(8%) Consider Program 11-2C on pages 435 and 436 Look at these lines of code fro

Question

(8%) Consider Program 11-2C on pages 435 and 436

Look at these lines of code from this program

T0CON=0x0;

TMR0H=0x35;

TMR0L=0x00;

T1CON=0x88;

TMR1H=0x35;

TMR1L=0x00;

Suppose XTAL = 10M Hz, compute how long in milliseconds (or microseconds) it takes for Timer 0 to interrupt. Compute also how long it takes for Timer 1 to interrupt.

Repeat part (a) with XTAL = 16M Hz now.

(12%) Consider now Program 11-3C on pages 437 and 438.

Some lines of code for timer configuration and count are as follows:

T0CON=0x08;

TMR0H=0;

TMR0L=0;

T1CON=0x06;

TMR1H=255;

TMR1L=-200;

Near the top of page 437, it is stated that program11-3C has two interrupts: (1) PORTC counts up every time Timer0 overflows and (2) a 1-Hz pulse is fed into Timer 1 used as a counter. When the count reaches 200, it will toggle pin RB6.

Explain why the code of program11-3C toggles pin RB6 when timer 1 count reaches 200. Note you may need to use 1 or 2 lines of code not shown above. You do not have to explain the chk_isr, T0_ISR, T1_ISR functions and also the My_HiPrio_Int function with the #pragma code etc.

With the code of program 11-3C, compute when Timer0 overflows (in milliseconds) and also when Timer1 overflows.

Is it possible to change code so that Timer 1 toggles RB6 when it reaches count 400? If so, how? If not, why?

Explanation / Answer

timer0 =0x3500 or 13568 in decimal

so now count tis incresed from 0x3500to 0xFFFF ---> 65535-13568=51877 counts

T0CON =0x0 hence it is in 16 bit mode with prescalar of 1:2

we know that The CPU clock speed is the oscillator speed divided by four. also it will be divided by prescalar

clock frequency = (XTAL/4)/2 =XTAL/8

time taken by timer0 to interrupt = 51877 /(XTAL/8)

= 415016/XTAL

now XTAL = 10 Mhz

time taken by timer0 to interrupt= 415016/10000000 s = 41.5016 ms

count tis incresed from 0x3500 to 0xFFFF ---> 65535-13568=51877 counts

T1CON =0x88

hence timer1 is started with 16bit mode and no prescalar

clock frequency = (XTAL/4) =XTAL/4

time taken by timer0 to interrupt = 51877 /(XTAL/4)

= 207508/XTAL

now XTAL = 10 Mhz

time taken by timer1 to interrupt= 207508/10000000 s = 20.7508ms

2 Xtal =16Mhz

time for timer0 = 415016/16000000 = 25.9385 ms

time for timer1 = 207508/16000000 = 12.96925 ms

2.

1.

Note that TMR1H =255 and TML1H =-200

-200 in hex =0x38

hence TIMER1 counts from 0xFF38 to 0xFFFF which is 199 counts nearly equal to 200 counts

When the timer1 overflows a interrupt is triggered. The program would jump to TIMER1 Interrupt Service Routine.where the PIN RB6 is toggled.

2

Timer 0 counts from 0 to 65535 = 65535 counts
T0CON=0x08; timer 0 16 bit mode prescalar off.

clock frequency = (XTAL/4) =XTAL/4

time taken by timer0 to interrupt = 65535/(XTAL/4)

= 262140/XTAL

Assuming XTAL = 10 Mhz for second question

time taken by timer1 to interrupt= 262140/10000000 s = 26.214 ms

timer 1 counts 200 counts

external sourc =1 HZ

time taken =200/1 = 200sec

3

yes it is possible we just want timer1 to have 400 counts hence timer1 = 65535-400 =65135 =0xFE6F in hex

hence

TMR1H=0xFE

TMR1L=0x6F

for 400 count

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