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Next: Butterworth filters Up: Chapter 6: Active Filter Previous: The Twin-T notch (band-stop)

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:

WienBridge2.png

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 $R_1=R_2=R$ and $C_1=C_2=C$, 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

WienRobinson2.png


\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}

or

\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 $A=-R_2R_3/R_4(R_2+R_3)=(R_2//R_3)/R_4$, stop-band $\omega_n=1/\tau$ and $Q=(R_2+R_3)/3R_3$.

Further reading for


next up previous
Next: Butterworth filters Up: Chapter 6: Active Filter Previous: The Twin-T notch (band-stop)
Ruye Wang 2018-07-23