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Stan’s Legacy The Stanley Meyer Archive

Capacitor discharge into a coil

Dump a charged capacitor into the coil: how big a current pulse, how fast, and does it ring?

The formula
Ipk=V0ωd·L⁢e−α·tpk⁢sinωd·tpk
Ipk
Peak current, A
V0
Charge voltage, V
ωd
Damped angular frequency √(ω₀² − α²), rad/s (the hyperbolic rate √(α² − ω₀²) when overdamped, and sin becomes sinh)
L
Coil inductance, H
α
Damping rate R ÷ 2L, per second
tpk
Time to peak, s
LaTeX
Ipk = \frac{V0}{ωd \cdot L} \, e^{-α \cdot tpk} \sin\left(ωd \cdot tpk\right)

Work it out

µF

The bank that is charged and then switched across the coil.

V

The capacitor's voltage at the moment the switch closes.

Ω

The coil's resistance plus the capacitor's ESR, the switch and the leads. Everything the current passes through.

mH

The drive coil, with whatever is in the tube.

Method

  1. Find the undamped angular frequency ω₀ = 1/√(LC) and the damping rate α = R/2L. Their ratio ζ = α/ω₀ names the regime.
  2. Underdamped (ζ < 1): the current is (V₀/ω_d L)·e^(−αt)·sin(ω_d t) with ω_d = √(ω₀² − α²). It peaks at t = atan(ω_d/α)/ω_d and then reverses, ringing at ω_d/2π and decaying at α.
  3. Overdamped (ζ > 1): the current is (V₀/βL)·e^(−αt)·sinh(βt) with β = √(α² − ω₀²) — the same shape with sinh for sin — and peaks at ln(s₂/s₁)/(s₁ − s₂) where s₁,₂ = −α ± β. It never reverses.
  4. Critically damped (ζ = 1): the current is (V₀ t/L)·e^(−αt), peaking at t = 1/α with V₀/(αL)·e^(−1).
  5. The stored energy is ½CV₀², and since the capacitor ends up empty and the coil ends up with no current, every joule of it is dissipated in the circuit's resistance — mostly the coil's copper.

Assumptions

  • Constant L, R and C throughout, and no initial current in the coil. A medium that saturates during the pulse lowers L as the current rises; an electrolytic bank's ESR rises with frequency. Both make the real peak a little different from this.
  • The switch closes instantly and stays closed. A thyristor opens itself at the first current zero, which in the underdamped case cuts the ring off after its first half-cycle and leaves the capacitor charged the other way; an IGBT or a contactor does not.
  • No freewheel or clamp diode. With a diode across the coil the reverse half-cycle is shorted out and the current decays through the diode at the coil's own L/R instead of reversing — which is what a pulse magnetiser does so the medium is not un-set by the swing back.
  • Nothing here is Meyer's. The estate material does not describe how the EPG coils were driven beyond the pulse timings; a capacitor dump is what a workshop would use to get a low-resistance coil to a high current, and the page shows what that costs and what it does.