Slew rate expressions and constant GBP

Slew rate

Slew rate (SR) is the maximum rate of change of output voltage per unit time that an op-amp can produce, SR = (dVo/dt)max, usually in V/µs (0.5 V/µs for the 741).

Cause: inside the op-amp, a limited current Imax charges the compensation capacitor C (30 pF in the 741). Since i = C·dv/dt:

SR = (dVo/dt)max = Imax / C
741: SR = 15 µA / 30 pF = 0.5 V/µs

Slew rate and sine waves: for vo = Vm sin(2πft),

dvo/dt = 2πf·Vm·cos(2πft)
(dvo/dt)max = 2πf·Vm
No distortion if  2πf·Vm ≤ SR
Full-power bandwidth: fmax = SR / (2π·Vm)
Maximum output peak:  Vm   = SR / (2πf)

If the required rate is larger than SR, the output becomes a triangular wave with smaller amplitude (slew-rate distortion). Example: 741 at 10 V peak gives fmax = 0.5 × 10⁶/(2π × 10) ≈ 7.96 kHz. For a step input the output rises as a ramp of slope SR, e.g. a 10 V step takes 20 µs.

Proof that gain-bandwidth product 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.