Recovery, restriking voltage, RRRV and derivation

(i) Recovery voltage

The normal-frequency (50 Hz) rms voltage that appears across the CB contacts after the final arc extinction and after the high-frequency transients have died out. Its peak is about V_m sin φ.

(ii) Restriking voltage

The transient, high-frequency voltage that appears across the contacts at or near current zero during arc interruption. It oscillates about the recovery voltage and can reach 2V_m.

(iii) RRRV

The rate of rise of restriking voltage, dv/dt, in kV/μs. If the voltage across the gap rises faster than the gap regains dielectric strength, the arc restrikes. It is the main factor in arc extinction.

Expression for restriking voltage

Consider a short circuit at the CB terminals of a generator with inductance L and stray capacitance C across the contacts (resistance neglected).

  e = V_m cos ωt
 (~)----^^^L^^^----+----o/ o----X fault
  |                |
 earth            === C
                   |
                  earth

1. Before interruption, C is shorted by the fault. The fault current lags e by 90°, so at current zero e = V_m (taken constant during the short transient).
2. After the arc goes out at current zero, L and C form a series circuit driven by V_m. Let v be the voltage across C (= across the contacts) and i the current:

V_m = L di/dt + v,    i = C dv/dt
V_m = LC d²v/dt² + v

3. Taking the Laplace transform with v(0) = 0, i(0) = 0:

V_m/s = LC s² V(s) + V(s)
V(s)  = V_m / [ s (1 + s² LC) ]
      = V_m ω_n² / [ s (s² + ω_n²) ],  ω_n = 1/√(LC)
      = V_m [ 1/s − s/(s² + ω_n²) ]

4. Inverse transform:

v(t) = V_m (1 − cos ω_n t),   ω_n = 1/√(LC)

Results

f_n          = 1 / (2π √(LC))
Peak value   = 2 V_m   at t = π √(LC)
RRRV         = dv/dt = V_m ω_n sin ω_n t
RRRV (max)   = V_m / √(LC)  at t = (π/2) √(LC)

So a smaller capacitance or inductance raises the natural frequency and the RRRV, making interruption harder.