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 R1 is in series with the capacitor in Hay bridge. The unknown inductance Lx and its resistance part are represented by Lx and Rx respectively as shown in the schematic diagram. One of the ratio arm used capacitor C1 and a variable resistance R1. The other two arms have R2 and R3.

The balance condition is obtained as follows:

From the schematic diagram we find that:

Z1 = R1 – (j/ωC1)
Z2 = R2
Z3 = R3
and Z4 = Rx + jωLx


(R1 - (j/ωC1))( Rx + jωLx) = R2 R3

R2 R3 = R1 Rx + jωLxR1 + (Lx/C1) – (jRx/ ωC1)

Taking real terms:

R2 R3 = R1 Rx + (Lx/C1) ------------------------ 1

Taking imaginary terms we have:

Rx/ ωC1 = ωLx R1 ------------------- 2

Both the equations 1 and 2 contain Lx and Rx therefore we must solve these equations simultaneously.

Taking equations 1 and 2 and multiplying equation 1 with ωR1 we have:

ωR1R2R3 = ωR12Rx + Lx ω R1/C1 ------------------ 3

Dividing the equation 2 with C1

Rx/ ωC12 = ωLx R1/ C1--------------------- 4

Subtracting 4 from 3 we have,

ωR1R2R3 - Rx/ ωC12 = ωR12Rx
ωR1R2R3 = ωR12Rx + Rx/ ωC12
ωR1R2R3 = Rx ((1 + ω2 C12R12 )/ ωC12)
Rx = ω2 C12R1R2R3/(1 + ω2 C12R12)

From equation 2
Lx = Rx/ ω2 C1R1
Lx = ω2 C12R1R2R3/(1 + ω2 C12R12) ω2 C1R1
Lx = R2R3 C1/(1 + ω2 C12R12)

The solution yields
Rx = ω2 C12R1R2R3/(1 + ω2 C12R12) ------------------- 5
Lx = R2R3 C1/(1 + ω2 C12R12) ------------------- 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 C1R1 substituting the value of Q in the equation 6 he expression for Lx becomes:

Lx = R2R3 C1/(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:  Lx = R2R3C1------------------- 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 R1 to be appearing in the denominator. Hence for high Q coils value of R1 must be small. It shows that the bridge requires only a low value of resistor R1, 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.

Sreejith Hrishikesan

Sreejith Hrishikesan is a ME post graduate and has been worked as an Assistant Professor in Electronics Department in KMP College of Engineering, Ernakulam. For Assignments and Projects, Whatsapp on 8289838099.

Post a Comment

Previous Post Next Post