Electrical – How to calculate closed-loop bandwidth from open-loop parameters using feedback factor

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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.

Best Answer

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

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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.