Tow-Thomas high pass derivation; LPF design

Tow-Thomas high-pass biquad

The basic Tow-Thomas loop (lossy integrator OA1, integrator OA2, inverter OA3) gives LP and BP outputs. A high-pass output is obtained by feeding the input to OA1 through a capacitor $C_f$ instead of the resistor $R_4$; the output is taken at OA1.

            R1
          +--/\/\--+
          |   C1   |
          +---||---+
          |        |
Vin--Cf---+--[OA1]-+--> V1 (high-pass output)
          ^
          |  V1--R3-->[OA2 integrator, C2]--> V2
          |  V2--r--->[OA3 inverter, r/r]---> V3=-V2
          |                                    |
          +------------------R2----------------+

Derivation. KCL at OA1's inverting input (virtual ground):

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

From OA2 and OA3: $V_3 = -V_2 = \dfrac{V_1}{sR_3C_2}$. Substituting:

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

Multiplying by $s/C_1$:

$$\frac{V_1}{V_{in}} = \frac{-\dfrac{C_f}{C_1}s^2}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}$$

This is a second-order high-pass function with $\omega_0 = 1/\sqrt{R_2R_3C_1C_2}$, $\omega_0/Q = 1/(R_1C_1)$ and high-frequency gain $-C_f/C_1$.

Low-pass Tow-Thomas used for the design

With the resistor $R_4$ at the input (figure above with $C_f$ replaced by $R_4$), the same analysis gives at $V_2$:

$$\frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \quad \omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}},\ \frac{\omega_0}{Q} = \frac{1}{R_1C_1},\ K = \frac{R_2}{R_4}$$

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 450$, $\omega_d = 893.03$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 450^2 + 893.03^2 = 1{,}000{,}002.6 \;\Rightarrow\; \omega_0 \approx 1000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{1000}{900} = 1.111 \end{aligned}$$

Required transfer function (dc gain $K = 1.5$):

$$T(s) = \frac{1.5\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{1.5 \times 10^{6}}{s^2 + 900\,s + 10^{6}}$$

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. Choose $C_1 = C_2 = C = 0.1\ \mu$F (a standard value that gives resistors in the k$\Omega$ range).

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{1000 \times 0.1\times10^{-6}} = 10\ \text{k}\Omega \\ R_1 &= QR = 1.111 \times 10\ \text{k}\Omega = 11.11\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{10\ \text{k}\Omega}{1.5} = 6.667\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 1.5.

Answer: $C = 0.1\ \mu\text{F}$, $R_1 = 11.11\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).