STM8S207RB STMicroelectronics, STM8S207RB Datasheet - Page 96

• Manufacturer Part Number: STM8S207RB
• Description: Performance line, 24 MHz STM8S 8-bit MCU, up to 128 Kbytes Flash, integrated EEPROM
• Manufacturer: STMicroelectronics

Specifications of STM8S207RB

• Max Fcpu: up to 24 MHz, 0 wait states @ fCPU≤ 16 MHz
• Program: up to 128 Kbytes Flash; data retention 20 years at 55 °C after 10 kcycles
• Data: up to 2 Kbytes true data EEPROM; endurance 300 kcycles
• Ram: up to 6 Kbytes
• Advanced Control Timer: 16-bit, 4 CAPCOM channels, 3 complementary outputs, dead-time insertion and flexible synchronization

Package characteristics
11.2.2
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Selecting the product temperature range
When ordering the microcontroller, the temperature range is specified in the order code (see
Figure 49: STM8S207xx/208xx performance line ordering information scheme(1) on
page
The following example shows how to calculate the temperature range needed for a given
application.
Assuming the following application conditions:
●
●
●
●
●
Using the values obtained in
calculated as follows for LQFP64 10 x 10 mm = 46 °C/W:
T
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.
Jmax
Maximum ambient temperature T
I
Maximum eight standard I/Os used at the same time in output at low level with I
mA, V
Maximum four high sink I/Os used at the same time in output at low level with I
mA, V
Maximum two true open drain I/Os used at the same time in output at low level with
I
P
P
This gives: P
P
Thus: P
DDmax
OL
99).
INTmax =
IOmax =
Dmax
= 82 °C + (46 °C/W x 443 mW) = 82 °C + 20 °C = 102 °C
= 20 mA, V
OL
OL
= 82.5 mW + 360 mW
= 15 mA, V
Dmax
= 2 V
= 1.5 V
(10 mA x 2 V x 8 ) + (20 mA x 2 V x 2) + (20 mA x 1.5 V x 4) = 360 mW
15 mA x 5.5 V = 82.5 mW
INTmax
= 443 mW
OL
= 2 V
DD
= 82.5 mW and P
= 5.5 V
Table 57: Thermal characteristics on page 95
Doc ID 14733 Rev 12
Amax
IOmax
= 82 °C (measured according to JESD51-2)
360 mW:
J
< 105 °C).
STM8S207xx, STM8S208xx
T
Jmax
is
OL
OL
= 20
= 10