We discus
on my earlier post that each instruction passes through different combinations
of Fetch, Memory Read, and Memory Write cycles at http://amienbcafe.blogspot.in/2013/08/video-tutorial-on-how-to-draw-timing.html.

Now if
you know the combinations of cycles then you can calculate how long such an
instruction would require to complete.

To know how many T-States an
instruction requires, and keeping in mind

that a T-State is one clock cycle long, we can calculate the time using the following formula:

that a T-State is one clock cycle long, we can calculate the time using the following formula:

**Delay = No. of T-States / Frequency**

Now let we take a example to understand easily.

If the speed of your microprocessor is 2 MHz then for a “MVI”
instruction uses 7 T-States. Therefore the instruction would require 3.5 μ Seconds
to complete.

Now we can use a loop to
produce a certain amount of time delay in a program.

See an example of a delay
loop:

MVI C, FFH 7 T-States

LOOP: DCR C 4 T-States

JNZ LOOP 10 T-States

Here see the first
instruction initializes the loop counter and is executed only once requiring
only 7 T-States.

The following two
instructions which is inside a loop that requires 14 T-States to execute and is
repeated 255 times until C becomes 0.

We need to keep in mind
though that in the last rotation of the loop, the JNZ instruction will not jump
to the address and so it requires only 7 T-States rather than the 10.

•Therefore, we must
deduct 3 T-States from the total delay to get an accurate delay calculation.

•To calculate the delay, we
use the following formula:

T

_{delay = }T_{O + }T_{L}
T

_{delay}= total delay
T

_{O}= delay outside the loop
T

_{L}= delay of the loop
Now using these formulas, we can calculate the time delay for the above
mansion example:

•T

_{O}= 7 T-States Delay of the MVI instruction
•T

_{L}= (14 X 255) -3 = 3567 T-States (14 T-States for the 2 instructions repeated 255 times).
Now your
microprocessor speed is 2 MHz. so we know

T (time for one T
state) = 1/ F

= ½
X 10

^{-6 }sec
=
.5 µ sec

So now we have 3567 T-state so we will get
3567 X .5 µ sec =
1.7835 m sec of delay.

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