# Combinational Logic Circuit to convert Excess 3 code into BCD code

#### hitmen

Joined Sep 21, 2008
159
I know that -3 = 0000
-2 = 0001
-1 = 0010
and that 10, 11, 12 are dont care conditions.

But how do I implement the K-map?

#### beenthere

Joined Apr 20, 2004
15,819
Make a table of the excess 3 codes and their BCD equivalents. Work from that.

#### hitmen

Joined Sep 21, 2008
159
Table ? I dont exactly understand.

-3 = 0000
-2 = 0001
-1 = 0010 (These 3 dont care)

0 = 0011
1 = 0100
2 = 0101
3 = 0110
4 = 0111
5 = 1000
6 = 1001
7 = 1010
8 = 1011
9 = 1100
10 onwards not possible dont care

Where do I go from here?

#### beenthere

Joined Apr 20, 2004
15,819
Think of your excess 3 codes being applied as select bits to some multiplexers.

#### vvkannan

Joined Aug 9, 2008
138
Hello hitmen,

Set up a 4 variable k-map . The input variables A,B,C,D are the
excess-3 code bits(starting from 0011 ,which corresponds to A'B'CD) . This is for the first bit of all the output values(which is BCD ).

Similarly do it for the other 3 bits which means you will get 4 K-maps in all

#### hitmen

Joined Sep 21, 2008
159
Hello hitmen,

Set up a 4 variable k-map . The input variables A,B,C,D are the
excess-3 code bits(starting from 0011 ,which corresponds to A'B'CD) . This is for the first bit of all the output values(which is BCD ).

Similarly do it for the other 3 bits which means you will get 4 K-maps in all
Hi vvkannan,

I understand how to do the question already. But how do the examiners know whether your answer is right or wrong since there are so many dont care conditions? Also, how can a student verify his answer?

#### vvkannan

Joined Aug 9, 2008
138
But how do the examiners know whether your answer is right or wrong since there are so many dont care conditions? Also, how can a student verify his answer?
First of all i am an undergraduate student and i don't have any experience in correction . I studied this subject last semester.

But yes you are right ,we can have many answers for these type of questions.
We try to get the simplified form which uses the minimum number of gates and incase of dontcares just give them the value which reduces the total number of inputs and gates.
But i think you can have many answers