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Digital Electronics - Code Converters and Multiplexers - Discussion

Discussion Forum : Code Converters and Multiplexers - General Questions (Q.No. 5)
Code Converters and Multiplexers
5.
From the following list of input conditions, determine the state of the five output leads on a 74148 octal-to-binary encoder.
I0 = 1 I3 = 1 I6 = 1
I1 = 1 I4 = 0 I7 = 1
I2 = 1 I5 = 1 EI = 0
GS = L, A0 = L, A1 = L, A2 = H, EO = H
GS = L, A0 = H, A1 = L, A2 = L, EO = H
GS = L, A0 = L, A1 = H, A2 = L, EO = H
GS = L, A0 = H, A1 = H, A2 = L, EO = H
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
3 comments Page 1 of 1.
Ashish said:   1 decade ago
I guess the answer should be a) as if I4=0 then output in binary is 100, NOTE- inputs of 74148 are low active hence giving a zero to I4 gives binary equivalent of 4 i.e 100, A2=1,A1=0,A0=0.
Chinnu said:   1 decade ago
If any one knows please tell the answer, I try to solve but I don't know how to solve this.
Tariku Kajela said:   1 decade ago
explain it
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