Designing Embedded Hardware, 2nd Edition

Timing of Instructions

On some processors, particularly CISC chips, different instructions take varying lengths of time to complete. Sometimes it is important to know how long a given section of code will take to execute, particularly if the software is interacting with a time-critical external system. The execution time can be calculated by looking up the number of cycles each instruction takes and adding them together.

Tip

A "cycle" relates to the internal clock that is driving the processor. This is not necessarily the same speed as the crystal. The datasheet will tell you how the crystal frequency relates to the internal clock for a given processor.

The following example shows some 68HC11 assembly language and the number of cycles associated with each instruction:

      LDAA #$0F ; 2 cycles
again DECA      ; 2 cycles
      BNE again ; 3 cycles
exit  JMP $1000 ; 3 cycles

From examining the code, we can see that the loop is executed 15 (0x0F) times. This loop comprises five cycles in total. Therefore, the above program will take:

2 + (15 * 5) + 3 cycles = 80 cycles

For a 2 MHz 68HC11, a cycle is 500 ns. Hence, this code fragment will take 40,000 ns (or 40 μsecs) to complete.

Knowing the timing of instructions is often very useful. For example, suppose we need to have a program that will provide a delay of approximately 9 msec. To do this, we need to load a value into a register and count it down to zero. This will provide the delay.

So a program that would do this is simply:

 LDX #some_number ; load X ...