Power-Electronics Thermal Management, Extended-Surface Fin Efficiency & Heat Transfer
Extruded Heatsink Fin Efficiency TANH(mL)/(mL) Formula
Calculate Dimensionless Fin Parameter m*L, Single-Fin Efficiency eta_f = TANH(mL)/(mL) (%), Overall Finned-Array Thermal Resistance R_sa (°C/W), and Heatsink Base Temperature Ts (°C).
Making heatsink fins taller yields diminishing returns—conduction temperature drop toward the fin tip follows TANH(mL) / (mL)!
Treating every square centimeter of a tall, thin aluminum heatsink fin as isothermal at Base Temperature `T_s` under-predicts IGBT/MOSFET operating temperatures! Along a thin rectangular fin of length `L` (m), thickness `t` (m), thermal conductivity `k` (`~205` W/(m·K) for 6063-T5 aluminum), and convection coefficient `h` (`W/(m²·K)`), axial conduction competes against two-sided surface convection via Dimensionless Fin Parameter `m * L = SQRT((2 * h) / (k * t)) * L`. Adiabatic-tip differential conduction yields Single-Fin Efficiency `eta_f = TANH(m*L) / (m*L)` (expressed universally with `EXP` as `((EXP(2*mL) - 1) / (EXP(2*mL) + 1)) / mL`), Overall Finned-Surface Efficiency `eta_o = 1 - (A_fin / (A_fin + A_base)) * (1 - eta_f)`, Sink-to-Ambient Thermal Resistance `R_sa = 1 / (eta_o * h * (A_fin + A_base))` (°C/W), and Baseplate Temperature `T_s = T_amb + Q * R_sa` (°C)!
Verify on a Forced-Air 6063-T5 Inverter Heatsink—Fin_Height `L` = 32.0 mm (B2), Fin_Thickness `t` = 1.80 mm (B3), Forced_Air_Convection `h` = 38.0 W/(m²·K) (B4), Aluminum_Conductivity `k` = 205.0 W/(m·K) (B5 -> `m*L = 0.4592` in E2 & Single-Fin Efficiency `eta_f = 93.52%` in E3), Total_Finned_Area `A_f` = 0.085 m² (B6), Unfinned_Root_Area `A_b` = 0.015 m² (B7), Power_Dissipation `Q` = 45.0 W (B8), and Ambient_Air `T_a` = 35.0 °C (B9): Sink-to-Ambient Resistance `R_sa` equals `0.279 °C/W` (E4) and Heatsink Base Temperature settles at `47.53 °C` (E5)!
| Extended-Surface Heatsink KPI (L=32 mm, t=1.8 mm, h=38 W/m²K, Af=0.085 m², Q=45 W) | Formula | Verified Output |
|---|---|---|
| Dimensionless Fin Parameter m*L | SQRT((2 * h) / (k * (t_mm/1000))) * (L_mm/1000) | 0.4592 (0.2235 @ fan-fail h=9) |
| Single-Fin Efficiency eta_f (%) | (TANH(mL) / mL) * 100 | 93.52% (98.37% @ fan-fail h=9) |
| Sink-to-Ambient Resistance Rsa | 1 / (eta_o * h * (Af_m2 + Ab_m2)) | 0.279 °C/W (1.127 °C/W @ fan-fail!) |
| Heatsink Baseplate Temp Ts (°C) | Ambient_Ta_C + Power_W * Rsa_C_W | 47.53 °C (85.70 °C @ cooling-fan fail!) |
Hyperbolic-Tangent Fin Efficiency & Overall Heatsink Thermal Resistance Formula
Validate `Fin_Length L > 0` mm, `Fin_Thick t > 0` mm, `Convection h > 0` W/(m²·K), `Conductivity k > 1` W/(m·K), `Fin_Area > 0` m², `Base_Area >= 0` m², and `Power >= 0` W. Expressing `TANH(x)` via `(EXP(2*x) - 1) / (EXP(2*x) + 1)` guarantees identical evaluation across every spreadsheet engine. In a cloud collaboration thermal budget, teammates using the in-app Pro Assistant (metered monthly Arte AI credits) or testing the Arte Sheets ChatGPT `0.1.0` connector (`In review` / `Not published`; `read_range` inspects up to `200` cells of raw formulas and `revision` without consuming Arte AI credits) can audit fan-failure derating margins side by side.
=LET(ln_m, B2 / 1000, th_m, B3 / 1000, hc_w, B4, kc_w, B5, af_m, B6, ab_m, B7, pw_w, B8, ta_c, B9, IF(OR(ln_m <= 0, th_m <= 0, hc_w <= 0, kc_w <= 1, af_m <= 0, ab_m < 0, pw_w < 0), "Invalid Heatsink Input", LET(ml_p, SQRT((2 * hc_w) / (kc_w * th_m)) * ln_m, th_v, (EXP(2 * ml_p) - 1) / (EXP(2 * ml_p) + 1), ef_f, th_v / ml_p, at_m, af_m + ab_m, eo_f, 1 - (af_m / at_m) * (1 - ef_f), rs_a, 1 / (eo_f * hc_w * at_m), ROUND(ta_c + pw_w * rs_a, 2))))Cooling-Fan Failure Natural-Convection Fallback Check (Set Convection h B4 = 9.0 W/(m²·K))
Lose the axial cooling fan so `h` B4 drops from `38.0` down to natural convection `9.0` W/(m²·K): although Fin Efficiency E3 rises to `98.37%` (`m*L = 0.2235` in E2), Sink Thermal Resistance `R_sa` E4 quadruples from `0.279` to `1.127 °C/W`, driving Heatsink Base Temperature E5 from `47.53 °C` up to `85.70 °C`!
=IF(ISTEXT(E5), E5, LET(ml_p, SQRT((2 * B4) / (B5 * (B3 / 1000))) * (B2 / 1000), ef_f, ((EXP(2 * ml_p) - 1) / (EXP(2 * ml_p) + 1)) / ml_p, at_m, B6 + B7, eo_f, 1 - (B6 / at_m) * (1 - ef_f), ROUND(1 / (eo_f * B4 * at_m), 3)))Try the example
Use the original values or change them to test your understanding.