10.2 Overcurrent Protection (50/51), Time-Current Curves (TCC) & Relay Coordination

10.2 Overcurrent Protection (50/51), Time-Current Curves (TCC) & Relay Coordination

Overcurrent protection is the most universally applied form of power system protection. It operates on the fundamental premise that short circuits and severe overloads produce current magnitudes substantially higher than normal full-load operating levels. Achieving selective coordination ensures that the protective device closest to the fault opens first, isolating only the faulted branch while leaving the remainder of the electrical distribution system energized and operational.

On the NCEES PE Electrical and Computer: Power examination, overcurrent coordination problems require precise mathematical evaluation of IEEE/IEC inverse curve equations, determination of Time Dial Settings ($TDS$), calculation of Coordination Time Intervals ($CTI$), and validation against transformer damage curves (ANSI/IEEE C57.109) and inrush constraints.

---

1. Overcurrent Relaying Mechanics: 50 vs. 51

Modern microprocessor overcurrent relays integrate both instantaneous and time-delayed overcurrent elements:

+---------------------------------------------------------------------------------------------------+
|                         INSTANTANEOUS (50) vs. TIME-OVERCURRENT (51)                              |
+---------------------------------------------------------------------------------------------------+
| Characteristic        | Instantaneous Element (Device 50)     | Time-Overcurrent Element (Device 51) |
| :---                  | :---                                  | :---                                 |
| **Operating Time**    | No intentional delay ($<1-2\text{ cycles}$) | Inverse time-delayed ($0.05-10.0\text{ s}$) |
| **Pickup Parameter**  | $I_{pickup,50}$ (Amperes primary/sec) | $I_{pickup,51}$ (Amperes primary/sec)|
| **Setting Basis**     | Set above maximum through-fault at    | Set at $1.25\times$ to $1.50\times$ full-load|
|                       | downstream bus ($>1.25\times I_{f,max}$)| continuous rating ($FLA$).           |
| **Primary Function**  | Severe close-in short-circuit clearing | Overload protection & coordinated     |
|                       | with minimum arc flash hazard.        | backup fault clearing.               |
| **Coordination Method**| Coordinate by current magnitude       | Coordinate by time-current curves    |
|                       | (definite current separation).        | (CTI time separation).               |
+---------------------------------------------------------------------------------------------------+

                       IDEALIZED TIME-CURRENT CURVE (TCC) PROFILE

    Operating Time (Seconds)
        ^
        |    \ (51 Inverse Time-Overcurrent Characteristic)
        |     \
   10.0 |      \  <-- Long trip time at low overload (e.g., 1.5x FLA)
        |       \
        |        \
    1.0 |         \
        |          \
        |           \               (50 Instantaneous Element Pickup)
    0.1 |------------+ - - - - - - - - - - *---------------------
        |                                  |  <-- High-speed trip (<0.03 s)
   0.01 |                                  |      for faults > I_pickup,50
        +----+-------+---------------------+---------------------> Current (Amperes)
            FLA   I_pickup,51            I_pickup,50

---

2. Standard IEEE / IEC Inverse Time Curve Equations

Electromechanical overcurrent relays utilize an induction disk whose rotational torque is proportional to $I^2$. Modern digital relays mathematically emulate or exceed these curves using standard parameterized equations defined by IEEE C37.112 and IEC 60255.

IEEE C37.112 Standard Equation

$t(I) = TDS \times \left[ \frac{A}{\left(\frac{I}{I_{pickup}}\right)^p - 1} + B \right] \quad [\text{seconds}]$

Where:

• $t(I)$ = Relay operating time in seconds.
• $TDS$ = Time Dial Setting (typically adjustable from $0.5$ to $15.0$).
• $I$ = Measured primary or secondary fault current magnitude $[\text{A}]$.
• $I_{pickup}$ = Relay current pickup setting $[\text{A}]$.
• $M = \frac{I}{I_{pickup}}$ = Plug Setting Multiplier (PSM) / Multiple of Pickup.
• $A, B, p$ = Dimensionless curve constants defining the degree of inverseness.

IEC 60255 Standard Equation

$t(I) = TMS \times \left[ \frac{A}{\left(\frac{I}{I_{pickup}}\right)^p - 1} \right] \quad [\text{seconds}]$

Where $TMS$ is the Time Multiplier Setting (analogous to $TDS$). Note that standard IEC equations do not include the additive constant $B$.

+---------------------------------------------------------------------------------------------------+
|                     STANDARD IEEE & IEC INVERSE CURVE CONSTANTS                                   |
+---------------------------------------------------------------------------------------------------+
| Curve Family                | Standard Code | Constant A | Constant B | Exponent p | Typical Use  |
| :---                        | :---          | :---       | :---       | :---       | :---         |
| **IEEE Moderately Inverse** | IEEE MI       | $0.0515$   | $0.1140$   | $0.02$     | Utility feeds|
| **IEEE Very Inverse**       | IEEE VI       | $19.61$    | $0.4910$   | $2.00$     | Long lines   |
| **IEEE Extremely Inverse**  | IEEE EI       | $28.20$    | $0.1217$   | $2.00$     | Fuse coord.  |
| **IEEE Short-Time Inverse** | IEEE STI      | $0.00342$  | $0.00262$  | $0.02$     | Motor locked |
| **IEC Standard Inverse**    | IEC Class A   | $0.1400$   | $0.0000$   | $0.02$     | European grid|
| **IEC Very Inverse**        | IEC Class B   | $13.500$   | $0.0000$   | $1.00$     | Feeder lines |
| **IEC Extremely Inverse**   | IEC Class C   | $80.000$   | $0.0000$   | $2.00$     | XFMR & Fuses |
+---------------------------------------------------------------------------------------------------+

[!TIP] Curve Selection Rule:

• Use Extremely Inverse (EI) curves when coordinating with downstream power fuses or protecting transformers, because the $I^2 t$ shape ($p = 2.0$) closely matches the thermal melting profile of fuses and the thermal damage envelope of transformers.
• Use Moderately Inverse (MI) curves when fault current variations between minimum and maximum generation scenarios are large, ensuring consistent operating times.

---

3. Fuse Protection Characteristics: MMT vs. TCT

Power fuses are non-adjustable, direct-acting thermal interrupting devices. Their performance is defined by two standardized published time-current curves:

1. Minimum Melting Time (MMT) Curve: The lowest boundary of the time-current band. Represents the minimum time required for the fusible element to melt at a given current without arcing. Damage occurs if current flows beyond this time.
2. Total Clearing Time (TCT) Curve: The upper boundary of the band. Represents the maximum total time required for the fusible element to melt, establish an arc, and fully extinguish the arc at current zero ($TCT = MMT + t_{arc}$).

                   FUSE TIME-CURRENT CHARACTERISTIC BAND

    Time (s)
        ^
        |       Total Clearing Time (TCT): Maximum fault clearing time
        |       .-----------------------
        |      /                       /
        |     /   Arcing Region       /
        |    /  (Melting + Arcing)   /
        |   /                       /
        |  /                       /
        | .-----------------------.
        | Minimum Melting Time (MMT): Threshold of irreversible thermal damage
        +----------------------------------------------> Current (A)

+---------------------------------------------------------------------------------------------------+
|                         FUSE COORDINATION CRITERIA & RULES                                        |
+---------------------------------------------------------------------------------------------------+
| Rule 1: Fuse-to-Fuse Coordination:                                                                |
|   --> Downstream Fuse Total Clearing Time (TCT) MUST be ≤ 75% of Upstream Fuse Minimum Melting   |
|       Time (MMT) across the entire fault range:                                                   |
|                                                                                                   |
|             TCT_downstream(I_f) ≤ 0.75 * MMT_upstream(I_f)                                        |
|                                                                                                   |
|   --> The 25% safety margin prevents pre-damaging or annealing the upstream fuse link during a    |
|       downstream through-fault.                                                                   |
|                                                                                                   |
| Rule 2: Relay-to-Fuse Coordination:                                                               |
|   --> Upstream Relay Curve must clear above downstream Fuse TCT by at least 0.15 to 0.20 seconds. |
|   --> Downstream Relay Curve must clear below upstream Fuse MMT by at least 0.20 seconds.         |
+---------------------------------------------------------------------------------------------------+

---

4. Time-Current Coordination & Coordination Time Interval (CTI)

The Coordination Time Interval ($CTI$) is the intentional vertical time buffer between the operating curves of two series-connected protective devices plotted on a common TCC. It guarantees that the downstream interrupting device completely isolates the fault before the upstream device's timing circuit reaches its trip threshold.

                      COORDINATION TIME INTERVAL (CTI) STACK

    Time (s)
        ^
        |        [ Upstream Relay Curve (51) ]  t_upstream
        |                   ^
        |                   |  <-- Safety Margin (~0.05 s)
        |                   |  <-- CT / Relay Tolerance (~0.05 - 0.10 s)
        |             CTI   |  <-- Relay Overtravel / Coasting (0.00 s digital / 0.10 s disk)
        |                   |  <-- Downstream Breaker Clearing Time (3-5 cycles = 0.05-0.08 s)
        |                   v
        |        [ Downstream Relay / Breaker Curve ]  t_downstream
        +-------------------------------------------------------------> Current (A)

Mathematical Breakdown of CTI Components

$CTI = t_{breaker} + t_{overtravel} + t_{tolerance} + t_{margin}$

+---------------------------------------------------------------------------------------------------+
|                     STANDARD COORDINATION TIME INTERVALS (CTI)                                    |
+---------------------------------------------------------------------------------------------------+
| Coordinating Device Pair                    | Recommended CTI Range | Key Factor Driving Setting   |
| :---                                        | :---                  | :---                         |
| **Digital Relay to Digital Relay**          | **0.20 to 0.25 s**    | Zero disk overtravel;        |
|                                             | ($12-15\text{ cycles}$)| high microprocessor accuracy.|
| **Digital Relay to Electromechanical Relay**| **0.25 to 0.30 s**    | Disc overtravel on upstream. |
| **Electromechanical to Electromechanical**  | **0.35 to 0.40 s**    | Induction disk inertia coasting|
|                                             | ($21-24\text{ cycles}$)| and mechanical calibration.  |
| **Digital Relay to Downstream Low-Volt CB** | **0.15 to 0.20 s**    | Direct-acting trip clearing.  |
+---------------------------------------------------------------------------------------------------+

---

5. Transformer Damage Curves (ANSI/IEEE C57.109) & Inrush Constraints

Properly coordinating primary overcurrent protection for a liquid-immersed or dry-type power transformer requires plotting two critical boundary points alongside the relay TCC:

1. The ANSI/IEEE C57.109 Transformer Damage Curve (Upper thermal/mechanical withstand bound).
2. The Transformer Magnetizing Inrush Point (Lower operating limit to avoid nuisance tripping).

                  TRANSFORMER PROTECTION COORDINATION WINDOW

    Time (s)
        ^
   10.0 |          [ ANSI/IEEE C57.109 Damage Curve ] (Upper Bound)
        |                     \            /
    1.0 |                      \  [Relay] /
        |                       \  [TCC] /
    0.1 |---[ Inrush Point ]-----\------/ (Lower Bound)
        |    (8-12x FLA @ 0.1s)   \
   0.01 |                          \
        +----+----------------------+---------------------------------> Current (A)
            FLA                   I_fault,max

ANSI/IEEE C57.109 Transformer Categories

Transformers are categorized by three-phase kVA ratings to establish their thermal and mechanical through-fault withstand limits:

• Category I: $15\text{ to }500\text{ kVA}$ (3-phase)
• Category II: $501\text{ to }5,000\text{ kVA}$ (3-phase)
• Category III: $5,001\text{ to }30,000\text{ kVA}$ (3-phase)
• Category IV: $>30,000\text{ kVA}$ (3-phase)

Frequent vs. Infrequent Fault Withstand Curves

• Infrequent Fault Withstand Curve: Applies where faults occur rarely (e.g., secondary feeds encapsulated in metal-enclosed bus duct). Follows a pure $I^2 t = K$ thermal limit line:

$t = \frac{1250}{I_{pu}^2} \quad (\text{for Category II transformers}, 501-5000\text{ kVA})$

• Frequent Fault Withstand Curve: Applies where transformers feed overhead distribution circuits subject to recurring lightning, tree contacts, and recloser operations. The curve transitions from thermal $I^2 t$ slope to a horizontal mechanical withstand limit to prevent cumulative mechanical winding deformation.

Transformer Magnetizing Inrush Point

When a transformer is energized, core flux saturation produces a transient inrush current surge that flows only into the primary winding:

• Standard Rule-of-Thumb Inrush Point: $8\times$ to $12\times$ Full Load Amps ($FLA$) for $0.10\text{ seconds}$ ($6\text{ cycles}$).
• First Half-Cycle Crest Inrush: Up to $25\times FLA$ for $0.01\text{ s}$.

$\text{Primary Protection Criterion: } t_{relay}(12 \times FLA) > 0.10\text{ s}$

$\text{ANSI Protection Criterion: } t_{relay}(I_{fault}) < t_{ANSI\_Damage}(I_{fault})$

---

5a. Overhead Distribution Devices: Reclosers, Sectionalizers & Fuse Coordination

The NCEES specification lists protective devices (e.g., fuses, breakers, reclosers) as its own sub-topic. Reclosers deserve separate treatment because they coordinate on a fundamentally different principle from the relay/breaker pairs above.

Why reclosers exist. Roughly 70% to 80% of overhead distribution faults are temporary — a branch contact, an animal, a lightning flashover. The arc self-extinguishes once current is interrupted, so the circuit can be safely re-energized. A recloser (IEEE device 79) is a self-contained interrupter with its own control that trips, waits a dead time, and recloses automatically, typically up to four trips before lockout.

Fuse Saving vs. Fuse Clearing

This trade-off is the classic recloser exam item, and both schemes are defensible engineering — which one is correct depends on what the question says the utility is optimizing.

Modern practice on feeders serving power-quality-sensitive customers has shifted toward fuse clearing, precisely because momentary interruptions carry a real economic cost.

The two fuse-coordination inequalities you must satisfy, using NEC- and IEEE-recognized factors from the fuse manufacturer's time-current curves:

1. To save the fuse: $t_{recloser,,fast} \times k ; < ; MMT_{fuse}$, where $k$ (about 1.2 to 1.35) accounts for cumulative fuse heating across successive fast operations — the fuse does not fully cool between trips, so the second fast operation sees a partially damaged fuse element.
2. To ensure selectivity on permanent faults: $TCT_{fuse} < t_{recloser,,delayed}$ at the maximum available fault current on the lateral.

Sectionalizers

A sectionalizer is not a fault interrupter — it has no interrupting rating and cannot break fault current. It counts the number of times an upstream recloser has interrupted fault current and opens during the dead time, while the line is de-energized, after a preset count (typically 2 or 3). Because it does not need a time-current curve, it can be applied where two devices in series would otherwise be impossible to coordinate on time alone. Applying a sectionalizer as though it were a fuse or recloser — expecting it to clear a fault — is a frequent distractor.

---

6. Step-by-Step Selective Coordination Methodology (Load to Utility)

+---------------------------------------------------------------------------------------------------+
|                    RADIAL SYSTEM SELECTIVE COORDINATION WORKFLOW                                  |
+---------------------------------------------------------------------------------------------------+
| Step 1: Base Data Gathering:                                                                      |
|   - Obtain continuous full-load currents (FLA) for all loads and transformers.                    |
|   - Compute maximum and minimum available short-circuit currents (3Φ and SLG) at every bus.       |
| Step 2: Establish Common Voltage Base:                                                            |
|   - Select a reference voltage base (typically the primary service voltage) and reflect all      |
|     downstream currents across transformer turns ratios: I_pri = I_sec * (V_sec / V_pri).         |
| Step 3: Set Downstream Load/Branch Devices First:                                                 |
|   - Select branch circuit breakers or fuses based on NEC 240 / 430 load and motor rules.          |
| Step 4: Set Upstream Phase Overcurrent Relay Pickup (51P):                                        |
|   - Set pickup I_pickup = 1.25 to 1.50 * FLA (above maximum anticipated continuous load).         |
|   - Ensure pickup is sensitive enough to detect minimum end-of-line fault: I_pickup < 0.5 * I_f,min.|
| Step 5: Coordinate Time Dial Setting (TDS):                                                       |
|   - Identify the Maximum Coordination Current (I_coord): The highest through-fault current seen   |
|     simultaneously by both devices (typically the maximum bolted fault on the downstream bus).    |
|   - Determine downstream clearing time t_downstream at I_coord.                                   |
|   - Target upstream operating time: t_upstream = t_downstream + CTI.                               |
|   - Solve IEEE/IEC curve equation for required TDS.                                                |
| Step 6: Set Instantaneous Element (50P):                                                          |
|   - Set I_pickup,50 ≥ 1.25 to 1.30 * I_fault,downstream_bus to prevent misoperation during          |
|     downstream faults (or disable 50 element if selective reach cannot be achieved).              |
| Step 7: Validate Asset Protection & Inrush Security:                                               |
|   - Verify relay curve clears BELOW transformer ANSI damage curve and ABOVE the 12x FLA @ 0.1s inrush.|
+---------------------------------------------------------------------------------------------------+

---

7. Step-by-Step Worked Mathematical Example

Problem Statement

A $2500\text{ kVA}$, 3-phase, $60\text{ Hz}$ step-down transformer rated at $13.8\text{ kV}$ (Delta) to $480\text{ V}$ (Grounded Wye) has a nameplate impedance of $Z = 5.75%$ ($X/R = 8.0$). The transformer feeds a $480\text{ V}$ main switchboard equipped with a secondary low-voltage power circuit breaker ($R_2$).

The primary side is protected at $13.8\text{ kV}$ by a microprocessor time-overcurrent relay ($R_1$) utilizing an IEEE Very Inverse (IEEE VI) curve ($A = 19.61, B = 0.491, p = 2.0$) connected to $200:5\text{ A}$ CTs ($CTR = 40$).

System data:

• Primary Full-Load Amps: $I_{FLA,pri} = \frac{2500\text{ kVA}}{\sqrt{3} \times 13.8\text{ kV}} = 104.59\text{ A}$
• Secondary Full-Load Amps: $I_{FLA,sec} = \frac{2500\text{ kVA}}{\sqrt{3} \times 0.480\text{ kV}} = 3007.03\text{ A}$
• Available short-circuit duty on the $480\text{ V}$ bus (bolted 3-phase fault): $I_{f,sec} = 48.0\text{ kA}$ at $480\text{ V}$
• Downstream breaker $R_2$ total clearing time at $48.0\text{ kA}$: $t_{downstream} = 0.060\text{ s}$ ($3.6\text{ cycles}$)
• Required Coordination Time Interval: $CTI = 0.25\text{ s}$
• Primary relay pickup is set at $150\text{ A}$ primary ($I_{pickup} = 150\text{ A}$, which is $1.43\times FLA_{pri}$)

Calculate:

1. The maximum $480\text{ V}$ fault current reflected to the primary $13.8\text{ kV}$ voltage base.
2. The required operating time for upstream relay $R_1$ at the maximum coordination fault current.
3. The exact Time Dial Setting ($TDS$) for relay $R_1$.
4. Verify that relay $R_1$ will not nuisance-trip on transformer magnetizing inrush ($12\times FLA_{pri}$ for $0.10\text{ s}$).

=========================================================================================
CALCULATION WORKFLOW & DETAILED STEP-BY-STEP SOLUTION:
=========================================================================================

Step 1: Reflect Secondary Fault Current to 13.8 kV Primary Base
  Turns Ratio: a = V_pri / V_sec = 13,800 V / 480 V = 28.75
  
  Primary Reflected Fault Current (I_coord):
    I_coord = I_f,sec / a = 48,000 A / 28.75
            = 1,669.57 A ≈ 1,670 A on 13.8 kV base

Step 2: Determine Required Upstream Relay Operating Time (t_upstream)
    t_upstream = t_downstream + CTI
               = 0.060 s + 0.250 s
               = 0.310 seconds at I_coord = 1,669.57 A

Step 3: Calculate Plug Setting Multiplier (M) and Solve for TDS
  Plug Setting Multiplier at fault current:
    M = I_coord / I_pickup = 1,669.57 A / 150.0 A = 11.1305

  Evaluate the IEEE Very Inverse bracket term with A = 19.61, B = 0.491, p = 2.0:
    Bracket = [ A / (M^p - 1) ] + B
            = [ 19.61 / ((11.1305)^2 - 1) ] + 0.491
            = [ 19.61 / (123.887 - 1) ] + 0.491
            = [ 19.61 / 122.887 ] + 0.491
            = 0.15958 + 0.49100
            = 0.65058

  Solve for TDS:
    t(I) = TDS * 0.65058 = 0.310 s
    TDS  = 0.310 / 0.65058 = 0.4765

  Select next highest standard setting: TDS = 0.50

  Recompute actual operating time at TDS = 0.50:
    t_actual = 0.50 * 0.65058 = 0.3253 seconds
    Actual CTI = 0.3253 - 0.060 = 0.2653 s ≥ 0.25 s (COORDINATION ACHIEVED!)

Step 4: Verify Security Against Transformer Magnetizing Inrush
  Inrush Current Magnitude: I_inrush = 12 * FLA_pri = 12 * 104.59 A = 1,255.08 A
  Inrush Duration: t_inrush = 0.100 seconds

  Multiplier at Inrush:
    M_inrush = I_inrush / I_pickup = 1,255.08 A / 150.0 A = 8.3672

  Relay Trip Time at Inrush Current (with TDS = 0.50):
    Bracket_inrush = [ 19.61 / ((8.3672)^2 - 1) ] + 0.491
                   = [ 19.61 / (70.010 - 1) ] + 0.491
                   = [ 19.61 / 69.010 ] + 0.491
                   = 0.28416 + 0.491 = 0.77516

    t_trip(I_inrush) = 0.50 * 0.77516 = 0.3876 seconds

  Security Verification:
    Since t_trip (0.3876 s) > t_inrush (0.100 s), the relay will NOT nuisance-trip
    during transformer energization. The scheme is robustly secure.
=========================================================================================

---

8. Common PE Exam Traps & Tactical Pitfalls

• Omitting Voltage Ratio Scaling in TCC Coordination: Coordinating an upstream $13.8\text{ kV}$ relay directly against a secondary $480\text{ V}$ breaker's physical amperes without dividing by the turns ratio ($a = 28.75$). All device characteristics plotted on a single TCC sheet must be scaled to a single common reference voltage base.
• Confusing IEEE and IEC Formula Structures: Omitting constant $B$ when computing IEEE curve operating times, or erroneously adding constant $B$ to IEC formulations. IEEE curves approach a horizontal asymptotic floor of $TDS \times B$ as $I \to \infty$, whereas IEC curves decay to zero.
• Setting TDS Based on Normal Load Current: Solving for $TDS$ at nominal full-load amps ($FLA$) instead of the maximum through-fault coordination current ($I_{coord}$). $TDS$ is determined exclusively by the required $CTI$ at the point of maximum through-fault current.
• Neglecting the 75% Fuse Pre-Damage Rule: Aligning downstream fuse clearing time equal to upstream fuse melting time ($TCT_{down} = MMT_{up}$). Because the upstream fuse accumulates irreversible thermal heat during the downstream fault, the downstream fuse $TCT$ must never exceed $75%$ of upstream $MMT$ ($0.75 \times MMT_{up}$).