接着又有怪问题了...研究下资料后，我发现那个150mA灌电流又有些迷茫了，比如资料中有下面例子：
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
Thus: PDmax = 447 mW
Using the values obtained in Table 70 TJmax is calculated as follows:
– For LQFP100, 46 °C/W
TJmax = 82 °C + (46 °C/W × 447 mW) = 82 °C + 20.6 °C = 102.6 °C
This is within the range of the suffix 6 version parts (–40 < TJ < 105 °C).
按上面的灌电流算的话：50mA + 20 × 8mA + 8 × 20mA = 370mA，带来的是447mW的功耗（对应20.6 °C温升），早就超过那150mA了，那么是不是只要温升+环境温度不超过芯片所能承受的温度范围就没事？所以又萌发新的想法：
我所用的是LQFP144封装，对应温升是1W/30 °C，产品所在环境温度最大40 °C，最大工作温度范围-40 °C ~ +85 °C，取85 °C，那么（85 - 40） °C = 35 °C就是允许的温升，取30 °C算就是允许有1W功耗在IC上，芯片供电如之前所说3.3V/70mA，VOL按0.4V算，接着：
1W = 3.3V × 70mA + 0.4V × 最大允许电流
则最大允许灌电流有1.9225A？！
|
楼主
