Advanced Proximity Effect Loss Calculator

Analyze round wires or layered windings with practical engineering assumptions. Review losses, ratios, and depth. Size conductors wisely before heat and efficiency problems appear.

Calculator Inputs


Use the round model for wire bundles in external fields. Use the Dowell model for layered windings or foil conductors.

Example Data Table

Case Model Current (A) Frequency (Hz) Main geometry Observed design takeaway
Transformer lead pair Round wire 12 20,000 0.8 mm wire, 3.5 mm spacing, 2 neighbors Comfortable spacing keeps field coupling moderate.
Tight choke winding Round wire 18 80,000 1.2 mm wire, 2.0 mm spacing, 4 neighbors High frequency and crowding sharply increase AC loss.
Planar secondary Dowell 25 150,000 10 mm width, 0.2 mm thickness, 5 layers More layers can dominate total copper heating.

Formula Used

This calculator offers two engineering approaches because proximity loss depends strongly on conductor shape and winding arrangement.

Common equations

Angular frequency: ω = 2πf

Temperature-adjusted resistivity: ρT = ρ20 × [1 + α(T − 20)]

Skin depth: δ = √[2ρT / (ωμ0μr)]

Round wire transverse-field model

Cross-sectional area: A = π(d/2)2

DC resistance: RDC = ρTL / A

Peak field intensity: Hpk ≈ k × n × Ipk / (2πs)

Peak flux density: Bpk = μ0μrHpk

Approximate proximity loss: Pprox ≈ (π2Bpk2d2f2V) / (16ρT)

Dowell multilayer winding model

Dimension ratio: x = t / δ

Skin factor: Fskin = (x/2) × (sinh x + sin x) / (cosh x − cos x)

Proximity factor: Fprox = ((m2 − 1)/3) × x × (sinh x − sin x) / (cosh x + cos x)

Effective AC resistance ratio: RAC / RDC = Fskin + Fprox

The round model is useful for early wire-spacing studies. The Dowell model is widely used for transformer and inductor winding estimates.

How to Use This Calculator

  1. Select the model that best matches your conductor arrangement.
  2. Enter RMS current, operating frequency, conductor length, temperature, and material values.
  3. For round wire, add diameter, center spacing, nearby conductor count, and geometry factor.
  4. For layered windings, add width, thickness, and the number of stacked layers.
  5. Press Calculate to display the result block beneath the header.
  6. Review proximity loss, total AC loss, AC to DC resistance ratio, and skin depth.
  7. Use the CSV or PDF buttons to save the computed summary.
  8. Compare multiple cases by changing spacing, thickness, or layer count.

FAQs

1. What is proximity effect loss?

It is extra conductor loss caused by magnetic fields from nearby conductors. Those fields push current toward limited regions, increasing current density and raising AC resistance.

2. Why does frequency matter so much?

Higher frequency reduces skin depth. As current crowds into thinner regions, both skin and proximity effects intensify, which increases copper loss and temperature rise.

3. When should I use the round-wire model?

Use it for quick estimates of circular conductors exposed to nearby alternating fields. It works well for comparing spacing, neighbor count, and wire diameter during early design screening.

4. When is the Dowell model more appropriate?

Use Dowell for transformer or inductor windings with layered rectangular conductors or foil. It is especially helpful when thickness and layer count strongly affect AC resistance.

5. Does this replace finite-element analysis?

No. This calculator is intended for engineering estimates and comparisons. Final validation for critical hardware should still use detailed field simulation or laboratory measurements.

6. Why is temperature included?

Resistivity increases with temperature. A hotter conductor has higher resistance, so both DC loss and AC-related loss predictions should be adjusted for realistic operating conditions.

7. How can I reduce proximity effect loss?

Increase spacing, reduce conductor thickness, split current among strands, lower layer count, use litz wire where suitable, and optimize winding arrangement to weaken transverse magnetic fields.

8. What does the AC to DC resistance ratio show?

It shows how much effective resistance rises under AC conditions. A value of 2 means the conductor behaves like it has twice the DC resistance at that frequency.

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Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.