Morris Mano Numerical

Question

• A. How many 128 × 8 RAM chips are needed to provide a memory capacity of 2048 bytes?
• B. How many lines of the address bus must be used to access 2048 byte of memory? How many of these lines will be common to all chips?
• C. How many lines must be decoded for chip select? Specify the size of the decoders?

Answer 1

i)

128 X 8 RAM provides a memory capacity of 128*8=1024 bits

Number of chips required = $\frac{2048*8}{1024}$ = 16

ii)

Number of address line to access 2048 bytes = log 2048 = 11

Since there are 16 chips, so to distinguish among these chips bits required = log 16 = 4

So common lines = 11-4 =7

iii)

since 7 lines are common to each chip, chip select will have = log 7 =3 lines

since there are 16 chips we will need 4 X 16 decoder

Comments

anyone please give the diagrammatic representation for (ii) please.

thanks in advance

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since 7 lines are common to each chip, chip select will have = log 7 =3 lines

chip select lines 4 because there are 16 chips.

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How you came to know that no of cs bits are 5.

Answer 2

Under the assumption of byte addressable memory 1 word=8bits.
1. no of RAM chips=2048*8/128*8=16
2. size of memory is 2048 B=2¹¹ so we need 11 bits 2 uniqually determine a Byte so address BUS requires 11 lines
Now Understand the purpose of these 11 bits we knows that we need 4 bits to identify a RAM chip bcoz we have 16
so 11 is divided into 2 parts 4+7  " first 4 bits 2 select a RAM chip out of 16 and next  bits 2 select 8bits inside that RAM chip " so i think no of common address lines to all chips =7.
3. we have 16 chips, we need 4*16 decoder to select the right chip. so decoder size =4*16