Wien bridge

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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:

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

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

$\begin{displaymath} 1-\omega^2R_1R_2C_1C_2=0,\;\;\;\;\;\mbox{i.e.,}\;\;\;\;\; \omega=\frac{1}{\sqrt{R_1R_2C_1C_2}} \end{displaymath}$

and the equation above becomes

$\begin{displaymath} \frac{R_3}{R_4}=\frac{R_1C_1+R_2C_2}{R_1C_2}=\frac{C_1}{C_2}+\frac{R_2}{R_1} \end{displaymath}$

In particular, if

and

, we have:

$\begin{displaymath} \omega=\frac{1}{\sqrt{R_1R_2C_1C_2}} =\frac{1}{\sqrt{R^2C^2}}=\frac{1}{RC} \end{displaymath}$

and

$\begin{displaymath} \frac{R_3}{R_4}=\frac{C_1}{C_2}+\frac{R_2}{R_1}=1+1=2, \;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;R_4=2R_3 \end{displaymath}$

Wien-Robinson Filter

$\begin{displaymath} \frac{V_{in}}{R_4}+\frac{V_{out}}{R_3}+\frac{V_1}{R_2}=0\;\;\;\;\;\;\;\;(1) \end{displaymath}$
$\begin{displaymath} V_2=\frac{R+1/j\omega C}{R+1/j\omega C+R/j\omega C/(R+1/j\o... ... =\frac{(j\omega\tau+1)^2}{(j\omega\tau+1)^2+j\omega\tau}V_1, \end{displaymath}$

where $\tau=RC$ , i.e.,

$\begin{displaymath} V_1=\left(1+\frac{j\omega\tau}{(j\omega\tau+1)^2}\right)V_2,\;\;\;\;\;\;(2) \end{displaymath}$
$\begin{displaymath} \frac{V_1-V_2}{R_1}+\frac{V_{out}-V_2}{2R_1}=0, \;\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\; V_{out}=3V_2-2V_1,\;\;\;\;\;\;(3) \end{displaymath}$

$\begin{displaymath} V_{out}=3V_2-2V_1=3V_2-2\left(1+\frac{j\omega\tau}{(j\omega\... ...u+1)^2}\;V_2 =\frac{(j\omega\tau)^2+1}{(j\omega\tau+1)^2}\;V_2 \end{displaymath}$

$\begin{displaymath} V_2=\frac{(j\omega\tau+1)^2}{(j\omega\tau)^2+1}V_{out} \end{displaymath}$

Substituting this into (2) we get

$\begin{displaymath} V_1=\left(1+\frac{j\omega\tau}{(j\omega\tau+1)^2}\right)V_2 ... ...ac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\;V_{out} \end{displaymath}$

Substituting this into (1) we get

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

We rearrange to get

$\begin{displaymath} \frac{V_{in}}{R_4}=-\left(\frac{1}{R_3} +\frac{(j\omega\tau)... ...j\omega\tau+1}{(j\omega\tau)^2+1}\;\frac{1}{R_2}\right)V_{out} \end{displaymath}$

$\begin{displaymath} \frac{R_2}{R_4}=-\left(\frac{R_2}{R_3} +\frac{(j\omega\tau)^... ...\omega\tau+1}{(j\omega\tau)^2+1}\right) \frac{V_{out}}{V_{in}} \end{displaymath}$

$\begin{displaymath} H(j\omega)=\frac{V_{out}}{V_{in}} =-\frac{R_2R_3}{R_4(R_2+R_... ...\omega\tau)^2+1}{(j\omega\tau)^2+3j\omega\tau R_3/(R_2+R_3)+1} \end{displaymath}$

$\begin{displaymath} H(j\omega)=A\;\frac{(j\omega)^2+\omega_n^2}{(j\omega)^2+\Del... ...\omega_n=1/\tau \ A&\omega\rightarrow\infty\end{array}\right. \end{displaymath}$

This is a band-stop filter with passband gain

, stop-band $\omega_n=1/\tau$ and