I've seen the assumption of gain-bandwidth product being constant for some amplifiers. More specifically for a compensated op-amp.

Is there any mathematical proof to establish why the product is constant?

  • \$\begingroup\$ Constant relative to what? Frequency, temperature, closed loop gain, voltage, process variation, luck? \$\endgroup\$
    – bobflux
    Commented Apr 26, 2021 at 22:45
  • \$\begingroup\$ From batch to batch I have seen differences in GBWP for op-amps. My sample size in in the hundred of thousands of parts over 12 years for one product. \$\endgroup\$
    – qrk
    Commented Apr 26, 2021 at 23:42
  • \$\begingroup\$ Why is the product constant? Is this what you are looking for? en.wikipedia.org/wiki/Gain%E2%80%93bandwidth_product#Examples \$\endgroup\$
    – Mattman944
    Commented Apr 26, 2021 at 23:59
  • 2
    \$\begingroup\$ Does this answer your question? Why is an op amp's bandwidth higher at lower gains? \$\endgroup\$
    – Null
    Commented Apr 27, 2021 at 11:32
  • \$\begingroup\$ Constant across frequencies. \$\endgroup\$
    – Curiosity
    Commented Apr 27, 2021 at 23:03

2 Answers 2


The GBW product of an op-amp is set by internal resistances or current sources, and an internal capacitor. These are all made on silicon, and will be quite variable. The one datasheet that I checked lists the GBW product as "typical", without even giving a minimum and maximum.

Bottom line: if you're depending on the GBW product to be constant, you probably want to revisit the decisions you made designing that circuit.


Is there any mathematical proof to establish why the product is constant?

As I understand your question, you are not asking about constancy (repeatability) from one device to another, but about the constant 6 dB per octave slope of the gain-bandwidth function. If so, then ...


It is the math of an integrator, a single-pole, R-C lowpass filter, which is derived from the charge equation of a capacitor.



These sections spell out the math that show the impedance of a capacitor to be inversely proportional to the frequency of the signal going through it. If you plot the ratios of the impedance at various frequencies against the ratios of those frequencies, the log-log plot is a straight line.

  • \$\begingroup\$ and the fact that it's dominant-pole can come from "pole splitting" so OP could look that up \$\endgroup\$
    – Pete W
    Commented Apr 27, 2021 at 1:02

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