Design of binary subtractor using MSI ICs

IC 74LS83 is a 4-bit parallel binary adder chip. It adds/subtracts a four-bit number (nibble) with another 4-bit number. The block symbol for the IC is shown in Fig.1. This IC has two sets of 4-bit inputs along with a carry input C0. It performs binary subtraction on the A & B inputs and the carry input C0. It generates a 4-bit Difference and a Borrow out C4.


Fig. 1 Block Symbol of four-bit Subtractor IC

A circuit that can subtract 4-bit numbers can be designed using a control input and additional EX-OR IC 74LS86  For this we use the EX-OR gate as a “Controlled Inverter”. The explaination for this concept can be easily understood from Fig.2. The four bit input B4, B3, B2 & B1 can be passed through the controlled inverter IC74LS83 and the A4, A3, A2 & A1 are connected directly to A inputs of IC 74LS83 as shown in Fig 3.

Fig.2. Exclusive-OR gate used as a Controlled Inverter



Fig.3. 4-bit Binary Subtractor

Four-bit Subtraction: When Control input is set = 1, the Carry –in input C0 = 1. In this situation, the Ex-OR gates will provide 1,’s complement of the Input-2 to the B-inputs of Adder IC. Moreover as C0 = 1, the addition of 1 to the 1’s complement of B gives 2’s complement of B.Now the IC74LS83 adds Input-1 i.e. A4,A3, A2,A1 to the 2’s complemet of B and produces the Carry and Sum output on C4 & the lines Σ4, Σ3, Σ2, & Σ1.

Example1: Let Input-1 = A4 A3 A2 A1 = 1001 & Input-2 = 0111 & control input be set to 1.1’s complement of 0111 = 1000. Since carry input C0 = 1, the input B becomes,
B4 B3 B2 B1 = 1000 + 1 = 1001.
Now IC 74LS83 performs addition of A & 2,s complement of B and produces the output.



Since a Carry is generated discard the Carry and the Sum is the final output of subtraction operation.
The result is Σ4 Σ3 Σ2 & Σ1 = 0 0 1 0.

Example2: Let Input-1 = A4 A3 A2 A1 = 0111 & Input-2 = 1001 & control input be set to 1.
1’s complement of Input-2 i.e. 1001 = 0110  Since carry input C0 = 1, the input B becomes,
B4 B3 B2 B1 = 0110 + 1 = 0111.
Now IC 74LS83 performs addition of A & 2’s complement of B and produces the output.


In this case no Carry is generated during addition. Hence the answer can be obtained by taking the 2’s complement of the Sum output and attaching a negative sign.So the 2’s complement of 1110 = 0010 and the final result of subtraction is Σ4 Σ3 Σ2 & Σ1 = - 0 0 1 0.   

Numerical:

For subtracting two number, the 2's complement of one number is calculated and then added to other number

To calculate 2's complement of a number, the 1's complement of number is calculated i.e. number is represented in binary format, and then inverted. And then add 1 to it.

For eg.:        Complement of 2 is:

                             2                  =>     0010

                   1's complement     =>     1101

                   2's complement     =>     1101  + 1     => 1110

          Hence 2's complement of 2 is 1110.

                  

Examples:

1.         1.  4 - 0

 

2's Complement of   0  =  1 0000

 

 

4

 

0

1

0

0

+

2's complement of 0

 1

0

0

0

0

 

4

1

0

1

0

0

 

         (The MSB bit is complemented to achieve the Sign-bit)

          Sign Bit = 0 (Positive Number)

 

 

2.         2.  4 - 2

 

2's Complement of   2  =  0 1110

 

 

4

 

0

1

0

0

+

2's complement of 2

0

1

1

1

0

 

2

1

0

0

1

0

 

         Sign Bit = 0 (Positive Number)

 

3.         3.  2 - 4

 

2's Complement of   4  =  1100

 

 

2

 

0

0

1

0

+

2's complement of 4

 0

1

1

0

0

 

14

 0

1

1

1

0

 

         Sign Bit = 1 (Negative Number)

 

         As the output is negative number, it is in the 2's complement format. Hence the 2's complement of Answer need to be taken, to achieve the right answer:

         Answer => 1110

         1's Complement => 0001

         2's Complement => 0001 + 1 = 0010

         Hence, final answer = 0010 with Sign bit = 1