Question

Design a 8-to-3 binary encoder, with priority in binary sequence (i.e. input line 0 has higher priority than input line 1 etc). There should also be an "active" output which is 1 when any input line is active, and "0" when all input lines are 0. If all input lines are 0, the output lines other than the "active" output are "don't care".

Answer 1

The truth table for the 8-to-3 priority encoder will be

D7	D6	D5	D4	D3	D2	D1	D0	Q2	Q1	Q0	O
0	0	0	0	0	0	0	0	x	x	x	0
x	x	x	x	x	x	x	1	0	0	0	1
x	x	x	x	x	x	1	0	0	0	1	1
x	x	x	x	x	1	0	0	0	1	0	1
x	x	x	x	1	0	0	0	0	1	1	1
x	x	x	1	0	0	0	0	1	0	0	1
x	x	1	0	0	0	0	0	1	0	1	1
x	1	0	0	0	0	0	0	1	1	0	1
1	0	0	0	0	0	0	0	1	1	1	1

In the above truth table, D0 to D7 are 8 input lines; Q0 to Q2 are 3 output lines and O is active output.

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  Steps to write the equations:

• For a particular output, select the rows with output value 1 only. For example, Q0 has output value 1 for rows highlighted in green above.
• Make the expression of that particular output using those rows only. For example, Q0 will result in the expression:

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Using the above truth table, we will derive the expression for Q0 to Q2 and O.

We are using the property to reduce the expressions below:

Verification:

provide input to lines 2 and 3.

Q0 = 0

Q1 = 1

Q2 = 0

We got the answer 010 which is 2.

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For the bonus part, the truth table will be

The priority order is 3, 4, 0, 2, 1, 5, 7, 6.

D7	D6	D5	D4	D3	D2	D1	D0	Q2	Q1	Q0	O
0	0	0	0	0	0	0	0	x	x	x	0
x	x	x	0	0	x	x	1	0	0	0	1
x	x	x	0	0	0	1	0	0	0	1	1
x	x	x	0	0	1	x	0	0	1	0	1
x	x	x	x	1	x	x	x	0	1	1	1
x	x	x	1	0	x	x	x	1	0	0	1
x	x	1	0	0	0	0	0	1	0	1	1
0	1	0	0	0	0	0	0	1	1	0	1
1	x	0	0	0	0	0	0	1	1	1	1

Now use this above truth table to find the equation for Q0 to Q2 and simplify them using boolean algebra properties.