What is the gain bandwidth product of an amplifier? (What does it mean?) How can I find the gain bandwidth product for an amplifier which has a gain of -100?

  • \$\begingroup\$ Do you really mean a "has gain of -100", and not "has gain of 100"? \$\endgroup\$
    – gbulmer
    Sep 13, 2014 at 17:43
  • \$\begingroup\$ It should be given in the datasheet of the Op Amp. If you use a traditional inverting amplifier (with 2 resistors) with gain of 100, and if the component's G.BW=100Mhz, your amplifier should keep its x100 gain up to around 1MHz. If it is a component with fixed ampification, you can calculate the G.BW from the maximum frequency indicated in the datasheet. \$\endgroup\$
    – Grabul
    Sep 13, 2014 at 17:45
  • \$\begingroup\$ Yes, that's how the question is framed on my textbook. Gain of -100 and yes I am using a traditional inverting amplifier with 2 resistors. \$\endgroup\$
    – Timmy
    Sep 13, 2014 at 18:00
  • \$\begingroup\$ @gbulmer a gain of -100 is easily understood as a inverting gain of 100 without any loss of generality. \$\endgroup\$ Sep 13, 2014 at 18:35
  • \$\begingroup\$ @placeholder - I thought I needed to check as the OP also used '-' as punctuation, as do I. \$\endgroup\$
    – gbulmer
    Sep 13, 2014 at 18:43

1 Answer 1


GBP is the to do with the open loop gain of the op-amp. If you have a closed-loop circuit then GBP can help you find where the flatness of the frequency response starts to be eroded.

GBP - if it has an open loop DC gain of 1 million and unity gain at 1MHz then the GBP is said to be 1 million.

This is useful to know because if you have an op-amp that has a GBP of 1 million and you have a manufactured a closed loop gain of 100 then I would expect the gain to remain flat up to about 10kHz then roll-off gently at 6 dB / octave: -

enter image description here

Here is an extract from the data sheet for the AD8606 op-amp and I've drawn four red lines on it at 10kHz, 100kHz, 1MHz and 10MHz. The line at 10MHz is important because this is the unity gain point of the op-amp i.e. it has a GBP of 10,000,000. If this was all we knew we could predict the open loop gain at 10kHz by dividing 10,000,000 by 10,000 to get 1,000 (this would be the open loop gain at 10kHz and of course a gain of 1,000 is 60 dB - exactly as seen in the graph.

  • \$\begingroup\$ It might be worth adding that the GBP is typically specified in an OpAmps datasheet. \$\endgroup\$
    – gbulmer
    Sep 13, 2014 at 17:42
  • \$\begingroup\$ +1 Even better! Small point "GBP is the to do with" might be clearer as "GBP is to do with" \$\endgroup\$
    – gbulmer
    Sep 13, 2014 at 18:38
  • \$\begingroup\$ It is hardly irrelevant - the excess gain is what - amongst many other things - determines the accuracy of circuit in closed loop configurations. Determines the maximum BW possible in the configuration (which you refer to) and also helps reduce offset error. \$\endgroup\$ Sep 13, 2014 at 18:39
  • \$\begingroup\$ @placeholder yeah "irrelevant" is a bit strong. I'll amend that. \$\endgroup\$
    – Andy aka
    Sep 13, 2014 at 19:22
  • 1
    \$\begingroup\$ The GBP is said to be 1MHz, not one million. It isn't unit-free, it is gain times -3dB bandwidth, whose units are Herz, and it coincides with unity-gain bandwidth in a one-pole amplifier. \$\endgroup\$
    – user207421
    Sep 13, 2014 at 21:06

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