# Wien Bridge

**The Wien bridge is a circuit that uses a bridge topology to accurately measure capacitance or frequency.**

In terms of construction, the Wien bridge is very similar to the Wheatstone bridge, with the addition of one capacitor in series and another in parallel. Both use the concept of **balancing a bridge circuit** to perform a measurement.

The basic Wien bridge has one capacitor (C_{1}) in series with resistor R_{1} and one capacitor (C_{2}) in parallel with resistor R_{2}.

## Balanced Wien Bridge Circuit

In order to understand the functionality of the Wien bridge, we have to look at how the circuit is balanced.

Recall that the **Wheatstone bridge** circuit balances when the ratio of the resistances of the first and second arms (R_{1}/R_{2}) is equal to the ratio of the third and fourth arms (R_{2}/R_{3}):

\frac{R_1}{R_2}=\frac{R_3}{R_4}\,(Wheatstone\,Bridge)

The Wien bridge circuit extends this by adding two capacitors. Balancing the circuit now requires including the **impedance** introduced by the capacitors (i.e. capacitive reactance).

The balancing condition can be found by trading the branch resistances of the Wheatstone bridge, for the total impedances of the new Wien bridge circuit:

\frac{Z_1}{Z_2}=\frac{Z_3}{Z_4}\,(Wien\,Bridge)

Like the Wheatstone bridge, the Wien bridge circuit is balanced when the ratio of the impedances of the first and second arms (Z_{1}/Z_{2}) is equal to the ratio of the impedances of the third and fourth arms (Z_{3}/Z_{4}). The primary difference is that impedance only includes resistance in the Wheatstone bridge circuit.

Let’s solve for each value of impedance in the Wien Bridge circuit.

#### Impedance of Arm 1

The impedance of the first arm (Z_{1}) is determined by resistor R_{1} and the capacitive reactance X_{C1} of C_{1}. The two terms add, with the reactance being differentiated by the inclusion of the unit imaginary number *j* (which is equal to the square root of negative one).

Z_1=R_1-jX_{C1}=R_1-\frac{j}{\omega C_1}

#### Impedance of Arm 2

The impedance of the second arm, like the first, depends on the contributions of both the resistor and capacitor.

This time, however, the capacitor is in parallel with the resistor. This means that we have to use the equation for parallel impedances in order to find Z_{2}.

Z_2=\frac{1}{\frac{1}{R_2}+\frac{1}{jX_{C2}}}

We can plug in the value of *X*_{C2} and simplify:

Z_2=[\frac{1}{R_2}+\frac{1}{j(\frac{1}{\omega C_2})}]^{-1}=[\frac{1}{R_2}+j\omega C_2]^{-1}

#### Impedances of Arms 3 and 4

The impedances of arms 3 and 4 are equal to the value of the corresponding resistors:

Z_3=R_3

Z_4=R_4

#### Balancing the Wien Bridge

Now let’s combine these results with the conditions for a balanced circuit:

\frac{Z_1}{Z_2}=\frac{Z_3}{Z_4} => Z_1Z_4=Z_2Z_3

To go further, we’ll need to calculate Z_{1}Z_{4} and Z_{2}Z_{3}:

Z_1Z_4=(R_1-\frac{j}{\omega C})R_4

Z_2Z_3=[\frac{1}{R_2}+j\omega C_2]^{-1}R_3=\frac{R_3}{\frac{1}{R_2}+j\omega C_2}

Setting Z_{1}Z_{4} equal to Z_{2}Z_{3} (per the balancing conditions):

(R_1-\frac{j}{\omega C})R_4=\frac{R_3}{\frac{1}{R_2}+j\omega C_2}

\omega ^2=\frac{1}{C_1C_2R_1R_2}

\frac{C_4}{C_2}=\frac{R_4}{R_3}-\frac{R_2}{R_1}