Engineering Guide CDA3000
3-11
3 Selection of inverter module
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EN
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A
Example: Heat transfer via a cooler
The inverter module has a
temperature evaluation facility
as standard.
The current temperature at
point 1 of the heat sink is
displayed by way of parameter
427-KTEMP in subject area
_VAL (actual values).
•
Inverter module CDA34.014
•
Power stage clock frequency 4 kHz
1. Power loss discharged by way of the mounting plate of the inverter
module.
The CDA34.014 has a power loss of 180 W (table 3.2).
75% of the power loss is discharged via the mounting plate (active cool-
ing area) and 25% as radiated heat via the housing (table 3.9)
P
Mountingplate
= 180 W x 0.75 = 135 W
2. Calculate temperature difference between mounting plate and cool-
ing plate.
∆ϑ
= P
Mounting plate
x R
th
1)
= 135 W x 0.02 K/W = 2.7 K
1)
See table 3.9
3. Maximum temperature at point 2 and on the cooler
ϑ
Point 2
=
ϑ
Point 1
-DJ = 85 °C - 2.7 °C = 82.3 °C
4. Calculation of the cooler:
•
At point 2 the max. temperature of 82.3 °C must not be exceeded.
•
135 W of power loss must be discharged by way of the cooler.
•
The exact solution depends on the cooler used, e.g. heat sink to air
or water, heat exchanger etc.
Heat transfer
compound
Mounting plate
CDA34.014
Cooler
Point 1: 85˚C, see table 3.9
Point 2: To be ascertained (max.
temperature on cooling plate)
