Skip to content
Stan’s Legacy
Sign in

Damping and ringdown, two ways

The same calculation run twice, side by side. Change anything on either side and the difference is shown output by output. This page is a link: both sets of numbers are in the address bar.

Side A open alone
mH

Everything inductive in the series loop.

nF

The cell.

Ω

Everything resistive in the loop — choke wire, cell ESR, wiring. This is what takes the energy out.

Side B open alone
mH

Everything inductive in the series loop.

nF

The cell.

Ω

Everything resistive in the loop — choke wire, cell ESR, wiring. This is what takes the energy out.

Swap sides

What changed

The two sides are the same. Change something on either.

Output A B B against A
ζ Damping ratio 0.05 0.05 same
τ Decay time constant 20 µs 20 µs same
d Ringing frequency 159 kHz 159 kHz same
m Cycles to 1 % energy 7.329 7.329 same

Side A, with its numbers

0.05=100 Ω21 nF1 mH20 µs=2·1 mH100 Ω159 kHz=12π11 mH·1 nF−120 µs27.329=ln10π·1100 Ω1 mH1 nF
LaTeX 
0.05 = \frac{100\,\mathrm{Ω}}{2} \sqrt{\frac{1\,\mathrm{nF}}{1\,\mathrm{mH}}} \qquad 20\,\mathrm{µs} = \frac{2 \cdot 1\,\mathrm{mH}}{100\,\mathrm{Ω}} \qquad 159\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{1\,\mathrm{mH} \cdot 1\,\mathrm{nF}} - \frac{1}{20\,\mathrm{µs}^{2}}} \qquad 7.329 = \frac{\ln 10}{\pi} \cdot \frac{1}{100\,\mathrm{Ω}} \sqrt{\frac{1\,\mathrm{mH}}{1\,\mathrm{nF}}}

Side B, with its numbers

0.05=100 Ω21 nF1 mH20 µs=2·1 mH100 Ω159 kHz=12π11 mH·1 nF−120 µs27.329=ln10π·1100 Ω1 mH1 nF
LaTeX 
0.05 = \frac{100\,\mathrm{Ω}}{2} \sqrt{\frac{1\,\mathrm{nF}}{1\,\mathrm{mH}}} \qquad 20\,\mathrm{µs} = \frac{2 \cdot 1\,\mathrm{mH}}{100\,\mathrm{Ω}} \qquad 159\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{1\,\mathrm{mH} \cdot 1\,\mathrm{nF}} - \frac{1}{20\,\mathrm{µs}^{2}}} \qquad 7.329 = \frac{\ln 10}{\pi} \cdot \frac{1}{100\,\mathrm{Ω}} \sqrt{\frac{1\,\mathrm{mH}}{1\,\mathrm{nF}}}