Transfer function of op-amp with Rs ∥ C feedback

This is an inverting amplifier with a feedback impedance $Z_f = R_s \parallel C$; $R_2$ on the non-inverting input only compensates bias current.

Assumptions (ideal op-amp)

• No current into the input terminals, so no drop across $R_2$ and $V_+ = 0$.
• Virtual ground: $V_- = V_+ = 0$.

Impedances

$$Z_1 = R_1, \qquad Z_f = R_s \parallel \frac{1}{sC} = \frac{R_s\cdot\frac{1}{sC}}{R_s + \frac{1}{sC}} = \frac{R_s}{1 + sR_sC}$$

Derivation

Current through $R_1$ equals current through $Z_f$:

$$\frac{V_{in}(s) - 0}{R_1} = \frac{0 - V_{out}(s)}{Z_f}$$ $$\frac{V_{out}(s)}{V_{in}(s)} = -\frac{Z_f}{Z_1}$$

Transfer function

$$\frac{V_{out}(s)}{V_{in}(s)} = -\frac{R_s}{R_1}\cdot\frac{1}{1 + sR_sC}$$

This is a first-order lag (inverting low-pass) circuit:

• DC gain $= -R_s/R_1$
• pole at $s = -\dfrac{1}{R_sC}$, time constant $\tau = R_sC$

For example, with $R_1 = 10\ \text{k}\Omega$, $R_s = 100\ \text{k}\Omega$, $C = 1\ \mu\text{F}$, the gain is $-10/(1 + 0.1s)$. If $R_s \to \infty$ the circuit becomes a pure integrator $-1/(sR_1C)$.