Electrolytic capacitor heating - QuickField simulation example
Joule losses and operating temperature of an electrolytic capacitor heated by ripple current superimposed on its normal DC operating voltage.
Engineering question
How to calculate the Joule heat loss and the temperature of the capacitor?
Answer Calculate ripple-current Joule loss from the temperature-dependent ESR, use it as a volumetric heat source, and solve the steady-state thermal model for the capacitor temperature.
Given Voltage V = 40 VDC, capacitance C = 2200 uF
Rated ripple current I = 7.2 A, AC frequency f = 100 kHz.
ESR vs. Temperature dependence
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).
Thermal conductivities of materials, W/K·m
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.
