Digital Electronics - Code Converters and Multiplexers - Discussion

Discussion Forum : Code Converters and Multiplexers - General Questions (Q.No. 29)
29.
For the following conditions on a 7485 magnitude comparator, what will be the state of each of the three outputs?
A0 = 0B0 = 1IA < B = 0
A1 = 1B1 = 0IA = B = 1
A2 = 1B2 = 0IA > B = 0
A3 = 0B3 = 0 
A = B = 0, A < B = 0, A > B = 1
A = B = 0, A < B = 1, A > B = 0
A = B = 1, A < B = 0, A > B = 0
A = B = 0, A < B = 0, A > B = 0
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
4 comments Page 1 of 1.

Sukriti said:   9 years ago
Bit position:

i.e 15 14 13 12 ...................................... 1 0
The LSB (least significant bit) is 0.
MSB (most significant bit ) is 15.

So,
A3 A2 A1 A0 => 0110
B3 B2 B1 B0=> 0001

The decimal Values for A is 6 & B is 1

Now compare for each state.

State 1: A=B
=> No A!=B(A is not equal to B)
=> So, A=B=0

State 2: A<B
=> No A!<B(A is not less than B)
=> So, A<B=0 ()

State 3: A>B
=> A>B(A is greater than B)
=> So, A>B=1 (6>1)

Therefore option (A) is correct.

Karan said:   9 years ago
@Avy.

Why 0001?

Avy said:   9 years ago
A --> 0110 = 6.
B --> 0001 = 1.
A>B.

Kelsy21 said:   1 decade ago
Can anybody please explain me how the answer A came?

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