Electrical tools
ElectricalTools.
Five calculators for motor current, cable sizing, power supply loading, panel heat dissipation and short-circuit current rating (SCCR). Everything runs in your browser, nothing is uploaded, and every result exports to CSV.
Tatva Logix / Engineering Tools
Motor Current Calculator
Full load current, starting current and switchgear sizing for AC and DC motors, with NEC 430 table values or IEC 60947 selection.
Nameplate values give the most accurate result. Leave the defaults for a typical 4-pole induction motor.
Full load current
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Electrical
Protection & switchgear sizing
Formula reference: how this result is derived
Every value on the right comes from one of the expressions below. P is rated shaft output, V the line voltage, pf the power factor and η the efficiency as a decimal.
Full load current, 3-phase AC
I = (P × 1000) ÷ (√3 × V × pf × η)
Line current. √3 = 1.7321. For 7.5 kW at 415 V, 0.86 pf, 90% efficiency: 13.5 A.
Full load current, 1-phase AC
I = (P × 1000) ÷ (V × pf × η)
V is the phase-to-neutral voltage for a single-phase load.
Full load current, DC
I = (P × 1000) ÷ (V × η)
No power factor term on a DC supply.
Power unit conversion
P[kW] = HP × 0.746
HP = P[kW] ÷ 0.746
1 HP = 745.7 W. Nameplate HP is output, not input.
Apparent and reactive power
S[kVA] = √3 × V × I ÷ 1000
P[kW] = S × pf
Drop √3 for single phase. Input kW = output kW ÷ η.
Starting current
Istart = IFL × k
k: 6 to 8 direct on line, 2.2 star-delta, 3 soft starter at 300% limit, 1.5 on a VFD.
Overload relay setting
IOL = 1.15 to 1.25 × Inameplate
IEC 60947-4-1 class 10 for a standard duty cycle. NEC 430.32 sets 115 to 125% of nameplate FLA.
Conductor and protection
Icable ≥ 1.25 × IFL
NEC 430.22. Short-circuit protection to NEC 430.52 (inverse-time breaker up to 250% of FLC); under IEC follow the manufacturer type 2 coordination chart.
Torque and speed
T[N·m] = 9550 × P[kW] ÷ n[rpm]
nsync = 120 × f ÷ poles
Slip is the difference between synchronous and nameplate speed.
NEC table current
FLC from NEC Table 430.250
In NEC mode conductors and protection are sized from the table FLC, not the nameplate current. The nameplate value is used only for the overload element.
Indicative results for preliminary sizing and reference only. Verify against the applicable standard, product data and site conditions before design or procurement.
Tatva Logix / Engineering Tools
Cable Size Calculator
Minimum conductor size from current-carrying capacity and voltage drop, with ambient and grouping derating applied to IEC 60364-5-52 or NEC Table 310.16 values.
Recommended conductor
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Sizing basis
Derating applied
Formula reference: how this result is derived
A cable has to satisfy two independent checks. The larger of the two sizes wins.
Check 1: current-carrying capacity
Iz = It × Kamb × Kgroup ≥ Ib
It is the tabulated ampacity for the insulation and installation method, Ib the design current. Every derating factor multiplies.
Check 2: voltage drop, IEC
ΔU = (mV/A/m) × Ib × L ÷ 1000
The mV/A/m figure already contains the √3 for a three-phase circuit. Single phase uses the two-conductor value.
Check 2: voltage drop, NEC
ΔV = √3 × I × R × L ÷ 1000
(2 × I × R × L ÷ 1000 single phase)
R is the conductor resistance in ohms per 1000 ft from NEC Chapter 9 Table 8.
Voltage drop as a percentage
ΔU% = ΔU ÷ V × 100
Common limits: 3% on a final circuit and 5% total from the origin (IEC 60364-5-52 Annex G, NEC 210.19 and 215.2 informational notes).
Utilisation
Utilisation% = Ib ÷ Iz × 100
Above 95% there is no room for load growth or harmonic heating.
Protective device coordination
Ib ≤ In ≤ Iz
IEC 60364-4-43. The device rating sits between the design current and the derated cable capacity.
Aluminium conductors
It,Al ≈ 0.78 × It,Cu
Estimate used in IEC mode. Resistance is about 1.64 times copper for the same section. Confirm against the manufacturer table.
Termination temperature limit
NEC 110.14(C)(1)
For equipment rated 100 A or less, ampacity is generally taken from the 60 °C column even where the conductor is rated 90 °C, unless the terminations are listed higher.
Cable sizing tables: IEC 60364-5-52 and NEC 310.16
Indicative results for preliminary sizing and reference only. Verify against the applicable standard, product data and site conditions before design or procurement.
Tatva Logix / Engineering Tools
Power Supply Calculator
Size a 24 VDC control supply from the connected device list, with spare capacity, thermal derating and backup sizing.
Connected load
| Device | Qty | mA each |
|---|
Enter steady-state consumption from the device datasheet. Inrush is handled separately below.
Recommended supply
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Load summary
Selection notes
Formula reference: how this result is derived
Connected load
Itotal = Σ (qty × mA) ÷ 1000
Steady-state current from each device datasheet, at the supply voltage.
Design load
Idesign = Itotal × (1 + spare%) ÷ Ktemp
Spare capacity covers future I/O. The supply is then chosen as the next standard rating at or above this figure.
Thermal derating Ktemp
1.0 up to 45 °C
0.9 at 50, 0.8 at 55, 0.7 at 60 °C
Typical DIN-rail derating curve. Always confirm the curve for the model actually specified.
Headroom on the selected unit
Headroom% = (Irated − Itotal) ÷ Irated × 100
Below 20% leaves no room for panel growth.
Heat rejected into the panel
Q[W] = Pout × (1 − η) ÷ η
At 90% efficiency Q is about 0.111 × the delivered watts. Carry this figure into the panel heat calculation.
Battery or UPS backup
Ah = Iload × t[min] ÷ 60
Before charge efficiency, end-of-discharge voltage and ageing allowance. Add 25 to 40% for a usable figure.
Branch protection
Itrip,branch ≤ 0.5 × Irated,supply
Keeps one shorted branch from collapsing the whole 24 V bus before the branch device clears.
Inrush allowance
Ipeak = 2 to 6 × Irated for 10 to 50 ms
Capacitive loads (drives, HMIs, switches) draw well above steady state at power-up. Use a supply with a stated peak-power reserve.
Indicative results for preliminary sizing and reference only. Verify against the applicable standard, product data and site conditions before design or procurement.
Tatva Logix / Engineering Tools
Panel Heat Dissipation Calculator
Effective cooling surface, natural dissipation and surplus heat load for a control enclosure, with filter fan and cooling unit sizing to IEC 60890 surface factors.
Internal heat load
| Source | Qty | Watts |
|---|
For a VFD, enter the drive loss - roughly 3% of rated motor power. For a transformer or power supply, enter rated output x (1 − efficiency).
Cooling required
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Thermal balance
Equipment selection
Formula reference: how this result is derived
Temperature rise is the balance between the heat generated inside the enclosure and the heat its surface can shed. ΔT is the internal target minus the maximum ambient.
Effective surface area (IEC 60890)
Ae = Σ (Asurface × b)
Each face is counted with a factor b for its exposure: 1.4 free top, 0.7 covered or floor-standing base, 0.5 an obstructed side. A blocked face sheds almost nothing.
Natural dissipation
Qout = k × Ae × ΔT
k is the surface heat transfer coefficient: 5.5 W/m²K painted sheet steel, 3.7 stainless, 12 bare aluminium.
Temperature difference
ΔT = Tinternal − Tambient
Keep the internal target at or below the lowest component rating, typically 40 to 45 °C for drives and PLCs.
Net heat to remove
Qnet = Qinternal + Qsolar − Qout
Positive means forced cooling is required. Negative means natural convection is enough.
Filter fan airflow
V[m³/h] = 3.1 × Qnet[W] ÷ ΔT[K]
Sea level. Add about 15% per 1000 m of altitude. A fan can only ever hold the panel above ambient, never below.
Cooling unit capacity
Pcool ≥ Qinternal − k × Ae × ΔT
Required when the target internal temperature is at or below ambient. Rated capacity is quoted at L35/L35: derate for the real ambient.
Panel heater
Pheat ≥ k × Ae × (Tmin,req − Tmin,amb)
Sized with the panel de-energised, so no internal load is counted. Add a hygrostat where condensation is the concern.
Typical component losses
VFD ≈ 3% of rated kW
PSU = Pout × (1 − η) ÷ η
Transformer ≈ 5% of kVA
Use datasheet losses where available. Contactors, relays and fuses are a few watts each but add up in a large panel.
Indicative results for preliminary sizing and reference only. Verify against the applicable standard, product data and site conditions before design or procurement.
Tatva Logix / Engineering Tools
SCCR Calculator
Short-circuit current rating of an industrial control panel by the weakest-link method of UL 508A Supplement SB, checked against the available fault current at the point of installation.
Main device IR is the interrupting rating of the panel main protective device. Available fault current is the figure stated by the utility or the plant study at the panel terminals.
Power circuit components
| Component | Circuit | Marked SCCR |
|---|
List only components in the power circuit. Control circuit devices downstream of a transformer rated 10 kVA or less are excluded by SB4.1. Components with an unlimited rating (bus bar, terminal blocks, wiring, marked distribution blocks) do not limit the panel.
Estimate the available fault current
Infinite primary method: Isc = IFL / Zpu. Conservative, and it ignores primary impedance and motor contribution. A plant short-circuit study is the figure to design to.
Panel SCCR
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Rating build-up
Limiting components
If the limit is addressed
Guide: how a panel SCCR is determined
UL 508A Supplement SB determines the rating of an assembly from the ratings of its parts. The panel is only as strong as its weakest power-circuit component, and the marked rating must equal or exceed the fault current available where the panel is installed (NEC 409.110 and 110.10).
The method, step by step
- Identify every component in the power circuit: from the incoming terminals through the main device, distribution, branch protection, starters, drives and to the load terminals.
- Record each component's marked SCCR. Where a component carries no marking, use the default from Table SB4.1 below.
- Components listed as unlimited in SB4.1 (bus bar, wiring, terminal blocks, marked power distribution blocks, current transformers) are skipped: they do not limit the rating.
- The feeder circuit rating is the lowest rating among the feeder components.
- Each branch circuit rating is the lowest rating among that branch's components. A branch may be raised where the branch protective device is current-limiting and the combination is tested and marked for the higher rating.
- The panel SCCR is the lowest of the feeder rating and every branch rating, and it can never exceed the interrupting rating of the main protective device.
- Mark the panel with the result and confirm it is at or above the available fault current.
Default ratings for unmarked components, Table SB4.1
| Component | Default SCCR |
|---|---|
| Bus bar, wiring, terminal block, wire connector | Unlimited |
| Power distribution block, spacings observed and marked | Unlimited |
| Current transformer, current shunt, meter with external CT | Unlimited |
| Fuse holder (the fuse carries the rating) | Unlimited |
| Motor controller, 2 hp (1.5 kW) or less and 300 V or less | 1 kA |
| Motor controller, above 2 hp up to 50 hp | 5 kA |
| Motor controller, 51 to 200 hp | 10 kA |
| Motor controller, 201 to 400 hp | 18 kA |
| Motor controller, 401 to 600 hp | 30 kA |
| Overload relay and other auxiliary devices | 5 kA |
| Disconnect switch, switch unit | 5 kA |
| Circuit breaker, unmarked | 5 kA |
| Miniature and miscellaneous fuse | 10 kA |
| Meter socket base | 10 kA |
Reproduced for guidance only, abbreviated. Work from the current edition of UL 508A and the component markings on the actual bill of material.
Raising a low rating
Specify a higher marked rating
100 kA drives and starters exist
Cheapest at design time. Most drive and starter families publish an SCCR table conditional on a specific input fuse or breaker type: honour that condition on the drawing and the bill of material.
Current-limiting protection
Class J, CC, T or RK1 fuses
Lets a tested combination stand at the higher rating (SB4.2.2). The fuse class, maximum amp rating and manufacturer must match the tested combination exactly.
Reduce the fault current
Impedance in front of the panel
A smaller transformer, a longer feeder or a series reactor lowers the available fault current. The panel then only needs a rating above that lower figure.
Split the load
Move a weak branch out
Feeding a low-rated component from a separate transformer of 10 kVA or less takes it out of the power-circuit calculation.
Fault current estimate
Transformer full load current
IFL = kVA × 1000 ÷ (√3 × V)
Drop √3 for a single-phase transformer.
Available fault current
Isc = IFL ÷ (Z% ÷ 100)
Infinite primary assumption. A 500 kVA 480 V transformer at 5% impedance gives about 12 kA at its terminals.
Indicative results for preliminary engineering and reference only. Verify against the applicable standard, product data and site conditions before design, procurement or issue to a customer.