# How to make 4 bit binary adder using IC 7483?

Discussion in 'Homework Help' started by Shreyash, Oct 5, 2016.

1. ### Shreyash Thread Starter New Member

Oct 5, 2016
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0
its my project.
we have the following: IC7483, IC 7400(NAND), IC 7432(OR GATE), BREADBOARD, WIRES. two of each and leds.
how to implement it?

2. ### AlbertHall Well-Known Member

Jun 4, 2014
2,275
449
Have you read the datasheet? Which bits did you not understand?

Oct 5, 2016
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4. ### Shreyash Thread Starter New Member

Oct 5, 2016
3
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why do we need the a1,a2,a3,a4 and b1-4 again below.
is just the upper half not enough?why does this have only one set of ai's and bi's (i=1to4)

• ###### Capture.PNG
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Last edited: Oct 5, 2016
5. ### AlbertHall Well-Known Member

Jun 4, 2014
2,275
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A four bit adder adds two four bit numbers to a four bit sum and a carry.

6. ### ci139 Member

Jul 11, 2016
341
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we belive you, but what you actually have to do with those chips?
HD74HC83RPEL

7. ### ci139 Member

Jul 11, 2016
341
39
you list NAND 7400 but #3 uses 7408 AND - ? so you need to rewire this thing for NAND
and i don't think #3 works as BCD adder coz the carry is set at outputs (12 to 15) or (6,7,14,15) or (16 to 31)
wy to se t carry at 6,7 ?
you need to set carry extra to 16 and beyond at 10 to 15 that'd be
10 1010
11 1011 -- AND(S4,S2) // the lower AND in Fig #3 is connected wrong or shown wrong or both
12 1100 -- OR AND(S4,S3)
13 1101
14 1110
15 1111
-- suppose we connect the lower and "right"
and our carry is now correct
so for nubers >=10 the sum is extra incremented by 6 as 10=10*((carry=1) OR 1)+((10+6) MOD 16)
10 + i = 10*carry + (((10+i) mod 16 + 6*carry) MOD 16) =10*carry + i
--- so remake it for NAND instead of AND
your carry is currently carry = C10 = CO or (S4 and S2) or (S4 and S3)

Last edited: Oct 5, 2016
8. ### ci139 Member

Jul 11, 2016
341
39
the carry cant be converted using boolean algebra
but paying attention to function where NAND=0 if both inputs are 1 otherwise it's 1
so we need to avoid settin our NAND-s while SUM <=9 ... obvously we need 1 NAND to invert a bit for that
0 0000 0
1 0001 0
2 0010 0
3 0011 0
4 0100 0
5 0101 0
6 0110 0
7 0111 0
8 1000 0
9 1001 0
A 1010 1
B 1011 1
C 1100 1
D 1101 1
E 1110 1
F 1111 1
it likely is more likely if part of the sum is formed in one adder and part in an other - but it's not shure it can be done so
it may also be your teacher was drunk? giving such a task ...

9. ### ci139 Member

Jul 11, 2016
341
39
i dont think it has solution using the components you listed with partial summing eighter coz if we bit mask one operand to 1001 0110 then
the condition that has to correct the final modulus would had to change it's input causing indefinit/strobing output if we bit mask the mod 10 e.g. 1010 0101 we'd be adding hopefully 0 or10 to our sum but also 8 or 2 the least likely can't be set to form what is required
change the school ? "We teach here everything that can't be done!" -- weird motto

10. ### ci139 Member

Jul 11, 2016
341
39
• Do NOT just provide solutions to student's problems. This is known as "cheating".
so it turns out you don't have to change the school
new carry = OR( c16 , NANDasNOT( NAND( S4 , OR( S3 , S2 ) ) ) )

it was tricky to see

11. ### ci139 Member

Jul 11, 2016
341
39

confirmed it with (if there's no more flaws ...)

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