---
title: "Tow-Thomas band pass biquad design"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:tow-thomas-bandpass-design"
source: https://www.bctengineer.com/q/semester_5/filter_design/tow-thomas-bandpass-design
site: BCT Engineering
---

# Tow-Thomas band pass biquad design

Filter Design · Chapter 6: Active Filters

- Asked in: 2078 Bhadra, 5+4 marks
- Video: [Tow-Thomas Biquad Filters (2): Op-Amp RC Circuit](https://www.youtube.com/watch?v=OBJmKUgmBNc) — Semiconductor Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/tow-thomas-bandpass-design
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

Derive the transfer function of Tow-Thomas band pass biquad and design it with centre frequency of 1000 rad/sec, bandwidth of 200 rad/sec and a maximum gain of 1.

## Short answer

For a Tow-Thomas band-pass filter with ω0=1000 rad/s, bandwidth 200 rad/s, and peak gain 1, use equal capacitors C=0.1 μF and equal R2=R3=R. R=1/(ω0C)=10 kΩ, R1=1/(BW·C)=50 kΩ, R4=R1/gain=50 kΩ, r=10 kΩ. The band-pass output is at V1.

## Answer

#### Tow-Thomas band-pass biquad

The band-pass output of the Tow-Thomas biquad is taken at the output of the lossy integrator OA1.

```text
            R1 (sets Q)
          +--/\/\--+
          |   C1   |
          +---||---+
          |        |
Vin--R4---+--[OA1]-+--> V1 (band-pass)
          ^
          |  V1--R3-->[OA2 integrator, C2]--> V2 (LP)
          |  V2--r--->[OA3 inverter, r/r]---> V3=-V2
          |                                    |
          +------------------R2----------------+
```

All op-amp (+) inputs are grounded. OA1 is a lossy integrator ($R_1 \parallel C_1$ in feedback), OA2 an ideal integrator ($C_2$ in feedback), OA3 a unity-gain inverter.

**Derivation.** KCL at the inverting input of OA1 (virtual ground):

$$
\frac{V_{in}}{R_4} + \frac{V_3}{R_2} + V_1\left(\frac{1}{R_1} + sC_1\right) = 0
$$

OA2 integrates and OA3 inverts:

$$
V_2 = -\frac{V_1}{sR_3C_2}, \qquad V_3 = -V_2 = \frac{V_1}{sR_3C_2}
$$

Substituting $V_3$:

$$
V_1\left(sC_1 + \frac{1}{R_1} + \frac{1}{sR_2R_3C_2}\right) = -\frac{V_{in}}{R_4}
$$

Multiplying both sides by $s/C_1$:

$$
\frac{V_1}{V_{in}} = \frac{-\dfrac{s}{R_4C_1}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}
$$

Comparing with the standard band-pass form $T(s) = \dfrac{-H(\omega_0/Q)s}{s^2 + (\omega_0/Q)s + \omega_0^2}$:

$$
\omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}}, \qquad \text{BW} = \frac{\omega_0}{Q} = \frac{1}{R_1C_1}, \qquad Q = \omega_0R_1C_1
$$

and the gain at the centre frequency $s = j\omega_0$ is

$$
|T(j\omega_0)| = \frac{1/(R_4C_1)}{1/(R_1C_1)} = \frac{R_1}{R_4}
$$

So $\omega_0$ is set by $R_2, R_3$, the bandwidth by $R_1$ and the peak gain by $R_4$, independently.

#### Design: $\omega_0 = 1000$ rad/s, BW = 200 rad/s, maximum gain 1

**Step 1: Parameters.**

$$
Q = \frac{\omega_0}{\text{BW}} = \frac{1000}{200} = 5, \qquad H = \frac{R_1}{R_4} = 1
$$

Required: $T(s) = \dfrac{-200s}{s^2 + 200s + 10^6}$.

**Step 2: Element values.** Choose $C_1 = C_2 = C = 0.1\ \mu$F and $R_2 = R_3 = R$.

$$
\begin{aligned}
R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{1000 \times 0.1\times10^{-6}} = 10\ \text{k}\Omega \\
R_1 &= \frac{1}{\text{BW}\cdot C} = \frac{Q}{\omega_0 C} = 5 \times 10 = 50\ \text{k}\Omega \\
R_4 &= \frac{R_1}{H} = 50\ \text{k}\Omega \\
r &= 10\ \text{k}\Omega\ \text{(inverter)}
\end{aligned}
$$

**Check:** $1/(R_1C) = 1/(5\times10^4 \times 10^{-7}) = 200$ rad/s (bandwidth), $1/(R^2C^2) = 1/(10^4\times10^{-7})^2 = 10^6$, and $1/(R_4C) = 200$, so $T(s) = -200s/(s^2+200s+10^6)$ with $|T(j1000)| = 1$.

**Answer:** $C_1 = C_2 = 0.1\ \mu$F, $R_1 = R_4 = 50$ k$\Omega$, $R_2 = R_3 = 10$ k$\Omega$, $r = 10$ k$\Omega$; output at $V_1$ (OA1).

## Check yourself

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

1. For centre frequency 1000 rad/s and bandwidth 200 rad/s, what is Q?
   A. 0.2
   B. 5
   C. 200
   D. 1000
   Answer: B. Q = w0/BW = 1000/200 = 5. (Key point: Q from specs)

2. In the Tow-Thomas band-pass, which components set the bandwidth and the peak gain?
   A. R1 sets BW (1/(R1C1)) and R4 sets the peak gain (R1/R4)
   B. R2 sets BW and R3 sets gain
   C. R4 sets BW and R1 sets gain
   D. C2 alone sets both
   Answer: A. BW = 1/(R1C1) and |T(jw0)| = R1/R4, so R1 and R4 control them independently of w0. (Key point: Which parameter sets what)

3. With R1 = 50 kΩ and maximum gain 1, what is R4?
   A. 10 kΩ
   B. 5 kΩ
   C. 500 kΩ
   D. 50 kΩ
   Answer: D. Peak gain = R1/R4 = 1, so R4 = R1 = 50 kΩ. (Key point: R4 for unity gain)

4. With C = 0.1 µF and w0 = 1000 rad/s, R2 = R3 = 1/(w0C) equals:
   A. 10 kΩ
   B. 1 kΩ
   C. 100 kΩ
   D. 50 kΩ
   Answer: A. 1/(1000 x 10^-7) = 10 kΩ. (Key point: Element R)

## Related in Filter Design

- [Tow-Thomas band pass biquad transfer function](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-bandpass-derivation) — Filter Design, chapter 6
- [Tow-Thomas LPF, poles 1000 ± j8994.03, gain 1.89](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-lpf-1000-8994) — Filter Design, chapter 6
- [Tow-Thomas high pass derivation; LPF design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-highpass-derive-lpf-design) — Filter Design, chapter 6, asked 3×
- [Band pass (10000 rad/s, 1000 rad/s) from given ladder](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abandpass-10000-1000-from-ladder) — Filter Design, chapter 3
- [Tow-Thomas LPF, poles -450 ± j893.03, gain 1.5](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-lpf-450-893) — Filter Design, chapter 6, asked 3×
- [Tow-Thomas LPF for ECG (gain 1.3)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-ecg-lpf-design) — Filter Design, chapter 6