STM32L476RG datasheet(227/232 Pages) STMICROELECTRONICS

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STM32L476xx

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As applications do not commonly use the STM32L476xx 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 20 I/Os used at the same time in output at low

level with IOL = 8 mA, VOL= 0.4 V and maximum 8 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 = 20 × 8 mA × 0.4 V + 8 × 20 mA × 1.3 V = 272 mW

This gives: PINTmax = 175 mW and PIOmax = 272 mW:

PDmax = 175 + 272 = 447 mW

Using the values obtained in Table 113 TJmax is calculated as follows:

–

For LQFP64, 45 °C/W

TJmax = 82 °C + (45 °C/W × 447 mW) = 82 °C + 20.115 °C = 102.115 °C

This is within the range of the suffix 6 version parts (–40 < TJ < 105 °C) see Section 8: Part

numbering.

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

numbering).

Note:

With this given PDmax we can find the TAmax allowed for a given device temperature range

(order code suffix 6 or 7).

Suffix 6: TAmax = TJmax - (45°C/W × 447 mW) = 105-20.115 = 84.885 °C

Suffix 7: TAmax = TJmax - (45°C/W × 447 mW) = 125-20.115 = 104.885 °C

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 = 100 °C (measured according to JESD51-2),

IDDmax = 20 mA, VDD = 3.5 V, maximum 20 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 = 20 × 8 mA × 0.4 V = 64 mW

This gives: PINTmax = 70 mW and PIOmax = 64 mW:

PDmax = 70 + 64 = 134 mW

Thus: PDmax = 134 mW

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