4 bit binary adder

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Kaiser Ed Bayawa

Joined Mar 19, 2015
64
does anyone here already made a circuit called binary adder
i have here: 74ls83 and 74ls86 ic and a logic gate about the circuit.
my problem is i dont know how to make it work. i already have researched its data sheets to determine the correct pins but still not working.
 

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hp1729

Joined Nov 23, 2015
2,304
does anyone here already made a circuit called binary adder
i have here: 74ls83 and 74ls86 ic and a logic gate about the circuit.
my problem is i dont know how to make it work. i already have researched its data sheets to determine the correct pins but still not working.
The schematic you have looks like a good start. What part doesn't work? All you need is switches for inputs and LEDs for outputs.
Something like this ... I like your addition of the 7486.
 

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absf

Joined Dec 29, 2010
1,942
no sir i thought sed would be connected to the source.. should i put some resistors to the pins for safety?
No, all pins connected to SEL are inputs. You can make it either High or Low. No resistor needed unless you insist.
C1 is carry-in and C5 is carry-out (as in your diagram), of the adder.

Allen
 

hp1729

Joined Nov 23, 2015
2,304
sir what do you mean by vcc=5 and ground=12 should i use 2 sources? the 5 volts and the 12volts?
No, he is saying ground is pin 12. All your schematic shows looks good so your problem must be something not shown on the schematic. It just didn't show power pins. What are you using for inputs and outputs?
 

absf

Joined Dec 29, 2010
1,942
no sir i thought sed would be connected to the source.. should i put some resistors to the pins for safety?
What "source"? Do you mean +5V or logic High?

I think 'SEL' is making the B inputs of the adder negative. When 'SEL' is high, the XOR gates would complement the B inputs. Thus 0001 would become 1110 with C1 set high. So 0001 + 1110 + carry in becomes 0000 with carry out =1.

Allen
 

hp1729

Joined Nov 23, 2015
2,304
What "source"? Do you mean +5V or logic High?

I think 'SEL' is making the B inputs of the adder negative. When 'SEL' is high, the XOR gates would complement the B inputs. Thus 0001 would become 1110 with C1 set high. So 0001 + 1110 + carry in becomes 0000 with carry out =1.

Allen
No, the low into one side of the Ex-OR makes it non-inverting.
 
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