Can somebody explain to me how the gain-bandwidth product is constant in an amplifier or op-amp? I want to know the mathematical expression behind it if possible.

As said here: https://en.wikipedia.org/wiki/Gain%E2%80%93bandwidth_product

  • \$\begingroup\$ What resources have you already looked at and what specific issue are you trying to understand within that context? (In a number of amplifiers, it is not a constant). \$\endgroup\$ Apr 1, 2021 at 10:36
  • \$\begingroup\$ @PeterSmith en.wikipedia.org/wiki/Gain%E2%80%93bandwidth_product \$\endgroup\$
    – Sayan
    Apr 1, 2021 at 10:37
  • \$\begingroup\$ You may have to elaborate the question (not in comments) with what you have understood/found out so far. \$\endgroup\$
    – Mitu Raj
    Apr 1, 2021 at 10:42
  • \$\begingroup\$ Read the section AC imperfections: en.m.wikipedia.org/wiki/Operational_amplifier , if you are familiar with frequency response of "integrators", you should be able to figure out why. \$\endgroup\$
    – Mitu Raj
    Apr 1, 2021 at 10:51
  • 1
    \$\begingroup\$ It is a direct consequence of the open-loop gain having the form of a single-pole or an integrator (doesn't matter which). I.e. A_OL=GBW/jω .. Here's my answer to a similar question with a quick walk-thru of the math \$\endgroup\$
    – Pete W
    Apr 2, 2021 at 2:21

1 Answer 1


can somebody explain to me how the gain and bandwidth product is constant in an amplifier or OP AMP ?

Just look at the open loop gain of a typical op-amp: -

enter image description here

Mathematically this happens because an op-amp internally is equivalent to a DC gain stage with high gain followed by a single order low pass filter and, as we know with a single order low pass filter, the amplitude reduces above the 3 dB point at a rate proportional to frequency i.e. ten times the frequency means one-tenth the amplitude.

It's all embodied in a simple RC low-pass filter when we move beyond the cut-off frequency.

An RC low pass filter has a transfer function of: -

$$\dfrac{1}{1+j\omega RC}$$

And, when you get beyond the cut-off frequency the transfer function becomes asymptotic with \$\dfrac{1}{j\omega RC}\$ hence, gain is inversely proportional to frequency.

That slope has a constant gain-bandwidth product.

  • \$\begingroup\$ that is fine, but it is based on a graphical method which may not rigorous all the time, is there any mathematical derivation of the statement? also is it fair to expect that all the amplifier posing band width product constant would have the same kind of (geometrically similar at least) frequency response curve as the one you have provided? \$\endgroup\$
    – Sayan
    Apr 1, 2021 at 10:43
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    \$\begingroup\$ The roll-off of amplitude against frequency is totally based on the maths of a single order low pass filter (as I said in my answer). It's not based on a graph method but I used a graph to make things easier to understand i.e. a picture speaks a thousand words. Most op-amps will adhere to the 20 dB/decade slope characteristic graph (and math) as shown. There will be some exceptions i.e. those there are not general purpose and those that are not unity gain stable but, in the main, all general purpose op-amps will comply @Sayan. \$\endgroup\$
    – Andy aka
    Apr 1, 2021 at 10:48

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