---
title: "Sallen-Key LPF; obtaining HPF; 12 kHz design"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:sallen-key-highpass-12khz"
source: https://www.bctengineer.com/q/semester_5/filter_design/sallen-key-highpass-12khz
site: BCT Engineering
---

# Sallen-Key LPF; obtaining HPF; 12 kHz design

Filter Design · Chapter 6: Active Filters

- Asked in: 2074 Asoj, 4+2+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-highpass-12khz
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

Draw the circuit diagram of Sallen-Key lowpass biquad circuit and derive the transfer function. How can you obtain highpass filter from lowpass one? Design the second order lowpass Butterworth filter having half power frequency of 12 KHz using Sallen-Key biquad circuit. T2(s) = 1/(s² + √2 s + 1)

## Short answer

Use equal-resistor design: R1=R2=1 Ω, C1=C2=1 F, K=3–1/Q=1.586. Scaling to 12 kHz with C=0.01 µF gives R=1.33 kΩ. For unity overall gain, a resistive input divider scales gain to 1: R1a=2.10 kΩ, R1b=3.59 kΩ.

## 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 $= K$.

#### Obtaining a high-pass filter from the low-pass one
Use the **RC-CR transformation**: replace every resistor $R_i$ by a capacitor of value $1/R_i$ and every capacitor $C_j$ by a resistor of value $1/C_j$ (normalized values, $\omega_o = 1$). The op-amp gain resistors $R_A, R_B$ set only a ratio, so they stay unchanged. This replaces $s$ by $1/s$, so for the normalized function
$$
T_{LP}(s) = \frac{K}{s^2 + \frac{1}{Q}s + 1} \;\Rightarrow\; T_{HP}(s) = T_{LP}\!\left(\frac{1}{s}\right) = \frac{K s^2}{s^2 + \frac{1}{Q}s + 1}
$$
Frequency scaling by $\omega_o$ then gives $\dfrac{K s^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$. In the high-pass circuit, capacitors sit in the series positions and resistors in the feedback and grounded positions.

#### Design: Butterworth LPF at 12 kHz
$T_2(s) = \dfrac{1}{s^2 + \sqrt2 s + 1}$: $\omega_o = 1$, $Q = 0.7071$.

**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
$$
\frac{1}{Q} = 1 + 1 + (1 - K) = 3 - K \;\Rightarrow\; K = 3 - \frac{1}{Q}
$$
$$
K = 3 - 1.414 = 1.586, \quad R_B/R_A = 0.586
$$
**Scaling** (choose $C = 0.01\ \mu$F):
$$
\begin{aligned}
k_f &= 2\pi \times 12\,000 = 75\,398.2\ \text{rad/s} \\
k_m &= \frac{1}{75\,398.2 \times 10^{-8}} = 1326.3 \\
R_1 = R_2 &= 1.326\ \text{k}\Omega
\end{aligned}
$$
$R_A = 10\ \text{k}\Omega$, $R_B = 5.86\ \text{k}\Omega$.

**Answer:** $R_1 = R_2 = 1.33\ \text{k}\Omega$, $C_1 = C_2 = 0.01\ \mu$F, $R_A = 10\ \text{k}\Omega$, $R_B = 5.86\ \text{k}\Omega$; $f_c = 12$ kHz, DC gain 1.586.

If unity gain is needed (as in $T_2(s)$), use an input divider with $\alpha = 1/1.586 = 0.631$: $R_{1a} = 1.326/0.631 = 2.10\ \text{k}\Omega$ and $R_{1b} = 1.326/0.369 = 3.59\ \text{k}\Omega$.

## Check yourself

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

1. How is a high-pass obtained from the low-pass?
   A. Swap the inputs and outputs of the op-amp
   B. Double all capacitors and halve the resistors
   C. Swap R to C (1/R) and C to R (1/C); gain resistors RA, RB unchanged
   D. Replace every element including RA, RB by its inverse
   Answer: C. This replaces s by 1/s; gain-ratio resistors stay the same. (Key point: RC-CR transformation)

2. The resulting high-pass transfer function is:
   A. K wo^2 / (s^2 + (wo/Q)s + wo^2)
   B. K s / (s^2 + (wo/Q)s + wo^2)
   C. K / (s^2 + (wo/Q)s + 1)
   D. K s^2 / (s^2 + (wo/Q)s + wo^2)
   Answer: D. Under s -> 1/s the numerator becomes K s^2 with the same poles. (Key point: HP transfer function)

3. For Butterworth with R1=R2=C1=C2=1, K is:
   A. 1.586
   B. 1
   C. 1.414
   D. 2
   Answer: A. K = 3 - 1/Q = 3 - 1.414 = 1.586. (Key point: Gain for equal elements)

4. With C = 0.01 uF and fc = 12 kHz, R1 = R2 is about:
   A. 13.3 kΩ
   B. 1.33 kΩ
   C. 15.9 kΩ
   D. 133 Ω
   Answer: B. km = 1/(2π×12000×10^-8) = 1326, so R = 1.33 kΩ. (Key point: Resistor scaling)

## Related in Filter Design

- [Sallen-Key 2nd order Butterworth LPF at 10 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-butterworth-10khz) — Filter Design, chapter 6
- [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
- [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
- [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×
- [3rd order Butterworth HPF from LPF transfer function](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Athird-order-hpf-from-lpf-tf) — Filter Design, chapter 5

## Related in other subjects

- [LPF prototype conversion and second order HPF](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Alpf-prototype-hpf-conversion) — RF and Microwave Engineering, chapter 6