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Consider a processor which contains 32 time multiplexed pins used to carry address and data, and 8 address pins carry address only.Then calculate the size of memory and number of address bits needed to address memory?
in CO and Architecture by Boss (25.6k points) | 96 views

is it 32 address bits and size = 32$\times$240

Total pins for address =32+8 ,so address is 40 bits.

Word size here =4B

So if word addressable =  $2^{40} W=>40$

If byte addressable then =$2^40 *2^2 B => 2^{42} Bytes=>42 $

I dont have answer.Please check if i am correct.Why have you taken address bits as 32 only?
See Multiplexed pin used for a 1st clock for address and for remaining clock used to send data.

so address bits will be 32 only, after that data bits will be 32+8 = 40 bits

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