Hay Bridge Derivation
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 L x and
its resistance part are represented by L x and Rx respectively as shown in the schematic
diagram. One of the ratio arm used capacitor C 1 and a variable resistance R1. The other
two arms have R2 and R3.
Therefore
Rx/ ωC1 = ωLx R1 ------------------- 2
Both the equations 1 and 2 contain Lx and Rx therefore we must solve these equations
simultaneously.
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)
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:
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
1. This bridge gives very easy derivation for unknown inductance for high Q coils and is
appropriate for coils having Q greater than 10.
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 R1, unlike Maxwell bridge which required
impractically large value of resistance.
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.