Tow-Thomas LPF, poles -500 ± j2449.49

Tow-Thomas low-pass biquad

The Tow-Thomas biquad is a two-integrator-loop circuit using three op-amps: a lossy integrator, an ideal integrator and an inverter in a feedback loop. It gives low-pass and band-pass outputs at the same time, has low sensitivity, and lets $\omega_0$, $Q$ and gain be tuned independently.

            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 (ideal op-amps, so each inverting input is a virtual ground):

KCL at the inverting input of OA1:

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

OA2 (integrator) and OA3 (inverter):

$$V_2 = -\frac{V_1}{sR_3C_2}, \qquad V_3 = -V_2 = \frac{V_1}{sR_3C_2} \;\Rightarrow\; V_1 = sR_3C_2V_3$$

Substituting $V_1$ into the KCL equation:

$$\begin{aligned} sR_3C_2V_3\left(\frac{1}{R_1} + sC_1\right) + \frac{V_3}{R_2} &= -\frac{V_{in}}{R_4} \\ V_3\left(s^2R_3C_1C_2 + \frac{sR_3C_2}{R_1} + \frac{1}{R_2}\right) &= -\frac{V_{in}}{R_4} \end{aligned}$$

Dividing by $R_3C_1C_2$:

$$\frac{V_3}{V_{in}} = \frac{-\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \qquad \frac{V_2}{V_{in}} = -\frac{V_3}{V_{in}}$$

Comparing with $T(s) = \dfrac{\pm K\omega_0^2}{s^2 + (\omega_0/Q)s + \omega_0^2}$:

$$\omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}}, \qquad \frac{\omega_0}{Q} = \frac{1}{R_1C_1}, \qquad K = \frac{R_2}{R_4}\ \text{(dc gain)}$$

$V_2$ is a non-inverting low-pass output and $V_3$ an inverting one; $V_1$ is a band-pass output.

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 500$, $\omega_d = 2449.49$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 500^2 + 2449.49^2 = 6{,}250{,}001 \;\Rightarrow\; \omega_0 \approx 2500\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{2500}{1000} = 2.5 \end{aligned}$$

Required transfer function (dc gain $K = 2$):

$$T(s) = \frac{2\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{2 \times 6.25\times10^{6}}{s^2 + 1000\,s + 6.25\times10^{6}}$$

Design equations (choose $C_1 = C_2 = C$, $R_2 = R_3 = R$):

$$R = \frac{1}{\omega_0 C}, \qquad R_1 = QR = \frac{Q}{\omega_0 C}, \qquad R_4 = \frac{R_2}{K}, \qquad r = \text{any convenient value}$$

Step 2: Element values. Given $C_1 = C_2 = C = 0.1\ \mu$F.

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{2500 \times 0.1\times10^{-6}} = 4\ \text{k}\Omega \\ R_1 &= QR = 2.5 \times 4\ \text{k}\Omega = 10\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{4\ \text{k}\Omega}{2} = 2\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.1\ \mu\text{F}$, $R_1 = 10\ \text{k}\Omega$, $R_2 = R_3 = 4\ \text{k}\Omega$, $R_4 = 2\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 2.

Answer: $C = 0.1\ \mu\text{F}$, $R_1 = 10\ \text{k}\Omega$, $R_2 = R_3 = 4\ \text{k}\Omega$, $R_4 = 2\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).