Practical GBP and amplifier output resistance

Gain-bandwidth product of a practical op-amp

A practical (internally compensated) op-amp has a single dominant pole, so its open-loop gain is A(f) = A0/(1 + jf/fo). For the 741, A0 ≈ 2 × 10⁵ and fo ≈ 5 Hz. When negative feedback with factor β is applied:

Closed-loop gain: Af = A0/(1 + A0β)
Bandwidth:        fF = fo(1 + A0β)
Af × fF = A0 × fo = ft (constant)
741: GBP = 2 × 10⁵ × 5 Hz = 1 MHz

So the product of closed-loop gain and bandwidth is constant and equal to the unity-gain frequency ft. Reducing gain by a factor increases bandwidth by the same factor (gain 10 → BW 100 kHz; gain 100 → BW 10 kHz). This lets a designer find the bandwidth for a given gain: BW = GBP/gain (using the noise gain 1 + Rf/R1).

Output resistance of an amplifier using a practical op-amp

Model the op-amp by its open-loop gain A, input resistance Ri (very large, ignored) and output resistance Ro. To find the closed-loop output resistance, set the input source to zero, apply a test voltage vx at the output and find the current ix drawn.

                    Rf
      +-----------/\/\/-----------+
      |                           |
 v- --+-----(-)                   |
      |         >--(A·vd)---Ro----+----o vx (test source,
      R1   +---(+)                         ix flows in)
      |    |
     GND  GND

Feedback voltage at inverting input:
  v- = vx·R1/(R1 + Rf) = β·vx
  vd = v+ - v- = -β·vx
Current through Ro:
  io = (vx - A·vd)/Ro = vx(1 + Aβ)/Ro
Current through R1 + Rf:
  iF = vx/(R1 + Rf)   (very small, neglected)
  ix ≈ io = vx(1 + Aβ)/Ro
Rof = vx/ix = Ro / (1 + Aβ),   β = R1/(R1 + Rf)

Because the output voltage is sampled and fed back (voltage feedback), the output resistance is divided by (1 + Aβ). Example: 741 with Ro = 75 Ω, A = 2 × 10⁵ and β = 0.1 gives Rof = 75/(1 + 2 × 10⁴) ≈ 3.75 mΩ, practically zero. The same result holds for the inverting and non-inverting amplifiers, since with the input grounded both have the same feedback network.