STM32F105xx, STM32F107xx
Package characteristics
6.2.2
Selecting the product temperature range
When ordering the microcontroller, the temperature range is specified in the ordering
information scheme shown in Table 62: Ordering information scheme.
Each temperature range suffix corresponds to a specific guaranteed ambient temperature at
maximum dissipation and, to a specific maximum junction temperature.
As applications do not commonly use the STM32F103xx at maximum dissipation, it is useful
to calculate the exact power consumption and junction temperature to determine which
temperature range will be best suited to the application.
The following examples show how to calculate the temperature range needed for a given
application.
Example 1: High-performance application
Assuming the following application conditions:
Maximum ambient temperature T
= 82 °C (measured according to JESD51-2),
Amax
I
= 50 mA, V = 3.5 V, maximum 20 I/Os used at the same time in output at low
DDmax
DD
level with I = 8 mA, V = 0.4 V and maximum 8 I/Os used at the same time in output
OL
OL
at low level with I = 20 mA, V = 1.3 V
OL
OL
P
P
= 50 mA × 3.5 V= 175 mW
INTmax
= 20 × 8 mA × 0.4 V + 8 × 20 mA × 1.3 V = 272 mW
IOmax
This gives: P
= 175 mW and P
= 272 mW:
IOmax
INTmax
P
= 175 + 272 = 447 mW
Dmax
Thus: P
= 447 mW
Dmax
Using the values obtained in Table 61 T
is calculated as follows:
Jmax
–
T
For LQFP100, 46 °C/W
= 82 °C + (46 °C/W × 447 mW) = 82 °C + 20.6 °C = 102.6 °C
Jmax
This is within the range of the suffix 6 version parts (–40 < T < 105 °C).
J
In this case, parts must be ordered at least with the temperature range suffix 6 (see
Table 62: Ordering information scheme).
Example 2: High-temperature application
Using the same rules, it is possible to address applications that run at high ambient
temperatures with a low dissipation, as long as junction temperature T remains within the
J
specified range.
Assuming the following application conditions:
Maximum ambient temperature T
= 115 °C (measured according to JESD51-2),
Amax
I
= 20 mA, V = 3.5 V, maximum 20 I/Os used at the same time in output at low
DDmax
DD
level with I = 8 mA, V = 0.4 V
OL
OL
P
P
= 20 mA × 3.5 V= 70 mW
INTmax
= 20 × 8 mA × 0.4 V = 64 mW
IOmax
This gives: P
= 70 mW and P
= 64 mW:
IOmax
INTmax
P
= 70 + 64 = 134 mW
Dmax
Thus: P
= 134 mW
Dmax
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