Digital Logic | Code Converters – BCD(8421) to/from Excess-3

Prerequisite – Number System and base conversions

Excess-3 binary code is a unweighted self-complementary BCD code.
Self-Complementary property means that the 1’s complement of an excess-3 number is the excess-3 code of the 9’s complement of the corresponding decimal number. This property is useful since a decimal number can be nines’ complemented (for subtraction) as easily as a binary number can be ones’ complemented; just by inverting all bits.
For example, the excess-3 code for 3(0011) is 0110 and to find the excess-3 code of the complement of 3, we just need to find the 1’s complement of 0110 -> 1001, which is also the excess-3 code for the 9’s complement of 3 -> (9-3) = 6.

Converting BCD(8421) to Excess-3 –

As is clear by the name, a BCD digit can be converted to it’s corresponding Excess-3 code by simply adding 3 to it.
Let $A,:B,:C,:and:D$ be the bits representing the binary numbers, where $D$ is the LSB and $A$ is the MSB, and
Let $w,:x,:y,:and:z$ be the bits representing the gray code of the binary numbers, where $z$ is the LSB and $w$ is the MSB.
The truth table for the conversion is given below. The X’s mark don’t care conditions.
$egin{tabular}{||c|c|c|c||c|c|c|c||} hline multicolumn{4}{||c||}{BCD(8421)} & multicolumn{4}{|c||}{Excess-3}\ hline A & B & C & D & w & x & y & z \ hline hline 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \ hline 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \ hline 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \ hline 0 & 0 & 1 & 1 & 0 & 1 & 1 & 0 \ hline hline 0 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \ hline 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \ hline 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \ hline 0 & 1 & 1 & 1 & 1 & 0 & 1 & 0 \ hline hline 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 \ hline 1 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \ hline 1 & 0 & 1 & 0 & X & X & X & X \ hline 1 & 0 & 1 & 1 & X & X & X & X \ hline hline 1 & 1 & 0 & 0 & X & X & X & X \ hline 1 & 1 & 0 & 1 & X & X & X & X \ hline 1 & 1 & 1 & 0 & X & X & X & X \ hline 1 & 1 & 1 & 1 & X & X & X & X \ hline hline end{tabular}$
To find the corresponding digital circuit, we will use the K-Map technique for each of the Excess-3 code bits as output with all of the bits of the BCD number as input.

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Corresponding minimized Boolean expressions for Excess-3 code bits –
$w = A+BC+BD\ x = B^prime C + B^prime D +BC^prime D^prime\ y = CD + C^prime D^prime \ z = D^prime$
The corresponding digital circuit-

Converting Excess-3 to BCD(8421) –

Excess-3 code can be converted back to BCD in the same manner.
Let $A,:B,:C,:and:D$ be the bits representing the binary numbers, where $D$ is the LSB and $A$ is the MSB, and
Let $w,:x,:y,:and:z$ be the bits representing the gray code of the binary numbers, where $z$ is the LSB and $w$ is the MSB.
The truth table for the conversion is given below. The X’s mark don’t care conditions.
$egin{tabular}{||c|c|c|c||c|c|c|c||} hline multicolumn{4}{||c||}{Excess-3} & multicolumn{4}{|c||}{BCD}\ hline w & x & y & z & A & B & C & D \ hline hline 0 & 0 & 0 & 0 & X & X & X & X \ hline 0 & 0 & 0 & 1 & X & X & X & X \ hline 0 & 0 & 1 & 0 & X & X & X & X \ hline 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \ hline hline 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \ hline 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \ hline 0 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \ hline 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \ hline hline 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \ hline 1 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \ hline 1 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \ hline 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \ hline hline 1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \ hline 1 & 1 & 0 & 1 & X & X & X & X \ hline 1 & 1 & 1 & 0 & X & X & X & X \ hline 1 & 1 & 1 & 1 & X & X & X & X \ hline hline end{tabular}$
K-Map for D-

K-Map for C-

K-Map for B-

K-Map for A-
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Corresponding minimized boolean expressions for Excess-3 code bits –
$A = wx+wyz\ B = x^prime y^prime + x^prime z^prime +xyz\ C = y^prime z+ yz^prime \ D = z^prime$
The corresponding digital circuit –
Here $E_3,:E_2,:E_1,:and:E_0$ correspond to $w,:x,:y,:and:z$ and $B_3,:B_2,:B_1,:and:B_0$ correspond to $A,:B,:C,:and:D$ .