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Bridge for high-Q inductance measurement

Short answer

The Hay bridge measures high-Q inductors (Q>10) using a series R4–C4 arm. At balance, L1 ≈ R2R3C4, Q = 1/(ωC4R4). It is preferred because high Q requires only a small R4, while a Maxwell bridge would need an impractically large R4.

Which bridge is used for the measurement of inductance of coil having high quality factor? Explain how the measurement is done and why the bridge is preferred.

The Hay bridge is used to measure the inductance of coils with a high quality factor (Q > 10).

Circuit

          A
        /   \
  L1,R1(Z1)  R3 (Z4)
      /         \
     B---[D]----D
      \         /
    R2 (Z2)   R4 + C4 (series) (Z3)
        \     /
          C
   source across A-C, detector across B-D

Arm AB: unknown coil (L1 in series with R1). Arm BC: R2. Arm CD: R4 in series with a standard capacitor C4. Arm DA: R3. The detector is between B and D.

Measurement

Adjust R3 (or R2) and R4 until the detector shows zero. At balance, Z1·Z3 = Z2·Z4, with Z3 = R4 − j/(ωC4):

(R1 + jωL1)(R4 − j/ωC4) = R2R3

Equate the real and imaginary parts:

 R1R4 + L1/C4 = R2R3
 ωL1R4 − R1/(ωC4) = 0   ->  R1 = ω²L1R4C4

Solving:

 L1 = R2R3C4 / (1 + ω²C4²R4²)
 R1 = ω²C4²R2R3R4 / (1 + ω²C4²R4²)
 Q  = ωL1/R1 = 1/(ωC4R4)

For high Q (ωC4R4 ≪ 1), the denominator is nearly 1, so L1 ≈ R2R3C4.

Why it is preferred for high Q

  • Q = 1/(ωC4R4), so a high Q needs only a small R4, which is cheap and practical.
  • In the Maxwell bridge the same Q would need a very large parallel R4 (Q = ωC4R4), which is impractical.
  • For high Q the term ω²C4²R4² is negligible, so L1 ≈ R2R3C4 is almost independent of frequency.
  • The standard capacitor is nearly lossless and easy to obtain, which gives good accuracy.
  • Drawback: the exact balance equations contain ω, so the source frequency must be known and stable, and the bridge is not suitable for low-Q coils.

✓ Answer checked by Claude Opus 5.5 (high effort)

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