Overview
Selects and sizes current transformers (CTs) for protection applications per IEC 60044-1 and IEC 61869-2. Verifies that a selected CT is suitable for the installation by checking three criteria: secondary burden does not exceed rated VA, the system fault multiple is within the Accuracy Limit Factor (ALF), and the required secondary voltage under fault conditions remains below the CT knee-point voltage (Vk).
Use this calculator when specifying CTs for protection relays or metering installations, selecting CT ratios, verifying burden compatibility, or checking saturation risk under fault conditions.
Input Parameters
System Parameters
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
i_nominal |
Number | 50 A | 1 – 10,000 A | Nominal primary current (motor FLC, feeder current, etc.) |
i_fault |
Number | 5,000 A | 100 – 200,000 A | Maximum expected fault current at CT location (from fault study) |
voltage_ll |
Number | 415 V | 100 – 15,000 V | Line-to-line system voltage (for reference) |
frequency |
Enum: 50 / 60 Hz | 50 Hz | — | System frequency |
CT Parameters
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
ct_primary |
Number | 50 A | 1 – 10,000 A | CT primary winding rated current |
ct_secondary |
Enum: 1 / 5 A | 5 A | — | CT secondary winding rated current |
ct_class |
Enum | 5P10 | — | CT accuracy class (see table below) |
ct_rated_va |
Number | 10 VA | 1 – 100 VA | CT rated burden at secondary current |
ct_knee_point_v |
Number | — | 10 – 2,000 V | Knee-point voltage from CT datasheet (optional) |
ct_internal_resistance |
Number | 0 Ω | 0 – 10 Ω | CT secondary winding internal resistance (optional) |
Secondary Circuit Parameters
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
cable_length_m |
Number | 20 m | 1 – 500 m | One-way cable length from CT to relay/meter |
conductor_area_mm2 |
Enum | 2.5 mm² | 0.75 – 50 mm² | CT secondary cable conductor cross-sectional area |
relay_burden_va |
Number | 0.5 VA | 0.1 – 20 VA | Relay/meter burden at rated secondary current |
cable_temp_c |
Enum | 75 °C | 20 – 90 °C | Expected cable operating temperature |
CT Accuracy Classes
| Category | Classes | Composite Error | ALF |
|---|---|---|---|
| Protection | 5P5, 5P10, 5P20, 10P5, 10P10, 10P20 | 5% or 10% | 5, 10, or 20 |
| Special Protection | PX, TPX, TPS | — | — |
| Metering | 0.1, 0.2, 0.5, 1.0 | Class value | — |
Output Results
| Output | Unit | Description |
|---|---|---|
ct_ratio |
— | CT ratio string (e.g., "100:5") |
ct_ratio_factor |
— | Turns ratio factor \(n = I_{pri} / I_{sec}\) |
r_cable |
Ω | Secondary cable loop resistance |
r_relay |
Ω | Relay equivalent resistance |
r_total |
Ω | Total secondary circuit resistance |
va_actual |
VA | Actual VA burden at rated secondary current |
burden_ok |
Boolean | Whether actual burden ≤ rated VA |
burden_utilization |
% | Burden as percentage of CT rated VA |
fault_multiple |
× | Ratio of fault current to CT rated primary current |
i_sec_fault |
A | Secondary current under fault conditions |
v_required_fault |
V | Secondary voltage required to drive fault current through burden |
alf_check_ok |
Boolean | Whether fault multiple ≤ ALF |
vk_check_ok |
Boolean | Whether required voltage ≤ knee-point voltage |
saturation_risk |
text | "low", "medium", or "high" |
overall_suitable |
Boolean | Overall CT suitability (all checks pass) |
recommended_ratio |
— | Next standard CT ratio above nominal current |
status |
text | "SUITABLE" or "NOT SUITABLE" |
Formulas
CT Transformation Ratio
\[ n = \frac{I_{primary}}{I_{secondary}} \]
Where \(I_{primary}\) is the CT rated primary current and \(I_{secondary}\) is the rated secondary current (1 A or 5 A).
Cable Loop Resistance
\[ R_{cable} = \frac{\rho \times 2L}{A} \]
Where \(\rho\) is the temperature-adjusted copper resistivity (Ω·mm²/m), \(L\) is the one-way cable length (m), and \(A\) is the conductor area (mm²). The factor of 2 accounts for the go-and-return loop.
Temperature-Adjusted Resistivity
\[ \rho(T) = \rho_{20} \times \left(1 + \alpha \times (T - 20)\right) \]
Where \(\rho_{20} = 0.017241\) Ω·mm²/m (copper at 20 °C) and \(\alpha = 0.00393\) /°C (temperature coefficient for copper).
Relay Equivalent Resistance
\[ R_{relay} = \frac{VA_{relay}}{I_{sec}^2} \]
Converts the relay/meter burden in VA to an equivalent resistance at rated secondary current.
Total Secondary Circuit Resistance
\[ R_{total} = R_{cable} + R_{relay} + R_{CT,internal} \]
Actual VA Burden
\[ VA_{actual} = I_{sec}^2 \times R_{total} \]
The total burden imposed on the CT at rated secondary current. Must not exceed the CT rated VA.
Fault Multiple
\[ M_{fault} = \frac{I_{fault}}{I_{rated,primary}} \]
The ratio of maximum fault current to CT rated primary current (per IEC 60044-1 cl. 3.4.202). Must not exceed the Accuracy Limit Factor (ALF) for the CT class.
Secondary Fault Current
\[ I_{sec,fault} = \frac{I_{fault}}{n} \]
The current reflected to the CT secondary winding during a fault.
Required Secondary Voltage
\[ V_{required} = I_{sec,fault} \times R_{total} \]
The voltage the CT must develop across the secondary burden under fault conditions. Must not exceed the knee-point voltage \(V_k\) to avoid saturation.
Suitability Checks
1. Burden Check:
\[ VA_{actual} \leq VA_{rated} \implies \text{PASS} \]2. ALF Check (protection classes only):
\[ M_{fault} \leq ALF \implies \text{PASS} \]3. Knee Point Check (when \(V_k\) provided):
\[ V_{required} \leq V_k \implies \text{PASS} \]Standard CT Ratios
5 A Secondary
| Ratio | Primary (A) | Ratio | Primary (A) | Ratio | Primary (A) |
|---|---|---|---|---|---|
| 5:5 | 5 | 10:5 | 10 | 15:5 | 15 |
| 20:5 | 20 | 25:5 | 25 | 30:5 | 30 |
| 40:5 | 40 | 50:5 | 50 | 60:5 | 60 |
| 75:5 | 75 | 100:5 | 100 | 150:5 | 150 |
| 200:5 | 200 | 250:5 | 250 | 300:5 | 300 |
| 400:5 | 400 | 500:5 | 500 | 600:5 | 600 |
| 750:5 | 750 | 800:5 | 800 | 1000:5 | 1000 |
| 1200:5 | 1200 | 1500:5 | 1500 | 2000:5 | 2000 |
1 A Secondary
| Ratio | Primary (A) |
|---|---|
| 50:1 | 50 |
| 100:1 | 100 |
| 200:1 | 200 |
| 500:1 | 500 |
| 1000:1 | 1000 |
Worked Examples
The 9 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_protection; worked examples for every calculation).
| # | Case | CT | Burden (VA) | Fault multiple / ALF | Vreq / Vk (V) | Result |
|---|---|---|---|---|---|---|
| 1 | 400/5 5P20 feeder CT, all checks pass | 400/5 5P20 | 8.86 / 15 | 15.0 / 20 | 26.6 / 200 | PASS |
| 2 | 300/5 5P20 with winding resistance, fully passing | 300/5 5P20 | 10.62 / 30 | 13.3 / 20 | 28.3 / 500 | PASS |
| 3 | 50/5 5P10 at 100 × rated current | 50/5 5P10 | 8.89 / 10 | 100.0 / 10 | 177.7 / – | FAIL |
| 4 | 100/5 5P10, burden passes but ALF fails | 100/5 5P10 | 8.89 / 10 | 60.0 / 10 | 106.6 / – | FAIL |
| 5 | 200/5 motor-feeder CT: burden and ALF fail, Vk passes | 200/5 5P20 | 21.36 / 15 | 50.0 / 20 | 213.6 / 400 | FAIL |
| 6 | 600/5 10P10 with 100 m of leads at 90 °C | 600/5 10P10 | 27.98 / 15 | 8.3 / 10 | 46.6 / – | FAIL |
| 7 | 1000/1 5P20 on a 150 m lead run | 1,000/1 5P20 | 7.02 / 10 | 15.0 / 20 | 105.2 / 150 | PASS |
| 8 | Class PX 400/1 saturates at 40 kA | 400/1 PX | 3.78 / 5 | 100.0 / – | 377.7 / 200 | FAIL |
| 9 | Class 0.5 metering CT, leads at 20 °C | 200/5 0.5 | 4.95 / 5 | 25.0 / – | 24.7 / – | PASS |
Example 1 — 400/5 5P20 feeder CT, all checks pass
A 400/5 5P20 15 VA CT with 30 m of 4 mm² leads: burden, ALF and knee point all comfortable.
Given: 400/5 A CT, class 5P20, rated burden 15 VA, knee point Vk = 200 V, winding resistance Rct = 0 Ω; nominal current 300 A, maximum fault current 6,000 A; 30 m (one way) of 4 mm² copper leads at 75 °C; relay burden 1 VA (415 V, 50 Hz).
| CT ratio | n = 400 ÷ 5 = 80 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 30 ÷ 4 = 0.3145 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 1 ÷ 5² = 0.0400 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 0.3145 + 0.0400 + 0 = 0.3545 Ω |
| Burden | Isn² × Rtotal = 5² × 0.3545 = 8.863 VA (59.1 % of rating) → 8.863 vs limit 15.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 6,000 ÷ 400 = 15.00 × rated current against ALF 20 of class 5P20 → 15.00 vs limit 20.00 × In PASS |
| Secondary fault current | Is,f = If ÷ n = 6,000 ÷ 80 = 75.00 A |
| Knee point | Vreq = Is,f × Rtotal = 75.00 × 0.3545 = 26.59 V against Vk (saturation risk low) → 26.59 vs limit 200.00 V PASS |
| Ratio and lead length | Smallest standard ratio ≥ 300 A: 300:5; maximum lead length = ⌊(15 ÷ 5² − 0.0400 − 0) ÷ 0.010484 Ω/m⌋ = ⌊0.5600 ÷ 0.010484⌋ = 53 m |
| Result | 3/3 checks pass → PASS; SUITABLE, saturation risk low; governing: accuracy limit factor. Cable resistance dominates burden - consider shorter cable run or larger conductor |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 30 ÷ 4 = 0.3145 Ω; R_relay = 1 ÷ 5² = 0.0400 Ω; R_total = 0.3145 + 0.0400 + 0 = 0.3545 Ω; burden = 5² × 0.3545 = 8.863 VA ≤ 15 VA; fault multiple = 6000 ÷ 400 = 15.00 ≤ ALF 20; Is,fault = 6000 ÷ (400/5) = 75.00 A; V_req = 75.00 × 0.3545 = 26.59 V ≤ Vk 200 V; smallest standard ratio ≥ 300 A is 300:5; max lead = ⌊(15/5² − 0.0400 − 0) ÷ 0.010484 Ω/m⌋ = 53 m (validation case c0008-400-5-5p20-suitable) |
Example 2 — 300/5 5P20 with winding resistance, fully passing
Help-page example: 300/5 5P20 30 VA, Rct 0.3 Ω, Vk 500 V, 10 m of 4 mm² — every check passes.
Given: 300/5 A CT, class 5P20, rated burden 30 VA, knee point Vk = 500 V, winding resistance Rct = 0.3 Ω; nominal current 250 A, maximum fault current 4,000 A; 10 m (one way) of 4 mm² copper leads at 75 °C; relay burden 0.5 VA (415 V, 50 Hz).
| CT ratio | n = 300 ÷ 5 = 60 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 10 ÷ 4 = 0.1048 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 0.5 ÷ 5² = 0.0200 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 0.1048 + 0.0200 + 0.3 = 0.4248 Ω |
| Burden | Isn² × Rtotal = 5² × 0.4248 = 10.621 VA (35.4 % of rating) → 10.621 vs limit 30.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 4,000 ÷ 300 = 13.33 × rated current against ALF 20 of class 5P20 → 13.33 vs limit 20.00 × In PASS |
| Secondary fault current | Is,f = If ÷ n = 4,000 ÷ 60 = 66.67 A |
| Knee point | Vreq = Is,f × Rtotal = 66.67 × 0.4248 = 28.32 V against Vk (saturation risk low) → 28.32 vs limit 500.00 V PASS |
| Ratio and lead length | Smallest standard ratio ≥ 250 A: 250:5; maximum lead length = ⌊(30 ÷ 5² − 0.0200 − 0.3) ÷ 0.010484 Ω/m⌋ = ⌊0.8800 ÷ 0.010484⌋ = 83 m |
| Result | 3/3 checks pass → PASS; SUITABLE, saturation risk low; governing: accuracy limit factor. Cable resistance dominates burden - consider shorter cable run or larger conductor |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 10 ÷ 4 = 0.1048 Ω; R_relay = 0.5 ÷ 5² = 0.0200 Ω; R_total = 0.1048 + 0.0200 + 0.3 = 0.4248 Ω; burden = 5² × 0.4248 = 10.621 VA ≤ 30 VA; fault multiple = 4000 ÷ 300 = 13.33 ≤ ALF 20; Is,fault = 4000 ÷ (300/5) = 66.67 A; V_req = 66.67 × 0.4248 = 28.32 V ≤ Vk 500 V; smallest standard ratio ≥ 250 A is 250:5; max lead = ⌊(30/5² − 0.0200 − 0.3) ÷ 0.010484 Ω/m⌋ = 83 m (validation case c0008-300-5-5p20-fully-passing) |
Example 3 — 50/5 5P10 at 100 × rated current
A 5 kA fault on a 50/5 CT is 100 × rated current, ten times the 5P10 accuracy limit factor.
Given: 50/5 A CT, class 5P10, rated burden 10 VA, knee point not given, winding resistance Rct = 0 Ω; nominal current 50 A, maximum fault current 5,000 A; 20 m (one way) of 2.5 mm² copper leads at 75 °C; relay burden 0.5 VA (415 V, 50 Hz).
| CT ratio | n = 50 ÷ 5 = 10 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 20 ÷ 2.5 = 0.3355 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 0.5 ÷ 5² = 0.0200 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 0.3355 + 0.0200 + 0 = 0.3555 Ω |
| Burden | Isn² × Rtotal = 5² × 0.3555 = 8.887 VA (88.9 % of rating) → 8.887 vs limit 10.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 5,000 ÷ 50 = 100.00 × rated current against ALF 10 of class 5P10 → 100.00 vs limit 10.00 × In FAIL |
| Secondary fault current | Is,f = If ÷ n = 5,000 ÷ 10 = 500.00 A |
| Knee point | Vreq = Is,f × Rtotal = 500.00 × 0.3555 = 177.74 V; no Vk given, so the required voltage is banded (< 50 V low, < 150 V medium, else high) → saturation risk high |
| Ratio and lead length | Smallest standard ratio ≥ 50 A: 50:5; maximum lead length = ⌊(10 ÷ 5² − 0.0200 − 0) ÷ 0.016774 Ω/m⌋ = ⌊0.3800 ÷ 0.016774⌋ = 22 m |
| Result | 1/2 checks pass → FAIL; NOT SUITABLE, saturation risk high; governing: accuracy limit factor. Verify knee-point voltage from CT datasheet Cable resistance dominates burden - consider shorter cable run or larger conductor Use higher ALF class (e.g., 5P20) or increase CT ratio Option 2: Use CT with ALF ≥ 100 (e.g., 5P20) Option 3: Increase CT ratio to 100:5 (adjust relay settings) |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 20 ÷ 2.5 = 0.3355 Ω; R_relay = 0.5 ÷ 5² = 0.0200 Ω; R_total = 0.3355 + 0.0200 + 0 = 0.3555 Ω; burden = 5² × 0.3555 = 8.887 VA ≤ 10 VA; fault multiple = 5000 ÷ 50 = 100.00 > ALF 10; Is,fault = 5000 ÷ (50/5) = 500.00 A; V_req = 500.00 × 0.3555 = 177.74 V (no Vk: < 50 V low, < 150 V medium, else high); smallest standard ratio ≥ 50 A is 50:5; max lead = ⌊(10/5² − 0.0200 − 0) ÷ 0.016774 Ω/m⌋ = 22 m (validation case c0008-50-5-5p10-alf-fails) |
Example 4 — 100/5 5P10, burden passes but ALF fails
Help-page example: the 8.89 VA burden fits 10 VA but 6 kA is 60 × rated current against ALF 10; no knee point given.
Given: 100/5 A CT, class 5P10, rated burden 10 VA, knee point not given, winding resistance Rct = 0 Ω; nominal current 80 A, maximum fault current 6,000 A; 20 m (one way) of 2.5 mm² copper leads at 75 °C; relay burden 0.5 VA (415 V, 50 Hz).
| CT ratio | n = 100 ÷ 5 = 20 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 20 ÷ 2.5 = 0.3355 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 0.5 ÷ 5² = 0.0200 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 0.3355 + 0.0200 + 0 = 0.3555 Ω |
| Burden | Isn² × Rtotal = 5² × 0.3555 = 8.887 VA (88.9 % of rating) → 8.887 vs limit 10.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 6,000 ÷ 100 = 60.00 × rated current against ALF 10 of class 5P10 → 60.00 vs limit 10.00 × In FAIL |
| Secondary fault current | Is,f = If ÷ n = 6,000 ÷ 20 = 300.00 A |
| Knee point | Vreq = Is,f × Rtotal = 300.00 × 0.3555 = 106.64 V; no Vk given, so the required voltage is banded (< 50 V low, < 150 V medium, else high) → saturation risk medium |
| Ratio and lead length | Smallest standard ratio ≥ 80 A: 100:5; maximum lead length = ⌊(10 ÷ 5² − 0.0200 − 0) ÷ 0.016774 Ω/m⌋ = ⌊0.3800 ÷ 0.016774⌋ = 22 m |
| Result | 1/2 checks pass → FAIL; NOT SUITABLE, saturation risk medium; governing: accuracy limit factor. Verify knee-point voltage from CT datasheet Cable resistance dominates burden - consider shorter cable run or larger conductor Use higher ALF class (e.g., 5P20) or increase CT ratio Option 2: Use CT with ALF ≥ 60 (e.g., 5P20) Option 3: Increase CT ratio to 200:5 (adjust relay settings) |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 20 ÷ 2.5 = 0.3355 Ω; R_relay = 0.5 ÷ 5² = 0.0200 Ω; R_total = 0.3355 + 0.0200 + 0 = 0.3555 Ω; burden = 5² × 0.3555 = 8.887 VA ≤ 10 VA; fault multiple = 6000 ÷ 100 = 60.00 > ALF 10; Is,fault = 6000 ÷ (100/5) = 300.00 A; V_req = 300.00 × 0.3555 = 106.64 V (no Vk: < 50 V low, < 150 V medium, else high); smallest standard ratio ≥ 80 A is 100:5; max lead = ⌊(10/5² − 0.0200 − 0) ÷ 0.016774 Ω/m⌋ = 22 m (validation case c0008-100-5-5p10-alf-fails-no-vk) |
Example 5 — 200/5 motor-feeder CT: burden and ALF fail, Vk passes
Help-page example: 0.5 Ω of CT winding pushes the burden to 21.4 VA over a 15 VA rating and 50 × exceeds ALF 20, though the 400 V knee point is ample.
Given: 200/5 A CT, class 5P20, rated burden 15 VA, knee point Vk = 400 V, winding resistance Rct = 0.5 Ω; nominal current 150 A, maximum fault current 10,000 A; 30 m (one way) of 4 mm² copper leads at 75 °C; relay burden 1 VA (415 V, 50 Hz).
| CT ratio | n = 200 ÷ 5 = 40 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 30 ÷ 4 = 0.3145 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 1 ÷ 5² = 0.0400 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 0.3145 + 0.0400 + 0.5 = 0.8545 Ω |
| Burden | Isn² × Rtotal = 5² × 0.8545 = 21.363 VA (142.4 % of rating) → 21.363 vs limit 15.000 VA FAIL |
| Accuracy limit factor | If ÷ Ipn = 10,000 ÷ 200 = 50.00 × rated current against ALF 20 of class 5P20 → 50.00 vs limit 20.00 × In FAIL |
| Secondary fault current | Is,f = If ÷ n = 10,000 ÷ 40 = 250.00 A |
| Knee point | Vreq = Is,f × Rtotal = 250.00 × 0.8545 = 213.63 V against Vk (saturation risk low) → 213.63 vs limit 400.00 V PASS |
| Ratio and lead length | Smallest standard ratio ≥ 150 A: 150:5; maximum lead length = ⌊(15 ÷ 5² − 0.0400 − 0.5) ÷ 0.010484 Ω/m⌋ = ⌊0.0600 ÷ 0.010484⌋ = 5 m |
| Result | 1/3 checks pass → FAIL; NOT SUITABLE, saturation risk low; governing: accuracy limit factor. Cable resistance dominates burden - consider shorter cable run or larger conductor Increase CT rated VA to at least 25.6 VA Use higher ALF class (e.g., 5P20) or increase CT ratio Option 1: Reduce secondary burden (shorter cable, larger conductor, lower relay VA) Option 2: Use CT with ALF ≥ 50 (e.g., 5P40) Option 3: Increase CT ratio to 400:5 (adjust relay settings) |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 30 ÷ 4 = 0.3145 Ω; R_relay = 1 ÷ 5² = 0.0400 Ω; R_total = 0.3145 + 0.0400 + 0.5 = 0.8545 Ω; burden = 5² × 0.8545 = 21.363 VA > 15 VA; fault multiple = 10000 ÷ 200 = 50.00 > ALF 20; Is,fault = 10000 ÷ (200/5) = 250.00 A; V_req = 250.00 × 0.8545 = 213.63 V ≤ Vk 400 V; smallest standard ratio ≥ 150 A is 150:5; max lead = ⌊(15/5² − 0.0400 − 0.5) ÷ 0.010484 Ω/m⌋ = 5 m (validation case c0008-200-5-burden-and-alf-fail) |
Example 6 — 600/5 10P10 with 100 m of leads at 90 °C
Long 4 mm² leads at 90 °C make the lead resistance dominate: 28 VA against 15 VA, and the leads may be at most 52 m long.
Given: 600/5 A CT, class 10P10, rated burden 15 VA, knee point not given, winding resistance Rct = 0 Ω; nominal current 550 A, maximum fault current 5,000 A; 100 m (one way) of 4 mm² copper leads at 90 °C; relay burden 0.5 VA (415 V, 50 Hz).
| CT ratio | n = 600 ÷ 5 = 120 |
|---|---|
| Lead resistance | ρ(90 °C) = 0.017241 × (1 + 0.00393 × (90 − 20)) = 0.021984 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.021984 × 2 × 100 ÷ 4 = 1.0992 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 0.5 ÷ 5² = 0.0200 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 1.0992 + 0.0200 + 0 = 1.1192 Ω |
| Burden | Isn² × Rtotal = 5² × 1.1192 = 27.980 VA (186.5 % of rating) → 27.980 vs limit 15.000 VA FAIL |
| Accuracy limit factor | If ÷ Ipn = 5,000 ÷ 600 = 8.33 × rated current against ALF 10 of class 10P10 → 8.33 vs limit 10.00 × In PASS |
| Secondary fault current | Is,f = If ÷ n = 5,000 ÷ 120 = 41.67 A |
| Knee point | Vreq = Is,f × Rtotal = 41.67 × 1.1192 = 46.63 V; no Vk given, so the required voltage is banded (< 50 V low, < 150 V medium, else high) → saturation risk low |
| Ratio and lead length | Smallest standard ratio ≥ 550 A: 600:5; maximum lead length = ⌊(15 ÷ 5² − 0.0200 − 0) ÷ 0.010992 Ω/m⌋ = ⌊0.5800 ÷ 0.010992⌋ = 52 m |
| Result | 1/2 checks pass → FAIL; NOT SUITABLE, saturation risk low; governing: burden. Verify knee-point voltage from CT datasheet Cable resistance dominates burden - consider shorter cable run or larger conductor Increase CT rated VA to at least 33.6 VA Option 1: Reduce secondary burden (shorter cable, larger conductor, lower relay VA) |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(90 °C) = 0.017241 × (1 + 0.00393 × 70) = 0.021984 Ω·mm²/m; R_lead = 0.021984 × 2 × 100 ÷ 4 = 1.0992 Ω; R_relay = 0.5 ÷ 5² = 0.0200 Ω; R_total = 1.0992 + 0.0200 + 0 = 1.1192 Ω; burden = 5² × 1.1192 = 27.980 VA > 15 VA; fault multiple = 5000 ÷ 600 = 8.33 ≤ ALF 10; Is,fault = 5000 ÷ (600/5) = 41.67 A; V_req = 41.67 × 1.1192 = 46.63 V (no Vk: < 50 V low, < 150 V medium, else high); smallest standard ratio ≥ 550 A is 600:5; max lead = ⌊(15/5² − 0.0200 − 0) ÷ 0.010992 Ω/m⌋ = 52 m (validation case c0008-600-5-10p10-long-hot-leads) |
Example 7 — 1000/1 5P20 on a 150 m lead run
A 1 A secondary keeps 150 m of 2.5 mm² leads to 2.5 VA; with 4 Ω of winding the 15 kA fault needs 105 V against a 150 V knee point.
Given: 1,000/1 A CT, class 5P20, rated burden 10 VA, knee point Vk = 150 V, winding resistance Rct = 4 Ω; nominal current 800 A, maximum fault current 15,000 A; 150 m (one way) of 2.5 mm² copper leads at 75 °C; relay burden 0.5 VA (415 V, 50 Hz).
| CT ratio | n = 1,000 ÷ 1 = 1,000 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 150 ÷ 2.5 = 2.5161 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 0.5 ÷ 1² = 0.5000 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 2.5161 + 0.5000 + 4 = 7.0161 Ω |
| Burden | Isn² × Rtotal = 1² × 7.0161 = 7.016 VA (70.2 % of rating) → 7.016 vs limit 10.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 15,000 ÷ 1,000 = 15.00 × rated current against ALF 20 of class 5P20 → 15.00 vs limit 20.00 × In PASS |
| Secondary fault current | Is,f = If ÷ n = 15,000 ÷ 1,000 = 15.00 A |
| Knee point | Vreq = Is,f × Rtotal = 15.00 × 7.0161 = 105.24 V against Vk (saturation risk low) → 105.24 vs limit 150.00 V PASS |
| Ratio and lead length | Smallest standard ratio ≥ 800 A: 1000:1; maximum lead length = ⌊(10 ÷ 1² − 0.5000 − 4) ÷ 0.016774 Ω/m⌋ = ⌊5.5000 ÷ 0.016774⌋ = 327 m |
| Result | 3/3 checks pass → PASS; SUITABLE, saturation risk low; governing: accuracy limit factor. Cable resistance dominates burden - consider shorter cable run or larger conductor |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 150 ÷ 2.5 = 2.5161 Ω; R_relay = 0.5 ÷ 1² = 0.5000 Ω; R_total = 2.5161 + 0.5000 + 4 = 7.0161 Ω; burden = 1² × 7.0161 = 7.016 VA ≤ 10 VA; fault multiple = 15000 ÷ 1000 = 15.00 ≤ ALF 20; Is,fault = 15000 ÷ (1000/1) = 15.00 A; V_req = 15.00 × 7.0161 = 105.24 V ≤ Vk 150 V; smallest standard ratio ≥ 800 A is 1000:1; max lead = ⌊(10/1² − 0.5000 − 4) ÷ 0.016774 Ω/m⌋ = 327 m (validation case c0008-1000-1-5p20-long-leads) |
Example 8 — Class PX 400/1 saturates at 40 kA
A class PX CT has no ALF, so the knee point governs: 100 A secondary × 3.78 Ω needs 378 V against Vk = 200 V.
Given: 400/1 A CT, class PX, rated burden 5 VA, knee point Vk = 200 V, winding resistance Rct = 2 Ω; nominal current 350 A, maximum fault current 40,000 A; 100 m (one way) of 2.5 mm² copper leads at 75 °C; relay burden 0.1 VA (415 V, 50 Hz).
| CT ratio | n = 400 ÷ 1 = 400 |
|---|---|
| Lead resistance | ρ(75 °C) = 0.017241 × (1 + 0.00393 × (75 − 20)) = 0.020968 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.020968 × 2 × 100 ÷ 2.5 = 1.6774 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 0.1 ÷ 1² = 0.1000 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 1.6774 + 0.1000 + 2 = 3.7774 Ω |
| Burden | Isn² × Rtotal = 1² × 3.7774 = 3.777 VA (75.5 % of rating) → 3.777 vs limit 5.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 40,000 ÷ 400 = 100.00 × rated current; class PX has no accuracy limit factor, so no ALF check (verify from the datasheet) |
| Secondary fault current | Is,f = If ÷ n = 40,000 ÷ 400 = 100.00 A |
| Knee point | Vreq = Is,f × Rtotal = 100.00 × 3.7774 = 377.74 V against Vk (saturation risk high) → 377.74 vs limit 200.00 V FAIL |
| Ratio and lead length | Smallest standard ratio ≥ 350 A: 500:1; maximum lead length = ⌊(5 ÷ 1² − 0.1000 − 2) ÷ 0.016774 Ω/m⌋ = ⌊2.9000 ÷ 0.016774⌋ = 172 m |
| Result | 1/2 checks pass → FAIL; NOT SUITABLE, saturation risk high; governing: knee point voltage. CT class does not specify ALF - verify suitability from datasheet Cable resistance dominates burden - consider shorter cable run or larger conductor Use CT with higher knee-point voltage (PX/TPX class) Option 4: Use PX/TPX class CT with Vk ≥ 491 V |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(75 °C) = 0.017241 × (1 + 0.00393 × 55) = 0.020968 Ω·mm²/m; R_lead = 0.020968 × 2 × 100 ÷ 2.5 = 1.6774 Ω; R_relay = 0.1 ÷ 1² = 0.1000 Ω; R_total = 1.6774 + 0.1000 + 2 = 3.7774 Ω; burden = 1² × 3.7774 = 3.777 VA ≤ 5 VA; fault multiple = 40000 ÷ 400 = 100.00 (class PX: no ALF); Is,fault = 40000 ÷ (400/1) = 100.00 A; V_req = 100.00 × 3.7774 = 377.74 V > Vk 200 V; smallest standard ratio ≥ 350 A is 500:1; max lead = ⌊(5/1² − 0.1000 − 2) ÷ 0.016774 Ω/m⌋ = 172 m (validation case c0008-px-400-1-saturates) |
Example 9 — Class 0.5 metering CT, leads at 20 °C
A metering class has no ALF and no knee point given: only the burden is checked (4.95 VA of 5 VA) and the required voltage is banded as a saturation risk.
Given: 200/5 A CT, class 0.5, rated burden 5 VA, knee point not given, winding resistance Rct = 0 Ω; nominal current 180 A, maximum fault current 5,000 A; 10 m (one way) of 2.5 mm² copper leads at 20 °C; relay burden 1.5 VA (415 V, 50 Hz).
| CT ratio | n = 200 ÷ 5 = 40 |
|---|---|
| Lead resistance | ρ(20 °C) = 0.017241 × (1 + 0.00393 × (20 − 20)) = 0.017241 Ω·mm²/m; Rlead = ρ × 2L ÷ A = 0.017241 × 2 × 10 ÷ 2.5 = 0.1379 Ω (go-and-return loop) |
| Relay resistance | Rrelay = VA ÷ Isn² = 1.5 ÷ 5² = 0.0600 Ω |
| Loop resistance | Rtotal = Rlead + Rrelay + Rct = 0.1379 + 0.0600 + 0 = 0.1979 Ω |
| Burden | Isn² × Rtotal = 5² × 0.1979 = 4.948 VA (99.0 % of rating) → 4.948 vs limit 5.000 VA PASS |
| Accuracy limit factor | If ÷ Ipn = 5,000 ÷ 200 = 25.00 × rated current; class 0.5 has no accuracy limit factor, so no ALF check (verify from the datasheet) |
| Secondary fault current | Is,f = If ÷ n = 5,000 ÷ 40 = 125.00 A |
| Knee point | Vreq = Is,f × Rtotal = 125.00 × 0.1979 = 24.74 V; no Vk given, so the required voltage is banded (< 50 V low, < 150 V medium, else high) → saturation risk low |
| Ratio and lead length | Smallest standard ratio ≥ 180 A: 200:5; maximum lead length = ⌊(5 ÷ 5² − 0.0600 − 0) ÷ 0.013793 Ω/m⌋ = ⌊0.1400 ÷ 0.013793⌋ = 10 m |
| Result | 1/1 checks pass → PASS; SUITABLE, saturation risk low; governing: burden. CT class does not specify ALF - verify suitability from datasheet Verify knee-point voltage from CT datasheet |
| Reference | Hand calculation (IEC 61869-2 burden / ALF / knee-point): ρ(20 °C) = 0.017241 × (1 + 0.00393 × 0) = 0.017241 Ω·mm²/m; R_lead = 0.017241 × 2 × 10 ÷ 2.5 = 0.1379 Ω; R_relay = 1.5 ÷ 5² = 0.0600 Ω; R_total = 0.1379 + 0.0600 + 0 = 0.1979 Ω; burden = 5² × 0.1979 = 4.948 VA ≤ 5 VA; fault multiple = 5000 ÷ 200 = 25.00 (class 0.5: no ALF); Is,fault = 5000 ÷ (200/5) = 125.00 A; V_req = 125.00 × 0.1979 = 24.74 V (no Vk: < 50 V low, < 150 V medium, else high); smallest standard ratio ≥ 180 A is 200:5; max lead = ⌊(5/5² − 0.0600 − 0) ÷ 0.013793 Ω/m⌋ = 10 m (validation case c0008-metering-0-5-at-20c) |
Standards & References
- IEC 60044-1 — Instrument transformers — Part 1: Current transformers (superseded, still widely referenced)
- IEC 61869-2 — Instrument transformers — Part 2: Additional requirements for current transformers (current standard)
- IEC 61869-1 — Instrument transformers — Part 1: General requirements
- IEEE C57.13 — IEEE Standard Requirements for Instrument Transformers (North American practice)
- CT accuracy class designation: the first number is the composite error limit (%), "P" denotes protection class, and the second number is the Accuracy Limit Factor (e.g., 5P20 = 5% error at 20× rated current).
Tips & Limitations
Tips
- Modern digital relays typically have very low burdens (0.1–1.0 VA). Older electromechanical relays may present 5–15 VA — always check the relay datasheet.
- Cable resistance often dominates the total burden. Increasing conductor area from 2.5 mm² to 4 mm² reduces cable burden by 37%.
- If the ALF check fails, increasing the CT ratio is often more effective than changing the CT class, as it reduces both the fault multiple and the secondary fault current.
- For differential protection, use PX or TPX class CTs — standard P-class CTs do not guarantee performance for through-fault stability.
- A 1 A secondary carries one fifth of the current of a 5 A secondary, so the lead burden (I²R) is 1/25 and the lead voltage drop at fault current 1/5 of the 5 A figure. Use 1 A secondaries for long cable runs where burden must be minimised (the CT winding resistance and relay ohms are higher, so check the knee point).
- Always select a CT primary rating ≥ the expected maximum continuous current. The recommended ratio output shows the next standard ratio above nominal.
Limitations
- Steady-state sinusoidal only — DC offset and transient remanence during asymmetric faults are not modelled. For full transient analysis use TPX/TPY/TPZ class specifications.
- No derating for actual vs. rated burden — the effective ALF increases when actual burden is less than rated burden (per IEC 61869-2 clause), but this calculator does not apply that correction.
- Single CT core assumed — multi-core CTs (protection + metering on one unit) are not modelled. Analyse each core separately.
- Copper conductors only — cable resistance is calculated using copper resistivity. Aluminium secondary wiring is uncommon but would require different constants.
- Knee-point voltage is optional — when not provided, saturation risk is estimated heuristically from the required fault voltage (< 50 V = low, 50–150 V = medium, > 150 V = high).
Developer Guide
Architecture
File layout:
components/c0008_ct_selection/component.py— Component registration,PARAMETER_SCHEMA,CTSelectionCalculatorclass,equation_builder()components/c0008_ct_selection/calculations.py— Pure math functions, constants, and lookup tables (no Django dependencies)components/c0008_ct_selection/test_calculation.py— Unit tests for calculation logic
Key data structures:
STANDARD_CT_RATIOS— Dict mapping ratio strings (e.g.,'100:5') to(primary, secondary)tuplesCT_ACCURACY_CLASSES— Dict mapping class strings (e.g.,'5P10') to dicts withtype,error,alfkeys
The calculator class CTSelectionCalculator extends
ElectricalCalculator from
calcs.core.base_calculator. Registration follows the
standard component pattern via PARAMETER_SCHEMA and
equation_builder().
Code Paths
Key functions in calculations.py:
| Function | Purpose |
|---|---|
calculate_ct_selection(data) |
Main entry point — accepts input dict, returns full results dict with all checks and recommendations |
calculate_cable_resistance(length_m, area_mm2, temperature_c) |
Temperature-adjusted copper loop resistance |
calculate_relay_resistance(relay_va, i_secondary_rated) |
Convert relay VA burden to equivalent resistance |
recommend_ct_ratio(i_nominal, i_secondary) |
Find next standard ratio ≥ nominal current |
parse_ct_class(ct_class_str) |
Parse accuracy class string, extract error % and ALF |
Request flow:
CTSelectionCalculator.validate_parameters()— checks all required inputs and adds warnings for questionable combinations (e.g., CT primary < nominal current)CTSelectionCalculator.calculate()— builds the data dict and delegates tocalculate_ct_selection()equation_builder(results)— formats the results into the unified articles structure for the calculation report output
Testing
Running tests:
pytest calcs/components/c0008_ct_selection/test_calculation.py -vHow to extend:
- Add new CT accuracy classes by extending the
CT_ACCURACY_CLASSESdict incalculations.pyand updating thect_classenum inPARAMETER_SCHEMA. - Add new standard CT ratios by extending the
STANDARD_CT_RATIOSdict. - To support aluminium secondary wiring, add a resistivity
constant (
RHO_ALUMINIUM_20C) and a conductor material parameter. - To implement effective ALF correction (actual burden < rated burden), add the formula: \(ALF_{eff} = ALF_{rated} \times VA_{rated} / VA_{actual}\) per IEC 61869-2.
- Write tests covering new classes, ratios, and edge cases (zero internal resistance, maximum cable length, minimum conductor area).