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How many 256 X 4 RAM chips are required to organize a memory of capacity 32KB ? What is the size of decoder required in this implementation to select a row of chip?

Options :

(a) 128 , 7 X 128

(b) 256 , 7 X 128

(c) 512 , 7 X 128

(d) 256 , 8 X 256.

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2 Comments

Should be b
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I need the explanation.
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2 Answers

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Best answer
No of chips=32K*8/256*4=256

2 chips in one row

Total no of rows =128

So 7:128 decoder is required to select a row
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4 Comments

memory is assumed as byte addresable,
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If we consider 1 word equals 2 byte then how things will change??
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Here ask

What is the size of decoder required in this implementation to select a row of chip?

For select a row of chip need 8*256 decoder... Not 7*128...
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0 votes
0 votes
no. of chips required= 32k/(256*4)=256

Now we need 8 bits words so we will have to arrange at least 2 chips per row.
Since by constructing memory we will have 127 rows each having 2 chips. So to select a row we need at least log(127) = 7 bits.
So the decoder we need is 7x127 .

Hence (b) is the answer

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it will be 128 not 127
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