Engineering Design a logic circuit to add two 2-BCD decade numbers

[Fatima Hasan]

Homework Statement

Homework Equations

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The Attempt at a Solution

Here's my work :

When I added 0000 0001 to 0000 0000 , I didn't get the correct answer .
Could someone check where is my mistake please ?

 

[willem2]

It looks like you are adding 0001 0000 to 0000 0000 and the answer is correct..

 

[Fatima Hasan]

It looks like you are adding 0001 0000 to 0000 0000 and the answer is correct..

I added 0000 0001 not 0001 0000 and the answer was 0000 0XX1 0000 .
Why I didn't get 0000 0000 0001 ?

 

[willem2]

I added 0000 0001 not 0001 0000 and the answer was 0000 0XX1 0000 .
Why I didn't get 0000 0000 0001 ?

The only 1 input I can see is going into the adder on the bottom left, and this must be the adder for the most significant digit, since it has a carry in coming from the other adders. I have no idea where those XX come from. Could be unconnected inputs, inputs that are between a 1 and a 0 or inputs that are changing, and I don't see any of that.

 

[Fatima Hasan]

The only 1 input I can see is going into the adder on the bottom left, and this must be the adder for the most significant digit, since it has a carry in coming from the other adders. I have no idea where those XX come from. Could be unconnected inputs, inputs that are between a 1 and a 0 or inputs that are changing, and I don't see any of that.

I've solved it again and disconnected +5V from (A>B) :

I got the correct answer which is 0000 0001 0000 , but the question mentioned that the circuit will work if A and B are invalid BCD (A > 1001 and B > 1001 ) .So , should I connect +5V to A>B ( A > 1001 which is invalid BCD ) as I did in my first solution ?

 

[willem2]

I got the correct answer which is 0000 0001 0000 , but the question mentioned that the circuit will work if A and B are invalid BCD (A > 1001 and B > 1001 ) .So , should I connect +5V to A>B ( A > 1001 which is invalid BCD ) as I did in my first solution ?

Those are outputs, so tying them to +5 V will get you undefined behavior. You should use the outputs of the 2 comparators that check A to produce an error signal that is 1 whenever one of the digits of A is out of range.

 

[Fatima Hasan]

Those are outputs, so tying them to +5 V will get you undefined behavior. You should use the outputs of the 2 comparators that check A to produce an error signal that is 1 whenever one of the digits of A is out of range.

Here's when I connected the (A>B) of the 2 comparators to +5V :

 

[willem2]

Here's when I connected the (A>B) of the 2 comparators to +5V :

That is as I would expect. You are not using the outputs of the 4 comparators on the left to produce an error signal.
If you do use the comparators, you should connect all inputs. Do you understand what the inputs and outputs of the comparator are meant for? I think you use those. http://www.sycelectronica.com.ar/semiconductores/74LS85.pdf

 

[Fatima Hasan]

That is as I would expect. You are not using the outputs of the 4 comparators on the left to produce an error signal.
If you do use the comparators, you should connect all inputs. Do you understand what the inputs and outputs of the comparator are meant for? I think you use those. http://www.sycelectronica.com.ar/semiconductores/74LS85.pdf

I connected the cascade inputs of all the comparators as follows :

A<Bin and A>Bin = Ground.
A=Bin = Vcc
And the outputs as follows :
A>B =+5V to produce an error signal which is 1 if A or B are invalid BCD.
A=B and A<B = Ground .

But I still didn't get the correct answer .

 

[willem2]

And the outputs as follows :
A>B =+5V to produce an error signal which is 1 if A or B are invalid BCD.
A=B and A<B = Ground .

You should not tie any outputs to either 5V or ground. If you don't understand why that is, this circuit is too complicated for you. The XXX in the output is probably, because you shorted out the power supply, and your simulater doesn't know what to do. You should try to modify the circuit, so a lamp or LED gets lid whenever A or B has digits that are invalid. It's really much easier than the part of the circuit that modifies the digits after the addition.

 

[Fatima Hasan]

so a lamp or LED gets lid whenever A or B has digits that are invalid.

It lids when A = 1100 :