How to implement a Excess 3 to BCD Converter?

Thread Starter

The Rock

Joined Sep 25, 2011
3
I need to convert Excess 3 to BCD.
What are my inputs in the truth tables so that i can do the K Maps?
I understand that
-3 = 0000 ( Don't care)
-2 = 0001 ( Don't care)
-1 = 0010 ( Don't care )
0 = 0011
1 = 0100
etc n so on..until 10, 11, 12 are also don't cares.

Do i need to put the don't cares in the Excess 3 Code column? Plzz help! Thanks.
 

Georacer

Joined Nov 25, 2009
5,181
If your input and output are 4 bits long, you could make 4 4-input Boolean functions that will take the Excess 3 number and produce one digit of the output each. You will end up with a medium-sizes combinatorial circuit, which can be implemented either with simple gates or with MUXs.

Is that clear?

Post a conversion truth table if you need more help.
 

Thread Starter

The Rock

Joined Sep 25, 2011
3
Can u post the conversion truth table for me? Its still vague for me...
By the question, i think we need to implement it using K maps for each input used n construct the circuit as per the K map!
 

Georacer

Joined Nov 25, 2009
5,181
Your thinking this correct.

You started the truth table correctly yourself, you just need to finish it:
Rich (BB code):
Excess 3 | BCD (D3D2D1D0)
-----------------
  0000    | XXXX
  0001    | XXXX
  0010    | XXXX
  0011    | 0000
and so on.
You got four Boolean functions, D3, D2, D1 and D0 that will give you an output digit of the result depending on the 4 input digits.

Don't forget that BCD numbering doesn't go beyond 9.
 
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