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How many RAM chips needed & what is the size of decoder used?

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.

6Comments
Gagan Batra @gaganbatra
14 Mar 2016 11:33 pm

Option D should be correct.

target size = 32KB = 32K x 8 (1 Byte = 8 bits) = 215 x 8 bits

basic size = 256 x 4 = 28 x 4 bits

No of chips = target size/ basic size = 27 x 2 = 256

Size of decoder can be calculated from the row chips of basic size = 8 x 28 = 8 x 256

Note - If basic size is 2M x N then size of decoder is M x 2M

Jitendra Verma @jitendraverm
14 Mar 2016 11:48 pm

answer is B.

 

Gagan Batra @gaganbatra
14 Mar 2016 11:53 pm

Please explain.

Jitendra Verma @jitendraverm
14 Mar 2016 11:57 pm

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

1 chip 4 bits

And for byte (8 bits) we need 2 chips in row

Total no of rows =128

So 7:128 decoder is required to select a row

 

Charan @harisadu
27 Dec 2018 11:59 pm
Where is it written that it is Byte addressable? In this way we can take 16 bits for one row.
khdijah @khdijah
17 May 2017 11:21 am

the answer is a , 

we know that decoders used to distinguish between the RAMs so , first of all the Q said "memory of capacity 32KB " we need to convert from KB to Bytes , this gives whats 32*2^10 =32.768 Bytes . then we divdie those bytes on the size of RAMs to know how many of RAMs it can fit , 32.768\265= 128 which is the number of RAMs we need , now we just have to know the size of decoder and since there's 128 RAMs we should connect each one with a wire , but how we can differntiate between those RAMS ? 2^7 =128 , then all we need is 7 bits to differintaie , so the decoder should be 7*128 .