Offset/bias definitions; closed-loop input resistance

Definitions

• Input offset voltage (Vio): the dc voltage that must be applied between the two inputs to make the output zero when both inputs are at ground potential. It comes from mismatch of the input transistors. 741: 1 mV typical, 6 mV max.
• Input offset current (Ios): the difference between the currents flowing into the two inputs when the output is zero, Ios = |IB1 − IB2|. 741: 20 nA typical, 200 nA max.
• Input bias current (IB): the average of the currents flowing into the two input terminals when the output is zero, IB = (IB1 + IB2)/2. It is the base current of the input transistors. 741: 80 nA typical, 500 nA max.

Input resistance of the closed-loop op-amp

The answer depends on the configuration; both are shown.

Inverting amplifier (voltage-shunt feedback):

                Rf
       +------/\/\/-----+
       |                 |
 vin --R1--+--(-)        |
  -> i1    |        >----+---- vo
       GND-+--(+)

The input current i1 flows through R1 into node v− (the op-amp input current is negligible since Ri is very large).

vo = -A·v-          (v+ = 0)
Current through Rf:
  iF = (v- - vo)/Rf = v-(1 + A)/Rf
So, seen from node v-, Rf looks like
  Rf/(1 + A)        (Miller effect)
Node v- to ground: Ri ‖ Rf/(1 + A) ≈ Rf/(1 + A)
Input resistance:
  Rin = vin/i1 = R1 + Rf/(1 + A)
Since A is very large:
  Rin ≈ R1

So the input resistance of the inverting amplifier is set by R1, because the inverting input is a virtual ground. Example: R1 = 10 kΩ, Rf = 100 kΩ, A = 2 × 10⁵: Rin = 10 kΩ + 0.5 Ω ≈ 10 kΩ.

For the non-inverting amplifier (voltage-series feedback), vin drives the + input directly:

vd  = vin - β·vo ,   vo = A·vd ,   β = R1/(R1 + Rf)
vin = vd(1 + Aβ)
iin = vd/Ri
Rif = vin/iin = Ri(1 + Aβ)

Example: Ri = 2 MΩ, A = 2 × 10⁵, β = 0.1 gives Rif = 2 MΩ × 20 001 ≈ 40 GΩ, practically infinite.