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IEEE 48511 min read

Worked Example: A Duty-Cycle Battery Sizing That Passes Capacity but Fails Its Own Voltage Window

The IEEE 485 section method correctly sizes battery capacity for a four-period duty cycle — but the resulting cell count doesn't actually keep the string above its minimum discharge voltage.

Scenario

Duty cycle5 A for 15 min, then 35 A for 10 min, then 15 A for 75 min, then 45 A for 10 min
Correction factorsTemperature 1.0, aging 1.25, design margin 1.1
System DC voltage125 V nominal
Cell voltagesNominal 2.17 V, equalize 2.33 V, end-of-discharge 1.75 V
Acceptance window135–105 V for a 58-cell string

Step-by-step calculation

Step 1: Compute the required capacity section by section (IEEE 485 method)

Each later section's requirement isn't just its own current — it accounts for the capacity already 'used up' by earlier sections, scaled by how the battery's capacity-rating factor (Kt) changes with duration.

SectionCurrentCumulative timeKt factorRequired capacity
15 A15 min1.758.75 A
235 A25 min2.1575.25 A
315 A100 min4.1061.5 A
445 A110 min4.35195.75 A

Step 2: Apply temperature, aging and design-margin correction factors

final = maxSectionA x tempCF x agingCF x marginCF
195.75 x 1.0 x 1.25 x 1.1
final = 269.2 Ah required battery capacity

Step 3: Find the number of cells for the 125 V system

nCells = round(systemV / cellVnominal)
round(125 / 2.17)
58 cells

Step 4: Check the string voltage at equalize charge

vEqString = nCells x vEqCell
58 x 2.33
135.14 V — within the 140 V maximum acceptance limit

Step 5: Check the string voltage at end of discharge

vEodString = nCells x vEodCell
58 x 1.75
101.5 V — below the 105 V minimum acceptance limit

Result summary

CheckRequirementActualStatus
Required battery capacityn/a (this is the sizing result)269.2 Ah✓ PASS
String voltage at equalize charge≤ 140 V135.14 V✓ PASS
String voltage at end of discharge≥ 105 V101.5 V✗ FAIL
The duty-cycle capacity sizing itself is sound (269.2 Ah), but 58 cells at 1.75 V end-of-discharge only delivers 101.5 V — 3.5 V short of this system's 105 V minimum acceptance window. The cell count needs revisiting, not the capacity.

Key insight: Battery capacity sizing (how many amp-hours) and cell-count/voltage-window checking (how many cells, and does the string stay in an acceptable voltage range through the full discharge) are two genuinely separate calculations that can each pass or fail independently — a battery with plenty of capacity can still be the wrong string configuration if its end-of-discharge voltage falls outside what the connected DC equipment can tolerate.

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Frequently asked questions

What fixes the end-of-discharge voltage shortfall?

Adding one more cell (59 instead of 58) raises vEodString to 59 x 1.75 = 103.25 V — still short. The real fix here is usually a battery/cell chemistry with a higher end-of-discharge voltage point, or accepting a load-dependent cutoff earlier in the discharge (raising vEodCell), or revisiting the connected equipment's actual minimum operating voltage, since simply adding cells one at a time doesn't close a 3.5 V gap quickly without also pushing the equalize-charge voltage toward its own 140 V ceiling.

Why does the required capacity look at the largest single section instead of summing all four?

The section method reflects that later sections have to be supported by whatever capacity remains after earlier sections have already drawn on the battery — so Section 4's 195.75 A requirement already implicitly accounts for the capacity used in Sections 1-3 (that's what the ktPrev/kt ratio term in each section's formula does), rather than needing to be added to them separately. Simply summing all four sections' raw currents would badly oversize the battery.

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