UGC NET CSE | December 2014 | Part 2 | Question: 10

Question

The size of the $ROM$ required to build an 8-bit adder / subtractor with mode control, carry input, carry output and two's complement overflow output is given as 

• A. $2^{16} \times 8$
• B. $2^{18} \times 10$
• C. $2^{16} \times 10$ 
• D. $2^{18} \times 8$ 

Comments

result will be 8 bit so 8 vertical lines +( 1 for carry ) +1 ( for saying underflow) .

^^ i didnt get this

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Underflow occurs when result produced is smallest no. which can be represented.

Answer 1

total input to the rom decoder will be (8+8 ( two 8 bit number ) +1( mode ) +1( carry in))

so total number of words out of decoder will be 2^18 . result will be 8 bit so 8 vertical lines +( 1 for carry ) +1 ( for saying underflow) .

answer will be B.

a little more here .

 https://www.google.co.in/search?q=8-bit+adder%2F+subtractor&oq=8-bit+adder%2F+subtractor+&aqs=chrome..69i57.4118j0j7&sourceid=chrome&es_sm=93&ie=UTF-8#q=rom+required+8-bit+adder%2F+subtractor

Comments

Thank you sir.

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thanks mat bolo . bas ek jodi kapda sil dena. ;)

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@Chetana Tailor : Tag the source of this question. Or else here is a source you could update it with : UGC-NET December 2014 Paper II Qn 10.

Answer 2

Refer this