Hi. I have a question about ALU design. I have A and B variables which are 3-bit and ALU which has 4-bit inputs.My purpose is adding up 2A+3B but I can use ALU one time. How can I do it?

 

And later (F=2A+3B) save it 8-bit register.

 

Go to study 74HC181 then you will get the answer.

 

For example I have two variable A and B which are 4-bit and I want to make 2A plus B and save the answer in 8-bit register.

Solution:


Answer of my question will be like this but I can't do 2A plus 3B ( A and B 3-bits) and I need help

 

Is this homework or personal learning?

 

It is a question which has potatial to be in the exam

 

What is the maximum value that 2A + 3B can have? Can that be represented by the output, including carry, of a 4-bit adder?

How many bits are required to represent it?

You circuit diagram embodies some of what you need to do. Do you know which parts those are? Did you come up with that circuit diagram yourself? If so, think about why you made the connects you did and what parts of the math are taken care of and what parts aren't.

And remember that it IS a math function you are trying to implement, so play with the math.

For example, see if something like

Y = 2A + 3B = 2(A+B) + B

helps you out.

 

I think you should be able to use the S inputs, so that you can put A<<1 to Ain, B<<1 to Bin and B to Sin so that you get 2A+2B+1B.
(A<<1 being A shifted left by one bit)

 

I think put B to Select pins wrong choice because if I apply 000 to B , the F will be 0 although A is not 0

 

Oh sorry my mistake, i thought that S was the carry in, when in fact it is the mode select. Disregard my previous post.
My first idea was that +3B could be equal to -1B if the size of the processing busses was just right. I think this idea needs some practice becaues the ALU you described doesn't really cover all the possibilities of 4bit inputs and the according results.
So i suggest you try the A-B op code and see if it gets you anywhere.

 

Yes this ALU works unsigned numbers I understand(but I don't know ) so inputs must be positive therefore signed numbers can be confusing in this scenario.
My approximation like this

B0 is going to D0 directly and A0 and B0 are going to ALU's A0 and B0 respectively and B1 is going to Cn but I have hanged out there because A1+B1+B2 does not happen.Probably I should use ALU's A3 and B3 bits which are not used in the above event

 

can you use external gates as well, or only the ALU?

 

Is "Not Possible" a valid answer? Given that the ALU can only be used once, that would be my answer. Because the function as defined has three operations. 2A can be calculated without using the ALU. But 3B needs to use the ALU. Leaving no way (since the ALU can only be used once) to add the two terms.

 

Unfortinately only one ALU can be used

 

Is "Not Possible" a valid answer? Given that the ALU can only be used once, that would be my answer. Because the function as defined has three operations. 2A can be calculated without using the ALU. But 3B needs to use the ALU. Leaving no way (since the ALU can only be used once) to add the two terms.

I'm not quite convinced it's not possible yet. There is an easy way to get all but the msb of B of the odd term taken care of (so 2A + 2B + B%4). But I'm not seeing any way to get that last bit in there.