Microprogrammed-Control-Unit
Question 1 |
A microprogrammed control unit
Is faster than a hard-wired control unit. | |
Facilitates easy implementation of new instruction. | |
Is useful when very small programs are to be run. | |
Usually refers to the control unit of a microprocessor. |
Question 1 Explanation:
In micro-programmed control unit we can add new instruction by changing the content of control memory.
Question 2 |
Consider a CPU where all the instructions require 7 clock cycles to complete execution. There are 140 instructions in the instruction set. It is found that 125 control signals are needed to be generated by the control unit. While designing the horizontal microprogrammed control unit, single address field format is used for branch control logic. What is the minimum size of the control word and control address register?
125, 7 | |
125, 10 | |
135, 7 | |
135, 10 |
Question 2 Explanation:
Each instruction takes 7 cycles,
i.e., 140 instruction takes = 140 * 7 =980 cycles.
So, size of control address register = ⌈log2 980⌉
= 10 bit
Since horizontal microprogramming is used, so 125 control signals will require 125 bits.
Hence, size of control word = 125 + 10 = 135 bits
i.e., 140 instruction takes = 140 * 7 =980 cycles.
So, size of control address register = ⌈log2 980⌉
= 10 bit
Since horizontal microprogramming is used, so 125 control signals will require 125 bits.
Hence, size of control word = 125 + 10 = 135 bits
Question 3 |
The micro instructions stored in the control memory of a processor have a width of 26 bits. Each micro instruction is divided into three fields. a micro operation field of 13 bits, a next address field (X), and a MUX select field (Y). There are 8 status bits in the inputs of the MUX. How many bits are there in the X and Y fields, and what is the size of the control memory in number of words
10, 3, 1024 | |
8, 5, 256 | |
5, 8, 2048
| |
10, 3, 512 |
Question 3 Explanation:
The number of bits in Control memory =26.
From the given data each instruction divided into op field (13)+X(next address field)+Y(MUX)
8(23) status bits in the inputs of the MUX then three bits in the MUX select field.
No. of bits in control memory next address field=26-13-3 =10
size of the control memory in number of words is 210=1024 words
From the given data each instruction divided into op field (13)+X(next address field)+Y(MUX)
8(23) status bits in the inputs of the MUX then three bits in the MUX select field.
No. of bits in control memory next address field=26-13-3 =10
size of the control memory in number of words is 210=1024 words
There are 3 questions to complete.