--- title: "Gain enhancement in Sallen-Key circuit" subject: "Filter Design" chapter: 6 key: "semester_5/filter_design:sallen-key-gain-enhancement" source: https://www.bctengineer.com/q/semester_5/filter_design/sallen-key-gain-enhancement site: BCT Engineering --- # Gain enhancement in Sallen-Key circuit Filter Design · Chapter 6: Active Filters - Asked in: 2079 Baisakh, 4 marks; 2073 Shrawan, 5 marks; 2071 Shrawan, 5 marks (asked 3 times) - Video: [RESISTIVE GAIN ENHANCEMENT || Modified Form of Sallen-Key Circuit || 2nd Order Butterworth LPF](https://www.youtube.com/watch?v=__HBBTD7YM0) — Vivek Singh Rathore - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/sallen-key-gain-enhancement - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6 ## Question How can the gain enhancement be performed in a Sallen-Key circuit? Explain with necessary diagram. ## Short answer Gain enhancement is done by returning capacitor C1 to a tap of an output divider (β = Rd/(Rc+Rd)). The op‑amp gain μ is increased so that the effective gain βμ = K = 3−1/Q, preserving ω0 and Q, while the overall dc gain becomes H = K/β > K. Then μ = 1+RB/RA = H and the divider sets β = K/H. ## Answer In the equal-component Sallen-Key low-pass filter the amplifier gain is fixed by $Q$ ($K = 3 - 1/Q$), so the dc gain is fixed too. **Gain enhancement** raises the overall gain above $K$ while keeping $\omega_0$ and $Q$ unchanged. #### Basic idea The RC network only "sees" the voltage at the (+) input and the voltage fed back through $C_1$. Suppose the op-amp output is made larger by a factor $1/\beta$, but only the fraction $\beta V_o$ is fed back to $C_1$, with the effective gain $\beta\mu$ set equal to $K$. The RC network then works exactly as before, while the output is $1/\beta$ times larger. #### Circuit ```text C1 +---||-------------------+ | | Vin--R1---+A--R2--+B--(+) | | OA(mu)--+--|--- Vo C2 | | | Rc | GND +--+ Vo' = beta*Vo | Rd | GND OA(mu): non-inverting amp, mu = 1 + RB/RA ``` The op-amp output $V_o$ drives a divider $R_c$, $R_d$; capacitor $C_1$ is returned to the tap $V_o' = \beta V_o$ where $\beta = \dfrac{R_d}{R_c + R_d} < 1$ (keep $R_c\parallel R_d$ small compared with the impedance of $C_1$, or include it in the design). #### Derivation Let the non-inverting amplifier gain be $\mu = 1 + R_B/R_A$, so $V_o = \mu V_B$ and $V_o' = \beta\mu V_B$. The RC network now sees an "effective gain" $\beta\mu$ from node B to the point where $C_1$ returns. Writing the node equations exactly as for the normal circuit with $V_o$ replaced by $V_o'$: $$ \frac{V_o'}{V_{in}} = \frac{\dfrac{\beta\mu}{R_1R_2C_1C_2}}{s^2 + s\left(\dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-\beta\mu}{R_2C_2}\right) + \dfrac{1}{R_1R_2C_1C_2}} $$ For the required $Q$, set $\beta\mu = K = 3 - 1/Q$ (equal-component case). Then $\omega_0$ and $Q$ are the same as before, and the output is $$ \frac{V_o}{V_{in}} = \frac{1}{\beta}\cdot\frac{V_o'}{V_{in}} \;\Rightarrow\; \text{dc gain} = \frac{K}{\beta} > K $$ #### Design steps 1. Design the normal equal-component Sallen-Key for $\omega_0$ and $Q$; this gives $K = 3 - 1/Q$. 2. For a required dc gain $H > K$, choose $\beta = K/H$. 3. Set the op-amp gain $\mu = 1 + R_B/R_A = H$ and choose the divider $R_d/(R_c+R_d) = \beta$. **Example:** Butterworth ($Q = 0.707$, $K = 1.586$) with required gain $H = 5$: $\beta = 1.586/5 = 0.317$, $\mu = 5$ ($R_B = 4R_A$), and the divider is $R_c = 2.15R_d$ (e.g. $R_d = 1$ k$\Omega$, $R_c = 2.15$ k$\Omega$). For gain **reduction** (the opposite case) the input resistor $R_1$ is split into a voltage divider instead. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the basic idea of gain enhancement in a Sallen-Key filter? A. Make the op-amp output larger by 1/beta but feed back only beta*Vo to C1, so the RC network works as before B. Increase R1 and R2 to raise the gain C. Remove capacitor C2 to increase the output D. Cascade an inverting stage and ignore Q Answer: A. The RC network sees an effective gain beta*mu equal to K, while the actual output is 1/beta times larger. (Key point: Idea of gain enhancement) 2. What must be set so that w0 and Q stay unchanged? A. beta*mu = 1 B. beta*mu = K = 3 - 1/Q C. mu = K and beta = 1 D. mu = beta Answer: B. The effective gain beta*mu replaces K in the equal-component design, so it must equal 3 - 1/Q. (Key point: Condition for same Q) 3. For Butterworth (K = 1.586) with required gain H = 5, what are beta and mu? A. beta = 5/1.586 = 3.15, mu = 1.586 B. beta = 0.317, mu = 5 C. beta = 0.317, mu = 1.586 D. beta = 0.5, mu = 3.17 Answer: B. beta = K/H = 0.317 and the op-amp gain mu = H = 5 (RB = 4RA), giving dc gain K/beta = 5. (Key point: Resulting dc gain and example) ## Related in Filter Design - [Why gain enhancement is needed in Sallen-Key biquad](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-gain-enhancement-need) — Filter Design, chapter 6 - [Excess gain compensation in Sallen-Key circuit](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-excess-gain-compensation) — Filter Design, chapter 6, asked 3× - [Sallen-Key LPF with ωo 1000 rad/s, Q 0.866, gain 2](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-q-0866-gain) — Filter Design, chapter 6 - [Active filter advantages; Sallen-Key Butterworth at 4 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aactive-advantages-sallen-key-4khz) — Filter Design, chapter 6 - [4th order Butterworth LPF with two Sallen-Key biquads (10 kHz)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Acascade-sallen-key-fourth-order-10khz) — Filter Design, chapter 8 - [RC-CR transformation of Sallen-Key LPF to HPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Arc-cr-transformation-sallen-key) — Filter Design, chapter 6, asked 2×