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:

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

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

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

and the equation above becomes

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

In particular, if $R_1=R_2=R$ and $C_1=C_2=C$, we have:

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

and

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

Wien-Robinson Filter

• Apply KCL to $v^-$ of the first op-amp:
  
  $$\frac{V_{in}}{R_4}+\frac{V_{out}}{R_3}+\frac{V_1}{R_2}=0\;\;\;\;\;\;\;\;(1)$$
  
  
  Solve for $V_2$:
  
  $$V_2=\frac{R+1/j\omega C}{R+1/j\omega C+R/j\omega C/(R+1/j\o... ... =\frac{(j\omega\tau+1)^2}{(j\omega\tau+1)^2+j\omega\tau}V_1,$$
  
  
  where $\tau=RC$, i.e.,
  
  $$V_1=\left(1+\frac{j\omega\tau}{(j\omega\tau+1)^2}\right)V_2,\;\;\;\;\;\;(2)$$
  
  

• Apply KCL to $v^-=V_2$ of the second op-amp:
  
  $$\frac{V_1-V_2}{R_1}+\frac{V_{out}-V_2}{2R_1}=0, \;\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\; V_{out}=3V_2-2V_1,\;\;\;\;\;\;(3)$$
  
  

• Solve for $V_{out}$:
  
  $$V_{out}=3V_2-2V_1=3V_2-2\left(1+\frac{j\omega\tau}{(j\omega\... ...u+1)^2}\;V_2 =\frac{(j\omega\tau)^2+1}{(j\omega\tau+1)^2}\;V_2$$
  
  
  or
  
  $$V_2=\frac{(j\omega\tau+1)^2}{(j\omega\tau)^2+1}V_{out}$$
  
  

• Substitute this into (2) we get
  
  $$V_1=\left(1+\frac{j\omega\tau}{(j\omega\tau+1)^2}\right)V_2 ... ...ac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\;V_{out}$$
  
  
  Substituting this into (1) we get
  
  $$\frac{V_{in}}{R_4}+\frac{V_{out}}{R_3} +\frac{(j\omega\tau)^2+3j\omega\tau+1}{(j\omega\tau)^2+1}\;\frac{V_{out}}{R_2}=0$$
  
  

• Rearrange to get
  
  $$\frac{V_{in}}{R_4}=-\left(\frac{1}{R_3} +\frac{(j\omega\tau)... ...j\omega\tau+1}{(j\omega\tau)^2+1}\;\frac{1}{R_2}\right)V_{out}$$
  
  
  
  
  $$\frac{R_2}{R_4}=-\left(\frac{R_2}{R_3} +\frac{(j\omega\tau)^... ...\omega\tau+1}{(j\omega\tau)^2+1}\right) \frac{V_{out}}{V_{in}}$$
  
  

• Solve for $H(j\omega)=V_{out}/V_{in}$:
  
  $$H(j\omega)=\frac{V_{out}}{V_{in}} =-\frac{R_2R_3}{R_4(R_2+R_... ...\omega\tau)^2+1}{(j\omega\tau)^2+3j\omega\tau R_3/(R_2+R_3)+1}$$
  
  
  
  
  $$H(j\omega)=A\;\frac{(j\omega)^2+\omega_n^2}{(j\omega)^2+\Del... ...\omega_n=1/\tau A&\omega\rightarrow\infty\end{array}\right.$$
  
  

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

• Oscillator circuits
• Active filters