Transfer function of op-amp integrator-type circuit

The circuit is an inverting integrator (the input, through $R_1$, and the capacitor both go to the inverting input). $R_2$ only balances the input bias current; for an ideal op-amp it carries no current, so the non-inverting input is at 0 V.

Assumptions (ideal op-amp)

• Infinite input impedance: no current into the op-amp terminals.
• Virtual ground: $V_- = V_+ = 0$ (no current in $R_2$, so no drop across it).

Derivation

Current through $R_1$:

$$I_1(s) = \frac{E_{in}(s) - 0}{R_1}$$

This whole current flows through $C$ (impedance $1/sC$):

$$I_1(s) = \frac{0 - E_{out}(s)}{1/(sC)} = -sC\,E_{out}(s)$$

Equating:

$$\frac{E_{in}(s)}{R_1} = -sC\,E_{out}(s)$$

Transfer function

$$\frac{E_{out}(s)}{E_{in}(s)} = -\frac{1}{sR_1C}$$

In time domain, $e_{out}(t) = -\dfrac{1}{R_1C}\displaystyle\int e_{in}\,dt$. So the circuit is an integrator with time constant $R_1C$ and a sign inversion; it gives a pole at $s = 0$ and is used to realise the integral (I) action of a controller. Usually $R_2$ is chosen equal to $R_1$ to reduce output offset.