Site Power, Batteries & Heat

DC load and the current it really draws, conductors sized on what the equipment sees at the end of a discharge, and battery autonomy worked out at the rate you are actually discharging — not the twenty-hour rate on the label.

Two figures on a site are routinely optimistic A 100 Ah battery emptied in two hours does not give two hours. Rated capacity is quoted at a slow discharge, and Peukert's equation says a fast one gets nearly a third less — in the direction that leaves a site dark. And 3 kW at −48 V draws 63 A, not the 13 A the same load would draw at 230 V AC, so a volt lost in the cable is 2% of the supply rather than 0.4%. Both are computed here, alongside the naive answers, so the difference is visible.

The DC load

EquipmentQtyFigureUnit

Current is what sizes conductors, fuses and busbars; power is what sizes rectifiers and cooling. The figure to protect against is the current at the low-voltage disconnect, because that is the largest the load will ever draw.

Conductors

Battery autonomy

hours

Rectifiers, generator and heat

Why the battery figure is the one that catches people

A battery's rated capacity is quoted at a particular discharge rate, almost always the twenty-hour rate: a 100 Ah cell is one that delivers 5 A for twenty hours. Ask it for 50 A and it does not deliver two hours; the chemistry cannot keep up, and the usable charge falls.

Peukert's equation describes this. For a VRLA string with an exponent around 1.25, a two-hour discharge gets roughly 56% of the rated amp-hours — so the naive amp-hours-over-amps answer is out by nearly a factor of two. At the twenty-hour rate the correction vanishes entirely, which is the point: the error is a function of how hard you are pushing, not a constant to subtract.

Depth of discharge, temperature and age come off after that, and they compound. A four-year-old VRLA string at 0 °C, discharged fast, holding 80% of its rating and using 80% of what remains, delivers about a third of what the label suggests.

Why low voltage makes conductors interesting

Power is voltage times current, so at a twentieth of the voltage you have twenty times the current for the same load. Everything downstream of that scales: conductor size, lug size, fuse rating, busbar, and the heat dissipated in all of it.

Both conductors carry the current, so the resistive length of a run is twice its physical length. Halving that by forgetting the return path is a common and consistently optimistic error.

The test that matters is not whether the drop is a small percentage of 48 V. It is whether the equipment still sees enough voltage at the end of a battery discharge, when the supply is already down at the low-voltage disconnect. A drop of 2.7 V sounds trivial against 48 V and is fatal against the two volts of headroom between a 42 V disconnect and a 40 V equipment minimum.

What this does not do

This is engineering arithmetic, not a wiring-rules check. Conductor sizing here considers voltage drop only — not current-carrying capacity, grouping, installation method, ambient derating, fault-current withstand or protection coordination, all of which can demand a larger conductor than the drop alone. It is not a certification of anything, and the applicable electrical code governs.

Battery figures are model output. Vendor discharge curves are the authority, they differ between products, and a real string's condition is measured rather than calculated.

Everything is calculated in your browser. Equipment lists, site loads and battery details are not uploaded.