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Current and power under a pulsed drive

At the drive frequency, what current does the loop draw, what power does it dissipate, and what does that average out to over the pulse train?

The formula
X=2π⁢f·L−12π⁢f·CZ=R2+X2I=VZP=I2·RW=P·e100E=P·t
I
Peak current, A
X
Net reactance, Ω
Z
Loop impedance, Ω
P
Peak power, W
W
Average power, W
E
Energy per pulse, J
f
Drive frequency, kHz
L
Inductance, mH
C
Capacitance, nF
R
Series resistance, Ω
V
Drive amplitude, V
e
Effective duty, %
t
Pulse on-time, µs
LaTeX 
X = 2\pi f \cdot L - \frac{1}{2\pi f \cdot C} \qquad Z = \sqrt{R^{2} + X^{2}} \qquad I = \frac{V}{Z} \qquad P = I^{2} \cdot R \qquad W = P \cdot \frac{e}{100} \qquad E = P \cdot t

Work it out

Fill from a part the library

mH

Everything inductive in the series loop.

nF

The cell.

Ω

Everything resistive in the loop.

kHz

The pulse frequency the loop is driven at — not necessarily its resonance.

V

The peak voltage applied across the loop.

%

The fraction of all time the drive is on — pulse duty times gate duty.

µs

How long one pulse is high, for the energy each pulse costs.

or see a worked example with typical numbers

Method

  1. Find the two reactances at the drive frequency: X_L = 2πfL and X_C = 1/(2πfC), everything in SI. Their difference is the net reactance; it is zero only at resonance.
  2. The loop impedance is the resistance and the net reactance added in quadrature, Z = √(R² + X²) — they are 90° apart, so they do not simply add.
  3. The peak current is the drive amplitude over that impedance. Its phase relative to the voltage is atan(X/R): positive means the current lags, which is what an inductive loop does above resonance.
  4. The peak power is I²R — only the resistance dissipates anything; the reactances store and return.
  5. Multiply by the effective duty for the average over time, and by the on-time of one pulse for the energy each pulse costs.

Assumptions

  • Steady-state sinusoidal drive at the pulse frequency. The current here is the amplitude the loop settles to after many cycles; the first pulse of a burst draws less, and a square pulse has harmonics the loop answers differently. This is the fundamental, which carries most of the energy.
  • The peak current is the sinusoidal amplitude, so the power is I²R ÷ 2 averaged over a cycle if the current is sinusoidal; I²R is used here as the figure for the moment of peak current, and is the conservative one for sizing the drive.
  • The resistance is the only dissipation. Water loss and core loss add to the real figure; the comparison to a supply wattmeter is approximate.
  • The drive is a voltage source that holds its amplitude whatever the loop draws. A real transformer secondary sags under load, which limits the current when Z is small — that is, at resonance, where this calculation gives its largest number.