STM32F103C4H6ATR STMICROELECTRONICS [STMicroelectronics], STM32F103C4H6ATR Datasheet - Page 75

• Manufacturer Part Number: STM32F103C4H6ATR
• Description: Low-density performance line, ARM-based 32-bit MCU with 16 or 32 KB Flash, USB, CAN, 6 timers, 2 ADCs, 6 communication interfaces
• Manufacturer: STMICROELECTRONICS [STMicroelectronics]

6.2.2
Selecting the product temperature range
When ordering the microcontroller, the temperature range is specified in the ordering
information scheme shown in
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:
Thus: P
Using the values obtained in
This is within the range of the suffix 6 version parts (–40 < T
In this case, parts must be ordered at least with the temperature range suffix 6 (see
Table 55: Ordering information
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
specified range.
Assuming the following application conditions:
Thus: P
Maximum ambient temperature T
I
level with I
at low level with I
P
P
This gives: P
P
–
T
Maximum ambient temperature T
I
level with I
P
P
This gives: P
P
DDmax
DDmax
Jmax
INTmax
IOmax
Dmax
INTmax
IOmax
Dmax
Dmax
Dmax
For LQFP64, 45 °C/W
= 82 °C + (45 °C/W × 447 mW) = 82 °C + 20.115 °C = 102.115 °C
= 175 + 272 = 447 mW
= 70 + 64 = 134 mW
= 50 mA, V
= 20 mA, V
= 20 × 8 mA × 0.4 V + 8 × 20 mA × 1.3 V = 272 mW
= 20 × 8 mA × 0.4 V = 64 mW
= 447 mW
= 134 mW
= 50 mA × 3.5 V= 175 mW
= 20 mA × 3.5 V= 70 mW
OL
OL
INTmax
INTmax
= 8 mA, V
= 8 mA, V
OL
DD
DD
= 175 mW and P
= 70 mW and P
= 20 mA, V
= 3.5 V, maximum 20 I/Os used at the same time in output at low
= 3.5 V, maximum 20 I/Os used at the same time in output at low
OL
OL
Table 54
Table 55: Ordering information
Doc ID 15060 Rev 3
= 0.4 V and maximum 8 I/Os used at the same time in output
= 0.4 V
scheme).
OL
Amax
Amax
T
= 1.3 V
Jmax
IOmax
IOmax
= 82 °C (measured according to JESD51-2),
= 115 °C (measured according to JESD51-2),
is calculated as follows:
= 64 mW:
= 272 mW:
J
scheme.
< 105 °C).
J
remains within the
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