Q5Electronic Measurement & Instrumentation

Question

Show that the Wein frequency bridge will be balanced at only one frequency given by : $f = \frac{1}{2 π R _{1} R _{2} C _{1} C _{2}}$ Where $C_{1}, C_{2}, R_{1},$ and $R_{2}$ have their usual meanings.

Answer 1

The Wien bridge consists of a series RC network ( $R_{1}, C_{1}$ ) in arm 1, a parallel RC network ( $R_{2}, C_{2}$ ) in arm 2, and pure resistors ( $R_{3}, R_{4}$ ) in arms 3 and 4.

The general balance condition for an AC bridge is $Z_{1} Z_{4} = Z_{2} Z_{3}$ .

Substitute the impedances: $Z_{1} = R_{1} - j \frac{1}{ω C _{1}}$ $Z_{2} = \frac{R _{2} ( - j / ω C _{2} )}{R _{2} - j / ω C _{2}} = \frac{R _{2}}{1 + j ω R _{2} C _{2}}$

The balance equation becomes: $(R_{1} - j \frac{1}{ω C _{1}}) R_{4} = (\frac{R _{2}}{1 + j ω R _{2} C _{2}}) R_{3}$

Rearranging and expanding into real and imaginary parts: $\frac{R _{1} R _{4}}{R _{2}} - \frac{j R _{4}}{ω C _{1} R _{2}} + j ω R_{1} C_{2} R_{4} + \frac{C _{2} R _{4}}{C _{1}} = R_{3}$

To achieve balance, both the real and imaginary parts must equate independently. Setting the imaginary part to zero: $ω R_{1} C_{2} R_{4} - \frac{R _{4}}{ω C _{1} R _{2}} = 0$

Divide by $R_{4}$ and solve for $ω$ : $ω R_{1} C_{2} = \frac{1}{ω C _{1} R _{2}} ⟹ ω^{2} = \frac{1}{R _{1} R _{2} C _{1} C _{2}}$

Since $ω = 2 π f$ : $f = \frac{1}{2 π R _{1} R _{2} C _{1} C _{2}}$

Because this equation must be strictly satisfied to balance the imaginary part, the bridge only balances at this single unique frequency.

Q5Electronic Measurement & Instrumentation

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