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## Hay Bridge Derivation

**Hay Bridge Derivation**

The schematic diagram of the Hay bridge is shown in Figure below. It is a modified form of Maxwell bridge. The difference between Hay bridge and Maxwell bridge is that the resistor R

_{1}is in series with the capacitor in Hay bridge. The unknown inductance L_{x}and its resistance part are represented by L_{x}and R_{x}respectively as shown in the schematic diagram. One of the ratio arm used capacitor C_{1}and a variable resistance R_{1}. The other two arms have R_{2}and R_{3}.
The balance condition is obtained as follows:

From the schematic diagram we find that:

Z

_{1}= R_{1}– (j/ωC_{1})
Z

_{2}= R_{2}
Z

_{3}= R_{3}
and Z

_{4}= R_{x}+ jωL_{x}
Therefore

(R

_{1}- (j/ωC_{1}))( R_{x}+ jωL_{x}) = R_{2}R_{3}
R

_{2}R_{3}= R_{1}R_{x}+ jωL_{x}R_{1}+ (L_{x}/C_{1}) – (jR_{x}/ ωC_{1})
Taking real terms:

R

_{2}R_{3}= R_{1}R_{x}+ (L_{x}/C_{1}) ------------------------ 1
Taking imaginary terms we have:

R

_{x}/ ωC_{1}= ωL_{x}R_{1}------------------- 2
Both the equations 1 and 2 contain L

_{x}and R_{x}therefore we must solve these equations simultaneously.
Taking equations 1 and 2 and multiplying equation 1 with ωR

_{1}we have:
ωR

_{1}R_{2}R_{3}= ωR_{1}^{2}R_{x}+ L_{x}ω R_{1}/C_{1}------------------ 3
Dividing the equation 2 with C

_{1}
R

_{x}/ ωC_{1}^{2}= ωL_{x}R_{1}/ C_{1}--------------------- 4
Subtracting 4 from 3 we have,

ωR

_{1}R_{2}R_{3}- R_{x}/ ωC_{1}^{2}= ωR_{1}^{2}R_{x}
ωR

_{1}R_{2}R_{3}= ωR_{1}^{2}R_{x }+ R_{x}/ ωC_{1}^{2}
ωR

_{1}R_{2}R_{3}= R_{x}((1 + ω^{2}C_{1}^{2}R_{1}^{2}_{ })/ ωC_{1}^{2})
R

_{x}= ω^{2}C_{1}^{2}R_{1}R_{2}R_{3}/(1 + ω^{2}C_{1}^{2}R_{1}^{2})
From equation 2

L

_{x}= R_{x}/ ω^{2}C_{1}R_{1}
L

_{x}= ω^{2}C_{1}^{2}R_{1}R_{2}R_{3}/(1 + ω^{2}C_{1}^{2}R_{1}^{2}) ω^{2}C_{1}R_{1}
L

_{x}= R_{2}R_{3}C_{1}/(1 + ω^{2}C_{1}^{2}R_{1}^{2})
The solution yields

R

_{x}= ω^{2}C_{1}^{2}R_{1}R_{2}R_{3}/(1 + ω^{2}C_{1}^{2}R_{1}^{2}) ------------------- 5
L

_{x}= R_{2}R_{3}C_{1}/(1 + ω^{2}C_{1}^{2}R_{1}^{2}) ------------------- 6
From the equations 5 and 6, it is seen that they contain the angular velocity (ω). Therefore the frequency of the voltage source must be accurately known as it appears. However this is not true. As we known that Q = 1/ ω C

_{1}R_{1}substituting the value of Q in the equation 6 he expression for L_{x}becomes:
L

_{x}= R_{2}R_{3}C_{1}/(1 + (1/Q)^{2}) ---------------- 7
For values of Q greater than ten the term (1/Q)

^{2}will be smaller than 1/100 hence can be neglected.
Therefore: L

_{x}= R_{2}R_{3}C_{1}------------------- 8

**Hay's Bridge Advantages:**

1. This bridge gives very easy derivation for unknown inductance for high Q coils and is appropriate for coils having Q greater than 10.

2. The bridge also gives simple derivation for Q.

3. From the expression for Q we find the resistance R

_{1}to be appearing in the denominator. Hence for high Q coils value of R_{1}must be small. It shows that the bridge requires only a low value of resistor R_{1}, unlike Maxwell bridge which required impractically large value of resistance.

**Disadvantages of Hay's Bridge:**

This bridge is not suitable for measuring inductance values with Q values smaller than 10. The reason is that the factor (1/Q)

^{2}cannot be neglected in expression (7), in such cases.### This post was written by: Sreejith Hrishikeshan

Sreejith Hrishikesan Nair is a M-Tech graduate in Communication Systems. He completed B-tech Degree in Electronics and Communication.He is a person who wants to implement new ideas in the field of Technology.

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