If I build a resistor network where the op amp has a lower gain, it is able to maintain its gain for a larger bandwidth. Why?

  • \$\begingroup\$ because it needs to change from A to B with less 'effort' so it can change from A to B faster than if the change was larger (higher gain, effort, "distance travelled") \$\endgroup\$
    – KyranF
    Oct 15, 2014 at 14:21
  • \$\begingroup\$ Outside of feedback-controlled systems, think of it in terms of the slew rate required to produce a larger output signal. \$\endgroup\$ Oct 15, 2014 at 18:50

3 Answers 3


This is called constant gain-bandwidth product but it isn't true for every op amp. It is only true for voltage feedback op amps which use dominant pole compensation for stability. Such op amps can be approximated as a first order system since one pole dominates all others and the others can be ignored. (However, this is not true of current feedback op amps since current feedback op amps do not have a constant gain-bandwidth product.)

A first order system has a transfer function of the form

$$H(j\omega) = \frac{H_0}{j\omega\tau + 1} = \frac{H_0}{j\omega/\omega_c + 1}$$

where \$H_0\$ is the DC and passband gain, \$\tau\$ is the time constant of the dominant pole and \$\omega_c\$ is the cutoff frequency (bandwidth). The gain of this system is

$$|H(j\omega)| = \frac{H_0}{\sqrt{(\omega/\omega_c)^2+1}}$$

For \$\omega << \omega_c\$ the gain is approximately \$H_0\$ and the bandwidth does not come into play. If \$\omega >> \omega_c\$ the gain-bandwidth product can be approximated as

$$|H(j\omega)|\omega = \frac{H_0}{\sqrt{(\omega/\omega_c)^2+1}}\omega \approx \frac{H_0}{\sqrt{(\omega/\omega_c)^2}}\omega = H_0\omega_c$$

which is a constant. Since it is a constant, increasing the gain requires a decrease in the bandwidth while decreasing the gain allows an increase in the bandwidth.

  • \$\begingroup\$ Understood perfectly. The math checks out thx! \$\endgroup\$
    – Raaj
    Oct 21, 2014 at 5:10

Op amps are compensated with a dominant pole. That means the open loop gain rolls off at a constant 20dB/decade vs. frequency. Negative feedback increases the input impedance, decreases the output impedance and increases the bandwidth. Because of the single pole rolloff, the product of noise gain (or non-inverting gain) and bandwidth are constant. Another nice feature of dominant-pole compensation is that the amplifier will be stable at any closed-loop gain.

So if your amplifier has a dominant pole at 10Hz and an open loop gain of 100dB your gain*bandwidth will be 1MHz (10*100,000). So at a gain of 1000 you will have a 1KHz bandwidth.

  • \$\begingroup\$ I like to add that negative feedback not always does increase the input impedance. It depends on the feedback scheme: voltage-controlled current feedback (inverting opamp with decreased input impedance) or voltage-controlled voltage feedback (non-inverting operation with enlarged input impedance). \$\endgroup\$
    – LvW
    Oct 16, 2014 at 7:14
  • \$\begingroup\$ @LvW Good point- But it depends on how you define input impedance. The open loop impedance of an op-amp is measured from the non-inverting terminal, and by adding negative feedback you will increase the impedance from that terminal, (the "noise gain" input impedance), even though that is not the signal "input" terminal in an inverting op-amp. \$\endgroup\$
    – John D
    Oct 16, 2014 at 22:29
  • \$\begingroup\$ @JohnD-why do you define the open-loop impedance at the non-inv. terminal only? The opamp has two inputs with equal (similar) input impedances. Applying feedback we have an amplifier with a signal input which is used for defining the amplifier`s input impedance, am I wrong? \$\endgroup\$
    – LvW
    Oct 17, 2014 at 8:25
  • \$\begingroup\$ @LvW Well, we're discussing gain-bandwidth which only applies to the non-inverting gain. An inverting gain of -1 is still a gain of 2 for gain-bandwidth purposes. That's because in control theory for negative feedback you have some portion of the output fed back to the input (the "beta", which is our feedback network) that's fed to the inverting input of a summer (the diff pair in the op-amp) and compensated and gained up by the forward gain path. The non-inverting input of the summer is defined as the input to the control system. \$\endgroup\$
    – John D
    Oct 17, 2014 at 16:08
  • \$\begingroup\$ In the original question an opamp and a resistive feedback is mentioned - nothing else (neither inv. nor non-inv.). And I am sure you will agree that also the inverting gain is inverse prop. to the corresponding bandwidth. Hence, the question does not discriminate between inv. and non-inv. operation. And - the inv. input has an input impedance that is lowered due to feedback (principle of voltage-controlled current feedback always reduces input impedances). That`s all I wanted to mention. \$\endgroup\$
    – LvW
    Oct 17, 2014 at 20:27

In a first-order system, the product of the gain and bandwidth is constant. This is simply a consequence of the fact that the gain is proportional to R (typically it is some kind of gm time R) and the bandwidth is inversely proportional to R (the bandwidth is some variant of 1/RC).

So, increasing R makes the gain go up, but the bandwidth go down, in equal amounts.

Simple as that.

(please note this is only true for a first-order system or a system such as a closed-loop opamp amplifier that can be well approximated as a closed loop system).


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