# Combinational Logic

Discussion in 'Homework Help' started by pranavissmart, Mar 20, 2014.

1. ### pranavissmart Thread Starter New Member

Feb 23, 2014
5
0
How many don't care inputs are there in the BCD adder?

2. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Can you justify that assertion?

3. ### pranavissmart Thread Starter New Member

Feb 23, 2014
5
0
Let there be two numbers A and B. If A varies from 0 to 9 and B from 10 to 15 for the carries 0 or 1. Each digit of each carry would have one don't care input condition. Thus from multiplication rule, we get 10 x 6 x 2 = 120 don't cares.

I am unable to approach the rest of them.

4. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I don't think your problem statement is well formed. Are you concerned about "don't cares" in the sense that you don't care about the result with certain combinations of inputs, OR are you concerned about detecting the difference between valid inputs giving a valid result and invalid inputs giving a meaningless result?

As I count the inputs to a BCD Adder I get four for the first operand, call it A, and four for the second operand, call it B, and one for the Carry In for a total of nine inputs and none of them are "don't care".

Maybe you're thinking that with 9 inputs there 512 possible input combinations. There are 10 valid combinations for A and 10 valid combinations for b and two valid combinations for Carry In. That is 10 x 10 x 2 = 200 valid combinations. Now:

512 possible combinations - 200 valid combinations = 312 invalid combinations

Alternatively if you want to count the invalid combinations directly:

There are six invalid combinations for A times 10 valid combinations for B times 2 combinations for Carry In = 120
There are 10 valid combinations for A times six invalid combinations for B times 2 combinations for Carry In = 120
There are six invalid combinations for A times 6 invalid combinations for B times 2 combinations for Carry In = 72

120 + 120 +72 = 312

Last edited: Mar 20, 2014