The Drive That Overheated in the Cabinet
We installed a servo drive in a sealed electrical cabinet. The drive’s dissipation was 100 W. The cabinet was 600 × 600 × 300 mm, sealed (IP54), no fan. On the floor, the drive faulted with “overtemperature” after 20 minutes. The cabinet internal temperature climbed to 65°C. The drive’s derating kicks in at 50°C. The problem: 100 W of heat in a sealed cabinet with no cooling. The cabinet’s natural convection dissipated only about 50 W. The other 50 W raised the internal temperature. We added a filtered fan (20 CFM). The internal temperature dropped to 35°C. The drive ran fine. The mistake was not calculating the heat load and the cabinet’s cooling capacity.
Electrical cabinet thermal design is about matching the heat dissipated by the components to the cooling capacity of the cabinet. Too little cooling, and the drives and PLC overheat. Too much, and you waste money on oversized fans. This article runs the numbers.
Step 1: Calculate the Heat Dissipation
Every electrical component in the cabinet generates heat. The total heat load is the sum.
Q_total = Σ P_loss
Where P_loss is the power wasted as heat by each component.
Typical Heat Losses
- Servo drive: 3–5% of output power. A 1 kW servo drive dissipates 30–50 W (at full load).
- PLC and I/O: 10–30 W total.
- Power supply (24 V DC): 5–10% of output. A 24 V / 10 A supply dissipates about 12 W.
- Transformer: 3–5% of VA rating.
- Contactors, relays: 1–5 W each coil.
- HMI: 10–30 W.
Example: one servo drive (50 W), PLC (20 W), power supply (12 W), HMI (20 W), relays (5 W). Total Q = 107 W. Round up: 120 W heat load.
Step 2: Natural Convection (Sealed Cabinet)
A sealed cabinet (no fan) dissipates heat through the walls by natural convection (air outside flows over the cabinet). The cooling capacity depends on the cabinet surface area and the temperature difference.
Q_nat = h × A × ΔT
Where h is the heat transfer coefficient (about 5–10 W/m²·°C for natural convection on a vertical surface), A is the surface area (m²), and ΔT is the temperature rise (internal minus ambient, in °C).
For a 600 × 600 × 300 mm cabinet: surface area ≈ 1.5 m² (all six sides). At h = 8 W/m²·°C and ΔT = 15°C (ambient 25°C, internal 40°C): Q_nat = 8 × 1.5 × 15 = 180 W. That’s enough for our 120 W load.
But wait — our cabinet was sealed and overheated. Why? Because the cabinet was in a corner (the back and sides weren’t exposed to air flow). Effective area was only 0.8 m². Q_nat = 8 × 0.8 × 15 = 96 W. Our 120 W load exceeded it. The temperature kept rising.
| Cabinet Size | Surface Area | Natural Cooling (ΔT=15°C) | Handles Loads Up To |
|---|---|---|---|
| 400×400×200 | 0.6 m² | 72 W | ~50 W |
| 600×600×300 | 1.5 m² | 180 W | ~120 W |
| 800×600×300 | 1.8 m² | 216 W | ~150 W |
| 1200×800×400 | 3.0 m² | 360 W | ~250 W |
Step 3: Fan Cooling (Filtered Fan)
When natural convection isn’t enough, add a filtered fan. The fan blows outside air through the cabinet, exhausting the hot air.
Fan Sizing
The required air flow (CFM or m³/h) depends on the heat load and the allowable temperature rise.
Q (W) = 1.2 × CFM × ΔT
Where Q is the heat load (W), CFM is the air flow (cubic feet per minute), and ΔT is the temperature rise (°C). Rearranged: CFM = Q / (1.2 × ΔT).
For Q = 120 W, allowable ΔT = 10°C (ambient 25°C, internal 35°C): CFM = 120 / (1.2 × 10) = 10 CFM. A small 20 CFM fan (with margin) handles it. For a larger load (500 W): CFM = 500 / (1.2 × 10) = 42 CFM. A 50 CFM fan.
Fan Placement
- Intake filter (low): Outside air enters through a filter at the bottom of the cabinet.
- Exhaust fan (high): Hot air exits through a fan at the top. Hot air rises, so exhaust at the top.
- Air path: Air enters low, passes over the components, exits high. Don’t short-circuit (intake and exhaust next to each other).
Step 4: Air Conditioner (For Sealed or Hot Environments)
For washdown (IP65 sealed), outdoor, or high-ambient (40°C+), a filtered fan isn’t enough (it brings in humid/dirty air). Use an air conditioner (cabinet AC).
The AC is rated in watts (or BTU). It must exceed the heat load Q. For 120 W load, a 200 W AC (with margin) works. For 500 W, a 800 W AC.
The AC recirculates internal air and doesn’t bring in outside air. The cabinet stays sealed. But it’s expensive and needs maintenance (filter cleaning, refrigerant check).
| Method | Sealed? | Cooling Capacity | Cost | Best For |
|---|---|---|---|---|
| Natural convection | Yes (sealed) | 50–300 W | None | Low heat loads, clean environment |
| Filtered fan | No (filtered vent) | 100–1000 W | Low | Dry, clean environments, moderate loads |
| Cabinet AC | Yes (sealed) | 200–3000 W | High | Washdown, outdoor, high ambient, sealed cabinets |
| Heat exchanger | Yes (air-to-air) | 100–500 W | Medium | Sealed cabinets, ambient cooler than internal |
Step 5: Ambient Temperature Matters
The calculation assumes 25°C ambient. If the machine is in a hot plant (40°C), the cooling capacity drops.
- The allowable internal temperature is typically 50°C (drive derating starts). At 40°C ambient, ΔT is only 10°C. The cooling must handle the load with just 10°C rise.
- For 40°C ambient and 120 W load, fan CFM = 120 / (1.2 × 10) = 10 CFM (same math, but the internal is 50°C, not 35°C). The drive runs hot but within rating.
If the ambient is 45°C, the internal hits 55°C — over the drive limit. Need a bigger fan or AC. Check the plant ambient before sizing.
The thermal design workflow: 1) Sum all heat dissipation (Q in watts). 2) Estimate the cabinet’s natural cooling (h × A × ΔT). 3) If Q > natural cooling, add a fan. CFM = Q/(1.2 × ΔT). 4) For sealed cabinets or high ambient, use AC (sized to Q with margin). 5) Verify the internal temperature stays under the component limit (typically 50°C).
Component Placement: Heat Rises
Where the components go affects the thermal design.
- Hot components (drives, power supplies) at the bottom: Air passes over them first. The heated air rises and doesn’t pre-heat the cooler components above.
- Heat-sensitive components (PLC, HMI) at the top: They’re in the cooler (incoming) air, not under the drives’ exhaust.
- Leave gaps between components: Air flows through the gaps. Components packed tightly block the air path. Leave 50 mm between rows.
- Don’t block the fan intake: The filter intake must be unobstructed. Components mounted in front of the intake starve the fan.
Thermal Switches and Alarms
Add a temperature switch in the cabinet. If the internal temperature exceeds 45°C, the PLC gets an alarm. The operator knows the fan has failed or the filter is clogged.
A fan that fails (bearing seized) stops the air flow. The cabinet heats up. Without a temperature alarm, the drive overheats before anyone notices. The alarm gives early warning.
Heat Sinks for Individual Components
For a component that dissipates a lot of heat (a large servo drive), mount it on an external heat sink (through the cabinet wall). The drive’s fins are on the outside of the cabinet, exposed to ambient air. This dissipates the drive’s heat without heating the cabinet interior.
This is common for large drives (over 1 kW). The drive’s power stage bolts through the cabinet wall, and the heatsink is outside. The cabinet stays cooler because the biggest heat source doesn’t dump heat inside.
A Cabinet Cooling Checklist
- Sum all heat dissipation (Q in watts).
- What is the plant ambient temperature? (°C)
- What is the allowable internal temp? (Drive limit, typically 50°C)
- ΔT = allowable internal − ambient.
- Natural cooling: h × A × ΔT. Does it exceed Q?
- If not: fan CFM = Q/(1.2 × ΔT).
- Is the cabinet sealed (IP54+)? If yes, use AC or heat exchanger.
- Is the fan intake unobstructed?
- Are hot components at the bottom, sensitive at the top?
- Is there a temperature alarm switch?
- For large drives: external heat sink through wall?
- Fan filter maintenance schedule?
The Bottom Line
Electrical cabinet thermal design isn’t sealing the box and hoping. Sum the heat dissipation, calculate the natural convection, and add a fan (or AC) when needed. The drive that overheated in the cabinet wasn’t undersized — it was in a sealed box with no cooling for 100 W of heat. CFM = Q/(1.2 × ΔT) tells you the fan size. Place hot components at the bottom, leave gaps for air flow, and add a temperature alarm. The cabinet that stays at 35°C inside wasn’t lucky — it had the right fan for the heat load. Size the cooling to the watts, not the box size.
