I'm working through a past feedback analysis question for an exam and trying to calculate what the closed-loop bandwidth becomes given the open-loop gain \$|Av|=10^5\$, open-loop bandwidth \$B_{wo} = 20 \text{ Hz}\$ and a calculated feedback factor \$ \beta = 0.151515 \$. The class notes I'm given don't mention anything about how this works and neither does the textbook. From what I understand introducing negative feedback reduces the systems gain and widens its bandwidth? The answer given is that the closed-loop bandwidth becomes approximately \$303 \text{ kHz}\$. I initially thought to multiply the feedback factor with the open-loop bandwidth giving \$3.03 \text{Hz}\$, three orders of magnitude too small.

I have looked at this post Closed loop bandwidth vs open loop bandwidth
but I would like to know specifically how to calculate the closed-loop bandwidth using this feedback fraction.


3 Answers 3


Here's what you have in terms of open loop gain: -

enter image description here

You have an open-loop gain of 100,000 (100 dB) from DC to 20 Hz then it rolls off at 20 dB per decade until it reaches unity gain at 2 MHz.

With a feedback factor of 0.151515, the gain of the op-amp is the reciprocal i.e. 6.6. A gain of 6.6 is 16.4 dB hence, I've drawn an orange line across the graph at this point and it intersects the open loop gain a bit above 200 kHz.

Hopefully this makes sense now.

To get to 303 kHz you calculate the fraction of a decade above 200 kHz that the orange line intersects the open-loop line. I estimate it to be about 0.18 based on 1 - 16.4/20.

Take the antilog of 0.18 and multiply it by 200 kHz to get the real frequency number where the vertical orange line hits the base line. The answer I get is 302.7 kHz.


I'm too lazy to draw a new picture so I got a suitable one from the web:

enter image description here

Given is that \$A = 10^5\$

Open loop bandwidth: \$B_{wo} = 20 \text{ Hz}\$

Feedback: \$ \beta = 0.151515 \$

From this I can derive that the closed loop gain. As long as \$A\$ is much larger than the closed loop gain, I can use gain = \$ 1/\beta = 6.6 \$ Indeed that's much smaller than \$10^5\$ so that was a valid way to do it.

Since this will be a first order system (no other cutoff points are mentioned) above 20 Hz the open loop gain will decrease with 20 dB / decade and that means if the frequency increases with a factor \$x\$ then the gain will decrease with a factor \$x\$.

So for this kind of system the gain * bandwidth product remains constant. So at 20 Hz you'd get that \$10^5\$ times gain but at 200 Hz you'd get \$10^4\$ etc..

In this case the gain has decreased by a factor \$10^5 / 6.6 = 15151\$ so the bandwidth will increase with the same factor: \$15151 * 20 = 303 \$ kHz


Blargian - At first, I completely agree with Andy aka`s answer.

However, what will be your answer if during exam the question will arise WHY the closed-loop bandwidth will be at the frequency where the orange line intersects the open-loop gain of the opamp?

In case you know the answer, stop reading now. But if you are not sure about the answer continue the reading:

The key to the answer is a term that was not yet mentioned: LOOP GAIN LG(jw) (that is the gain of the complete feedback loop if it is open).

In your case it is simply LG(jw)=Av(jw)*beta.

The magnitude of the loop gain LG can also be found in the magnitude diagram for Av(w) (see the diagram from Andy aka) because of LG=Av(w)/(1/beta).

That means in dB: LG(w) in dB = Av(w) in dB - (1/beta) in dB.

(Comment: For a non-inverting amplifier 1/beta is identical to the closed-loop gain, see Andy aka`s nomenclature)

In the mentioned diagram, this function is shown, if you consider the orange line as a new abscissa (new frequency axis). This means: The loop gain LG (magnitude) is identical to the varying distance between the Av(w) curve and the orange line.

And it is important to realize that at the crossing point the loop gain is 0dB (equal to unity).

Why is this important? Analyze the closed-loop gain formula for negative feedback:


The magnitude of this complex function for Av(jw) will be 3 dB down (definition of closed-loop bandwidth) when the magnitude of the denominator will be SQRT(2). Hence, the closed-loop BW will be at that frequency where the magnitude of the loop gain is LG=1 (0 dB).

This is - as mentioned above - the crossing point between the orange line and the magnitude function for Av(jw).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.