifiers, and OTA based op-amps (like CCI and CCII class amplifiers) have different nuances to their limitations.


Open loop gain is determined by the gain characteristics of the internal devices and the internal circuit, and for an OP amp can be in the hundreds of thousands. Closed loop gain is determined by the external circuit, trivially the ratio of the input and feedback resistors.


open loop voltage gain of an operational amplifier is the gain obtained when no feed back is used in the circuit. open loop voltage gain usually exceedingly high.infact an applied an operational amplifier has infinite open loop volage gain.


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