GATE CSE 2004 | Question: 67

Question

The microinstructions stored in the control memory of a processor have a width of $26$ bits. Each microinstruction 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 input 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?

• A. $10, 3, 1024$
• B. $8, 5, 256$
• C. $5, 8, 2048$
• D. $10, 3, 512$

Comments

This question was asked in BARC 2019 (with the exact same values and options)

I couldn't add a tag to the question, I think it's because I don't have the question-editing privilege

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To get the clear Understanding on Mircroprogramming lec notes neptel

Link

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Answer to Why control memory 1024 ?

The X field tells us the address of the next microinstruction to execute.

Since: X=10 bits, the Control Address Register is 10 bits wide.

Therefore it can address: $2^{10}$ different locations in control memory.

So the control memory contains: $2^{10}$=1024 words​

Think of it exactly like normal memory addressing:

If an address field has:

• 8 bits → $2^{8}$=256 memory locations
• 10 bits → $2^{10}$=1024 memory locations

Here, X is the address used to access the control memory, so its 10 bits determine the number of microinstructions that can be stored.

Answer 1

$x + y + 13 = 26  \rightarrow (1)$
$y = 3$  $(y)$ is no of bits used to represent 8 different states of multiplexer $ \rightarrow (2)$
$x$ is no of bits required represent size of control memory
$x = 10$ from $(1)$ and $(2)$

$\therefore$ Size of control memory $= 2^x = 2^{10}= 1024$

Correct Answer: $A$

Comments

Both the video are same, ....please verify that only one video cover the topic completely

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any significance of the figure given?

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Can you provide a link to the full course of these lectures?

Answer 2

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