Tow-Thomas band pass biquad design

Tow-Thomas band-pass biquad

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

            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).