Assume the processor consumes $32$mA during normal operation, $8$mA at the idle mode, and

$30$uA at the sleep mode. Consider the following code:

main $()\{$

$\_$SWDTEN $=1$; $//$ Enable watchdog timer

while $(1)\{$

asm$($pwrsav#$0)$; $//$ idle mode

task$1()$;

$\}$

Assumptions :

$(1)$ The Watchdog Timer $($WDT$)$ prescaler N$1($WDTPRE$)$ divides by $32$ and postscaler N$2$

$($WDTPOST$)$ divides by $100.$

$(2)$ The time taken by the function task$1()$ is $10$ us$.$

$($a$)[2]$ Compute the WDT timeout

WDT timeout $=1/32768*$ N$1*$ N$2=(32*100)/32768=0\mathrm{.}09765625$ s

$($b$)[3]$ Compute the approximate percentage of time in which the processor is in the sleep

mode and normal mode

Normal mode time percentage $=100*10$us $/(10$us $+09765625)=0\mathrm{.}01\%$

Sleep mode time percentage $=100*1$s $/(10$us $+09765625)=99\mathrm{.}99\%$

$($c$)[3]$ Compute the average current consumption.

Average current $=30$uA $*0\mathrm{.}9999+32$mA $*0\mathrm{.}01=33\mathrm{.}2$ uA