STM32F303 datasheet(127/133 Pages) STMICROELECTRONICS

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STM32F302xx/STM32F303xx

Package characteristics

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7.2.2

Selecting the product temperature range

When ordering the microcontroller, the temperature range is specified in the ordering

information scheme shown in Section 8: Part numbering.

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 STM32F30xB/C 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 TAmax = 82 °C (measured according to JESD51-2),

IDDmax = 50 mA, VDD = 3.5 V, maximum 3 I/Os used at the same time in output at low

level with IOL = 8 mA, VOL= 0.4 V and maximum 2 I/Os used at the same time in output

at low level with IOL = 20 mA, VOL= 1.3 V

PINTmax = 50 mA × 3.5 V= 175 mW

PIOmax = 3 × 8 mA × 0.4 V + 2 × 20 mA × 1.3 V = 61.6 mW

This gives: PINTmax = 175 mW and PIOmax = 61.6 mW:

PDmax = 175 + 61.6 = 236.6 mW

Thus: PDmax = 236.6 mW

Using the values obtained in Table 79 TJmax is calculated as follows:

–

For LQFP64, 45°C/W

TJmax = 82 °C + (45°C/W × 236.6 mW) = 82 °C + 10.65 °C = 92.65 °C

This is within the range of the suffix 6 version parts (–40 < TJ < 105 °C).

In this case, parts must be ordered at least with the temperature range suffix 6 (see

Section 8: Part numbering).

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 TJ remains within the

specified range.

Assuming the following application conditions:

Maximum ambient temperature TAmax = 115 °C (measured according to JESD51-2),

IDDmax = 20 mA, VDD = 3.5 V, maximum 9 I/Os used at the same time in output at low

level with IOL = 8 mA, VOL= 0.4 V

PINTmax = 20 mA × 3.5 V= 70 mW

PIOmax = 9 × 8 mA × 0.4 V = 28.8 mW

This gives: PINTmax = 70 mW and PIOmax = 28.8 mW:

PDmax = 70 + 28.8 = 98.8 mW

Thus: PDmax = 98.8 mW

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