Skip to main content

Measurement of dielectric constant and loss factor

Short answer

Schering bridge measures Cx = Cs R4/R3 and tan δ = ω C4 R4; with guard-ring electrode, εr = Cx d/(ε0 A) and loss factor εr tan δ. Transformer ratio-arm bridge and Q-meter are alternatives. Tan δ rise with voltage indicates voids, with temperature indicates ageing.

Describe the methods for measurement of dielectric constant and loss factor.

The dielectric constant (εr\varepsilon_r) of an insulating material is the ratio of the capacitance of a specimen to its capacitance with vacuum (air) as dielectric. The loss factor is εrtan⁡δ\varepsilon_r\tan\delta, where tan⁡δ\tan\delta (dissipation factor) is the ratio of the loss (resistive) current to the charging (capacitive) current. Both are measured by bridge methods at power and higher frequencies; the most common is the high-voltage Schering bridge.

  Ix     Ic (leading by 90 deg)
   \     |
    \    |   delta = angle between
     \   |   Ix and Ic
      \d |
       \ |
  ------+--------> Ir (in phase with V)
tan⁡δ=IRIC,loss P=V2ωCtan⁡δ\tan\delta = \frac{I_R}{I_C}, \qquad \text{loss } P = V^2\omega C\tan\delta

1. High-voltage Schering bridge

              HV supply (V)
              /           \
      Arm 1  /             \ Arm 2
    Cx + Rx                 Cs (standard,
   (specimen)               loss-free)
          |                   |
          A------[ D ]--------B
          |                   |
      Arm 3: R3        Arm 4: R4 || C4
          \                  /
            ----- earth ----
  • Arm 1: test specimen, represented as CxC_x in series with RxR_x.
  • Arm 2: standard compressed-gas capacitor CsC_s (loss-free).
  • Arm 3: variable non-inductive resistance R3R_3.
  • Arm 4: fixed resistance R4R_4 in parallel with a variable capacitor C4C_4.
  • D: vibration galvanometer or electronic null detector.
  • Arms 3 and 4 are at low voltage (near earth), so adjustments are safe; protective gaps across them guard against a breakdown of CxC_x or CsC_s. Earthed shields (Wagner earth) remove errors due to stray capacitance.

Balance condition Z1Z4=Z2Z3Z_1 Z_4 = Z_2 Z_3:

(Rx+1jωCx)R41+jωC4R4=R3jωCs\left(R_x + \frac{1}{j\omega C_x}\right)\frac{R_4}{1 + j\omega C_4R_4} = \frac{R_3}{j\omega C_s}

Equating real and imaginary parts:

Cx=Cs R4R3Rx=R3 C4Cstan⁡δ=ωCxRx=ωC4R4\begin{aligned} C_x &= C_s\,\frac{R_4}{R_3} \\ R_x &= R_3\,\frac{C_4}{C_s} \\ \tan\delta &= \omega C_x R_x = \omega C_4 R_4 \end{aligned}

Usually R4R_4 is a fixed value such as 1000π Ω\frac{1000}{\pi}\ \Omega. At 50 Hz, ωR4=100π×1000π=105 Ω\omega R_4 = 100\pi\times\frac{1000}{\pi} = 10^5\ \Omega, so tan⁡δ=0.1 C4\tan\delta = 0.1\,C_4 (with C4C_4 in μF), and the C4C_4 dial can be calibrated directly in tan⁡δ\tan\delta. (With R4=104π ΩR_4 = \frac{10^4}{\pi}\ \Omega, tan⁡δ=C4\tan\delta = C_4 in μF.)

Dielectric constant: with specimen thickness dd and electrode area AA, C0=ε0A/dC_0 = \varepsilon_0 A/d, and

εr=CxC0=Cx dε0A,loss factor=εrtan⁡δ\varepsilon_r = \frac{C_x}{C_0} = \frac{C_x\,d}{\varepsilon_0 A}, \qquad \text{loss factor} = \varepsilon_r\tan\delta

A three-terminal (guard-ring) electrode system is used so that surface leakage and edge fringing do not affect the result.

2. Transformer ratio-arm bridge

  • Uses a precision ratio transformer (ratio arms) instead of resistive arms; balance is set by turns ratios Ns/NxN_s/N_x.
  • Cx=CsNs/NxC_x = C_s N_s/N_x and tan⁡δ\tan\delta from the conductance balance.
  • Advantages: very high accuracy, unaffected by stray capacitances to earth, can test earthed specimens, works over a wide frequency range.

3. Other methods

  • Resonance (Q-meter) method for high frequencies (kHz–MHz): change in resonance capacitance gives CxC_x, change in Q gives tan⁡δ\tan\delta.
  • Measurements versus voltage and temperature: tan⁡δ\tan\delta rising with voltage ("tip-up") indicates voids and partial discharge; rising with temperature indicates moisture and ageing.
QuantityFormulaTypical good value
tan⁡δ\tan\delta (oil-paper bushing)ωC4R4\omega C_4R_4< 0.005–0.007
εr\varepsilon_r (transformer oil)Cx/C0C_x/C_0about 2.2
εr\varepsilon_r (porcelain)Cx/C0C_x/C_0about 6

✓ Answer checked by Claude Opus 5.5 (high effort)

All High Voltage Engineering chapter 7 questions →