Checkpoint 1: Heatsink Thermal Resistance Calculated Critical
The heatsink thermal resistance (Rth_sa) determines how effectively it transfers heat from its base to the surrounding air. This must be calculated or selected to ensure the total thermal chain meets the junction temperature requirement.
Required Heatsink Rth Calculation
Start from the requirement and work backwards:
Rth_sa_required = (Tj_max - Ta_max) / P - Rth_jc - Rth_cs
Example: Power MOSFET in TO-220
Tj_max = 150°C (with 25°C margin: target = 125°C)
Ta_max = 50°C
P = 8W (conduction + switching)
Rth_jc = 1.5°C/W (TO-220, from datasheet)
Rth_cs = 0.5°C/W (thermal pad, Sil-Pad 400)
Rth_sa_required = (125 - 50) / 8 - 1.5 - 0.5
Rth_sa_required = 9.375 - 2.0 = 7.375°C/W
Need a heatsink with Rth_sa ≤ 7.4°C/W
Heatsink Performance Factors
Natural Convection Heatsink Rth (flat plate approximation):
Rth_sa ≈ 1 / (h × A_surface)
Where h = natural convection coefficient:
h_nat = 5-15 W/(m²·K) for vertical plates in still air
h_nat ≈ 10 W/(m²·K) is typical for electronics enclosures
For a finned heatsink with total fin area A_fins:
Rth_sa = 1 / (η_fin × h × A_fins)
η_fin = fin efficiency = tanh(mL) / (mL)
m = √(2h / (k_fin × t_fin))
Example: Aluminum extrusion, 10 fins, each 25mm tall, 50mm long, 1mm thick
A_fin_single = 2 × 25mm × 50mm = 2500 mm² = 0.0025 m²
A_total = 10 × 0.0025 = 0.025 m²
h = 10 W/(m²·K), k_Al = 200 W/(m·K), t = 1mm
m = √(2×10 / (200×0.001)) = √100 = 10 m⁻¹
mL = 10 × 0.025 = 0.25
η = tanh(0.25)/0.25 = 0.245/0.25 = 0.98 (very efficient fins)
Rth_sa = 1 / (0.98 × 10 × 0.025) = 4.08°C/W
Add base-to-air and radiation: Effective Rth_sa ≈ 3.5°C/W
Common Heatsink Selection Guide
| Heatsink Type | Size (L×W×H mm) | Rth_sa Natural (°C/W) | Rth_sa @ 2m/s (°C/W) | Suitable Power |
|---|
| BGA stamp (Aavid 375424) | 23×23×10 | 22 | 12 | 1-3W |
| TO-220 clip-on (Aavid 530002) | 19×14×10 | 18 | 10 | 2-5W |
| Extruded medium (Fischer SK104) | 50×30×20 | 8 | 4 | 5-15W |
| Extruded large (Fischer SK481) | 75×50×30 | 3.5 | 1.5 | 15-40W |
| Pin-fin (CUI HSF-1520) | 15×15×20 | 15 | 6 | 3-8W |
| Fan+heatsink combo (40mm) | 40×40×20 | N/A | 2-4 | 10-25W |
- Calculate required Rth_sa from your thermal budget (work backwards from Tj_max).
- Determine available space: height, footprint, and orientation constraints.
- Select natural convection vs. forced convection based on system architecture.
- Choose a heatsink from a catalog with Rth_sa less than or equal to your requirement (with 20% margin).
- Verify the heatsink's Rth is specified at your operating conditions (orientation, airflow, altitude).
- Confirm mechanical compatibility: mounting holes, clearance to nearby components, weight.
TO-220 MOSFET dissipating 6W: Required Rth_sa = (125-65)/6 - 1.5 - 0.4 = 8.1°C/W. Selected Aavid 531202B32G (30×25×13mm extrusion) with Rth_sa = 6.8°C/W natural, with mounting clip. Actual Tj = 65 + 6×(1.5+0.4+6.8) = 117.2°C. Margin = 7.8°C. Verified at worst-case altitude (Denver, 1600m) adds ~10% to Rth, still within margin.
Selected a 14×14×6mm BGA heatsink (Rth=25°C/W) for a 5W component because it "fit nicely." Tj = 50 + 5×(2+0.5+25) = 187.5°C. Component rated to 125°C. Heatsink is completely inadequate -- undersized by a factor of 3×.
Checkpoint 2: Airflow Requirements Determined Major
When natural convection is insufficient, forced airflow is needed. The required airflow velocity and volume must be determined to size fans and design air channels.
Airflow Velocity Impact on Heatsink Performance
Forced Convection Coefficient (flat plate, turbulent):
h_forced = 0.037 × (k_air/L) × Re^0.8 × Pr^(1/3)
Simplified for electronics cooling:
h_forced ≈ 3.9 × V^0.8 / L^0.2 (W/m²·K)
Where V = air velocity (m/s), L = length in flow direction (m)
Rough estimate:
h ≈ 10 W/(m²·K) at 0 m/s (natural convection)
h ≈ 25 W/(m²·K) at 1 m/s
h ≈ 40 W/(m²·K) at 2 m/s
h ≈ 55 W/(m²·K) at 3 m/s
h ≈ 80 W/(m²·K) at 5 m/s
Required Airflow Calculation:
Method 1: From heatsink datasheet curves
Most heatsink vendors provide Rth vs. airflow velocity curves.
Read the required velocity for your target Rth.
Method 2: From heat dissipation and temperature rise
Q = ṁ × Cp × ΔT_air
Where:
Q = total heat load (W)
ṁ = mass flow rate (kg/s)
Cp = air specific heat = 1005 J/(kg·K)
ΔT_air = allowable air temperature rise (°C)
Volume flow rate: V̇ = Q / (ρ × Cp × ΔT_air)
ρ_air = 1.18 kg/m³ at 25°C
Example: 50W total board dissipation, 10°C air rise allowed
V̇ = 50 / (1.18 × 1005 × 10) = 0.00422 m³/s = 4.22 L/s
Convert to CFM: 4.22 × 2.119 = 8.94 CFM
Need at least 9 CFM of airflow through the enclosure.
Air Velocity from CFM and Cross-Section
V = V̇ / A_cross_section
Example: 9 CFM through a heatsink with 50mm×20mm cross-section
V̇ = 9 CFM × 0.000472 m³/s per CFM = 0.00425 m³/s
A = 0.05 × 0.02 = 0.001 m²
V = 0.00425 / 0.001 = 4.25 m/s
At 4.25 m/s, heatsink Rth typically drops to 30-40% of natural convection value.
Industrial controller with 35W heat load in a sealed enclosure: Internal fan circulates air at 2 m/s across finned heatsink. External fins on enclosure wall dissipate heat to ambient. Air rise inside limited to 15°C. Heatsink sized for forced convection: Rth_sa = 2.5°C/W at 2 m/s. System operates with Tj_max = 105°C at 55°C ambient.
Assuming "some airflow will exist" inside a loosely sealed enclosure without any fans. In reality, internal natural convection is poor due to random component placement, and the air inside stratifies with hot air at the top reaching 30°C above ambient, far exceeding the 10°C assumption in the thermal analysis.
Checkpoint 3: Fan Selection (CFM, Static Pressure) Major
Fan selection requires matching the fan's performance curve to the system's impedance curve. The operating point is where these curves intersect, and it must provide sufficient airflow for cooling.
System Impedance and Fan Curves
System impedance (pressure drop):
ΔP = K × (V̇)² (parabolic relationship)
K depends on system geometry: filters, grilles, heatsink fins, PCBs, cables
Typical pressure drops:
Open PCB area: 0.01-0.05 inches H₂O per inch of travel
Heatsink (moderate fin density): 0.05-0.15 in H₂O
Filter (clean): 0.02-0.05 in H₂O
Filter (dirty): 0.1-0.3 in H₂O
Perforated panel (30% open): 0.05-0.1 in H₂O
Total system impedance example:
Inlet grille: 0.03 in H₂O
Filter: 0.05 in H₂O (clean), 0.15 (dirty)
PCB area: 0.04 in H₂O
Heatsink: 0.10 in H₂O
Exit grille: 0.03 in H₂O
Total (clean): 0.25 in H₂O
Total (dirty): 0.35 in H₂O
Fan Selection Process
- Calculate required airflow (CFM) from thermal load and allowable temperature rise.
- Estimate system impedance (pressure drop at required flow rate).
- Select fan size based on available space (common: 40mm, 60mm, 80mm, 92mm, 120mm).
- Find a fan whose pressure-flow curve passes through or above your operating point.
- Verify at the dirty-filter condition: Fan must still provide adequate flow.
- Check noise level (dBA) at operating point -- typically 25-40 dBA for office equipment.
- Consider redundancy: N+1 fans for mission-critical equipment.
Common Fan Specifications
| Fan Size (mm) | Max CFM | Max Pressure (in H₂O) | Noise (dBA) | Application |
|---|
| 25×25×10 | 3-5 | 0.1-0.2 | 20-30 | Spot cooling, small devices |
| 40×40×10 | 5-8 | 0.1-0.3 | 20-35 | Single IC/heatsink cooling |
| 60×60×25 | 15-25 | 0.2-0.4 | 25-40 | Small enclosures |
| 80×80×25 | 25-45 | 0.15-0.35 | 25-40 | Desktop equipment |
| 92×92×25 | 35-60 | 0.15-0.3 | 25-40 | Server/industrial |
| 120×120×25 | 45-80 | 0.1-0.25 | 20-35 | Low-impedance systems |
| 120×120×38 | 60-120 | 0.3-0.6 | 35-55 | High-impedance systems |
Fan Selection Example:
Required: 12 CFM at 0.25 in H₂O system impedance
Space: 60×60mm available
Option 1: Sunon MF60251V1 (60×60×25mm)
Free air: 23 CFM, Max pressure: 0.28 in H₂O
At 0.25 in H₂O: ~8 CFM (from curve) -- NOT ENOUGH!
Option 2: Delta AFB0612H (60×60×25mm, high speed)
Free air: 31 CFM, Max pressure: 0.42 in H₂O
At 0.25 in H₂O: ~18 CFM (from curve) -- ADEQUATE
Noise: 38 dBA -- May be too loud for office
Option 3: Two Sunon MF60251V1 in series
Combined: ~15 CFM at 0.25 in H₂O, 25 dBA each
Best solution for noise-sensitive application
- Using free-air CFM: Fan free-air rating is at zero pressure drop. Real systems have impedance, so actual flow is always less than free-air. Always use the fan curve.
- Altitude derating: Air density decreases ~12% per 1000m elevation. At 2000m, cooling capacity drops ~24%. Fan curves shift accordingly.
- Fan aging: Ball bearing fans lose 10-20% performance over their rated life. Sleeve bearing fans degrade faster, especially at high temperatures.
- Series vs. parallel fans: Series fans increase pressure capability (for high-impedance systems). Parallel fans increase flow (for low-impedance systems). Using the wrong configuration provides minimal benefit.
Checkpoint 4: Natural Convection Assessment Major
For products without fans (sealed enclosures, consumer devices, outdoor equipment), natural convection is the primary cooling mechanism. The design must be validated for this mode of heat transfer.
Natural Convection Fundamentals
Nusselt number for vertical flat plate:
Nu = 0.59 × (Gr × Pr)^0.25 (laminar, Ra < 10⁹)
Nu = 0.13 × (Gr × Pr)^0.33 (turbulent, Ra > 10⁹)
Where:
Gr = Grashof number = g × β × ΔT × L³ / ν²
Pr = Prandtl number ≈ 0.71 for air
Ra = Gr × Pr (Rayleigh number)
g = 9.81 m/s², β = 1/T_film (K⁻¹), ν = 15.9×10⁻⁶ m²/s (at 25°C)
h_natural = Nu × k_air / L
k_air = 0.026 W/(m·K) at 25°C
Example: Vertical heatsink, L = 50mm, ΔT = 40°C
T_film = (T_surface + T_air)/2 = (65+25)/2 = 45°C = 318K
β = 1/318 = 3.14×10⁻³ K⁻¹
Gr = 9.81 × 3.14×10⁻³ × 40 × 0.05³ / (15.9×10⁻⁶)²
Gr = 9.81 × 3.14×10⁻³ × 40 × 1.25×10⁻⁴ / 2.53×10⁻¹⁰
Gr = 6.1×10⁵
Ra = 6.1×10⁵ × 0.71 = 4.3×10⁵ (laminar)
Nu = 0.59 × (4.3×10⁵)^0.25 = 0.59 × 25.6 = 15.1
h = 15.1 × 0.026 / 0.05 = 7.85 W/(m²·K)
Natural Convection Design Rules
- Vertical orientation is always better than horizontal (30-50% improvement).
- Fin spacing for natural convection: minimum 6-10mm (closer fins choke the flow).
- Fin height: diminishing returns above 25-30mm for natural convection.
- Surface finish: anodized (black) surfaces radiate 2-3× more than bare aluminum.
- Enclosure venting: Provide inlet (bottom) and outlet (top) openings for chimney effect.
- Minimum chimney height for draft: ΔP = ρ × g × H × ΔT/T_avg.
Optimal Fin Spacing for Natural Convection:
S_opt ≈ 2.7 × L / Ra_L^0.25
For L = 50mm, Ra_L = 4.3×10⁵:
S_opt = 2.7 × 50 / (4.3×10⁵)^0.25 = 135 / 25.6 = 5.3mm
Practical: Use 6-8mm fin spacing for natural convection heatsinks.
Sealed outdoor sensor (5W internal dissipation): Aluminum enclosure with external fins (8mm spacing, 25mm height) on two sides. Enclosure oriented vertically. Black anodized for radiation. Total finned area = 0.04 m². Rth_enclosure_to_air = 1/(8×0.04) + radiation contribution = 3.1°C/W effective. Internal air temp rise above enclosure = 5°C. Total rise = 5 + 5×3.1 = 20.5°C above ambient. At 60°C ambient: 80.5°C internal. Acceptable.
Sealed plastic enclosure (ABS, k=0.17 W/m·K) with 8W internal dissipation. No external fins, smooth surfaces, horizontal orientation. The plastic acts as a thermal insulator. Internal temperature reaches 45°C above ambient. At 50°C ambient: 95°C inside -- components fail.
Checkpoint 5: Heatsink Mounting Method Verified Minor
The mounting method must provide adequate pressure for thermal interface performance while being reliable under vibration, thermal cycling, and assembly constraints.
Mounting Methods Comparison
| Method | Thermal Performance | Vibration Resistance | Ease of Assembly | Rework |
|---|
| Spring clip (TO-220) | Good (10-20 psi) | Excellent | Easy | Easy |
| Push-pin (BGA heatsink) | Moderate (5-10 psi) | Good | Easy | Easy |
| Screw/bolt | Excellent (controlled torque) | Excellent | Moderate | Easy |
| Thermal adhesive | Good (full contact) | Good | Easy | Difficult |
| Solder (direct) | Excellent | Excellent | Difficult | Very difficult |
| PCB-mounted tabs | Good | Excellent | Moderate | Moderate |
Screw Torque for Heatsink Mounting:
Target pressure: 20-50 psi (138-345 kPa) for most TIMs
Force = Pressure × Area
Example: 30mm × 30mm heatsink, target 30 psi
F = 30 psi × 30mm × 30mm × (1N / 145 psi·mm²) = 186N
For 2 screws: F_each = 93N
Torque = F × d / (2 × K_nut)
For M3 screw (K≈0.2): T = 93 × 0.003 / (2 × 0.2) = 0.7 N·m
Specify: "Tighten M3 screws to 0.7 N·m"
- Over-tightening screws: Excessive force can crack ceramic IC packages (especially BGAs) or cause PCB flexure that damages solder joints.
- Thermal adhesive on reworkable parts: If a component might need replacement, do not use permanent thermal adhesive. Use thermal tape or clips instead.
- Vibration loosening: In automotive/industrial applications, use thread-locking compound or spring washers to prevent screws from loosening.
- Uneven pressure: Single-point mounting creates tilting. Use multiple attachment points for large heatsinks to ensure even TIM compression.
Checkpoint 6: Thermal Grease/Pad Selection Major
The correct TIM must be specified with a specific manufacturer part number, application method, and quantity. Vague specifications like "use thermal paste" lead to inconsistent production and potential thermal failures.
TIM Selection Decision Tree
- Determine the gap to fill: If surfaces are flat and parallel within 0.05mm, use grease or phase-change. If gap is 0.5-5mm, use a thermal pad.
- Check if surfaces can be clamped together: Grease requires clamping pressure (10-50 psi). If no clamping, use thermal adhesive or pre-applied pad.
- Consider rework requirements: Grease is easy to clean and reapply. Adhesive is permanent. Pads are removable but may not be reusable.
- Match TIM to temperature range: Standard silicone grease works to 200°C. Thermal pads may be limited to 150°C. Check material specifications.
- Verify chemical compatibility: Some TIMs contain silicone (avoid near optical systems or relay contacts). Non-silicone options exist.
- Specify application: dot pattern, X-pattern, screen print, pre-cut pad with exact dimensions.
Recommended TIM Products
| Product | Type | k (W/m·K) | Use Case | Application |
|---|
| Shin-Etsu X23-7783D | Grease | 6.0 | High-performance IC cooling | Stencil or dispense |
| Dow TC-5022 | Grease | 4.0 | General heatsink mounting | Dispense, dot pattern |
| Bergquist GP5000S35 | Pad | 5.0 | Gap filling, 0.5-3mm | Die-cut, peel and stick |
| Laird Tflex 700 | Pad | 6.0 | High-performance gap fill | Die-cut, manual placement |
| Honeywell PTM7950 | Phase-change | 8.5 | Highest performance, reflow | Pre-applied to heatsink |
| 3M 8810 | Adhesive tape | 0.6 | Low-power, no clamp | Die-cut, peel and stick |
BOM specification: "Thermal pad: Bergquist Gap Pad 5000S35, 20×20×1.0mm, Part# GP5000S35-20-20-1.0. Apply centered on IC package top. Compress to 0.8mm minimum with heatsink mounting screws (0.5 N·m torque)." Assembly drawing shows exact placement with reference designator callout.
BOM line: "Thermal compound, generic, apply as needed." Production workers apply varying amounts -- some too little (poor contact), some too much (squeezes out onto nearby components, causing contamination). No torque specification means some heatsinks are barely touching and others crack the PCB.
