Understanding fault current calculations is essential for designing protective systems that ensure both operational safety and equipment protection. The methods presented here form the foundation for protection coordination studies in building services and industrial installations.
The Need for Fault Current Calculation
To determine the operational safety of a medium voltage design, the response time of protective equipment must be analyzed. This analysis requires realistic but prudent estimation of potential short-circuit currents throughout the distribution network.
For three-phase short-circuit protection design, calculating three-phase fault currents within the distribution network is essential. Additionally, when designing step and touch voltage protection or implementing separate earth-fault protection, earth-fault current calculations become necessary.
Three-Phase Fault Fundamentals
The calculation of three-phase fault current is straightforward because this represents a ‘balanced’—if unusual—load condition. The standard approach uses single-phase calculations to represent any (and all) of the three separate phase currents.
Using this technique, the maximum three-phase short-circuit current (per phase) can be derived from the following equation, which applies to any point in the network where a three-phase fault might occur:
I_fault = (c_max × U) / (√3 × (Z_s + Z_L))
Where:
- U is the line-to-line voltage at the point of fault
- Z_s is the line impedance of the supply transformer(s) feeding that section of the network (including upstream network impedance)
- Z_L is the line impedance of the distribution cable
- c_max is the voltage factor for maximum current as specified in BS EN 60909-0:2016 (1.1 for systems with ±10% voltage tolerance, 1.05 for systems with ±6% tolerance)
Important Note: Because quoted fault levels (MVA) are typically based on switchgear ratings rather than actual short-circuit values, calculated maximum short-circuit currents may err on the side of caution—a prudent approach for protection design.
Impedance Data Requirements
Typical transformer impedances are shown in the accompanying tables, while representative cable impedances are detailed in CIBSE AM18.2: Equipment (2022) and Annex C. For accurate assessment, both resistance and reactance must be included in calculations.
Practical Simplification: For quick check calculations, resistance values can often be omitted since they typically represent only a small fraction of overall reactance. Since impedance relates to the sum of squares of resistance and reactance, including the small resistance fraction changes the impedance value by only a few percent. This omission also designs the protection system to handle slightly higher fault currents than encountered in practice.
However, to establish peak fault currents, the X/R ratio at the fault point must be determined. This ratio can be calculated using industry standard power system software that accounts for both load losses and no-load losses of the transformer.
2.2 Per-Unit and Percentage Calculations
Why Use Per-Unit Methods?
Using the three-phase fault formula with impedance data allows calculation of both fault current and fault level for virtually any network point. However, such calculations often involve multiple voltage levels, and when undertaken with ohmic values, requires impedances to be ‘referred’ across transformer windings. In complex distribution networks, this becomes tedious—the per-unit method eliminates this problem.
The per-unit method offers three key advantages:
First, parameters become directly comparable across different systems. Component impedances, voltage drops, and power losses can be evaluated consistently regardless of voltage level or power rating. For example, in transformer impedance data tables, note how ohmic values change dramatically while per-unit values remain relatively similar across different transformer ratings.
Second, transformer impedances remain identical on both sides of the transformer, eliminating the need for impedance referral calculations. This is particularly valuable since MV/LV transformers commonly used in building services typically have impedances around 5-6%.
Third, all calculations use the same single-phase approach with phase voltage, simplifying the computational process for balanced three-phase systems.
Per-Unit Measurement Method
Transformer manufacturers typically measure per-unit impedance by applying a short-circuit to one transformer side—usually the low-voltage winding—with a voltmeter connected to the high-voltage winding and an ammeter measuring the shorted low-voltage winding current.
A very low voltage is applied to the high-voltage winding and gradually increased until normal full-load current flows in the low-voltage (short-circuited) winding. The applied voltage at this point, expressed as a fraction of normal rated voltage, becomes the transformer impedance voltage or per-unit impedance.
The underlying principle: if the applied voltage were increased to full rated value, the transformer short-circuit current would increase by the same ratio.
Notation Conventions
Critical Distinction: Per-unit values should strictly be expressed as decimal values (e.g., 0.0509), but engineers commonly use percentage values (5.09%) to reduce transcription errors, applying percentage conversion after completing data handling.
For clarity in this document:
- Decimal form variables use ‘pu’ suffix: X_pu, R_pu, Z_pu
- Percentage values use ‘(%)’ suffix: X(%), R(%), Z(%)
2.3 Calculation Methodology
Three fundamental equations govern the calculations:
Basic Equation 1: Fault Current Calculation
I_fault = (c_max × V_base) / (√3 × Z_per-unit × Z_base)
Note 1: Maximum short-circuit current can be determined by open-circuit voltage or by applying a multiplication voltage factor (c_max) of 1.1 at 11 kV and 1.05 at 400 V, whichever is larger (BS EN 60909-0:2016).
Note 2: Z_per-unit is expressed as a decimal value, not a percentage.
Basic Equation 2: Fault Level Calculation
In addition to using fault current magnitude for equipment rating, ‘fault MVA’ (also called fault level) is commonly used:
Fault MVA = √3 × V_line-to-line × I_fault
Where V_line-to-line is measured in kV and I_fault in kA.
Basic Equation 3: Base Conversion
To convert per-unit impedance from one base (kVA₁) to another base (kVA₂):
Z_pu,2 = Z_pu,1 × (kVA₂/kVA₁)
For calculation purposes, per-unit values are treated identically to ohmic resistance, reactance, or impedance values—combined through series or parallel impedance combination—with the advantage that impedance values represent ratios of full-load values.
Understanding Base Impedance
Consider a nominal 11 kV to 400 V, three-phase transformer rated at 500 kVA. The full-load current at 400 V nominal operating voltage is:
I_full-load = 500,000/(√3 × 400) = 721.7 A
To produce this secondary current magnitude, the transformer base impedance—the impedance value causing full-load secondary current flow—can be calculated using Ohm’s law:
Base impedance (per phase/balanced) = (400/√3)/721.7 = 0.32 ohms
This can be generalized as: Z_base = V_L²/S_3φ
Where V_L is line-to-line voltage and S_3φ is the transformer VA rating.
From this base (using 500 kVA rating as the base for all per-unit calculations), per-unit impedance expresses as a fraction of this value. For example:
- Load impedance of 0.16 ohms = 0.5 pu, producing double full-load current (1443 A)
- Load impedance of 0.032 ohms = 0.1 pu, producing ten times full-load current (7217 A)
Key Principle: Output current relates to full-load current and is inversely proportional to the per-unit value of total load impedance.
2.4 Practical Calculation Example
The following example demonstrates per-unit calculations for a typical building services installation. This configuration—featuring a 33kV supply, step-down transformers, and 11kV distribution—represents the complexity commonly encountered in large commercial or industrial facilities.
System Configuration
Consider the supply network shown in the system diagram. The complete network appears complex, but notice how opening the disconnectors where shown simplifies it to a radial configuration. This reduction technique is fundamental to fault analysis.
The methodology for calculating potential three-phase short-circuit current begins by drawing the equivalent single-phase impedance diagram. Rather than drawing horizontally, the diagram is typically rotated 90° so that the system neutral and fault star-point form the top and bottom of the diagram.
Step 1: Establish Common Base
The common volt-ampere base against which all per-unit impedances are referenced must be chosen. Any value works, but typically the rating of the local transformer feeding the final service point is selected. For this example, a base of 500 kVA (0.5 MVA) was chosen.
Important: When drawing impedance diagrams, start at the energy source and work toward the fault. For networks with multiple supply sources, sometimes it’s easier to start at the fault location and work backward toward the sources. Both techniques are valid.
Step 2: Include Supply Impedance
Starting with the incoming 33 kV supply, the first item is the impedance of the incoming supply. This is usually quoted by the Distribution Network Operator (DNO) as a fault level—for this example, 1000 MVA.
Understanding “Infinite Bus Bars”: An infinite bus bar does NOT mean the network can supply infinite current under fault conditions. It simply means the network behaves as a constant voltage source with zero internal impedance. The fault current magnitude won’t decay with time as it might from a single generator, because the fault represents an insignificant load on the system.
To convert the supply fault level to per-unit impedance: X_supply(%) = (Base MVA/Supply fault level MVA) × 100 X_supply(%) = (0.5/1000) × 100 = 0.05%
This impedance is considered wholly reactive, hence shown as X(%) rather than Z(%).
Step 3: Include Primary Transformer
The 10 MVA, 33/11 kV transformer has quoted reactance of 6.25%. However, this percentage relates to the transformer’s 10 MVA rating and needs conversion to the 0.5 MVA base:
X_transformer(%) = 6.25 × (0.5/10) = 0.3125%
Step 4: Include Cable Impedances
When considering 11 kV cables, recognize that:
- Generally, cable resistance exceeds its reactance
- Cable impedance information is quoted in ohms per kilometer, not per-unit values
Practical Note: Cable impedances often contribute minimally to final calculated fault current, leading some engineers to omit them for expediency. While this simplification is often acceptable for preliminary designs, include cable impedances for final protection coordination studies since their inclusion can only reduce calculated fault current.
Before including cable resistance and reactance values, convert from ohmic to per-unit values by dividing by the base impedance value:
Z_pu = (Z_ohmic × S_base)/(V_base²)
For the 11 kV system with 0.5 MVA base: Base impedance = (11,000)²/500,000 = 242 ohms
Step 5: Include Local Transformer
Include the impedance of the local transformer (adjacent to the fault). Since this transformer is rated at the same value as the chosen base (500 kVA), the impedance value can be inserted directly without conversion.
From typical data, a 500 kVA transformer has approximately 5.5% impedance.
Step 6: Combine Impedances
With all network impedances included, the equivalent impedance diagram reduces to a single impedance using traditional impedance combination methods. In this example, all per-unit impedances are in series, so similar types add algebraically—resistance with resistance, reactance with reactance.
The impedance values combine because they’re in series—this fundamental principle applies whether working in ohms or per-unit values.
Summary of impedances:
- Supply reactance: 0.05%
- Primary transformer reactance: 0.31%
- Cable resistance: 0.45%
- Cable reactance: 1.0%
- Local transformer: 5.5% (combined R and X)
Total impedance calculation:
- Total resistance: 0.45% + transformer resistance component
- Total reactance: 0.05% + 0.31% + 1.0% + transformer reactance component
- Combined impedance: √(R_total² + X_total²) ≈ 6.6%
Step 7: Calculate Final Fault Current
The final calculation uses Basic Equation 1 with the voltage factor from BS EN 60909-0:2016:
I_fault = (1.05 × 400)/(√3 × 0.066 × 0.32) = 11,073 A
Important: Convert the overall impedance (6.6%) back to decimal values (0.066) for the final calculation.
Key Observation: Note how dominant the local transformer impedance is compared to the overall impedance value. This explains why some designers, for quick checks, ignore supply network impedances and calculate maximum LV fault current using only the local supply transformer’s per-unit impedance.
2.5 Summary and Applications
This per-unit methodology provides a systematic approach to fault current calculation regardless of network complexity. By establishing a common base and working systematically from source to fault location, you can determine protection requirements for any point in the distribution system.
The method’s key strengths include:
- Consistency across voltage levels and power ratings
- Simplicity in handling transformer impedances
- Accuracy sufficient for protection coordination studies
- Flexibility for complex network configurations
Next Steps: The following sections extend this methodology to more complex scenarios involving multiple sources, motor contributions, and unbalanced fault conditions including single-phase-to-earth faults using symmetrical components.
Understanding these fundamental calculation principles enables engineers to design protection systems that ensure both safety and reliability in medium voltage distribution networks.