Substrate Current Capacity Calculator (AlN vs FR4)
Compares the maximum current a trace of the same geometry can carry on an aluminium-nitride versus an FR4 substrate using a 1D conduction model, and spells out voltage drop, drop ratio and dissipation — AlN’s high thermal conductivity raises the current capacity by an order of magnitude.
Key points
Substrate Current Capacity Calculator (AlN vs FR4): Compares the maximum current a trace of the same geometry can carry on an aluminium-nitride versus an FR4 substrate using a 1D conduction model, and spells out voltage drop, drop ratio and dissipation — AlN’s high thermal conductivity raises the current capacity by an order of magnitude. Formula: I = W·√(ΔT·k·T/(ρ·h))，ΔV = I·ρ·L/(W·t)，δ = √(ρ/(π·f·μ)). Enter parameters on the DD-CAT IDE online tool page to get results instantly, with a calculation-model diagram that updates with the inputs. 1D steady-state conduction model: all conductor heat is assumed to flow down through the substrate to an ideal heat sink, ignoring lateral spreading (conservative). AlN carries about √(170/0.3) ≈ 24× the current of FR4 — the same trace on a ceramic substrate handles an order of magnitude more current.
Inputs
I = W·√(ΔT·k·T/(ρ·h))，ΔV = I·ρ·L/(W·t)，δ = √(ρ/(π·f·μ))Calculation model (updates with parameters)
| AlN (aluminium nitride) current capacity | 117.5 A | k ≈ 170 W/(m·K) |
| FR4 current capacity | 4.936 A | k ≈ 0.3 W/(m·K) |
| Current ratio (AlN ÷ FR4) | 23.8 × | ≈ √(k_Aln / k_FR4) |
| Trace resistance R (same for both substrates) | 24.63 mΩ | ρ·L/(W·t), length 50 mm |
| voltage drop ΔV @ 10 A | 246.3 mV | ΔV = I·R = I·ρ·L/(W·t), Vs = 5.00 V |
| Voltage drop percentage | 4.93% | ΔV ÷ Vs; power rails are usually kept below 5% |
| Load-end voltage Vs − ΔV | 4.754 V | the voltage actually seen by the load |
| dissipation P @ 10 A | 2.463 W | P = I²·R; identical on both substrates |
| AlN full-load drop ΔV | 2.894 V | @ 117.5 A ｜ of Vs 57.9% |
| FR4 full-load drop ΔV | 121.6 mV | @ 4.936 A ｜ of Vs 2.4% |
| Copper current density (at AlN capacity) | 3357 A/mm² | high current needs thick copper — check fusing / electromigration |
| Skin depth δ | 66.08 µm | copper @20°C, μr ≈ 1 |
| Effective conductor thickness t_eff | 35.0 µm | copper thinner than skin depth — skin effect is negligible |
| AC / DC resistance ratio | 1.00 × | R_ac ≈ R_dc × (t / t_eff) |
Engineering notes
- 1D steady-state conduction model: all conductor heat is assumed to flow down through the substrate to an ideal heat sink, ignoring lateral spreading (conservative).
- AlN carries about √(170/0.3) ≈ 24× the current of FR4 — the same trace on a ceramic substrate handles an order of magnitude more current.
- Voltage drop ΔV = I·ρ·L/(W·t) is independent of the substrate material — only the copper geometry and length matter. The drop ratio ΔV/Vs on a power rail is usually kept below 5%.
- In width mode the AlN trace is narrower and therefore more resistive, so at the same current its drop is higher than FR4 — an unavoidable consequence of matching current with different widths. Check both drop and temperature rise.
- At AlN’s high currents the real limit is usually the copper: use thick copper / DBC (100–300µm) and check fusing and electromigration.
- Skin depth δ = √(ρ/(π·f·μ)): as frequency rises the current crowds toward the surface; when the copper is thicker than δ the AC resistance scales up by t/δ. Enter 0 for DC (δ is treated as infinite).
FAQ
- What is Substrate Current Capacity Calculator (AlN vs FR4) for?
- Compares the maximum current a trace of the same geometry can carry on an aluminium-nitride versus an FR4 substrate using a 1D conduction model, and spells out voltage drop, drop ratio and dissipation — AlN’s high thermal conductivity raises the current capacity by an order of magnitude.
- What is the formula for Substrate Current Capacity Calculator (AlN vs FR4)?
- Core formula: I = W·√(ΔT·k·T/(ρ·h))，ΔV = I·ρ·L/(W·t)，δ = √(ρ/(π·f·μ)). The tool returns results instantly using this formula and the selected parameters.
- What engineering notes should I keep in mind when using Substrate Current Capacity Calculator (AlN vs FR4)?
- 1D steady-state conduction model: all conductor heat is assumed to flow down through the substrate to an ideal heat sink, ignoring lateral spreading (conservative). AlN carries about √(170/0.3) ≈ 24× the current of FR4 — the same trace on a ceramic substrate handles an order of magnitude more current. Voltage drop ΔV = I·ρ·L/(W·t) is independent of the substrate material — only the copper geometry and length matter. The drop ratio ΔV/Vs on a power rail is usually kept below…
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