Electrolytic capacitor heating
QuickField simulation example
Electrolytic capacitor operates at DC voltage. The voltage ripple causes the ripple current that heats up the capacitor.
Problem Type
Axisymmetric problem of Steady state heat transfer.
Geometry
Case size 16*35 mm.
Given
Voltage V = 40 VDC, capacitance C = 2200 uF
Rated ripple current I = 7.2 A, AC frequency f = 100 kHz.
| Temperature | Equivalent series resistance |
| 20°C | 0.023 Ω |
| 125°C | 0.0099 Ω |
Ambient air temperature +20°C, convection coefficient 5 W/(K*m²).
| Aluminum (can and lid) | 237 |
| Rubber | 0.35 |
| Terminals (tinned copper) | 380 |
| Electrolyte | 0.21 |
| Capacitor body* | λx = 100 λy = 0.21 |
Task
Calculate the Joule heat loss and the temperature of the capacitor.
Solution
We can approximate ESR vs. temperature dependency with linear function:
ESR(T) = 0.025495 - 0.001247 * T
Joule heat losses are ESR(T) * I² = 1.32 - 0.00647 * T [W]
In QuickField we specify the loss density in W/m³. Capacitor body volume is 5386 mm³.
Joule heat losses density = 245000 - 1201 * T [W/m³].
There are also Ohmic losses in terminals. Terminal wire diameter is 1 mm, cross-section area is S = 0.785 mm², copper resistivity is 1.77e-8 Ohm*m. Joule heat losses density in the terminals is:
Resistivity * (I/S)² = 1.77e-8 * ( 7.2/0.785e-6 )² = 1'490'000 W/m³.
Results
Capacitor Joule heat losses are 0.72 W. Capacitor internal temperature is 93°C.
References:
* Sam G. Parler, Jr. Thermal Modeling of Aluminum Electrolytic Capacitors
** PEG227KKP4220ME1 KEMET Axial Aluminum Electrolytic Capacitors.
- Video: Electrolytic capacitor heating. Watch on YouTube
- Download simulation files (files may be viewed using any QuickField Edition).