I was reading wikipedia page of Gain-Bandwidth Product and got struck at one particular statement.

For devices such as operational amplifiers that are designed to have a simple one-pole frequency response, the gain–bandwidth product is nearly independent of the gain at which it is measured; in such devices the gain–bandwidth product will also be equal to the unity-gain bandwidth of the amplifier (the bandwidth within which the amplifier gain is at least 1).

I am unable to understand why only one-pole frequency response devices can have constant Gain-Bandwidth Product? Why can't a device with two pole have constant gain-bandwidth product? Can someone give me mathematical and intuitive explanation?

enter image description here

In the above picture, after looking at frequency response of an op-amp, it looks like it have two-poles rather than one-pole.

Also, the gain curve (with orange pen) is of op-amp with negative feedback. Does it mean negative feedback just moves the location of poles farther (away from each other) and hence, bandwidth is increased at expense of gain?

  • 1
    \$\begingroup\$ A one-pole system, given the gain is constant from DC out to the pole, gives 2:1 drop in open-loop-gain when the frequency is increased 2:1 which product (1/2:1 * 2:1 ) is constant. \$\endgroup\$ Commented Nov 28, 2018 at 9:52
  • \$\begingroup\$ You should be aware that that figure is just godawful, and it's the first time I've ever seen that sort of thing on a GBW discussion. That you have troubles with it is not surprising, and I suggest you find a different source of information on this particular subject. \$\endgroup\$ Commented Nov 28, 2018 at 14:30
  • \$\begingroup\$ I worked out the mathematical explanation in my answer to a related question. \$\endgroup\$
    – Null
    Commented Nov 28, 2018 at 15:02

2 Answers 2


Here's what a single-pole op-amp open-loop gain looks like (blue trace): -

enter image description here

Picture carved up from here.

I have marked two points with blue circles: -

  • G = 134 dB at 2 Hz
  • G = 0 dB at 10 MHz

If you convert 134 dB to real numbers it is a gain of 5,011,872 (or near enough 5 M). This means the GBWP is 10,000,000.

And at a frequency of 10 MHz, the gain is 0 dB or unity.

Any point along that sloping line will have a GBWP of ten million.

So, providing the open-loop response of the op-amp rolls-off as a single order low-pass filter, we can say GBWP is constant at ten million.

If the roll-off rate were doubled by introducing a second pole you would get the purple gain curve shown below: -

enter image description here

Now the GBWP is not constant at all because the rate at which the roll-off occurs is much steeper. The picture you have in your question is basically wrong - it doesn't show the open-loop gain of any op-amp I've ever seen because there is a high pass filter involved and that means the DC gain falls to zero.


In that diagram, there is one high frequency pole.

As the frequency doubles, the gain halves. As the frequency increases by a factor of n, the gain decreases by the same factor n. So the factor n cancels out when you multiply frequency by gain.

If the amplifier had two high frequency poles, the gain would fall by \$n^2\$ as the frequency increased by n, and the factors would not cancel.

Bonus comment -

There's a filter response which falls at 3dB per octave, what you would get if you had a 'half a pole' response. This has constant output power * bandwidth product, as the (voltage) gain falls by \$\sqrt{n}\$ for each increase of n in the frequency, so the power output (voltage squared) drops as n. This is often used in audio testing to shape white noise into 'pink noise', so that the power in each decade is constant.

  • \$\begingroup\$ @Nell_UK: In your bonus comment, do you refer to fractional-order techniques? \$\endgroup\$
    – LvW
    Commented Nov 28, 2018 at 10:43
  • \$\begingroup\$ @LvW I said, and watch my lips very carefully here, what you would get if you had a (note the use of quotes) 'half a pole' response. It's generally realised with a number of RC series connections all in parallel, either driven by another resistor or more usually in the feedback of an opamp, to approximate the response over a defined bandwidth (usually audio). The more RCs used, the better the approximation. Google for 'pink noise RC filter'. \$\endgroup\$
    – Neil_UK
    Commented Nov 28, 2018 at 10:56
  • \$\begingroup\$ I know about these techniqies based on many RC sections. It was my only intention to learn what you were referring to. Anyway, I think you have answerd the main question . \$\endgroup\$
    – LvW
    Commented Nov 28, 2018 at 11:24

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