find the time delay for the delay subroutine shown to the le

find the time delay for the delay subroutine shown to the left, if the system has an 8051 with frequency of 11.0592 MHz.

DELAY: MOV R5, #100

BACK: MOV R2, #200

AGAIN: MOV R3, #250

HERE: NOP

NOP

DJNZ R3, HERE

DJNZ R2, AGAIN

DJNZ R5, BACK

RET

Solution

For the 11.0592Mhz 8051 1 machine cycle=1 / 921.6Khz =1.085 micro seconds (11.0592 / 12 =921.6Khz).

In this program HERE loop time delay= 250*(2+1+1)* 1.085 X 10^-6=1085 X 10^-6 sec.

AGAIN loop repeats the HERE loop again 200 times ,so Time delay= 200*1085 X 10^-6 = 217 ms

BACK loop repeats 100 times ,so 100 X 217ms=21700ms

However, the instructions “MOV R3,#250 and “DJNZ R2, AGAIN” at the beginning and end of the AGAIN loop add (3 x 200 x 1.085 us) = 651 us to the time delay

Total time delay =21.700651 sec