Illustrative example only. Values are simplified for teaching and are not settings for any real installation. Real studies use manufacturer curves, verified field data, and utility-supplied source impedance.
1. The system
A commercial facility takes service at 13.2 kV from the utility and steps down through one transformer to a 480Y/277 V switchboard.
| Element | Data used in the study |
|---|---|
| Utility source at 13.2 kV | 8.0 kA available three-phase fault, X/R 10 |
| Service transformer | 2,500 kVA, 13.2 kV delta to 480Y/277 V, Z = 5.75% |
| Transformer primary protection | Relay with 51 and 50 elements, 200:5 CTs, IEEE very inverse curve |
| 480 V main | 3,000 A breaker, electronic trip (long time, short time, ground fault) |
| 480 V feeder | 800 A breaker, electronic trip (long time, short time, instantaneous) |
| Branch device | 225 A thermal-magnetic molded case breaker |
2. Short-circuit analysis
Coordination is checked across the full range of fault current, so the first step is bounding it. Using the MVA method and ignoring motor contribution for simplicity:
- Utility source: √3 × 13.2 kV × 8.0 kA = 182.9 MVA
- Transformer: 2.5 MVA / 0.0575 = 43.5 MVA
- Combined at the 480 V bus: 1 / (1/182.9 + 1/43.5) = 35.1 MVA
- Fault current at 480 V: 35.1 MVA / (√3 × 0.48 kV) = 42.3 kA
- Same fault seen at 13.2 kV: 42.3 kA × 0.48 / 13.2 = 1,537 A
Transformer full-load current is 109 A on the primary and 3,007 A on the secondary. A real study also calculates minimum fault current (weakest source, arcing faults, end-of-feeder faults), because pickups must still operate there.
3. Setting the 480 V devices, bottom up
Coordination is built from the load toward the source. Each device is set first to protect what it serves, then checked against the device above it.
Branch breaker, 225 A
Thermal-magnetic, with a fixed instantaneous trip near 10 times rating (about 2,250 A). It has no adjustment, so it sets the floor everything above it must clear.
Feeder breaker, 800 A
- Long-time pickup 800 A
- Short-time pickup 6 × = 4,800 A, short-time delay 0.1 s band
- Instantaneous 10 × = 8,000 A
For faults below 8,000 A downstream of the 225 A breaker, the 0.1 s short-time band lets the branch breaker clear first. Above 8,000 A both instantaneous elements can operate together. Selectivity in that overlap cannot be shown with time-current curves alone; it needs the manufacturer's published selectivity tables for that specific breaker pair, or the study documents the limit.
Main breaker, 3,000 A
- Long-time pickup 3,000 A
- Short-time pickup 4 × = 12,000 A, short-time delay 0.3 s band
- Instantaneous off, so the main stays selective with the feeder up to the full 42.3 kA
- Ground fault pickup 1,200 A, 0.3 s delay (ground fault protection is required on this service under NEC 230.95: solidly grounded wye, more than 150 V to ground, 1,000 A or larger)
The main's short-time band clears by about 0.4 s at maximum fault. That is the number the primary relay has to beat with margin.
4. Setting the transformer primary relay
51 time overcurrent
- Pickup: 160 A primary, about 146% of transformer full load. With 200:5 CTs that is 4.0 A secondary.
- Curve and time dial: IEEE very inverse, t = TD × (3.922 / (M² − 1) + 0.0982), where M is current divided by pickup.
- Target: at the maximum 480 V fault (1,537 A primary, M = 9.6), operate at least 0.3 s after the main breaker clears. Aiming for about 0.7 s gives TD = 0.7 / 0.141 = 5.0.
| Primary current | Multiple of pickup | 51 operate time (TD 5.0) |
|---|---|---|
| 320 A | 2.0 | 7.0 s |
| 800 A | 5.0 | 1.3 s |
| 1,537 A (max 480 V fault) | 9.6 | 0.71 s |
Margin over the main breaker at maximum fault: 0.71 s minus about 0.4 s = about 0.3 s, which covers relay tolerance, CT error, and breaker operating time.
50 instantaneous
The instantaneous element must not see a 480 V fault, or it would trip the whole service ahead of the main breaker. It also has to ride through transformer inrush, roughly 12 times full load (about 1,300 A) for 0.1 s. Setting it at 2,000 A primary (50 A secondary) clears both with margin while still operating fast for a fault on the 13.2 kV side, where 8.0 kA is available.
Checking transformer damage
A 2,500 kVA three-phase unit falls in IEEE C57.109 Category II. Its thermal through-fault limit (I²t = 1,250, per-unit current and seconds) allows about 6.3 s at the 14 per-unit through fault the relay sees here, against a relay time of 0.71 s. On a delta-wye transformer, a secondary line-to-ground fault appears on the primary at only 58% of its per-unit value, so the damage curve is shifted left to 58% before checking. The relay still clears inside the shifted curve, at about 1.2 s.
That same delta winding is why the primary relay cannot be relied on for 480 V ground faults: zero-sequence current does not pass through it. The main breaker ground fault element carries that job.
5. What changes when 500 kW of solar is added
Now the facility adds 500 kW of inverter-based solar, backfed through an 800 A breaker on the 480 V switchboard. The fault current changes are small, but they matter in the wrong places.
- Inverter fault contribution is limited. Inverters typically contribute about 1.1 to 1.5 times rated current for a short time (manufacturer data governs). Rated current is about 600 A, so the 480 V fault level barely moves.
- The primary relay cannot see a utility fault. For a fault out on the 13.2 kV feeder, the inverters push roughly 720 A at 480 V, which is only about 26 A at 13.2 kV. That is far below the 160 A pickup. Overcurrent elements will not detect it.
- Separation has to come from voltage and frequency. 27, 59, and 81 elements at the point of common coupling, plus inverter anti-islanding, become the elements that disconnect the plant. The utility may also require 32 (directional power) or direct transfer trip.
- The delta primary is not a ground source. If the feeder islands with a line-to-ground fault, the unfaulted phases can rise toward line-to-line voltage. Utilities often require ground fault overvoltage detection (59N) on the 13.2 kV side, or a grounding transformer, for this configuration.
None of this is visible in the original study. That is the general rule: adding generation invalidates the assumption that fault current only flows one way, and the study has to be revisited. The voltage and frequency settings themselves are worked in the relay setting calculation example.
6. What the final report documents
- Source data and its date, including the utility fault current letter
- System model assumptions, such as motor contribution and minimum source cases
- Short-circuit results at every bus in scope
- Time-current curves for each protection chain, with damage curves shown
- A settings table for every device, ready for field implementation
- Selectivity limits that remain, stated plainly rather than hidden
A study written this way can be checked by a utility reviewer line by line. That is the standard we hold our protection coordination studies to.