Op-Amp Configurations — Worked Example

Non-inverting amplifier driving a resistive load

Given: $V_{\text{in}} = 1\,\text{V}_{\text{pp}}$ sine (0.5 V amplitude), $R_f = 100\,\text{k}\Omega$, $R_{\text{in}} = 10\,\text{k}\Omega$, $R_L = 10\,\text{k}\Omega$, $V_{CC} = \pm 15\,\text{V}$.

Find: voltage gain, output amplitude, rail check, output current, and input impedance.

1. Compute gain.
   $$A_v = 1 + \frac{R_f}{R_{\text{in}}} = 1 + \frac{100\,\text{k}\Omega}{10\,\text{k}\Omega} = 11\,\text{V/V}$$
   The non-inverting topology adds 1 to the feedback ratio.
2. Output amplitude.
   $$V_{\text{out}} = A_v \times V_{\text{in}} = 11 \times 0.5\,\text{V} = 5.5\,\text{V}_{\text{peak}}$$
   Peak-to-peak: $11\,\text{V}_{\text{pp}}$.
3. Rail check.
   $$|V_{\text{out,peak}}| = 5.5\,\text{V} < 15\,\text{V} \quad \checkmark$$
   Output stays well within the supply rails — no clipping.
4. Output current.
   $$I_{\text{out}} = \frac{V_{\text{out,peak}}}{R_L} = \frac{5.5\,\text{V}}{10\,\text{k}\Omega} = 0.55\,\text{mA}$$
   Well within typical op-amp output capability (5–25 mA).
5. Input impedance.
   $$R_{\text{in,stage}} \approx \infty$$
   The non-inverting input connects directly to $V^+$; the ideal op-amp draws zero input current.
6. Result confirmed.
   $$A_v = 11,\quad V_{\text{out}} = 11\,\text{V}_{\text{pp}},\quad \text{no clipping, minimal loading}$$
   The design meets all requirements.