Peak, rms and duty-cycle heat

A statement such as "90 % reduction in battery heating" can mean any of three things:

• the reduction in peak instantaneous dissipation, at the moment of the transient;
• the reduction in root-mean-square battery current over the cycle;
• the reduction in total resistive heat generated across the complete duty cycle.

The first is the largest and the easiest to compute, which is presumably why it is the one most often quoted. It is also the least useful, because a cell's temperature responds to energy over a thermal time constant of minutes, not to instantaneous power.

Instantaneous dissipation in a cell of internal resistance R carrying current i(t) is P = i²R. Over a cycle of period T the heat generated is

Q = R · I_rms² · T    where    I_rms² = (1/T) ∫ i(t)² dt

So cycle heat is set by the mean of the square of current. That is why a duty cycle whose current is concentrated in brief peaks generates more heat than its average current suggests — and it is precisely that excess a capacitor buffer removes.

There is a second term people leave out. The charge the capacitor delivered has to be put back, and it comes back through the battery, paying conversion loss both ways.

The peak-instantaneous figure and the duty-cycle figure come from the same clamp on the same profile, and they differ by a factor of 2.3 in the reduction they report. Both are correct. Only one of them describes what a thermocouple on the cell will show.

Note that the peak-current and terminal-voltage reductions are not subject to this correction. They are first-order, they determine whether a bus stays above an inverter's dropout threshold, and they are far easier to measure than a thermal difference.

The benefit has an optimum. Write the duty-cycle heat as a function of the peak-to-base current ratio, the clamp, the transient duty fraction and the buffer's round-trip efficiency, and the result passes through a minimum. For a peak-to-base ratio of 5 with the clamp at 1.5 times base current, the optimum sits near a transient duty fraction of 0.14. Below it the peaks contribute little heat to begin with; above it the recharge burden dominates.

There is a break-even. Above a transient duty fraction of about 43 %, with a 90 % round-trip path, the buffer makes the battery hotter than no buffer at all. In the lossless limit with full buffering that boundary is exactly one half — and independently of how peaky the load is.

The reason is almost amusing once seen. At 50 % duty with full buffering, the battery carries the high current during the recharge half instead of the transient half. You have swapped which half of the cycle is hard, and the quadratic penalty does not care which half.

1. Name the quantity. Peak current, peak instantaneous dissipation, rms current, or duty-cycle heat. If the reduction is quadratic in current, say which current.
2. Publish the duty cycle. A time series with its sampling rate, or a reference to a published one. A peak and an average do not determine the heat.
3. Include the return path. State the round-trip efficiency assumed and show the recharge current in the battery's waveform. Omitting it is not conservative; it is wrong in a known direction.
4. Sample fast enough to be right. A 1 Hz logger on a profile with 0.2-second peaks at 1.7 % duty under-reports the heat integral by about 29 %, and the error flatters whichever configuration has the sharper peaks.

I have quoted the 91 % figure myself. It belongs in a sentence containing the words "peak instantaneous", and nowhere else.