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  #1  
Old 09-21-2008, 06:05 PM
Yash_Arya Yash_Arya is offline
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Default A 3-bit and 4-bit binary adder.

Hello, I was trying to make a 4-bit binary Adder and I reffered to some sites here and there but I could'nt get my concepts clear.

Could anyone please tell me (show me?) how a circuit of a 4-bit binary adder would be? Using standard Gates and ICs only.

Also, how would it be different from a 3-bit adder?

Please help me with both the circuits, I will be greatful. If my question was not clear, ask me to clear it.

Thank you, Greetings.
Yash Arya.
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Old 09-21-2008, 06:35 PM
scubasteve_911 scubasteve_911 is offline
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The 3 and 4-bit adders originate from a single bit adder, just scaled up.

The following link shows a single bit adder, then, how it is hooked up to make 4-bits.

http://www.play-hookey.com/digital/adder.html

I'm pretty sure that the logic isn't optimized for low-gate count, but it will work fine. It is possible to write out karnaugh maps or do boolean algebra to reduce the number of gates needed, but that's another level of complexity.

Steve

Or, an allaboutcircuits reference is here too

http://www.allaboutcircuits.com/vol_4/chpt_9/3.html
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  #3  
Old 09-21-2008, 06:36 PM
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Get the data sheet on a 74XX83. It is a 4 bit full adder. The data sheet should include the internal logic diagram.
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Old 09-23-2008, 02:02 AM
harsimranjit singh harsimranjit singh is offline
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whats 3 and 4 bit adder////i mean binary
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Old 09-23-2008, 02:28 AM
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A 3 bit adder adds 3 bit values, the 4 bit adder adds 4 bit values.
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Old 09-23-2008, 12:39 PM
Yash_Arya Yash_Arya is offline
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Thank you for your help people, now I think I do not need any further guidance unless I myself read the resources that you kindly provided.

However, (obviously) if someone else needs to clear something, go right ahead.

Thank you,
Yash Arya.
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  #7  
Old 09-23-2008, 12:55 PM
Yash_Arya Yash_Arya is offline
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**I already have querries to be cleared**

I went through the page [http://www.play-hookey.com/digital/adder.html] you provided me and I have the following querries.

(1) In the Full-Adder part, if we see the truth table, We have A,B (i/p) and Cin, Cout. Here, Is A itself a multi-bit number (and same for B)? Or do A, B, Cin and Cout together make a "multi-bit number"?

(2)If, A is itself a multi bit number, then how come in the truth table we give it only "0" or "1" in its coloumn? I mean, the number will be something like "A = 1011" for example.

Please clear my Querries. Thanks a lot in advance.

Thank you,
Yash Arya.
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Old 09-23-2008, 03:41 PM
veritas veritas is offline
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For each individual full adder, A and B are only 1 bit, and Cin/Cout are for those bits. For a multiple-bit adder, Cin goes to the least significant bit, and Cout comes from the most significant bit. The other carries are internal.
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Old 09-23-2008, 06:26 PM
Yash_Arya Yash_Arya is offline
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Quote:
Originally Posted by veritas View Post
For each individual full adder, A and B are only 1 bit, and Cin/Cout are for those bits. For a multiple-bit adder, Cin goes to the least significant bit, and Cout comes from the most significant bit. The other carries are internal.
Okay, so let us speak in terms of "order" of 'A' and 'B'. Suppose that 'A' and 'B' are bits of order 'n'. (i.e A[n] and B[n]) and the carry from previous bit is C[n-1].

Now to add A[n], B[n] and C[n-1], we need one full adder. Am I correct? So, for 4-bit adder, will I need four such full-adders? (Heck that will be complex)

Also, let us say that after the Addition, I recieve Sum (Sn) and Carry out(Cn).

Now what all should be taken as i/p for the next full-adder?

From what I understood, It will be A[n+1], B[n+1], C[n].

Am I correct?

Also, please tell me some suitable IC for the same. (The ICs I have in my syllabus are 7432, 7408, 7404, 7402, 7400, 7486, 74135) The ICs must be from the list only.

Thank you all for your co-operation.

Yash Arya.
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  #10  
Old 09-23-2008, 06:57 PM
veritas veritas is offline
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You are correct on all counts, I believe. If you have only logic gates, you will need to create logic for Sum[n] and Cout[n] from the inputs A[n], B[n], and Cin[n], and then duplicate you logic 4 times.

*edit* I would suggest making a truth table.
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