The table gives a ceiling, not a number. At 480 V, 125 percent of a 1,000 kVA primary is 1,503 amps, and the ladder offers 1,200 or 1,600. Field practice takes 1,600, which is 133 percent. That is legal. It is also why the primary breaker trips on the first cold start, and the table cannot tell you that.
That single gap is the whole subject. The code table answers one question, which is how large a device you are allowed to install. It does not answer how large a device will hold, and those are different questions with different answers. Once you separate them, most of the arguing about transformer protection stops being an argument.
Here is what follows. The table read correctly, including the 2026 change of address that has burned a few people. The reason 125 percent is a fifteen point wide region rather than a value. The arithmetic of inrush against magnetic trip tolerance, which is where field practice actually comes from. Four fixes and what each one costs you. Then the three checks that live outside the table entirely, which is usually where a drawing fails. A worked example, six submittal lines, and what to write on the RFQ.
The table is a ceiling, and in 2026 it moved
Article 450 protects the transformer. It does not protect the conductors on either side, it does not coordinate anything, and it does not care what is downstream. That narrow scope is why the numbers look generous. A 250 percent primary device sounds insane until you remember it is not trying to be an overload relay.
For units rated 1,000 volts and below, the rule lives in Table 450.3(B) in the 2023 edition and in Table 450.5(B) in the 2026 edition. That is not a typo and it is not a regional variant. The 2026 code added a Listing Requirements section at .2 and a Reconditioned Equipment section at .3 to a long list of articles, and in Article 450 both of those slots were created new immediately after 450.1. Everything behind them slid two places. Transformer overcurrent protection is at 450.5 now, and both of its tables moved with it.
| Edition | Section for transformer overcurrent protection | What sits at .2 | What sits at .3 |
|---|---|---|---|
| Through 2023 | 450.3 | Definitions | The rule itself, two subsections down from the top |
| 2026 | 450.5 | Listing Requirements | Reconditioned Equipment |
If you are holding a 2026 book and you look up 450.3, you will find a section titled Reconditioned Equipment with two reserved subsections. It is real code text, it reads normally, and it has nothing to do with transformers. That is the worst kind of wrong page, because nothing about it looks wrong.
The confusion runs both ways, which is worth knowing before you argue with anybody. In the editions through 2023 the number 450.5 is the autotransformer section for units over 600 volts. So the same three digits mean transformer overcurrent protection in one book and autotransformers in another, and a plan review that quotes 450.5 without saying which edition can be quoted back at you from the opposite direction. Say the edition out loud, or say the year, and the argument gets shorter.
One more thing about scope. This table stops at 1,000 volts. Above that, Table 450.3(A) in the old numbering works on impedance and on whether the location is supervised, with 300 and 600 percent numbers that have nothing to do with the low voltage ladder. If you are working on a 4,160 V primary, the percentages you remember from the 480 V world do not apply and the rounding rules do not either.
The four percentages, and the only one you design to
There are four numbers in circulation for low voltage: 167, 125, 250 and 300. Only two of them live in Table 450.3(B), and only one of them is a design target.
| Primary current | Primary device, maximum | Secondary device, maximum | Where it is used |
|---|---|---|---|
| Less than 9 A | 167% | 167% | Control and small single phase units |
| 9 A or more, primary only | 125% | Not required | The default for most dry types |
| 9 A or more, both sides | 250% | 125% | Where inrush has to be ridden through |
The 300 percent figure belongs to Table 450.3(A), which is over 1,000 volts, and it gets quoted into low voltage discussions constantly. It does not apply.
The 9 amp threshold sounds arbitrary until you convert it. Nine amps at 480 V three phase is 7.5 kVA. Nine amps at 208 V three phase is 3.2 kVA. Nine amps at 240 V single phase is 2.2 kVA. So the 167 percent row covers control transformers, buck-boost units, and almost nothing else in a commercial building. If you are arguing about a 300 kVA unit, that row is not on the table.
And the word “maximum” is doing real work in that table header. A device smaller than the maximum is compliant. A device at the maximum is compliant. There is no minimum in the table, because the table is not trying to make the transformer work, only to bound the damage.
125 percent is a fifteen point wide region, not a number
Here is the part that gets skipped. You compute 125 percent of the primary full load current and you get a number like 1,503.5 amps. The National Electrical Code does not make devices in that size. Article 240.6(A) publishes a ladder, and the ladder is coarse.
So you have two compliant moves. Go down to the largest standard rating at or below 125 percent. Or, where the value does not correspond to a standard rating, go up one step, which one of the table’s notes permits. Both are legal. They are not both sensible, and the distance between them is where field practice lives.
At 480 V three phase, the whole ladder looks like this. Full load current is kVA times 1.2028, and 125 percent of that is kVA times 1.5035.
| kVA | Primary FLA | 125% target | Largest standard at or below | Next standard above | Device as % of FLA |
|---|---|---|---|---|---|
| 45 | 54.1 | 67.7 | 60 | 70 | 129.3% |
| 75 | 90.2 | 112.8 | 110 | 125 | 138.6% |
| 112.5 | 135.3 | 169.1 | 150 | 175 | 129.3% |
| 150 | 180.4 | 225.5 | 225 | 250 | 138.6% |
| 225 | 270.6 | 338.3 | 300 | 350 | 129.3% |
| 300 | 360.8 | 451.1 | 450 | 500 | 138.6% |
| 500 | 601.4 | 751.8 | 700 | 800 | 133.0% |
| 750 | 902.1 | 1,127.6 | 1,000 | 1,200 | 133.0% |
| 1,000 | 1,202.8 | 1,503.5 | 1,200 | 1,600 | 133.0% |
| 1,500 | 1,804.2 | 2,255.3 | 2,000 | 2,500 | 138.6% |
| 2,000 | 2,405.6 | 3,007.0 | 3,000 | 4,000 | 166.3% |
| 2,500 | 3,007.0 | 3,758.8 | 3,000 | 4,000 | 133.0% |
Look at the last column. There are only four values in it across the entire ladder: 129.3, 133.0, 138.6 and 166.3. The rule that says 125 percent produces a device that is actually somewhere between 129 and 166 percent of full load current, and you do not get to choose where. The ladder chooses.
This is not a rounding detail. It is the reason two engineers can both be code compliant and disagree by one frame size, and it is why a submittal that says “protected at 125 percent” is almost always a false statement about the device that got installed. The device is at 133. The sentence is at 125. Nobody is lying, they are just describing the calculation instead of the hardware.
The rows where the table’s own number is useless
Two of the rows above have a floor that you cannot use.
At 1,000 kVA the primary is 1,202.8 amps and 125 percent is 1,503.5. The largest standard rating below that is 1,200 amps. It is compliant, it is 99.8 percent of full load current, and it will trip at nameplate load. So it exists in the table and it does not exist in practice. You are going to install a 1,600.
At 2,500 kVA the same thing happens. Primary is 3,007.0 amps, 125 percent is 3,758.8, and the largest standard rating below is 3,000 amps, which is 99.8 percent of full load. Unusable. You install a 4,000.
So at those two ratings the table’s permission to go down is theater. The only device that works is the one above, and at 1,000 kVA that is 133.0 percent.
The 33 percent cliff at 2,000 kVA
The 2,000 kVA row is the one that costs money. Primary current is 2,405.6 amps, 125 percent is 3,007.0 amps, and 3,000 amps is a standard rating. It sits seven amps below the target, so it is compliant at 124.7 percent. The next standard rating up is 4,000 amps, which is 166.3 percent.
Seven amps is the entire distance between a 3,000 amp frame and a 4,000 amp frame. Both are legal. One of them is 124.7 percent of full load, which is exactly what the rule intended. The other is 166.3 percent, which is a third more device than the calculation asked for, and on a 480 V service that is a different switchboard, a different available fault current study, and a different price.
If you have ever wondered why 2,500 kVA and 2,000 kVA units seem to attract oversized gear, that is why. The ladder has a hole in it right there. Between 3,000 and 4,000 amps there is nothing, and a 2,000 kVA transformer’s 125 percent value lands in the hole by seven amps.
You will see this seven amp miss twice if you work on these long enough. It is the same distance by which 2,500 kVA at 480 V misses 3,000 amps on full load current, since 3,007.0 is what both calculations produce. Two different questions, one answer, and both of them sit just outside a standard rating.
At 208 volts the drift is worse
The ladder is fixed in amps. The current scales as one over voltage. So the same rule applied to a 208 V secondary drifts further, because the gap between adjacent standard ratings is a larger fraction of the current at that voltage.
| kVA | Secondary FLA at 208 V | 125% target | Largest at or below | Next above | Device as % of FLA |
|---|---|---|---|---|---|
| 45 | 124.9 | 156.1 | 150 | 175 | 140.1% |
| 75 | 208.2 | 260.2 | 250 | 300 | 144.1% |
| 112.5 | 312.3 | 390.3 | 350 | 400 | 128.1% |
| 150 | 416.4 | 520.4 | 500 | 600 | 144.1% |
| 225 | 624.5 | 780.7 | 700 | 800 | 128.1% |
| 300 | 832.7 | 1,040.9 | 1,000 | 1,200 | 144.1% |
| 500 | 1,387.9 | 1,734.8 | 1,600 | 2,000 | 144.1% |
| 750 | 2,081.8 | 2,602.2 | 2,500 | 3,000 | 144.1% |
| 1,000 | 2,775.7 | 3,469.7 | 3,000 | 4,000 | 144.1% |
A 1,000 kVA unit with a 208Y/120 secondary lands on 4,000 amps, which is 144.1 percent of the secondary full load current. That is nineteen points above the rule. Not because anyone chose it, but because 3,469.7 has no home on the ladder and 4,000 is the next step.
Note also that the 3,000 amp option exists and is 108.1 percent. It is compliant. It will also trip when the building is fully occupied and the summer load arrives, because 108 percent is not a margin, it is a coincidence.
The 167 percent row, and where it lives
The 167 percent row gets skipped in most discussions because it seems to apply to nothing. Convert the 9 amp threshold and it stops being abstract.
| System | Current equal to 9 A | kVA at that current |
|---|---|---|
| 480 V three phase | 9 A | 7.5 kVA |
| 208 V three phase | 9 A | 3.2 kVA |
| 240 V single phase | 9 A | 2.2 kVA |
| 480 V single phase | 9 A | 4.3 kVA |
So the row covers control power transformers, buck-boost units, doorbell and signal transformers, and the small dry types inside equipment. It does not cover anything you would call a distribution transformer, which is why most people never use it and why the people who do use it are usually standing in front of a control cabinet.
The ladder down there is worse, not better. Take a 1 kVA control transformer with a 480 V single phase primary. Primary current is 2.08 amps. At 167 percent that is 3.48 amps. Standard fuse ratings jump 3, 6, 10, so the 3 amp fuse is 144 percent and the next one up is 6 amps at 288 percent. There is nothing between, and a 4 amp device does not exist in that part of the ladder.
That coarseness is why control circuit protection is so often wrong in the other direction. A 40 VA transformer on a 480 V primary draws 0.083 amps. It sits behind a 15 amp breaker, which is 180 times its rating. A dead short on the 24 V secondary might pull 3 amps on the primary side, which that breaker will never see. The transformer cooks and the breaker is happy. This is the case where a 1 or 2 amp fuse and nothing else is the right answer, and where a protective device sized to the panel rather than to the load is worse than no device at all, because it looks like protection.
Motor control circuit transformers get their own route through 430.72(C), which is where the very large percentages that circulate for tiny primaries come from. If you are looking at a control transformer and trying to make 450.3 fit, check 430.72 first. The percentages there are generous because the load is inherently current limited by the control wiring, and that reasoning does not transfer to a distribution transformer of any size.
Protection is not coordination
The table sets a maximum. Coordination sets an actual. Those are two different numbers and the installed device has to satisfy both, which means the real answer is the smaller of the two.
Here is why they diverge on a transformer. Take the 1,000 kVA unit with a 1,600 amp primary and a 4,000 amp secondary. A bolted fault at the secondary terminals is 48,273 amps. The secondary device sees that as 12.1 times its rating. The same fault, reflected through the transformer, is 20,918 amps at the primary, and the primary device sees that as 13.1 times its rating.
Twelve and thirteen. The two devices are at almost the same multiple of their own rating for the same event, which means their instantaneous elements are racing and neither is clearly faster.
| Device | Rating | Current it sees on a secondary bolted fault | Multiple of its own rating |
|---|---|---|---|
| Primary | 1,600 A | 20,918 A | 13.1x |
| Secondary | 4,000 A | 48,273 A | 12.1x |
That is why a fault on one feeder trips the whole building. Both devices are in their instantaneous band at the same time, and whichever one has the faster clearing time wins, which is a coin flip decided by manufacturing tolerance.
The fix is a deliberate time delay somewhere. Usually the primary gets a short time element set at 0.1 to 0.2 seconds with instantaneous off, so the secondary device clears first and the primary only operates if the secondary fails to. That works, and it costs let through energy into the transformer during a close in fault, which is the trade the code lets you make by putting the primary at 250 percent in the first place.
Note what this does to the earlier discussion. The primary could be a 1,600, a 2,000 or a 3,000 and all three are inside the table. Only one of them coordinates with a 4,000 amp secondary. The table did not pick it, and neither did the calculation. Coordination picked it.
Why field practice always takes the top of the window
Ask an electrician why the device is 1,600 instead of 1,200 on a 1,000 kVA unit and you will get a shrug and “that’s what the calc said.” The calc said 1,503.5. The calc did not say 1,600. Something else made that decision, and it is not in the table.
It is inrush. When you close the primary device on a transformer, the core has to establish flux from whatever residual state it happens to be in. If the breaker closes near the voltage zero crossing and the core is still magnetized from the last time it was energized in the opposite direction, the core saturates within the first half cycle and the winding looks like an air core inductor. Current spikes. Published figures for cast resin and dry type units run 8 to 12 times rated current for about 0.1 second, which is five to six cycles at 60 Hz. Liquid units and large cores sit at the top of that band.
That is the whole reason the 250 percent row exists. It is not there because transformers are tough. It is there because the device that protects a transformer from a fault is also the device that has to survive the transformer starting up, and those two jobs pull in opposite directions.
The number nobody multiplies by
Here is the arithmetic that turns this from folklore into a decision.
A thermal magnetic breaker’s instantaneous element is set at some multiple of its own rating, typically 8, 10 or 12 times. Under UL 489 the magnetic trip is allowed to operate anywhere within plus or minus 20 percent of that setting. So a breaker marked 10 times can trip at 8 times, and that is not a defective breaker, that is a breaker inside its band.
Now put that breaker on a transformer protected at 125 percent of full load. Express everything in multiples of transformer full load current instead of multiples of the device, because that is the unit the inrush lives in.
| Device as % of FLA | 8x inrush needs a nominal magnet of | 12x inrush needs a nominal magnet of | Does a 10x magnet hold? |
|---|---|---|---|
| 124.7% | 8.0x | 12.0x | 8x only, and exactly at the edge |
| 129.3% | 7.7x | 11.6x | 8x only |
| 133.0% | 7.5x | 11.3x | 8x only |
| 138.6% | 7.2x | 10.8x | 8x only |
| 166.3% | 6.0x | 9.0x | Yes, both |
Read the middle column. To ride through a 12 times inrush with a device at 125 percent of full load, the breaker’s nominal magnetic setting has to be at least 12 times the device rating, because the worst case tolerance takes 20 percent off it and 12 divided by 1.25 is 9.6, which divided by 0.8 is 12.0. A 10 times breaker cannot do it. It is not close. It is 2 multiples short.
And going up the ladder does not rescue you. At 138.6 percent, the most common drift on this ladder, you would need an 10.8 times magnet. A 10 times nominal breaker at minus 20 percent is 8 times, which against a 138.6 percent device is 11.1 times full load current. A 12 times inrush at 12.0 times full load current walks straight through it.
The only row where a 10 times magnet survives a 12 times inrush is 2,000 kVA, where the device is forced to 166.3 percent of full load. And that is not a design, that is the ladder hole in the previous section accidentally buying you inrush margin.
So the honest statement is this. Upsizing within the table does not fix nuisance tripping on a high inrush transformer. It changes a 2 multiple shortfall into a 1.2 multiple shortfall. The instinct that says “give it a little more room” is measuring the wrong quantity. What you need is a different trip characteristic, not a bigger number.
What actually decides it
Two variables move together and one of them is invisible on the nameplate.
The visible one is the inrush multiple. A 45 kVA single phase unit with a small core might be at 8 times. A 1,500 kVA cast resin three phase unit is at 12. You cannot read this off the kVA, and manufacturers do not always publish it, so the safe assumption for anything three phase and cold started is the top of the band.
The invisible one is where the breaker happened to land inside its tolerance band. Two identical breakers from the same carton can differ by 40 percent on magnetic pickup. This is why an installation can run for two years and then nuisance trip after a utility outage. Nothing changed except the residual flux state of the core when the device reclosed, and the phase angle at which it closed.
That combination is why “it worked at the factory” is not evidence, and why the first energization after a long outage is the test that matters. It is also why the fix that field crews reach for first, holding the handle closed, is the one that ends with a welded contact.
Four fixes, ranked by what they actually cost
There are four honest options and one dishonest one. The dishonest one is a bigger device, covered above. The other four each trade something different, and none of the four is a number.
| Fix | What it buys | What it costs |
|---|---|---|
| Time delay fuse | Full inrush ride through at 125% | Single use, no remote trip, selectivity harder |
| Adjustable magnetic trip | Tunable window, resettable | Frame cost, and it must be set and locked |
| The 250% path | Inrush margin on paper | A second device, and the primary stops seeing overload |
| Full primary and secondary | Fault coverage the primary cannot provide | The most expensive of the four |
A time delay fuse
This is the cheapest correct answer and the one most often skipped because the panel schedule already says breaker.
A Class RK5 or Class J time delay fuse has no magnetic element at all. It has a thermal element with mass, and a short circuit element that works by restricted cross section rather than by sensing anything. Mass takes time to heat. A 90 amp dual element fuse carries several times its rating for seconds before the overload element opens, which is far more time than a 0.1 second inrush needs, and once current reaches the current limiting range the same fuse clears a bolted fault in half a cycle. There is nothing in it to be fooled by a five cycle spike.
The trade is real and it is not about price. A fuse means a spare inventory, a person who knows which class is in the holder, and no ability to trip remotely or to be opened by a shunt trip from a fire alarm panel. On a main service that last item usually kills the idea. On a 75 kVA unit feeding a machine, the fuse is almost always right.
One caution that gets missed. A time delay fuse sized at 125 percent is protecting the transformer, and the fuse curve has a minimum melting band. Check that the fuse will actually clear a secondary arcing fault, not just a bolted one. At low fault levels the melting time can run long enough to let the winding cook.
An adjustable magnetic trip
On an electronic trip breaker you can separate the long time element from the instantaneous element, and that is exactly the separation this problem needs. Long time pickup stays at or below 125 percent of full load current. Instantaneous goes to the top of its range, or gets turned off entirely with short time pickup taking over.
Two things to get right.
First, an electronic trip breaker is adjustable, and the table’s permission to round up to the next standard rating is written for a fuse or a nonadjustable circuit breaker. The exact wording names those two and nothing else. So on an adjustable device you do not get the round up. Your long time pickup has to be set at or below 125 percent of full load current, which on a 1,000 kVA unit means setting it at 1,500 amps on a 1,600 amp frame, not leaving the factory setting of 1,600.
That is a real inspection finding, and it is one of the few where the installer is technically wrong while the arithmetic looks right. A 1,600 amp frame left at 1.0 long time pickup is a 1,600 amp device in effect, which is 133 percent of full load, which exceeds 125 percent on an adjustable device. Where the trip resolves in small steps, 1,500 amps is available on a 1,600 amp frame. Where it steps in tenths of frame, 0.9 gives you 1,440 amps, which is 119.7 percent and also compliant. Either setting works. The factory setting does not, and the next person to open that door will not know what it was set to, so write it on the schedule.
Second, turning instantaneous off entirely is not free. You have just made the short time element the fastest thing in the chain, and short time delays are usually set for coordination, which means 0.1 to 0.3 seconds of let through. On a transformer primary that is a lot of energy into a winding during a close in fault. Higher pickup with instantaneous on is usually better than lower pickup with it off.
The 250 percent path
The table permits the primary device at 250 percent of primary full load current provided the secondary is protected at 125 percent. Two things about that arithmetic are not widely understood.
First, the 250 percent figure does not round up. The next size up allowance is written for the 125 percent values. So 250 percent is a hard ceiling, and where the computed value falls just above a standard rating you go down, not up.
| kVA at 480 V | Primary FLA | 250% target | Largest standard at or below | What you actually get |
|---|---|---|---|---|
| 75 | 90.2 | 225.5 | 225 | 249.4% |
| 150 | 180.4 | 451.1 | 450 | 249.4% |
| 300 | 360.8 | 902.1 | 800 | 221.7% |
| 500 | 601.4 | 1,503.5 | 1,200 | 199.5% |
| 750 | 902.1 | 2,255.3 | 2,000 | 221.7% |
| 1,000 | 1,202.8 | 3,007.0 | 3,000 | 249.4% |
| 1,500 | 1,804.2 | 4,510.5 | 4,000 | 221.7% |
So the headline 250 percent is really anything from 199.5 to 249.4, and at 500 kVA you get half of what you asked for. If you were counting on 250 percent to solve an inrush problem, the 500 kVA unit delivers 199.5 percent, and you now have a device that is only 2.0 times full load instead of 2.5.
Second, and this is the part that should make you nervous, the primary device stops being an overload device.
| Secondary loading | Secondary amps on a 1,000 kVA unit | What the 3,000 A primary device sees | As % of the device |
|---|---|---|---|
| 100% | 2,775.7 | 1,202.8 | 30.1% |
| 150% | 4,163.6 | 1,804.2 | 45.1% |
| 200% | 5,551.4 | 2,405.6 | 60.1% |
| 250% | 6,939.3 | 3,007.0 | 75.2% |
| 300% | 8,327.2 | 3,608.4 | 90.2% |
A transformer running at triple its rating, which is a fire, puts the primary device at 90 percent of pickup. It never operates. The only thing that clears that condition is the secondary device, which is why the table requires one on that path. The 250 percent row is not a trade you make to get inrush margin. It is a trade you make when you have already decided to protect the secondary and you need the primary to stop being the sensitive element.
If you want inrush margin and you want the primary to still see overload, the answer is a fuse or a different trip characteristic, not a percentage.
Full primary and secondary protection
The default good answer for anything industrial or commercial, and the reason is not legal. It is that a low level secondary arcing fault, the kind that sits at 1.5 times rated current and burns a cleat off a busbar, is invisible to a primary device at any percentage.
If you add a secondary device at 125 percent, you also add a device that has to interrupt the available fault current at the secondary terminals, and that number is bigger than most people expect.
| kVA, 480 to 208Y/120, Z 5.75% | Secondary FLA | Available fault current | Device and gear rating needed |
|---|---|---|---|
| 75 | 208.2 | 3,621 | 10 kA |
| 150 | 416.4 | 7,241 | 10 kA |
| 225 | 624.5 | 10,862 | 22 kA |
| 300 | 832.7 | 14,482 | 22 kA |
| 500 | 1,387.9 | 24,137 | 42 kA |
| 750 | 2,081.8 | 36,205 | 42 kA |
| 1,000 | 2,775.7 | 48,273 | 65 kA |
A 1,000 kVA transformer with a 208Y/120 secondary delivers 48,273 amps at the secondary terminals on an infinite bus, which puts you in the 65 kA product line. That is not the same switchboard as the 22 kA one the drawing showed.
This table is the reason a “we will just add a secondary main” decision in month nine blows a budget. The device is not the expensive part. The gear it sits in is.
Splitting the secondary six ways
One of the table’s notes allows the secondary protection to consist of up to six breakers or six sets of fuses grouped in one location, rather than a single main. Almost nobody uses it and it solves a specific cost problem.
On the 1,000 kVA unit with 2,775.7 amps of secondary, one 4,000 amp main is 144.1 percent and sits in 65 kA gear. Six devices instead, each carrying one sixth of the load:
| Arrangement | Amps per device | 125% of that | Standard device chosen | Total | As % of FLA |
|---|---|---|---|---|---|
| One main | 2,775.7 | 3,469.7 | 4,000 A | 4,000 A | 144.1% |
| Four ways | 693.9 | 867.4 | 1,000 A each | 4,000 A | 144.1% |
| Six ways | 462.6 | 578.3 | 600 A each | 3,600 A | 129.7% |
Six 600 amp breakers total 3,600 amps, which is closer to 125 percent than the single 4,000 amp main, and six 600 amp frames cost less than one 4,000 amp frame even though they add up to more amps. The six way split also means a fault on one feeder takes out one sixth of the building instead of all of it, which is a resilience argument nobody puts in the submittal.
The constraint is that all six have to be in one location, and the sum cannot exceed what one device would have been allowed. Read the exact wording in your edition before you design around it, because this note has been reworded more than once.
What the table does not protect
Three separate checks live outside Article 450, and on a real drawing they are done by three different people on three different days, which is why they get missed.
The conductors, 240.21(C)
A transformer secondary is a separately derived source, and the conductors leaving it need protection at their ampacity or a recognized tap rule. Article 450 does not do that for you. On a 1,000 kVA unit the secondary device is 4,000 amps, and the 25 foot tap rule wants conductor ampacity of at least one third of that, which is 1,333 amps.
| Conductor set | Ampacity at 75 C | Meets 1,333 A? |
|---|---|---|
| 3 x 600 kcmil | 1,260 | No, 73 amps short |
| 3 x 750 kcmil | 1,425 | Yes |
| 4 x 500 kcmil | 1,520 | Yes |
| 4 x 600 kcmil | 1,680 | Yes |
Three sets of 600 kcmil is a plan people draw, and it misses by 73 amps. That is the same kind of miss as the seven amp ladder hole, and it has the same cause, which is that somebody compared against full load current instead of against the device.
There is also an exemption worth knowing and worth not abusing. Where the secondary is two wire single phase or three wire delta to delta, the primary device can protect the secondary conductors, because the current ratio is fixed and there is no neutral to carry unbalance. A 208Y/120 secondary is four wire, so the exemption does not reach it. Most commercial work is four wire, so most commercial work does not get the exemption, and the drawing that assumes otherwise comes back.
The panelboard, 408.36(B)
A panelboard needs overcurrent protection not exceeding its rating, and that protection normally sits at or in the panelboard. Feed a panelboard through a transformer with primary only protection and you have to either provide a secondary device or show that the panelboard is protected some other recognized way.
This is the check that produces the “we need a main breaker in the panel” conversation after the panel is already built, which is the most expensive time to have it.
The interrupting rating, 110.9
Every device has to have a short circuit current rating at least equal to the available fault current where it is installed. That is separate from the amp rating, separate from the table, and it is the one that fails the loudest.
| Point in the system | Available fault current | Common gear rating | Verdict |
|---|---|---|---|
| 1,000 kVA primary at 480 V, Z 5.75% | 20,918 A | 22 kA | Just inside, 5.2% margin |
| 1,000 kVA secondary at 208 V, Z 5.75% | 48,273 A | 22 kA | Fails badly |
| 1,000 kVA secondary at 208 V, Z 5.75% | 48,273 A | 65 kA | Correct |
Two devices on the same transformer, one at 22 kA and one at 65 kA, and both numbers are right. The 22 kA figure is the primary, where the current is small. The 65 kA figure is the secondary, where it is 2.3 times larger. Anybody who carries the primary number over to the secondary has just underbuilt the switchboard by a factor of three.
The order to do the checks in
Most of the mistakes in this article come from doing these in the wrong order, or from stopping after step five. The order matters because each step can change the input to the next one.
- Read the nameplate. kVA, primary and secondary voltage, phase count, percent impedance, tap range, and whether there is a second kVA rating for forced air. Write all of it down. Half the arguments later are about which number is “rated.”
- Compute primary full load current. kVA times 1,000 divided by the square root of three divided by primary volts for three phase, or divided by volts for single phase. A 45 kVA unit at 480 V three phase is 54.1 amps and not 93.75, and the 93.75 comes from leaving out the 1.732.
- Decide the topology. Primary only, or primary and secondary. The code allows either. The load decides, because a low level secondary arcing fault is invisible to a primary device at any permitted percentage.
- Apply the percentage. 125 percent for primary only. 250 percent primary with 125 percent secondary for the two device path. 167 percent if primary current is under 9 amps.
- Walk the ladder twice. Largest standard rating at or below the target, and the next standard rating above it. Compute both as a percentage of full load current. If one of them is below 100 percent, strike it. Now you have a real choice instead of a number.
- Check the device type against the rounding rule. A fuse or a nonadjustable breaker may take the next size up on the 125 percent values. An adjustable trip breaker may not, so its long time pickup gets set, not selected.
- Check inrush. Multiply 15 by full load current and divide by the device rating. That is the minimum nominal magnetic multiple that will hold a 12 times inrush assuming the worst case tolerance. If the answer is above 12, no standard thermal magnetic breaker will do it and you need a fuse or an electronic trip.
- Check the secondary fault current and confirm every device and enclosure you just added is rated for it. This is the step that changes the switchboard, not the device.
- Check the three outside the table. Conductor ampacity at 240.21(C), the panelboard at 408.36(B), the interrupting rating at 110.9.
Step seven is the one that is missing from every checklist I have seen, and it is the one that decides whether the building comes back up on the first try after an outage.
One transformer, four compliant designs
Take a 1,000 kVA unit, 480 V delta primary, 208Y/120 secondary, 5.75 percent impedance, feeding a mixed commercial load. Every design below is code compliant. They are not close in cost or in behavior.
Design A, primary only at 125 percent. The primary device is 1,600 amps, which is 133.0 percent. There is no secondary device, so 240.21(C) tap rules govern the secondary conductors, and the panelboard question from 408.36(B) has to be answered. Available fault current at the primary is 20,918 amps, so 22 kA gear works. This is the cheapest design and the one with the least fault coverage. A secondary arcing fault at twice rated current is invisible.
Design B, primary only, with a time delay fuse. Same 1,600 amp rating but in a Class L or RK5 time delay fuse instead of a breaker. The inrush problem disappears because there is no magnetic element to fool. You give up remote trip and resettability, and you buy a spare inventory. On a unit this size the fuse and its holder are usually cheaper than the breaker, which surprises people.
Design C, primary and secondary at 125 percent. Primary 1,600 amps, secondary 4,000 amps. Now the secondary device has to interrupt 48,273 amps, so the switchboard and every device in it move to 65 kA. This is the design most industrial projects end up with, and it is the one where the gear cost is triple what the schedule assumed.
Design D, the 250 percent path. Primary 3,000 amps at 249.4 percent, secondary 4,000 amps. You have bought inrush margin on paper and given up primary overload protection entirely, as the table in the previous section showed. You are also now carrying a 3,000 amp frame to do the job a 1,600 amp fuse did in Design B.
| Design | Primary device | Secondary device | Secondary gear | Primary sees overload | Inrush margin |
|---|---|---|---|---|---|
| A | 1,600 A breaker, 133.0% | None | n/a | Yes | Poor at 12x |
| B | 1,600 A time delay fuse | None | n/a | Yes | Good |
| C | 1,600 A breaker, 133.0% | 4,000 A, 144.1% | 65 kA | Yes | Poor at 12x |
| D | 3,000 A breaker, 249.4% | 4,000 A, 144.1% | 65 kA | No | Marginal |
Design B wins on almost every axis for this transformer, and it is the one that appears least often on drawings. The reason is not technical. It is that the schedule template has a column for breaker frame size and no column for fuse class.
Three things that change the answer
A second kVA rating on the nameplate. Fan cooled dry types are often sold with a self cooled rating and a forced air rating. If a 1,000 kVA unit carries 1,250 kVA with the fans running, which one is “rated” for the purpose of the table? The safe read is the self cooled number, because a failed fan must not leave the protection oversized. Protecting to 1,250 kVA gives a primary current of 1,503.5 amps, 125 percent of that is 1,879, and the ladder takes you to 2,000, which against the self cooled 1,202.8 is 166.3 percent. That is the same band the 2,000 kVA ladder hole produced earlier, and it is a lot of device to protect a winding that is only rated for 1,000 kVA when the fan stops.
Taps. A unit set on the minus 5 percent tap delivers rated secondary voltage from 95 percent primary voltage, so at full load the primary current is 5.3 percent higher than nameplate. Whether the table percentage applies to the tapped current is argued, and the conservative answer is yes. On the 1,000 kVA unit that is 1,266.1 amps instead of 1,202.8, so 125 percent becomes 1,582.6, which still lands on 1,600. Check it, record it, and expect it to change the arithmetic without changing the frame. It changes the frame when the target is already within 5 percent of a ladder step, which is exactly the seven amp situation described above.
Harmonics. The table limits current, and harmonic current heats a winding more per amp than fundamental current does because the losses scale with frequency in the windings and the eddy paths. A K-rated transformer is designed for that. A standard unit carrying a high harmonic load is not, and no percentage in Article 450 accounts for it. If the load is mostly rectifiers, the current limit and the thermal limit come apart, and the protection that satisfies the table will not satisfy the winding.
Six lines I see on submittals
- “Primary protected at 125 percent.” It is not. It is at 133.0, because 1,503.5 became 1,600. Say 133 and say why, or the reviewer will do it for you and assume you did not notice.
- A 1,200 amp device on a 1,000 kVA unit. Compliant, 99.8 percent of full load, trips at nameplate. Somebody found the floor of the ladder and stopped without checking whether the floor was above the load.
- 3,007 amps rounded to 3,000 and called good. At 2,000 kVA that is 124.7 percent and it is fine. At 1,000 kVA the same 3,007 is the 250 percent value, and 3,000 is 249.4 percent, and that is also fine. Two different questions, one number, and both of them sit seven amps outside a standard rating. Check which question you are answering.
- A 1,600 amp frame left at factory long time pickup. On an adjustable trip device the round up does not apply, so 1,600 is 133 percent and it is over the limit. Set it at or below 1,503.5 and record the setting.
- A 250 percent primary device with no secondary device. The table does not allow that combination. The 250 percent number exists only because a secondary device at 125 percent is holding the other end.
- Primary fault current carried onto the secondary gear. 20,918 amps becomes 22 kA gear on the primary and gets copied to a secondary that actually sees 48,273. The number looks familiar so nobody rechecks it.
Every one of these is a five minute check. None of them is a calculation error. They are all cases of reading a table that answers a narrower question than the one being asked.
What to put on the RFQ
The table tells you the ceiling. The RFQ is where you tell the factory what you actually need, and four items do most of the work.
State the inrush assumption. Ask for the published inrush multiple and its duration at rated voltage. If they will not give you a number, ask for the first cycle peak in amps. A supplier who builds these units knows the figure and a supplier who is reselling them does not, which is useful information by itself.
State the impedance, not just the kVA. Everything downstream depends on it. At 5.75 percent a 1,000 kVA unit delivers 20,918 amps at 480 V. At 4.0 percent it delivers 30,070. Same transformer by name, different switchboard by a product line.
State the device type and the setting. Breaker or fuse, frame size, and for an adjustable trip the long time pickup in amps. Writing “protected at 125 percent” transfers the decision to whoever builds the lineup, and they will use the frame rating.
Ask for the compliance table row. Efficiency requirements for distribution transformers live in 10 CFR 431.196 as amended, with a compliance date of April 23, 2029, and the table is indexed by kVA and by basic impulse level. Ask which row the unit is being built to. A custom kVA has no row, which is one more reason the ladder is the ladder.
If you want the primary protection worked through against your actual secondary fault current and your actual gear, the fastest path is to send the single line and the utility available fault current to our engineering desk and let somebody check the three numbers that are not on the table.
FAQ
Does the 125 percent rule apply to the primary or the secondary?
Both, and they are separate devices. Table 450.3(B) caps the primary at 125 percent of primary full load current and the secondary at 125 percent of secondary full load current. On a 1,000 kVA 480 to 208Y/120 unit that is 1,503.5 amps on the primary, which lands on 1,600, and 3,469.7 on the secondary, which lands on 4,000. Neither number tells you anything about the other.
Can I always round up to the next standard breaker size?
Only on the 125 percent values, and only for a fuse or a nonadjustable circuit breaker. The next size up allowance is written for those two device types. It does not apply to the 250 percent primary value, and it does not apply to an adjustable electronic trip breaker, which means the long time pickup has to be set at or below 125 percent rather than left at the frame rating.
Why does my primary breaker trip when I energize the transformer?
Inrush. A cold core draws 8 to 12 times rated current for roughly 0.1 second, and the breaker’s magnetic element is allowed to be 20 percent below its marked setting. A 10 times breaker on a device sized at 125 percent of full load sees the trip point at 10 times full load, which is below a 12 times inrush. The device is not defective and it is not undersized in the code sense.
Does a bigger breaker fix nuisance tripping?
No, and this is the most expensive wrong answer in the subject. Going from 125 to 138.6 percent of full load reduces the shortfall from 2 multiples to about 1.2 multiples. What you need is a device with a different trip characteristic, either a time delay fuse or an adjustable magnetic trip set high, not a larger frame carrying the same characteristic.
What is the 250 percent primary protection actually for?
Riding through inrush where a fuse or a high magnetic setting is not acceptable. It only exists when the secondary is protected at 125 percent, and it turns the primary device into a short circuit device. A 1,000 kVA unit on that path has a 3,000 amp primary device, and the secondary would have to reach 300 percent of rating before that device sees 90 percent of pickup.
Can the 250 percent value be rounded up?
No. The next size up allowance applies to the 125 percent values. Where 250 percent of primary current falls just above a standard rating you go down, so a 500 kVA unit gets 1,200 amps, which is 199.5 percent rather than 250. On this ladder the achievable value runs from 199.5 to 249.4 percent depending on the kVA.
Is 450.3 still the right section number?
In the 2026 edition it is 450.5, and both tables moved with it. The 2026 code added a Listing Requirements section at 450.2 and a Reconditioned Equipment section at 450.3, which pushed transformer overcurrent protection two places down. The percentages did not change. Only the address did.
Why is the secondary fault current so much higher than the primary?
Because current scales as one over voltage while kVA stays the same. A 1,000 kVA unit at 5.75 percent impedance delivers 20,918 amps at 480 V and 48,273 amps at 208 V, a ratio of 2.3. The device on the secondary needs a 65 kA rating while the primary device is comfortable at 22 kA, and the two numbers belong to different product lines.
Do I need a secondary device if the primary is at 125 percent?
The table does not require one. Article 450 protects the transformer and nothing else, so the conductors still need 240.21(C), the panelboard still needs 408.36(B), and the interrupting ratings still need 110.9. In practice most industrial and commercial work ends up with a secondary device because a low level arcing fault at 1.5 times rated current is invisible to any primary device.
What is the six breaker allowance on the secondary?
One of the table’s notes permits the secondary protection to be up to six breakers or six sets of fuses grouped in one location rather than a single main, with the total not exceeding what one device would have been allowed. On a 1,000 kVA 208 V secondary, six 600 amp devices total 3,600 amps, which is 129.7 percent of full load against 144.1 percent for a single 4,000 amp main.
Should I use a fuse or a breaker on the primary?
A fuse if the inrush is high and remote operation is not required, because it has no magnetic element to be fooled and it costs less at the same rating. A breaker if you need a shunt trip, remote indication, or easy reset. On units below about 300 kVA the fuse is usually the better answer and the schedule template usually prevents it.
Does any of this change above 1,000 volts?
Yes, completely. Table 450.3(A) in the old numbering works from impedance and from whether the location is supervised, with different percentages for fuses and for breakers, and the 125 and 250 numbers from the low voltage table do not carry up. A 4,160 V primary is a different table, a different rounding rule, and usually a protective relay rather than a molded case device.
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