Skin Depth Calculator

Skin Depth Calculator Tutorial: Step-by-Step Analysis of Skin Depth Problems

1. How the Skin Depth Calculator Works

Calculator inputs and outputs

Skin depth equations

An electromagnetic wave entering a conductor decays exponentially. Skin depth δ is the distance over which its amplitude, and the current density it drives, falls by a factor of e.^[1][2][3] For a good conductor (σ ≫ ωε), skin depth is:

The calculator does not stop at Equation 1. It evaluates the exact attenuation length 1/α of a plane wave in a general lossy medium, which includes the displacement current ωε.^[2][3] That is why it asks for ε_r:

The ratio p = ωε/σ decides which equation you need. For metals, p is about 10⁻⁸ even at 10 GHz, so Equation 2 collapses to Equation 1 and ε_r has no effect. For seawater, soil or carbon-loaded materials, p approaches 1 and Equation 1 underestimates δ (Tutorial 4).

What skin depth tells you

Current density decays exponentially with depth z below the surface. Integrating it gives the fraction of the total current that flows within a thickness t:^[1][4]

A conductor several skin depths thick behaves like a sheet of thickness δ carrying uniform current. Its AC resistance per square is the surface resistance:^[1][4]

Because δ is proportional to 1/√f, one calculator result can be scaled to any other frequency:

2. Tutorial 1: Copper Thickness for a Microstrip Line

Problem: A 10 GHz microstrip filter uses 1 oz copper foil (34.8 µm nominal).^[11] Is the foil thick enough, what is the surface resistance, and how much does copper roughness matter?

1. Set the medium properties. Leave ε_r = 1.0 and μ_r = 1.0. Copper is non-magnetic, and its permittivity has no effect (p ≈ 9.6×10⁻⁹ at 10 GHz).
2. Select the conductivity. Select Annealed Copper (5.8E7), the 100% IACS reference value.^[10] Rolled or electrodeposited PCB foil is close to this; use the Copper (5.98E7) preset only for high-purity bulk copper.
3. Enter the frequency. Type 1E10 (or 10,000,000,000) for 10 GHz.
4. Read the skin depth. The calculator returns δ = 0.6609 µm.
5. Verify the result by hand. Substitute into Equation 1:
   Equation 6 — Copper check at 10 GHz $$\delta =\frac{1}{\sqrt{\pi \left({10}^{10}\right)(4\pi \times {10}^{-7})(5.8\times {10}^7)}}=0.661\text{µm}$$ δ = 1/√[π(10¹⁰)(4π×10⁻⁷)(5.8×10⁷)] = 1/√(2.2897×10¹²) = 6.609×10⁻⁷ m = 0.661 µm
   The shortcut in Equation 5 agrees: 66.09 / √10,000 = 0.661 µm.
6. Compare the foil thickness with the skin depth. 34.8 µm / 0.661 µm = 52.7δ. The 3δ to 5δ rule needs only 2.0 to 3.3 µm, so 1 oz foil carries essentially all of the current. Even ¼ oz foil (about 9 µm) would be 13.6δ thick.
7. Calculate the surface resistance. From Equation 4: R_s = 1 / (5.8×10⁷ × 0.661×10⁻⁶) = 26.1 mΩ/sq. Conductor attenuation of the line is directly proportional to R_s.^[1]
8. Check the effect of surface roughness. Skin depth at 10 GHz is comparable to the RMS roughness Δ of PCB copper, so current follows a longer path. The Hammerstad–Jensen model estimates the loss multiplier:^[6]
   Equation 7 — Hammerstad–Jensen roughness correction [6] $$K_{\mathrm{rough}}=1+\frac{2}{\pi }{\tan }^{-1}\left[1.4{\left(\frac{\Delta }{\delta }\right)}^2\right]$$ K_rough = 1 + (2/π)·arctan[1.4·(Δ/δ)²], Δ = RMS surface roughness; conductor loss ≈ K_rough × smooth-conductor loss

   Roughness loss multiplier for copper at 10 GHz (δ = 0.661 µm)

   RMS roughness Δ (µm)	Δ/δ	Krough
   0.3 (very low profile foil)	0.45	1.18
   0.5	0.76	1.43
   1.0	1.51	1.81
   2.0 (standard foil)	3.03	1.95

9. Repeat at other frequencies. Change only the frequency field. The calculator returns the values below for annealed copper.

   Calculator output: annealed copper (σ = 5.8×10⁷ S/m), ε_r = μ_r = 1

   Frequency	Entry	δ (µm)	1 oz foil (34.8 µm)
   1 MHz	1E6	66.09	0.53δ — current-limited
   10 MHz	1E7	20.90	1.67δ — 81% of current
   100 MHz	1E8	6.609	5.3δ — OK
   1 GHz	1E9	2.090	16.7δ — OK
   10 GHz	1E10	0.6609	52.7δ — OK
   28 GHz	2.8E10	0.3949	88δ — OK
   77 GHz	7.7E10	0.2382	146δ — OK

Result: At 10 GHz, foil thickness is not the limit; surface roughness is. Choosing a smooth foil (Δ ≈ 0.3 µm) instead of standard foil (Δ ≈ 2 µm) cuts conductor loss by about 40% (1.18 vs. 1.95). Below about 100 MHz the opposite is true: 1 oz foil is less than 5δ thick, and DC-to-AC resistance calculations must account for the finite thickness.

3. Tutorial 2: Plating Thickness for a Waveguide

Problem: An aluminum WR-90 waveguide assembly (8.2 to 12.4 GHz) will be silver plated to reduce loss. How thick must the silver be, what does it gain, and what happens if a nickel underlayer is used?

1. Find the lowest operating frequency. Skin depth is largest at the lowest frequency, so it sets the plating thickness: f_min = 8.2 GHz.
2. Calculate the skin depth of the plating metal. Set ε_r = 1, μ_r = 1, select Silver (6.3E7 S/m) and enter 8.2E9. The calculator returns δ_Ag = 0.7002 µm. At 12.4 GHz it returns 0.5694 µm.
3. Apply the 3δ to 5δ rule.
   Equation 8 — Minimum plating thickness [1], [4] $$t_{\min }=N\delta \left(f_{\min }\right),3\le N\le 5$$ t_min = N·δ(f_min), 3 ≤ N ≤ 5 → silver at 8.2 GHz: 3δ = 2.10 µm, 5δ = 3.50 µm
   A specification of 3.5 to 5 µm silver meets 5δ with margin for plating thickness variation inside corners and flanges.
4. Compare with the bare base metal. Select Aluminum (3.5E7 S/m) at 8.2 GHz: δ_Al = 0.9395 µm. Using Equation 4, R_s falls from 30.4 mΩ/sq (aluminum) to 22.7 mΩ/sq (silver). Conductor loss scales with R_s, so silver plating cuts wall loss by 25% (1 − √(3.5/6.3)) and raises cavity Q by up to 1.34×.
5. Size a gold finish the same way. Select Gold (4.52E7 S/m) at 8.2 GHz: δ_Au = 0.8267 µm, so 5δ = 4.13 µm. A 0.1 µm flash of immersion gold is only 0.12δ, so nearly all the current flows in whatever is underneath it.
6. Check a nickel underlayer with μ_r. Select Nickel (electrolytic) (1.46E7 S/m) at 10 GHz. With μ_r = 1, δ = 1.317 µm and R_s = 52.0 mΩ/sq, already twice copper. Nickel is ferromagnetic. Its effective microwave permeability depends on frequency, field and magnetization, so treat μ_r as a sensitivity input:

   Nickel at 10 GHz: effect of the μ_r input (illustrative values)

   μr entered	δ (µm)	Rs (mΩ/sq)	Rs / Rs,Cu
   1	1.317	52.0	2.0
   5	0.589	116	4.5
   20	0.295	233	8.9

   Because δ falls as 1/√μ_r, a magnetic underlayer is both thinner-skinned and far lossier. Keep the top plating at least 3δ to 5δ thick so the current never reaches the nickel.

Design note: Plating thickness is set at f_min, but surface finish matters at every frequency. A rough or porous plating behaves like the roughness case in Tutorial 1 and can erase the conductivity advantage of silver.

4. Tutorial 3: Shield Thickness and Absorption Loss

Problem: An enclosure wall is 1.0 mm thick. How much does it attenuate magnetic fields by absorption at 60 Hz, 10 kHz and 1 MHz if it is aluminum, and how does steel compare?

1. Calculate the aluminum skin depth at each frequency. Set ε_r = 1, μ_r = 1, select Aluminum (3.5E7 S/m), and enter 60, 1E4 and 1E6 in turn. The calculator returns 10,982.7, 850.7 and 85.07 µm.
2. Convert thickness to absorption loss. A field passing through a wall of thickness t is reduced by e^−t/δ, or 8.686 dB per skin depth:^[5]
   Equation 9 — Shield absorption loss [5, Ch. 6] $$A=20{\log }_{10}{e}^{t/\delta }=8.686\frac{t}{\delta }\text{dB}$$ A = 20·log10(e^(t/δ)) = 8.686·t/δ dB
3. Repeat for steel, entering a permeability. Select Steel (Cast) (6.21E6 S/m). Steel's low-frequency permeability is high, so enter μ_r = 300 as a representative value (check the supplier's data for the actual grade).^[5]

   Absorption loss of a 1.0 mm wall (calculator δ, Equation 9)

   Frequency	Al δ (µm)	Al A (dB)	Steel δ, μr = 300 (µm)	Steel A (dB)
   60 Hz	10,982.7	0.8	1,505.4	5.8
   10 kHz	850.7	10.2	116.6	74.5
   1 MHz	85.07	102.1	11.66*	745*

   *Steel's permeability falls with frequency, so values above about 100 kHz with μ_r = 300 are optimistic.
4. Solve for the thickness that meets a target. Rearrange Equation 9 for 30 dB of absorption at 10 kHz: t = 30δ/8.686 = 3.454δ. Aluminum needs 2.94 mm; steel (μ_r = 300) needs 0.40 mm.

Result: At 60 Hz, 1 mm of aluminum is only 0.09δ thick and absorbs less than 1 dB. Low-frequency magnetic shielding needs a high-permeability material. Above about 1 MHz, any practical metal wall gives large absorption, and seams, apertures and cable penetrations limit shielding instead. Total shielding effectiveness also includes reflection loss, which this calculator does not compute.^[5]

5. Tutorial 4: Poor Conductors with a User-Defined Conductivity

Problem: How deep do 1 MHz, 100 MHz and 1 GHz fields penetrate seawater (σ ≈ 4 S/m, ε_r ≈ 81), and when does the good-conductor formula fail?

1. Enter the medium's permittivity. Enter ε_r = 81 and μ_r = 1. Unlike a metal, seawater's displacement current is significant at radio frequencies, so ε_r now changes the answer.
2. Select a user-defined conductivity. Select User Defined Conductivity and type 4 in the User Defined Conductivity field. The same step works for alloys, plated finishes or measured values.
3. Enter each frequency and read the result. Enter 1E6, 1E8 and 1E9 in turn.

   Seawater (σ = 4 S/m, ε_r = 81): calculator (Eq. 2) vs. good-conductor formula (Eq. 1)

   Frequency	p = ωε/σ	Calculator δ	Equation 1 δ	Error of Eq. 1
   1 MHz	0.0011	251.8 mm	251.6 mm	−0.06%
   100 MHz	0.113	26.62 mm	25.16 mm	−5.5%
   1 GHz	1.127	12.91 mm	7.96 mm	−38%

   The calculator reports microns: 251,787.8 µm = 251.8 mm.
4. Verify the 1 GHz result with Equation 2.
   Equation 10 — Seawater check at 1 GHz (Eq. 2) $$p=\frac{2\pi \left({10}^9\right)\left(81\right)(8.854\times {10}^{-12})}{4}=1.127,\delta =7.958\text{mm}\times \sqrt{\sqrt{1+{1.127}^2}+1.127}=12.91\text{mm}$$ p = 2π(10⁹)(81)(8.854×10⁻¹²)/4 = 1.127; δ = 7.958 mm × √(√(1 + 1.127²) + 1.127) = 7.958 × 1.623 = 12.91 mm

Result: Use Equation 1 only when p = ωε/σ is below about 0.01. Above that, rely on the calculator's Equation 2. Seawater's σ and ε_r also vary with frequency, temperature and salinity, so treat the 1 GHz value as an estimate.

6. Common Input Mistakes

• Entering frequency in MHz or GHz. The frequency field is in hertz. Typing 10 for 10 GHz returns 20.9 mm for copper instead of 0.661 µm. Use E notation: 1E10.
• Misreading the output units. The result is in microns. Divide by 1,000 for millimeters; multiply by 0.03937 for mils.
• Leaving a non-zero user-defined value with a preset selected. The calculator adds the user-defined field to the preset conductivity. Keep the user-defined field at 0 when using a preset.
• Using the dielectric's ε_r for a metal. The ε_r and μ_r inputs describe the conducting medium itself. For a copper trace on FR-4, leave ε_r = 1; the FR-4 value of 4.4 has no physical role inside the copper.
• Ignoring permeability. Nickel, steel and other ferromagnetic metals need μ_r > 1, or the calculated δ will be too large and the predicted loss too small.
• Sizing plating at the center frequency. Thickness must be set at the lowest operating frequency, where δ is largest.
• Using room-temperature conductivity for hot hardware. Copper resistivity rises about 0.39% per °C, so δ grows about 0.2% per °C.^[9]

7. Skin Depth Equation Summary

Need help with conductor loss, plating specifications, shielding or RF heating penetration depth? Contact SiberSci, LLC. Related tools: Circular Waveguide Calculator, Rectangular Waveguide Calculator, Coax TEM Mode Calculator and Dielectric Heating Calculator.

8. References

1. D. M. Pozar, Microwave Engineering, 4th ed. Hoboken, NJ: Wiley, 2012. Ch. 1, "Electromagnetic Theory" (plane waves in good conductors, skin depth and surface resistance), and Ch. 2–3 (conductor loss in transmission lines and waveguides).
2. C. A. Balanis, Advanced Engineering Electromagnetics, 2nd ed. Hoboken, NJ: Wiley, 2012. Ch. 4, "Wave Propagation and Polarization" (attenuation and phase constants in lossy media; good-conductor approximation).
3. D. K. Cheng, Field and Wave Electromagnetics, 2nd ed. Reading, MA: Addison-Wesley, 1989. Sec. 8-3, "Plane Waves in Lossy Media."
4. H. A. Wheeler, "Formulas for the skin effect," Proceedings of the IRE, vol. 30, no. 9, pp. 412–424, Sept. 1942.
5. H. W. Ott, Electromagnetic Compatibility Engineering. Hoboken, NJ: Wiley, 2009. Ch. 6, "Shielding" (skin depth and absorption loss).
6. E. O. Hammerstad and Ø. Jensen, "Accurate models for microstrip computer-aided design," in IEEE MTT-S International Microwave Symposium Digest, 1980, pp. 407–409.
7. National Institute of Standards and Technology, "CODATA Value: vacuum magnetic permeability," μ₀ ≈ 1.25663706×10⁻⁶ N/A². https://physics.nist.gov/cgi-bin/cuu/Value?mu0
8. National Institute of Standards and Technology, "CODATA Value: vacuum electric permittivity," ε₀ ≈ 8.8541878×10⁻¹² F/m. https://physics.nist.gov/cgi-bin/cuu/Value?ep0
9. J. R. Rumble, Ed., CRC Handbook of Chemistry and Physics, 104th ed. Boca Raton, FL: CRC Press, 2023. "Electrical Resistivity of Pure Metals."
10. International Electrotechnical Commission, IEC 60028: International Standard of Resistance for Copper (100% IACS annealed copper, σ = 58.0 MS/m at 20 °C).
11. IPC, IPC-4562A: Metal Foil for Printed Board Applications. Bannockburn, IL: IPC, 2008 (nominal foil thickness by weight).

Tutorial by Dr. Scott Best, SiberSci, LLC. Values computed with the calculator's own equation (Equation 2), μ₀ = 4π×10⁻⁷ H/m and ε₀ = 8.8541853×10⁻¹² F/m.