Digital Electronics - Code Converters and Multiplexers - Discussion
Discussion Forum : Code Converters and Multiplexers - General Questions (Q.No. 5)
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 |
Discussion:
3 comments Page 1 of 1.
Tariku Kajela said:
1 decade ago
explain it
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.
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.
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