---
title: "Sallen-Key 2nd order Butterworth LPF at 10 kHz"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:sallen-key-butterworth-10khz"
source: https://www.bctengineer.com/q/semester_5/filter_design/sallen-key-butterworth-10khz
site: BCT Engineering
---

# Sallen-Key 2nd order Butterworth LPF at 10 kHz

Filter Design · Chapter 6: Active Filters

- Asked in: 2072 Kartik, 5+4 marks
- Video: [Sallen-Key Lowpass filter design and transfer function with programmable gain](https://www.youtube.com/watch?v=LYARYxwUJgc) — Engineering Prof.
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/sallen-key-butterworth-10khz
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

Draw the circuit diagram of Sallen-Key low pass filter and derive its transfer function. Design second order butterworth low pass filter having half power frequency of 10 KHz using Sallen Key biquad. In your final design the value of capacitors must be 0.01μF and feedback resistors should also be equal (Refer table 1).

## Short answer

With equal feedback resistors (K=2) and C=0.01µF, use R1=1.125kΩ, R2=2.25kΩ, RA=RB=10kΩ. This yields Butterworth response (Q=0.707) with half-power frequency 10kHz.

## Answer

#### Sallen-Key low-pass circuit and transfer function
The **Sallen-Key low-pass biquad** uses one op-amp wired as a non-inverting amplifier of gain $K = 1 + R_B/R_A$, two resistors $R_1, R_2$ in series, a capacitor $C_1$ fed back from the output to node A, and a grounded capacitor $C_2$ at node B.

```text
              C1 (feedback)
        +-----||-------------+
        |                    |
Vin -R1-+-(A)- R2 -+-(B)--|+\ |
                   |      |  >+---- Vo
                  === C2 +|-/ |
                   |     |    |
                  GND    +-RB-+
                         |
                         RA
                         |
                        GND
```

**Derivation** (ideal op-amp, so $V_o = K V_B$ and no current enters the + input):

Node B: the current through $R_2$ flows into $C_2$:
$$
\frac{V_A - V_B}{R_2} = sC_2 V_B \;\Rightarrow\; V_A = V_B(1 + sR_2C_2)
$$
Node A (KCL):
$$
\frac{V_{in} - V_A}{R_1} = \frac{V_A - V_B}{R_2} + sC_1 (V_A - V_o)
$$
Substituting $V_A$ and $V_B = V_o/K$ and simplifying:
$$
T(s) = \frac{V_o}{V_{in}} = \frac{\dfrac{K}{R_1R_2C_1C_2}}{s^2 + s\left(\dfrac{1}{R_1C_1} + \dfrac{1}{R_2C_1} + \dfrac{1-K}{R_2C_2}\right) + \dfrac{1}{R_1R_2C_1C_2}}
$$
Comparing with $T(s) = \dfrac{K\omega_o^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$:
$$
\begin{aligned}
\omega_o &= \frac{1}{\sqrt{R_1R_2C_1C_2}} \\
\frac{\omega_o}{Q} &= \frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}
\end{aligned}
$$
DC gain $= K$.

#### Design: 2nd-order Butterworth, $f_c = 10$ kHz
From Table 1, the normalized 2nd-order Butterworth function is
$$
T(s) = \frac{1}{s^2 + \sqrt{2}\,s + 1} \;\Rightarrow\; \omega_o = 1,\; Q = \frac{1}{\sqrt{2}} = 0.7071
$$
For a Butterworth filter the half-power frequency equals $\omega_o$.

**Equal-element design (Design 1):** take normalized $R_1 = R_2 = 1\ \Omega$, $C_1 = C_2 = 1$ F. Then $\omega_o = 1$ rad/s and
$$
\frac{1}{Q} = 1 + 1 + (1 - K) = 3 - K \;\Rightarrow\; K = 3 - \frac{1}{Q}
$$
$$
K = 3 - \sqrt{2} = 1.586 \;\Rightarrow\; \frac{R_B}{R_A} = K - 1 = 0.586
$$

**Denormalization:**
$$
\begin{aligned}
k_f &= 2\pi \times 10\,000 = 62\,831.85\ \text{rad/s} \\
C &= \frac{C_n}{k_f k_m} \Rightarrow k_m = \frac{1}{62\,831.85 \times 0.01\times10^{-6}} = 1591.55 \\
R_1 = R_2 &= k_m \times 1 = 1591.55\ \Omega \approx 1.59\ \text{k}\Omega
\end{aligned}
$$
Choose $R_A = 10\ \text{k}\Omega$, so $R_B = 0.586 \times 10 = 5.86\ \text{k}\Omega$.

| Element | Value |
|---|---|
| $R_1 = R_2$ | 1.59 kΩ |
| $C_1 = C_2$ | 0.01 μF |
| $R_A$ | 10 kΩ |
| $R_B$ | 5.86 kΩ |
| DC gain $K$ | 1.586 (4 dB) |

**Answer:** $R_1 = R_2 = 1.59\ \text{k}\Omega$, $C_1 = C_2 = 0.01\ \mu\text{F}$, $R_A = 10\ \text{k}\Omega$, $R_B = 5.86\ \text{k}\Omega$; this gives $f_c = 10$ kHz with $Q = 0.707$.

If unity overall gain is needed, use an input divider of gain $\alpha = 1/1.586 = 0.631$: $R_{1a} = 1.59/0.631 = 2.52\ \text{k}\Omega$ and $R_{1b} = 1.59/0.369 = 4.31\ \text{k}\Omega$.

**Other reading of "equal feedback resistors":** if it means $R_A = R_B$ (so $K = 2$) with $C_1 = C_2 = 0.01\ \mu$F, the coefficient of $s$ reduces to $1/(R_1C)$, so $Q = \sqrt{R_1/R_2}$. For $Q = 0.7071$ this needs $R_2 = 2R_1$, and $\omega_o = 1/(\sqrt{2}\,R_1C)$ gives $R_1 = 1.125\ \text{k}\Omega$, $R_2 = 2.25\ \text{k}\Omega$, $R_A = R_B = 10\ \text{k}\Omega$. The passband gain is then 2.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the Sallen-Key low-pass, the op-amp gain K is set by:
   A. K = R1/R2
   B. K = 1 + RB/RA
   C. K = RB/RA
   D. K = 1 + C1/C2
   Answer: B. The op-amp is a non-inverting amplifier, so K = 1 + RB/RA. (Key point: Amplifier gain)

2. With R1 = R2 and C1 = C2, 1/Q = 3 - K. For Q = 0.7071, K equals:
   A. 3 - sqrt2 = 1.586
   B. 3 + sqrt2 = 4.41
   C. 2
   D. 1.414
   Answer: A. K = 3 - 1/Q = 3 - 1.414 = 1.586. (Key point: Equal-element Q)

3. With C = 0.01 µF and w0 = 2π x 10^4, R1 = R2 = 1/(w0C) is about:
   A. 15.9 kΩ
   B. 159 Ω
   C. 1.59 kΩ
   D. 10 kΩ
   Answer: C. 1/(62832 x 10^-8) = 1.59 kΩ. (Key point: Resistors R1 = R2)

4. With RA = 10 kΩ and K = 1.586, RB is:
   A. 15.86 kΩ
   B. 5.86 kΩ
   C. 1.586 kΩ
   D. 10 kΩ
   Answer: B. RB = (K - 1) RA = 0.586 x 10 kΩ = 5.86 kΩ. (Key point: Feedback resistors)

## Related in Filter Design

- [Sallen-Key 2nd order Butterworth LPF design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-butterworth-equal-resistors) — Filter Design, chapter 6, asked 3×
- [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
- [4th order Butterworth LPF with two Sallen-Key biquads (1 kHz)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Acascade-sallen-key-fourth-order-1khz) — Filter Design, chapter 8
- [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
- [Sallen-Key LPF; obtaining HPF; 12 kHz design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-highpass-12khz) — Filter Design, chapter 6
- [Tow-Thomas 2nd order Butterworth LPF at 5 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-butterworth-5khz) — Filter Design, chapter 6

## Related in other subjects

- [Transfer function of RC low-pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Arc-lowpass-transfer-function) — Signals and Systems, chapter 5, asked 2×
- [Transfer function of DT low pass filter](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-lowpass-transfer-function) — Signals and Systems, chapter 6