Wien bridge

The Wien bridge is a particular type of the Wheatstone bridge of which two of the four arms are composed of a capacitor as well as a resistor in parallel and series:

For this bridge to balance, the ratios of the left and right branches should be the same:

$\displaystyle \frac{R_3}{R_4}=\frac{R_2+1/j\omega C_2}{R_1\vert\vert 1/j\omega ... ... R_1C_2} =\frac{1-\omega^2R_1R_2C_1C_2+j\omega(R_1C_1+R_2C_2)}{j\omega R_1C_2}$ (50)

For this equation to hold, the right-hand side needs to be real, i.e.,

$\displaystyle 1-\omega^2R_1R_2C_1C_2=0,\;\;\;\;\;\mbox{i.e.,}\;\;\;\;\; \omega=\frac{1}{\sqrt{R_1R_2C_1C_2}}$ (51)

and the equation above becomes

$\displaystyle \frac{R_3}{R_4}=\frac{R_1C_1+R_2C_2}{R_1C_2}=\frac{C_1}{C_2}+\frac{R_2}{R_1}$ (52)

In particular, if $R_1=R_2=R$

and

, we have:

$\displaystyle \omega=\frac{1}{\sqrt{R_1R_2C_1C_2}} =\frac{1}{\sqrt{R^2C^2}}=\frac{1}{RC}$ (53)

and

$\displaystyle \frac{R_3}{R_4}=\frac{C_1}{C_2}+\frac{R_2}{R_1}=1+1=2, \;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;R_4=2R_3$ (54)

Wien-Robinson Filter

$\displaystyle \frac{V_{in}}{R_4}+\frac{V_{out}}{R_3}+\frac{V_1}{R_2}=0\;\;\;\;\;\;\;\;(1)$ (55)
$\displaystyle V_2=\frac{R+1/j\omega C}{R+1/j\omega C+R/j\omega C/(R+1/j\omega C)}V_1 =\frac{(j\omega\tau+1)^2}{(j\omega\tau+1)^2+j\omega\tau}V_1,$ (56)
where $\tau=RC$ , i.e.,
$\displaystyle V_1=\left(1+\frac{j\omega\tau}{(j\omega\tau+1)^2}\right)V_2,\;\;\;\;\;\;(2)$ (57)
$\displaystyle \frac{V_1-V_2}{R_1}+\frac{V_{out}-V_2}{2R_1}=0, \;\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\; V_{out}=3V_2-2V_1,\;\;\;\;\;\;(3)$ (58)

$\displaystyle V_{out}=3V_2-2V_1=3V_2-2\left(1+\frac{j\omega\tau}{(j\omega\tau+1... ...au}{(j\omega\tau+1)^2}\;V_2 =\frac{(j\omega\tau)^2+1}{(j\omega\tau+1)^2}\;V_2$ (59)

$\displaystyle V_2=\frac{(j\omega\tau+1)^2}{(j\omega\tau)^2+1}V_{out}$ (60)

Substituting this into (2) we get

$\displaystyle V_1=\left(1+\frac{j\omega\tau}{(j\omega\tau+1)^2}\right)V_2 =\le... ...+1}V_{out} =\frac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\;V_{out}$ (61)

Substituting this into (1) we get

$\displaystyle \frac{V_{in}}{R_4}+\frac{V_{out}}{R_3} +\frac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\;\frac{V_{out}}{R_2}=0$ (62)

We rearrange to get

$\displaystyle \frac{V_{in}}{R_4}=-\left(\frac{1}{R_3} +\frac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\;\frac{1}{R_2}\right)V_{out}$ (63)

$\displaystyle \frac{R_2}{R_4}=-\left(\frac{R_2}{R_3} +\frac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\right) \frac{V_{out}}{V_{in}}$ (64)

$\displaystyle H(j\omega)=\frac{V_{out}}{V_{in}} =-\frac{R_2R_3}{R_4(R_2+R_3)}\;\frac{(j\omega\tau)^2+1}{(j\omega\tau)^2+3j\omega\tau R_3/(R_2+R_3)+1}$ (65)

$\displaystyle H(j\omega)=A\;\frac{(j\omega)^2+\omega_n^2}{(j\omega)^2+\Delta\om... ...a=0\\ 0&\omega=\omega_n=1/\tau \\ A&\omega\rightarrow\infty\end{array}\right.$ (66)

This is a band-stop filter with passband gain $A=-R_2R_3/R_4(R_2+R_3)=(R_2//R_3)/R_4$

, stop-band $\omega_n=1/\tau$ and $Q=(R_2+R_3)/3R_3$