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Question 3.5: Find the size of the delay in following program, if the crys...

Find the size of the delay in following program, if the crystal
frequency is 11.0592MHz.

Machine Cycle

DELAY:        MOV   R2,#200                                                                                                                                           1
AGAIN:        MOV   R3,#250                                                                                                                                            1
HERE:          NOP                                                                                                                                                              1

NOP                                                                                                                                                              1
DJNZ   R3,HERE                                                                                                                                         2
DJNZ   R2,AGAIN                                                                                                                                       2
RET^{\nearrow ^{\text{Notice in nested loop, as in all other time delay loops, the time is approximate since we have ignored the first and last instructions in the subroutine.} } }              2
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For HERE loop, we have (4×250) x1.085 μs＝1085μs.
For AGAIN loop repeats HERE loop 200 times, so we have 200×1085 μs＝217000 μs. But “MOV R3,#250” and “DJNZ R2,AGAIN” at the start and end of the AGAIN loop add (3x200x1.805)=651 μs.
As a result we have 217000 + 651= 217651 μs.

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