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Why your PDU says 16 A and the breaker says 25

Watts to amps is one division. Choosing the breaker is three, because a continuous load may only use eighty percent of it — and the outlet count is a third calculation again.

By Alex Xu · 21 October 2026 · 5 min read

A rack draws 3.6 kW. The PDU’s outlets are rated 16 A. The feed is a 20 A breaker. All three numbers appear to agree, and the installation trips after about an hour at full load, in a warm rack, with no fault present. Nothing is broken: the arithmetic was never done, only estimated.

Sizing a rack feed is three calculations, in a fixed order, and skipping the second one is the usual mistake.

One: watts to amps

For a single-phase feed:

I = P / (V x PF)

3,600 W at 208 V, PF 0.95
I = 3600 / (208 x 0.95) = 3600 / 197.6 = 18.2 A

For a three-phase feed, the line-to-line voltage is the one on the nameplate, and the current is lower by sqrt(3):

I = P / (sqrt(3) x V_LL x PF)

3,600 W at 400 V three-phase, PF 0.95
I = 3600 / (1.732 x 400 x 0.95) = 3600 / 658.2 = 5.5 A

The sqrt(3) is not a fudge factor. Three phases deliver 3 x V_phase x I x PF, and V_LL = sqrt(3) x V_phase, so the line-to-line form carries the root of three. Entering the phase-to-neutral voltage instead — 230 V on a 400 V supply — overstates the current by a factor of sqrt(3), which is the difference between a 16 A and a 10 A breaker.

Power factor matters and is not 1. Modern 80 PLUS Gold and Titanium supplies sit around 0.95; 0.9 is the conservative assumption if the supply is unknown. Ignoring it understates the current by five to ten percent, which is exactly the margin the next step is about to consume.

Two: the eighty percent rule

A continuous load — three hours or more — may not exceed eighty percent of the overcurrent device protecting it. Equivalently: size the breaker at 125 % of the load. This is a requirement, not a safety margin someone invented for UPS sizing.

So the largest continuous load each standard breaker can carry is:

BreakerMax continuous loadMax at PF 0.95, 208 V
15 A12 A2,366 W
20 A16 A3,154 W
25 A20 A3,943 W
30 A24 A4,731 W
50 A40 A7,886 W

Return to the 18.2 A example. It needs a breaker of at least 18.2 x 1.25 = 22.8 A. A 20 A breaker is not adequate — it is at 114 % of its rating continuously. The smallest standard size that works is 25 A, and in practice a 30 A single-pole is what gets installed, because 25 A is less common in that form factor.

The 400 V three-phase version draws 5.5 A, needs 6.9 A of breaker, and lands on a 10 A device. Same rack, two feeds, a factor of three between the breaker sizes — which is why the voltage cannot be assumed when someone hands you a watt figure.

The failure this prevents is not a fire. It is a nuisance trip at thermal equilibrium: a breaker loaded at 90 % passes current happily at 20 °C ambient and opens after an hour in a rack that is at 35 °C. That reads as a hardware fault and gets escalated as one.

Three: the outlets

Connectors have their own rating and their own derating, and they are lower than the breaker:

ConnectorRatedDerated (80 %)At 230 VAt 208 V
C13 IEC 6032010 A8 A1,840 W1,664 W
C19 IEC 6032016 A12.8 A2,944 W2,662 W

So 3.6 kW on 230 V is two C13s at minimum (3,600 / 1,840 = 1.96, so two), or one C19 plus one C13 if the load can be split that way. At 208 V it is three C13s (3,600 / 1,664 = 2.16). The rack does not care about your breaker size; the connector is a separate bottleneck and it is usually the one that surprises people.

A 16 A nameplate next to a 20 A breaker is not permission to draw 16 A. It is a connector rating and a breaker rating, both of which are derated, and the smaller of the two derated figures is the real limit.

The two numbers people get wrong

  • Nameplate watts are not draw. A PSU’s nameplate is its maximum output, and the input draw is that divided by efficiency at the operating point. Sizing to the sum of nameplates over-sizes the feed, which is safe and wasteful; sizing to the sum of measured draw with no efficiency correction under-sizes it. Use the measured figure at the plug, or the vendor’s derated configuration.
  • Dual-corded gear does not halve. Two feeds both have to be able to carry the whole rack, because either one may be the only one. The breaker math runs twice, not once divided by two.

Do the three steps in order

The PDU power calculator takes a rack load in watts and does all three: line current for a single- or three-phase feed, the 80 % derating, the next standard breaker size, and the C13 and C19 count assuming a quarter of the load lands on C19. It reports the numbers rather than a recommendation, because the last decision — which breaker your facility actually stocks — is not one a calculator can make.

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