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    Transformer Calculator

    Three-phase cast-resin or oil-immersed transformer currents, winding quantities, Dyn5/Dyn11 vector groups, impedance and preliminary LV fault level.

    Before you calculate

    Understand three-phase transformer nameplates

    Calculate rated HV/LV currents, line and winding ratios, Dyn5/Dyn11 phase displacement, impedance magnitude and an illustrative transformer-limited LV three-phase fault estimate. Use actual nameplate data and a complete IEC 60909 study for design approval.

    Learn more — illustrated guide →
    Step 1 · Three-phase transformer

    Nameplate inputs

    Two-winding, balanced, three-phase MV/LV transformer. Use the nameplate ratings for power, voltages, vector group and uk%. No cos φ is required for rated currents.

    Default Dyn5 · HV/LV winding connections and clock displacement.
    6% is an editable example; use manufacturer/nameplate values.
    AN: natural air · AF: forced air. Ratings are not automatically boosted.
    Step 2 · Electrical results

    Transformer characteristics

    HV rated line current46.19 A
    LV rated line current2,309.40 A
    Transformer-limited LV 3φ fault38.49 kASimplified initial symmetrical RMS estimate at LV terminals; infinite upstream source
    Rated line-voltage ratio50.00 : 1Not always the physical winding turns ratio

    Winding quantities

    HV winding voltage20,000.00 V
    LV winding voltage230.94 V
    HV winding current26.67 A
    LV winding current2,309.40 A
    Ideal winding voltage / turns ratio86.60 : 1
    Secondary neutral brought outYes

    Impedance & fault level

    Z referred to LV side0.00600 Ω
    Z referred to HV side15.00000 Ω
    Transformer-limited LV fault power26.67 MVA
    uk nameplate6.00%

    Dyn5 · Winding configuration

    HV: Delta (Δ)
    LV: Star (Y) with brought-out neutral

    The LV phasor is at clock 5: 150° lagging the HV reference.

    Vector group changes the winding-level values and phase displacement, not the rated line currents when kVA and voltages are unchanged.
    121234567891011HV 12 o'clock · LV 5 o'clock
    HV referenceLV phasor
    Preliminary fault estimate — not IEC 60909 approval

    Calculated from transformer uk% assuming an ideal infinite upstream supply. Does not include grid source impedance, IEC 60909 voltage and transformer correction factors, motor or other transformer contributions, tap position, fault X/R, peak current, or downstream cable impedance. Earth faults and protection coordination are not evaluated. Manufacturer data and a site-specific study are required.

    Cooling: AN is descriptive only; no overload or additional fan-cooled kVA is assumed.

    1 · Three-phase rated line currents

    I=S3 ULLI=\frac{S}{\sqrt3\,U_{LL}}
    IHV=1,600×10003×20,000=46.19 AI_{HV}=\frac{1,600\times1000}{\sqrt3\times20,000}=46.19\,\mathrm A
    ILV=1,600×10003×400=2,309.40 AI_{LV}=\frac{1,600\times1000}{\sqrt3\times400}=2,309.40\,\mathrm A

    2 · Delta/star winding voltage & current

    Uw,Y=ULL3Uw,Δ=ULLU_{w,Y}=\frac{U_{LL}}{\sqrt3}\qquad U_{w,\Delta}=U_{LL}
    Iw,Y=ILIw,Δ=IL3I_{w,Y}=I_L\qquad I_{w,\Delta}=\frac{I_L}{\sqrt3}
    n≈Uw,HVUw,LV=20,000.00230.94=86.60n\approx\frac{U_{w,HV}}{U_{w,LV}}=\frac{20,000.00}{230.94}=86.60

    3 · Transformer equivalent impedance

    Zbase,LV=ULL,LV2SVAZ_{base,LV}=\frac{U_{LL,LV}^{2}}{S_{VA}}
    ∣ZT,LV∣=uk100⋅Zbase,LV=0.00600 Ω\left|Z_{T,LV}\right|=\frac{u_k}{100}\cdot Z_{base,LV}=0.00600\,\Omega

    Only the magnitude of the equivalent impedance is obtained from uk%. Separating R and X requires additional data such as load losses.

    4 · Simplified transformer-only LV fault

    Ik,approx′′=ILV,nuk/100I''_{k,\mathrm{approx}}=\frac{I_{LV,n}}{u_k/100}
    Ik,approx′′=2,309.400.0600×11000=38.49 kAI''_{k,\mathrm{approx}}=\frac{2,309.40}{0.0600}\times\frac{1}{1000}=38.49\,\mathrm{kA}

    The transformer-only result in A is converted to kA by dividing by 1,000.