Transformer Secondary Conductor Size Screening Calculator — Instant Results from Full-Load Current

This article provides a precise screening calculator for transformer secondary conductor sizing and verification requirements.

Instant results derive from full load current inputs, adjustments, corrections, and short-circuit considerations environmental factors

Transformer Secondary Conductor Size Screening Calculator (from Full-Load Current)

Advanced options

You may upload a clear transformer nameplate or wiring diagram photo to suggest typical values (processed by external AI tools).

Enter transformer kVA, secondary voltage and phase to obtain minimum secondary conductor ampacity and a suggested conductor size.
Engineering basis and formulas used
  • Transformer full-load current (single-phase): I_FL = (S_kVA × 1000) / V_sec [A], where S_kVA is transformer apparent power in kVA and V_sec is secondary voltage in V.
  • Transformer full-load current (three-phase): I_FL = (S_kVA × 1000) / (√3 × V_sec) [A].
  • Design current for conductor ampacity screening: I_design = I_FL × (Design multiplier [%] / 100).
  • Temperature correction factor (75 °C insulation, based on typical ampacity tables): 21–25 °C → 1.08; 26–30 °C → 1.00; 31–35 °C → 0.94; 36–40 °C → 0.88; 41–45 °C → 0.82; 46–50 °C → 0.75; 51–55 °C → 0.67; 56–60 °C → 0.58. Values outside this range are limited to the nearest bound.
  • Grouping correction factor versus number of current-carrying conductors (typical): 1–3 → 1.00; 4–6 → 0.80; 7–9 → 0.70; 10–20 → 0.50; >20 → 0.45.
  • Total correction factor: F_total = F_temp × F_group.
  • Required base ampacity from standard tables: I_table_required = I_design / F_total.
  • The calculator selects the smallest standard copper or aluminum conductor size whose nominal ampacity at 75 °C and 30 °C ambient (3 current-carrying conductors) is greater than or equal to I_table_required.
Conductor sizeApprox. area (mm²)Copper ampacity 75 °C (A)Aluminum ampacity 75 °C (A)
14 AWG2.0820
12 AWG3.3125
10 AWG5.2635
8 AWG8.375040
6 AWG13.36550
4 AWG21.18565
3 AWG26.710075
2 AWG33.611590
1 AWG42.4130100
1/0 AWG53.5150120
2/0 AWG67.4175135
3/0 AWG85.0200155
4/0 AWG107.2230180
250 kcmil126.7255205
300 kcmil152.0285230
350 kcmil177.3310250
400 kcmil202.7335270
500 kcmil253.3380310

Technical FAQ about this transformer secondary conductor screening calculator

Does this calculator produce code-compliant conductor sizes?
No. The tool performs an engineering screening based on typical ampacity references and simple correction factors. Final conductor sizing must follow the applicable electrical code (for example NEC or IEC) and manufacturer data.
Which conductor temperature rating is assumed?
The screening uses typical 75 °C ampacity values for copper and aluminum conductors at 30 °C ambient with up to three current-carrying conductors. This generally reflects common termination ratings for low-voltage power feeders.
Why is there a design multiplier on full-load current?
The design multiplier allows you to include margins such as 125% of transformer full-load current, which is a common requirement for transformer secondary conductors in several installation codes, instead of sizing exactly at 100% full-load.
How should I choose the ambient temperature and number of current-carrying conductors?
Use the worst-case values expected in the installation. For typical indoor panels, 30–35 °C and three current-carrying conductors are common. For crowded raceways or high ambient temperatures, higher correction may drive a larger conductor size.

Overview of Transformer Secondary Conductor Screening

Selecting transformer secondary conductors requires rapid, precise screening to translate transformer kVA and system voltage into conductor requirements. This article presents a technical workflow and formulas that yield instant results from full load current, including correction and derating factors.We treat transformer secondary conductors as a system-design problem: convert transformer rating to full load current, apply continuous-load and grouping multipliers, correct for ambient temperature and insulation ratings, check voltage drop, and verify short-circuit and protection coordination. The objective is a deterministic screening calculator that outputs candidate conductor sizes or the number of parallel runs required for compliance with design targets and standards.

Key Inputs and Outputs for an Instant Screening Calculator

  • Inputs:
    • Transformer apparent power (kVA) and secondary nominal voltage (V).
    • Connection type: single-phase or three-phase; grounded wye or delta.
    • Maximum allowable voltage drop (percentage or volts) and circuit length (m or ft).
    • Conductor material (copper or aluminum) and insulation temperature rating (e.g., 75°C, 90°C).
    • Ambient temperature and cable grouping (number of conductors in conduit or tray).
    • Expected continuous percentage of load and motor starting or short-duration loads.
  • Outputs:
    • Full Load Current (A).
    • Required ampacity after continuous or demand factors (A).
    • Candidate conductor size(s) and number of parallels.
    • Estimated voltage drop and percentage.
    • Short-circuit prospective checks and time-current coordination notes.

Fundamental Electrical Formulas

Three-phase full load current (A):

I = (kVA × 1000) / (√3 × V)

Single-phase full load current (A):

Transformer Secondary Conductor Size Screening Calculator Instant Results From Full Load Current
Transformer Secondary Conductor Size Screening Calculator Instant Results From Full Load Current
I = (kVA × 1000) / V

Required conductor ampacity for continuous load (NEC-style 125% rule):

A_required = I × 1.25 × F_group × F_temp

Voltage drop for three-phase circuits (approximate):

V_drop = √3 × I × (R × cosφ + X × sinφ) × L

Or approximate resistive-only voltage drop percentage:

VD% = (√3 × I × R_total / V) × 100

Variable explanations and typical values

  • kVA: Apparent power rating of transformer in kilovolt-amperes (typical values: 15 kVA to 2000 kVA+).
  • V: Line-to-line RMS voltage for three-phase systems (typical values: 208 V, 400 V, 480 V, 600 V).
  • √3: Square root of three ≈ 1.732.
  • I: Full load current in amperes (A).
  • A_required: Minimum conductor ampacity necessary after continuous load multiplier and corrections.
  • F_group: Grouping factor (≤1) from cable bundling tables (e.g., 0.7–1.0 depending on number of conductors and installation method).
  • F_temp: Temperature correction factor for conductor insulation rating (e.g., 0.82 at 40°C for 75°C rated insulation; consult standard tables).
  • R, X: Circuit resistance and reactance per unit length (Ω/m or Ω/ft) for chosen conductor and installation.
  • L: One-way conductor length (m or ft) between transformer secondary and load or point of interest.
  • cosφ: Power factor (typical 0.8–1.0 for general balanced loads; motor loads often 0.8).

Lookup Table — Transformer kVA to Full Load Current (Three-Phase)

Transformer kVAI at 208 V (A)I at 240 V (A)I at 400 V (A)I at 480 V (A)I at 600 V (A)
1541.636.121.618.014.4
3083.172.143.336.128.9
75207.7180.4108.390.172.1
150415.5360.8216.5180.2144.1
250692.5602.5361.6301.8241.5
5001385.01205.1723.2603.6483.0
7502077.51807.61084.9905.4724.5
10002770.02410.21446.41207.2966.0

Notes: Values shown are computed using I = (kVA × 1000) / (√3 × V). Use these as immediate screening results for conductor sizing.

Typical Conductor Ampacity Tables (Representative Values)

The following tables present typical continuous ampacity values for commonly used cable cross-sectional areas (copper and aluminum). These are representative and must be confirmed against the code (e.g., NEC Table 310.16) and installation method.

Copper cross-section (mm²)Typical ampacity (A) @ 75°C
1676
25101
35125
50158
70198
95237
120269
150308
185348
240417
300515
400676
500810
630999
Aluminium cross-section (mm²)Typical ampacity (A) @ 75°C
35105
50135
70175
95215
120250
150290
185330
240390
300460
400600
500760

These ampacities are for estimating and screening only. Actual allowable ampacities per regulation depend on installation method, insulation type, and ambient conditions.

Screening Algorithm: Step-by-step Workflow

  1. Compute full load current (I) using the formula for single‑ or three‑phase systems.
  2. Decide if the load is continuous. If continuous, multiply I by 1.25 (125%).
  3. Apply grouping (F_group) and ambient temperature correction (F_temp) factors: A_required = I × 1.25 × F_group × F_temp.
  4. Select candidate conductor(s) whose tabulated ampacity ≥ A_required for the conductor insulation temperature column appropriate to the termination rating.
  5. Check voltage drop against design limit (commonly 1.5% for branch circuits, 3% for feeder + branch total). Compute V_drop and VD%.
  6. If VD% > limit, increase conductor size or use parallel conductors, or reduce length.
  7. Verify short-circuit withstand: conductor short-circuit temperature and protective device clearing times; consult IEC/IEEE/NEC rules.
  8. Finalize and document assumptions: ambient, grouping, conductor type, parallel runs, and reference normative clauses.

Correction and Derating Factors — Practical Values

  • Temperature correction factor (F_temp): Typical values:
    • At 30°C: ~1.00 for 75°C-rated conductor.
    • At 40°C: ~0.91–0.95 depending on table.
    • At 50°C: ~0.82 for 75°C-rated conductors.
  • Grouping factor (F_group): Depends on number of current-carrying conductors and conduit/tray configuration. Common screening values: 1.0 (single), 0.8–0.9 (small group), 0.7 (large bundle).
  • Altitude correction: above 1000 m altitude may require derating (refer to local code).
  • Terminal temperature limitation: Terminations may be limited to 75°C maximum even if conductor ampacity table is for 90°C. Use the lower column when necessary.

Voltage Drop Calculation Details

For instantaneous screening, use a resistive approximation sufficient for copper and aluminum where reactance is small or where conductor length is modest.

Resistive voltage drop (three‑phase) in volts, assuming R_total is roundtrip conductor resistance per meter and one‑way length L (m):

V_drop = √3 × I × R_total × L

Where R_total (Ω/m) is the conductor resistance per meter for the chosen cross-section at operating temperature. To compute percentage voltage drop:

VD% = (V_drop / V) × 100

Typical DC resistance at 20°C (example assumptions for screening):

  • Copper 500 mm²: R ≈ 0.00008 Ω/m
  • Copper 300 mm²: R ≈ 0.00012 Ω/m
  • Aluminium 300 mm²: R ≈ 0.00020 Ω/m

These resistance values are representative; use manufacturer cable tables for exact R and temperature correction.

Example 1 — Detailed Case (Copper conductors)

Problem statement:

  • Transformer rating: 500 kVA three-phase
  • Secondary voltage: 480 V (line-to-line)
  • Load condition: Continuous (100% continuous)
  • Conductor material: Copper, insulation rating 75°C
  • Ambient temperature: 30°C (no temperature correction assumed for screening)
  • Circuit length: 40 meters (one-way)
  • Allowed voltage drop: 3% (overall feeder + branch permitted)
  • Cable grouping: single run per phase in conduit (F_group = 1.0)
Step 1 — Compute full load current:

Use the three-phase formula:

I = (kVA × 1000) / (√3 × V)
Substitute: I = (500 × 1000) / (1.732 × 480) = 500000 / 831.36 = 601.5 A (rounded)
Step 2 — Apply continuous load multiplier:
A_required_initial = I × 1.25 = 601.5 × 1.25 = 751.9 A
Step 3 — Apply grouping and temperature corrections:
F_group = 1.0, F_temp ≈ 1.00 at 30°C → A_required = 751.9 A
Step 4 — Select candidate conductor:

From the copper ampacity table, a 500 mm² copper conductor has a typical ampacity ≈ 810 A @ 75°C, which exceeds 751.9 A. Therefore, a single 500 mm² copper per phase is acceptable from an ampacity perspective and termination rating permitting.

Step 5 — Voltage drop estimate (simplified resistive):

Assume resistive DC R ≈ 0.00008 Ω/m for 500 mm² copper (representative). For three-phase:

V_drop = √3 × I × R × L = 1.732 × 601.5 × 0.00008 × 40
Compute: V_drop ≈ 1.732 × 601.5 × 0.0032 = 1.732 × 1.9248 ≈ 3.334 V
VD% = (3.334 / 480) × 100 = 0.695% ≈ 0.7%
Result summary:
  • Full load current: 601.5 A
  • Required ampacity after continuous multiplier: 751.9 A
  • Selected conductor: Copper 500 mm² (ampacity ≈ 810 A)
  • Estimated voltage drop: 3.334 V (0.7%), well under 3% limit
Design notes:
  • Confirm 500 mm² termination compatibility on transformer and switchgear.
  • Validate exact conductor R and X from manufacturer; refine VD calculation with reactance if longer runs or harmonic-rich loads.
  • Verify short-circuit withstand and protective device settings with supplier curves.

Example 2 — Detailed Case (Aluminium conductors and paralleled cables)

Problem statement:

  • Transformer rating: 150 kVA three-phase
  • Secondary voltage: 208 V (line-to-line)
  • Load condition: 100% continuous
  • Conductor material: Aluminium, limited to standard mm² sizes
  • Ambient temperature: 40°C (apply temperature correction)
  • Circuit length: 30 m one-way
  • Allowed voltage drop: 3% total
Step 1 — Compute full load current:
I = (150 × 1000) / (1.732 × 208) = 150000 / 360.256 ≈ 416.3 A
Step 2 — Continuous multiplier:
A_required_initial = 416.3 × 1.25 = 520.4 A
Step 3 — Temperature correction (assume F_temp = 0.91 at 40°C for aluminium insulation reference):

A_required = 520.4 / 0.91 = 572.0 A (we divide by factor if tabular ampacity is reduced — alternative formulation: A_required = 520.4 × (1/F_temp) when using table values that must be multiplied by F_temp; many tables use multiplication less than 1: to be consistent, treat A_required = 520.4 / F_temp).

Step 4 — Candidate selection from aluminium ampacity table:

From table, a single 500 mm² aluminium conductor shows typical ampacity ≈ 760 A (representative). However, consider practical availability: 300 mm² Al ≈ 460 A (insufficient), 400 mm² Al ≈ 600 A (marginally sufficient depending on exact tabulated values and F_temp). To be conservative and allow for future margin, one may choose either:

  1. Single 400 mm² aluminium per phase if manufacturer ampacity at 75°C and correction yields ≥ 572 A; or
  2. Two parallel 240 mm² aluminium conductors per phase (each 240 mm² ≈ 390 A typical; two parallels share current → combined rating ≥ 780 A and provide redundancy).
Step 5 — Voltage drop check for option with two parallel 240 mm² Al conductors:

Assume R ≈ 0.00020 Ω/m for 240 mm² aluminium (per conductor). When two conductors are paralleled per phase, effective resistance halves: R_effective = 0.00010 Ω/m.

V_drop = √3 × I × R_effective × L = 1.732 × 416.3 × 0.00010 × 30
Compute: V_drop ≈ 1.732 × 416.3 × 0.003 = 1.732 × 1.2489 ≈ 2.162 V
VD% = (2.162 / 208) × 100 ≈ 1.04%
Result summary:
  • Full load current: 416.3 A
  • Required ampacity after continuous and temperature correction: ≈ 572 A
  • Single 400 mm² Al may be marginal; two parallel 240 mm² Al per phase provide combined ampacity > 572 A and voltage drop ≈1.04%.
Design notes:
  • Parallel conductor design must follow code rules for paralleled conductors (equal length, same material, same type and insulation, same lay, and same manufactured type).
  • Terminate paralleled conductors appropriately and verify busbar/terminal sizes.
  • Confirm exact aluminium ampacity tables and temperature correction factors from the regulating authority or manufacturer data.

Short-Circuit and Protective Device Considerations

  • Ensure conductor short-circuit rating is adequate for the maximum prospective fault current and clearing time. Use adiabatic heating equation or manufacturer short-circuit ratings:

    k = I_fault × √t ≤ k_s where k_s is conductor-specific constant, or explicitly use tables providing I^2t withstand.

  • Coordinate protective devices so that conductor clearing time does not allow excessive heating. Reference IEEE and IEC protective device coordination principles.
  • Verify fuse/CB interrupting capacity and transformer inrush characteristics when choosing upstream protection.

Standards, Normative References and Authoritative Links

  • NFPA 70 (National Electrical Code) — conductor ampacity, ampacity correction and continuous-load rules: https://www.nfpa.org/nec
  • IEC 60076 — Power transformers — general standard for transformer ratings and testing: https://www.iec.ch
  • IEEE C57 series — Transformer standards and ratings: https://standards.ieee.org
  • IEC 60287 — Electric cables — calculation of the continuous current rating (ampacity): https://www.iec.ch/
  • British Standard BS 7671 (IET Wiring Regulations) — for UK wiring practices: https://www.theiet.org/publishing/bs-7671/

These references should be consulted for specific local regulatory requirements and the authoritative ampacity and derating tables applicable to your jurisdiction.

Implementation Notes for a Screening Calculator

A practical instant-results screening calculator must:

  • Accept kVA and voltage inputs and compute FLC instantly using the formulas above.
  • Provide configurable parameters: continuous (%) flag, ambient temperature, grouping, conductor material and insulation rating, accepted voltage drop limit, and conductor length.
  • Use internal lookup tables for conductor ampacity and resistance per unit length by size and material, referencing manufacturer tables and normative tables.
  • Automatically evaluate parallel runs if single conductor size cannot meet ampacity or if a voltage drop constraint forces upsizing.
  • Output candidate solutions with rationale and display the governing formulas and variable values for auditability.

Validation and Practical Recommendations

  • Always validate screening results against local code tables and manufacturer data. Screening calculators are an engineering aid, not a substitute for final design checks.
  • Document all assumptions: ambient, grouping, insulation rating, calculation method used for VD (resistive or full R+jX), and any rounding rules.
  • For installations with motors or non-linear loads, consider inrush currents, harmonic heating, and verify cable thermal behavior under harmonic distortion using standards such as IEC 60287 and IEC 60502.
  • When paralleling conductors, follow code provisions strictly including identical conductor types and lengths.

Appendix — Quick Reference Formulas and Constants

QuantityFormula / Typical value
Three-phase full load currentI = (kVA × 1000) / (√3 × V)
Single-phase full load currentI = (kVA × 1000) / V
Continuous-load multiplier1.25 (125%) typical for continuous loads per many codes
Typical √31.732
Voltage drop three-phaseV_drop = √3 × I × R × L
Example R for Cu 500 mm²~0.00008 Ω/m (representative)

Final Engineering Checklist

  1. Confirm full load currents per transformer nameplate and actual operating voltage.
  2. Identify continuous load percentage and apply appropriate multipliers.
  3. Select conductor sizes from authoritative ampacity tables for the installation method and conductor temperature rating.
  4. Apply ambient and grouping correction factors using normative tables.
  5. Perform voltage drop calculation and iterate conductor size or parallel runs until limits are met.
  6. Check short-circuit heating I^2t against conductor limits and protective device timing.
  7. Document references, assumptions, and selected solution for compliance and review.
References
  • NFPA 70, National Electrical Code (NEC) — https://www.nfpa.org/nec
  • IEC 60076 — Power transformers — https://www.iec.ch
  • IEC 60287 — Electric cables — calculation of current ratings — https://www.iec.ch
  • IEEE Standards — Transformer and switchgear standards — https://standards.ieee.org
  • IET Wiring Regulations (BS 7671) — https://www.theiet.org/publishing/bs-7671/
Note: This article provides a technical screening methodology and representative values for rapid decision-making. Final conductor selection must reference the exact local code tables, manufacturer cable datasheets, transformer terminal temperature limits, and must be reviewed by a licensed electrical engineer prior to construction.