Microinstructions Format

9. Microinstructions Format

It should be noted that micro programmed control could adapt easily to changes in the system design. We can easily add new instructions without changing hardware. In this lecture will be devoted to study the format of micro- instruction
Like instructions, the micro- instruction consists of flieds and The standard instruction format contains opcode field and address(s),The micro- instruction format:

The length of a microinstruction is determined based on the following parameters:

1-number of micro operations specified in the microinstructions:
2- The way the control bits will be interpreted.

A microinstruction may specify one or more micro-operations that will be activated simultaneously. The length of the microinstruction will increase as the number of parallel micro-operations per microinstruction increases. Furthermore, when each control bit in the microinstruction corresponds to exactly one control line, the length of microinstruction could get bigger.

Examples:



1- Suppose that the size of CM is 256×24, the computer executes as maximum 32 micro operations and no more 4 micro operations at the same time. There are two fields : branch and subroutine and no more 4 conditions for each. The micro instruction has one address. Based on previous information find the length of micro instruction and then draw it.


Sol:




256×24 ---> 8 bit for address field.

4 micro operations at the same time---> 4 Groups (F1,F2,F3,F4).

No. of bits for each Group= ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

= 32/4= 8---> 23--->3 bit for each F
no more 4 conditions for BR and CD ---> 4= 22--->2 bit for each
F1 F2 F3 F4 CD BR Address
3 bit 3 bit 3 bit 3 bit 2 bit 2 bit 8 bit
The length of instruction comes from the summation of bits of the fields = 24.


2- Suppose that the size of Control Memory is 512×m, the computer executes as maximum 40 micro operations and no more 5 micro operations at the same time. There are two equal fields for branching and subroutine to assign 4 states for each. The micro instruction has one address. Find m and draw the microinstruction

Sol:
512×m ---> 9 bit for address field.

5 micro operations at the same time---> 5 Groups (F1,..,F5).

No. of bits for each Group= ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

= 40/5= 8---> 23--->3 bit for each F
no more 4 conditions for BR and CD ---> 4= 22--->2 bit for each
F1 F2 F3 F4 CD BR Address
3 bit 3 bit 3 bit 3 bit 2 bit 2 bit 9 bit
M = 28, how?

-What is the size of CDR? Why?


Answer: ----------------------------------------------------------------------------



-What is the size of CAR? Why?

Answer: ----------------------------------------------------------------------------

Question for you…
If the size of Control Memory is n×32, the computer executes as maximum 48 micro operations and there are 6 micro operations executed at the same time. There are two equal fields for branching and subroutine to assign 4 states for each. The micro instruction has one address. Find n.
Answer:
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