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AN4043 Datasheet(PDF) 46 Page - STMicroelectronics

Part # AN4043
Description  In recent years the variable speed motor control market has required high performance
PDF  60 Pages
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Manufacturer  STMICROELECTRONICS [STMicroelectronics]
Direct Link  http://www.st.com
Logo STMICROELECTRONICS - STMicroelectronics

AN4043 Datasheet(HTML) 46 Page - STMicroelectronics

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Power losses and dissipation
AN4043
46/60
Doc ID 022726 Rev 2
4.3
Thermal impedance overview
During operation, power losses generate heat which elevates the temperature in the
semiconductor junctions contained in the SLLIMM-nano, limiting its performance and
lifetime. To ensure safe and reliable operation, the junction temperature of power devices
must be kept below the limits defined in the datasheet, therefore, the generated heat must
be conducted away from the power chips and into the environment using an adequate
cooling system.
The SLLIMM-nano was designed to drive electric motors up to 100 W without any heatsink.
Therefore, the thermal aspect of the system is one of the key factors in designing high
efficiency and high reliability equipment. In this environment the package and its thermal
resistance play a fundamental role.
Thermal resistance quantifies the capability of a given thermal path to transfer heat in
steady-state and it is generically given as the ratio between the temperature increase above
the reference and the relevant power flow:
Equation 29
The thermal resistance specified in the datasheet is the junction-ambient Rth(j-a) which is
commonly used with natural and forced convection air cooled systems and it is defined as
the difference in temperature between junction and ambient reference divided by the power
dissipation per device:
Equation 30
Figure 28.
Typical switching waveforms of the STGIPN3H60
V
CE
E
on=23.5μJ(*)
IC
t
on = 270ns
V
LIN = 2V/Div
I
C = 500mA/Div
V
CE = 100V/Div
t = 100ns/Div
V
LIN
Turn on
STGIPN3H60
Low side
Tj=100
°C
Turn off
STGIPN3H60
Low side
Tj=100
°C
toff = 905ns
(*) E
on and Eoff are the areas under the red plots
E =
∫ (V
CE · IC) dt
V
CE
E
off=5.1μJ(*)
I
C
V
LIN
V
LIN = 2V/Div
I
C = 200mA/Div
V
CE = 100V/Div
t = 200ns/Div
AM11816v1
P
T
Rth
Δ
Δ
=
D
amb
j
a)
th(j-
P
T
T
R
−
=



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