Gain-bandwidth relationship; GBP is constant

Relationship between gain and bandwidth

The open-loop gain of an op-amp is very high (about 2 × 10⁵ for the 741) only at low frequencies. Because of the internal compensation capacitor it falls at −20 dB/decade above a low break frequency fo (about 5 Hz), reaching unity at ft (about 1 MHz). When negative feedback reduces the gain, the bandwidth increases by the same factor. So gain and bandwidth are inversely proportional, and their product is constant.

Gain (dB)
 106 |-------.        open loop
     |        \
  40 |---------\---.  closed loop, gain 100
     |          \  :
   0 +-----------\-:-----------> f (log)
          fo      fF  ft = 1 MHz

Proof that GBP is constant

The open-loop gain of an internally compensated op-amp (e.g. 741) has a single dominant pole at fo:

A(f) = A0 / (1 + j f/fo)
A0 = dc open-loop gain,  fo = open-loop break frequency

Apply negative feedback with feedback factor β (for a non-inverting amplifier β = R1/(R1 + Rf)):

Af(f) = A(f) / (1 + β A(f))
      = [A0/(1 + jf/fo)] / [1 + βA0/(1 + jf/fo)]
      = A0 / (1 + βA0 + jf/fo)
      = [A0/(1 + βA0)] / [1 + j f / (fo(1 + βA0))]

This is again a single-pole response, with

Closed-loop dc gain : Af0 = A0 / (1 + βA0)
Closed-loop BW      : fF  = fo (1 + βA0)
Product             : Af0 × fF = A0 × fo = constant

The factor (1 + βA0) that divides the gain multiplies the bandwidth, so gain × bandwidth = A0·fo = ft, the unity-gain bandwidth, which is fixed for a given op-amp. For the 741, A0 ≈ 2 × 10⁵ and fo ≈ 5 Hz, so GBP ≈ 1 MHz: with a gain of 10 the bandwidth is 100 kHz, with a gain of 100 it is 10 kHz.