### Digital logic Circuits - Unit 1 Notes - BOOLEAN ALGEBRA AND COMBINATIONAL CIRCUITS

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UNIT I – BOOLEAN ALGEBRA AND COMBINATIONAL CIRCUITS
Boolean Variables & Truth Tables
Boolean algebra differs in a major way from ordinary algebra in that Boolean constants and variables are allowed to have only two possible values, 0 or 1.
Boolean 0 and 1 do not represent actual numbers but instead represent the state of a voltage variable, or what is called its logic level.
Some common representation of 0 and 1 is shown in the following diagram.
In Boolean algebra, there are three basic logic operations: AND ,OR, and NOT.
These logic gates are digital circuits constructed from diodes, transistors, and resistors connected in such a way that the circuit output is the result of a basic logic operation (OR, AND, NOT) performed on the inputs.
Truth Table
A truth table is a means for describing how a logic circuit's output depends on the logic levels present at the circuit's inputs.
In the following two-input logic circuit, the table lists all possible combinations of logic levels present at inputs A and B along with the corresponding output level X.

When either input A OR B is 1, the output X is 1. Therefore the "?" in the box is an OR gate.

OR Operation
The expression X = A + B reads as "X equals A OR B". The + sign stands for the OR operation, not for ordinary addition.
The OR operation produces a result of 1 when any of the input variable is 1.
The OR operation produces a result of 0 only when all the input variables are 0.
An example of three input OR gate and its truth table is as follows:
With the OR operation, 1 + 1 = 1, 1+ 1 + 1 = 1 and so on.
AND Operation
The expression X = A * B reads as "X equals A AND B".
The multiplication sign stands for the AND operation, same for ordinary multiplication of 1s and 0s.The AND operation produces a result of 1 occurs only for the single case when all of the input variables are 1.The output is 0 for any case where one or more inputs are 0
An example of three input AND gate and its truth table is as follows:
With the AND operation, 1*1 = 1, 1*1*1 = 1 and so on.
NOT Operation
The NOT operation is unlike the OR and AND operations in that it can be performed on a single input variable. For example, if the variable A is subjected to the NOT operation, the result x can be expressed as x = A' where the prime (') represents the NOT operation. This expression is read as:
x equals NOT A
x equals the inverse of A
x equals the complement of A
Each of these is in common usage and all indicate that the logic value of x = A' is o           pposite to the logic value of A. The truth table of the NOT operation is as follows:
1'=0 because NOT 1 is 0
0' = 1                because  NOT  0 is 1
The NOT operation is also referred to as inversion or complementation, and these terms are used interchangeably.
NOR Operation
NOR and NAND gates are used extensively in digital circuitry. These gates combine the basic operations AND, OR and NOT, which make it relatively easy to describe then using Boolean algebra.
NOR gate symbol is the same as the OR gate symbol except that it has a small circle on the output. This small circle represents the inversion operation. Therefore the output expression of the two input NOR gate is:
X = (A + B)'
An example of three inputs OR gate can be constructed by a NOR gate plus a NOT gate:
NAND Operation
NAND gate symbol is the same as the AND gate symbol except that it has a small circle on the output. This small circle represents the inversion operation. Therefore the output expression of the two input NAND gate is:
X = (AB)'
Describing Logic Circuits Algebraically
Any logic circuit, no matter how complex, may be completely described using the Boolean operations, because the OR gate, AND gate, and NOT circuit are the basic building blocks of digital systems.
This is an example of the circuit using Boolean expression:
If an expression contains both AND and OR operations, the AND operations are performed first (X=AB+C: AB is performed first), unless there are parentheses in the expression, in which case the operation inside the parentheses is to be performed first (X= (A+B) +C: A+B is performed first).
Circuits containing Inverters
Whenever an INVERTER is present in a logic-circuit diagram, its output expression is simply equal to the input expression with a prime (') over it.
Evaluating Logic Circuit Outputs
Once the Boolean expression for a circuit output has been obtained, the output logic level can be determined for any set of input levels.
These are two examples of the evaluating logic circuit output:
Let A=0, B=1, C=1, D=1
 X = A'BC (A+D)' = 0'*1*1* (0+1)' = 1 *1*1* (1)' = 1 *1*1* 0 = 0
Let A=0, B=0, C=1, D=1, E=1
 X = [D+ ((A+B)C)'] * E = [1 + ((0+0)1 )'] * 1 = [1 + (0*1)'] * 1 = [1+ 0'] *1 = [1+ 1 ] * 1 = 1

In general, the following rules must always be followed when evaluating a Boolean expression:

1. First, perform all inversions of single terms; that is, 0 = 1 or 1 = 0.
2. Then perform all operations within parentheses.
3. Perform an AND operation before an OR operation unless parentheses indicate otherwise.
4. If an expression has a bar over it, perform the operations of the expression first and then invert the result.
Determining Output Level from a Diagram
The output logic level for given input levels can also be determined directly from the circuit diagram without using the Boolean expression.
Implementing Circuits from Boolean Expression
If the operation of a circuit is defined by a Boolean expression, a logic-circuit diagram can he implemented directly from that expression.
Suppose that we wanted to construct a circuit whose output is y = AC+BC' + A'BC. This Boolean expression contains three terms (AC, BC', A'BC), which are ORed together. This tells us that a three-input OR gate is required with inputs that are equal to AC, BC', and A'BC, respectively.
Each OR-gate input is an AND product term, which means that an AND gate with appropriate inputs can be used to generate each of these terms. Note the use of INVERTERs to produce the A' and C' terms required in the expression.

Boolean Theorems
Investigating the various Boolean theorems (rules) can help us to simplify logic expressions and logic circuits.

Multivariable Theorems
The theorems presented below involve more than one variable:
 (9) x + y = y + x (commutative law) (10) x * y = y * x (commutative law) (11) x+ (y+z) = (x+y) +z = x+y+z (associative law) (12) x (yz) = (xy) z = xyz (associative law) (13a) x (y+z) = xy + xz (13b) (w+x)(y+z) = wy + xy + wz + xz (14) x + xy = x [proof see below] (15) x + x'y = x + y

Proof of (14)
 x + xy = x (1+y) = x * 1 [using theorem (6)] = x [using theorem (2)]
DeMorgan's Theorem
DeMorgan's theorems are extremely useful in simplifying expressions in which a product or sum of variables is inverted. The two theorems are:
(16) (x+y)' = x' * y'
Theorem (16) says that when the OR sum of two variables is inverted, this is the same as inverting each variable individually and then ANDing these inverted variables.
(17) (x*y)' = x' + y'
Theorem (17) says that when the AND product of two variables is inverted, this is the same as inverting each variable individually and then ORing them.
Example
 X = [(A'+C) * (B+D')]' = (A'+C)' + (B+D')' [by theorem (17)] = (A''*C') + (B'+D'') [by theorem (16)] = AC' + B'D
Three Variables DeMorgan's Theorem
(18) (x+y+z)' = x' * y' * z'
(19) (xyz)' = x' + y' + z'

Universality of NAND & NOR Gates
It is possible to implement any logic expression using only NAND gates and no other type of gate. This is because NAND gates, in the proper combination, can be used to perform each of the Boolean operations OR, AND, and INVERT.

In a similar manner, it can be shown that NOR gates can be arranged to implement any of the Boolean operations.
Alternate Logic Gate Representations
The left side of the illustration shows the standard symbol for each logic gate, and the right side shows the alternate symbol. The alternate symbol for each gate is obtained from the standard symbol by doing the following:
1. Invert each input and output of the standard symbol. This is done by adding     bubbles (small circles) on input and output lines that do not have bubbles, and by removing bubbles that are already there.
2. Change the operation symbol from AND to OR, or from OR to AND. (In the special case of the INVERTER, the operation symbol is not changed.)

Several points should be stressed regarding the logic symbol equivalences:
1. The equivalences are valid for gates with any number of inputs.
2. None of the standard symbols have bubbles on their inputs, and all the alternate symbols do.
3. The standard and alternate symbols for each gate represent the same physical circuit: there is no difference in the circuits represented by the two symbols.
4. NAND and NOR gates are inverting gates, and so both the standard and alternate symbols for each will have a bubble on either the input or the output. AND and OR gates are noninverting gates, and so the alternate symbols for each will have bubbles on both inputs and output.
Concept of Active Logic Levels:
When an input or output line on a logic circuit symbol has no bubble on it, that line is said to be active-HIGH. When an input or output line does have a bubble on it, that line is said to be active-LOW. The presence or absence of a bubble, then, determines the active-HIGH/active-LOW status of a circuit's inputs and output, and is used to interpret the circuit operation.
Boolean Function

A Boolean function is an algebraic expression consists of binary variables, the
constants 0 & 1, and the Boolean operators.For a set of given values of the variables,
the function is evaluated to either 0 or 1
e.g. f = x • y + x • z’
The Boolean function f has 3 binaryvariables x, y and z
The function is 1 if x and y are both 1 or if x is 1 and z is 0. Otherwise, f = 0

Operator Precedence

The operator precedence is:
1. Parentheses
2. NOT
3. AND
4. OR
e.g. f = x • y + x • z’
Precedence: z’, x • y, x • z’, x • y + x • z’

e.g. f = (a +b) • (c+d’)
Precedence: a+b, d’, c+d’, (a +b) • (c+d’)
The parentheses precedence is the same as in normal algebra

Boolean Function Truth Table

Boolean function can be represented by truth table as well.If the function has n variables, its truth table will have 2n rows
e.g. f = x • y + x • z’
f has 3 variables so 23 combinations
f is 1 when the expression is evaluated to 1 otherwise it is 0.
Minterm

In a Boolean function, a binary variable (x) may appear either in its normal form (x) or in its complement form (x’).Consider 2 binary variables x and y and an AND operation, there are 4 and only 4 possible combinations: x’•y’, x’•y, x•y’ & x•y
Each of the 4 product terms is called a MINTERM or STANDARD PRODUCT

By definition, a Minterm is a product which consists of all the variables in the normal form or the complement form but NOT BOTH.
e.g. for a function with 2 variables x and y:
x•y’ is a minterm but x’ is NOT a minterm
e.g. for a function with 3 variables x, y andz:
x’yz’ is a minterm but xy’ is NOT a minterm
Maxterm

Consider 2 binary variables x and y and an OR operation, there are 4 and only 4
possible combinations: x’+y’, x’+y, x+y’, x+y.Each of the 4 sum terms is called a
MAXTERM or STANDARD SUM.By definition, a Maxterm is a sum in which each variable appears once and only once either in its normal form or its complement
form but NOT BOTH.

Minterms and Maxterms for 3 Variables
Minterm Boolean Expression

Boolean functions can be expressed with minterms,
e.g.f1(x,y,z) = m1 + m4 + m6 = Î£m(1, 4, 6)
f2(x,y,z)  = m2 + m4 + m6+ m7
= Î£m(2, 4, 6, 7)

Maxterm Boolean Expression

Boolean functions can also be expressed with maxterms,
e.g.f1’ = x’y’z’+x’yz’+x’yz+xy’z+xyz
f1  = (x’y’z’+x’yz’+x’yz+xy’z+xyz)’
= (x+y+z)(x+y’+z)(x+y’+z’)(x’+y+z’)(x’+y’+z’)
= M0•M2•M3•M5•M7
= Î  M(0, 2, 3, 5, 7)
f2  = M0•M1•M3•M5
= Î  M(0, 1, 3, 5)

Literal

A Literal is a variable in a product or sum term
xy’ is a 2-literal product term
x’yz has 3 literals
x’ + xy’ + x’yz is an expression of sum of products with 3 product terms.The 3 product terms have 1, 2 and 3 literals respectively
x’(x+y’)(x’+y+z) is an expression of product of sums.The 3 sum terms have 1, 2 and 3 literals

Express Boolean Functions in Minterms

If product terms in a Boolean function are not minterms, they can be converted to minterms
e.g. f(a,b,c) = a’ + bc’ + ab’c
Function f has 3 variables, therefore, each minterm must have 3 literals
Neither a’ nor bc’ are minterms.They can be converted to minterm.ab’c is a minterm

Conversion to Minterms

e.g. f(a,b,c) = a’ + bc’ + ab’c
To convert a’ to a minterm, the 2 variables (b, c) must be added, without changing its
functionality .Since a’=a’•1 & 1 = b+b’, a’= a’(b + b’) = a’b + a’b’
Similarly, a’b = a’b(c + c’) = a’bc  +  a’bc’ and  a’b’ = a’b’(c+c’) = a’b’c + a’b’c’
bc’ = bc’(a+a’) = abc’ + a’bc’
f = a’bc+a’bc’+a’b’c+a’b’c’+abc’+a’bc’+ab’c

Express Boolean Functions in Maxterms

By using the Distribution Law: x+yz = (x+y)(x+z), a Boolean function can
be converted to an expression in product of maxterms
e.g. f(a,b,c) = a’+bc’
= (a’+b)(a’+c’) {not maxterms}
= (a’+b+cc’)(a’+c’+bb’) {cc’=0}
= (a’+b+c)(a’+b+c’)(a’+c’+b)(a’+c’+b’)
= (a’+b+c)(a’+b+c’)(a’+c’+b’)

Boolean Function Manipulation

Boolean functions can be manipulated with Boolean algebra.Manipulation can transform logic expressions, but still keep the same logic functionality.Manipulation can reduce the complexity, hence, easier to be implemented in hardware, i.e. fewer logic gates

Boolean Function Manipulation Example

f = xy’ + xyz + x’z
= x(y’ + yz) + x’z {common factor}
= x[(y’+y)(y’+z)] + x’z {Distribution law}
= x(y’+z) + x’z {y’ + y = 1}
= xy’ + xz + x’z {Distribution law}
= xy’ + (x + x’)z {common factor}
= xy’ + z {x + x’ = 1}
Simplify f1=abc+a’b+abc’ and f2=(a+b)’(a’+b’) to the minimum literals

f1 = abc+a’b+abc’ = ab(c+c’) + a’b = ab + a’b = (a+a’)b = b
f2 =(a+b)’(a’+b’) = a’b’(a’+b’) {DeMorgan}
= a’b’a’+a’b’b’
= a’b’ + a’b’ = a’b’

QUINE-McCLUSKEY MINIMIZATION
Quine-McCluskey minimization method uses the same theorem to produce the solution as the K-map method, namely X(Y+Y')=X
Minimization Technique
• The expression is represented in the canonical SOP form if not already in that form.
• The function is converted into numeric notation.
• The numbers are converted into binary form.
• The minterms are arranged in a column divided into groups.
• Begin with the minimization procedure.
• Each minterm of one group is compared with each minterm in the group immediately below.
• Each time a number is found in one group which is the same as a number in the group below except for one digit, the numbers pair is ticked and a new composite is created.
• This composite number has the same number of digits as the numbers in the pair except the digit different which is replaced by an "x".
• The above procedure is repeated on the second column to generate a third column.
• The next step is to identify the essential prime implicants, which can be done using a prime implicant chart.
• Where a prime implicant covers a minterm, the intersection of the corresponding row and column is marked with a cross.
• Those columns with only one cross identify the essential prime implicants. -> These prime implicants must be in the final answer.
• The single crosses on a column are circled and all the crosses on the same row are also circled, indicating that these crosses are covered by the prime implicants selected.
• Once one cross on a column is circled, all the crosses on that column can be circled since the minterm is now covered.
• If any non-essential prime implicant has all its crosses circled, the prime implicant is redundant and need not be considered further.
• Next, a selection must be made from the remaining nonessential prime implicants, by considering how the non-circled crosses can be covered best.
• One generally would take those prime implicants which cover the greatest number of crosses on their row.
• If all the crosses in one row also occur on another row which includes further crosses, then the latter is said to dominate the former and can be selected.
• The dominated prime implicant can then be deleted.
Example
Find the minimal sum of products for the Boolean expression,
f=(1,2,3,7,8,9,10,11,14,15), using Quine-McCluskey method.
Firstly these minterms are represented in the binary form as shown in the table below. The above binary representations are grouped into a number of sections in terms of the number of 1's as shown in the table below.
Binary representation of minterms
 Minterms U V W X 1 0 0 0 1 2 0 0 1 0 3 0 0 1 1 7 0 1 1 1 8 1 0 0 0 9 1 0 0 1 10 1 0 1 0 11 1 0 1 1 14 1 1 1 0 15 1 1 1 1
Group of minterms for different number of 1's
 No of 1's Minterms U V W X 1 1 0 0 0 1 1 2 0 0 1 0 1 8 1 0 0 0 2 3 0 0 1 1 2 9 1 0 0 1 2 10 1 0 1 0 3 7 0 1 1 1 3 11 1 0 1 1 3 14 1 1 1 0 4 15 1 1 1 1
Any two numbers in these groups which differ from each other by only one variable can be chosen and combined, to get 2-cell combination, as shown in the table below.
2-Cell combinations
 Combinations U V W X (1,3) 0 0 - 1 (1,9) - 0 0 1 (2,3) 0 0 1 - (2,10) - 0 1 0 (8,9) 1 0 0 - (8,10) 1 0 - 0 (3,7) 0 - 1 1 (3,11) - 0 1 1 (9,11) 1 0 - 1 (10,11) 1 0 1 - (10,14) 1 - 1 0 (7,15) - 1 1 1 (11,15) 1 - 1 1 (14,15) 1 1 1 -
From the 2-cell combinations, one variable and dash in the same position can be combined to form 4-cell combinations as shown in the figure below.
 Combinations U V W X (1,3,9,11) - 0 - 1 (2,3,10,11) - 0 1 - (8,9,10,11) 1 0 - - (3,7,11,15) - - 1 1 (10,11,14,15) 1 - 1 -
The cells (1,3) and (9,11) form the same 4-cell combination as the cells (1,9) and (3,11). The order in which the cells are placed in a combination does not have any effect. Thus the (1,3,9,11) combination could be written as (1,9,3,11).
From above 4-cell combination table, the prime implicants table can be plotted as shown in table below.
Prime Implicants Table
 Prime Implicants 1 2 3 7 8 9 10 11 14 15 (1,3,9,11) X - X - - X - X - - (2,3,10,11) - X X - - - X X - - (8,9,10,11) - - - - X X X X - - (3,7,11,15) - - - - - - X X X X - X X - X X - - - X -
The columns having only one cross mark correspond to essential prime implicants. A yellow cross is used against every essential prime implicant. The prime implicants sum gives the function in its minimal SOP form.
Y = V'X + V'W + UV' + WX + UW
Logic
         Combinational logic blocks have the outputs depending on the combinations of the current inputs. Sequential logic blocks have the outputs depending on the current inputs as well as any previous inputs.

Binary Adder
         Binary Adder is for binary number addition
                     Logic Circuit to be discussed:
§  Half Adder
§  Full Adder
§  Ripple Adder
§  Carry Look Ahead Adder
§
Half Adder

• Half adder is for addition of 2 single bits
• It has two 1-bit inputs and two 1-bit outputs
• The inputs are the 2 bits to be added (a, b)
• The outputs are 1-bit sum (s) & 1-bit carry (c)

The logic is:

Binary Addition

The half adder adds 2 single-bit inputs
It cannot complete a full addition

To complete a full addition, the adder needs to take in 3 inputs: a, b and the carry from the previous bit.

Full Adder

To carry the addition, an adder with 3 inputs is required. A Full Adder takes in 3 inputs (a, b and ci) and  produces 2 outputs (s, co) a & b are the 2 bits to be added, ci is the carry input (carry over from the  previous bit) and co is the carry output (to the next bit)

Logic for Full Adder
Logic equations derived from the truth table:

s = a Ã… b Ã… ci
co = ab + bci + aci

Full Adder
The below implementation shows implementing the full adder with AND-OR gates, instead of using XOR gates. The basis of the circuit below is from the above Kmap.
Circuit-SUM
Circuit-CARRY

Full adder can be built from 2 half adders
s = a Ã… b Ã… ci
co = ab+bci+aci
= ab+(a’bci+abci)+(abci+ab’ci)
= ab + abci + ci (a’b+ab’) = ab + ci (a Ã… b)

n-bit Ripple Adder
To perform an addition of 2 n-bit numbers An-1…A1A0 & Bn-1…B1B0, where An-1 & Bn-1 are theMSB & A0B0 are the LSB, we need a n-bit adder,which can be built from ‘n ‘fulladders

Ripple Adder: Carry ripples through the chain
Carry-Look-Ahead Adder
The delay generated by an N-bit adder is proportional to the length N of the two numbers X and Y that are added because the carry signals have to propagate from one full-adder to the next. For large values of N, the delay becomes unacceptably large so that a special solution needs to be adopted to accelerate the calculation of the carry bits. This solution involves a "look-ahead carry generator" which is a block that simultaneously calculates all the carry bits involved. Once these bits are available to the rest of the circuit, each individual three-bit addition (Xi+Yi+carry-ini) is implemented by a simple 3-input XOR gate. The design of the look-ahead carry generator involves two Boolean functions named Generate and Propagate. For each input bits pair these functions are defined as:
Gi = Xi . Yi
Pi = Xi + Yi
The carry bit c-out(i) generated when adding two bits Xi and Yi is '1' if the corresponding function Gi is '1' or if the c-out(i-1)='1' and the function Pi = '1' simultaneously. In the first case, the carry bit is activated by the local conditions (the values of Xi and Yi). In the second, the carry bit is received from the less significant elementary addition and is propagated further to the more significant elementary addition. Therefore, the carry_out bit corresponding to a pair of bits Xi and Yi is calculated according to the equation:
carry_out(i) = Gi + Pi.carry_in(i-1)
For a four-bit adder the carry-outs are calculated as follows
carry_out0 = G0 + P0 . carry_in0
carry_out1 = G1 + P1 . carry_out0 = G1 + P1G0 + P1P0 . carry_in0
carry_out2 = G2 + P2G1 + P2P1G0 + P2P1P0 . carry_in0
carry_out3 = G3 + P3G2 + P3P2G1 + P3P2P1G0 + P3P2P1 . carry_in0
The set of equations above are implemented by the circuit below and a complete adder with a look-ahead carry generator is next. The input signals need to propagate through a maximum of 4 logic gate in such an adder as opposed to 8 and 12 logic gates in its counterparts illustrated earlier.

Pi is called Carry Propagate

Gi is called Carry Generate

With Pi and Gi, we obtain the sum & carry for the full adder:

Ci+1= Gi + PiCi
C1  = G0 + P0C0
C2  = G1 + P1C1
= G1 + P1(G0 + P0C0)
= G1 + P1G0 + P1P0C0
C3   = G2 + P2C2
= G2 + P2(G1 + P1G0 + P1P0C0)
= G2 + P2G1 + P2P1G0 + P2P1P0C0
Carry no longer depend on its previous stage

Look-Ahead Carry Generator

Speed: 2 gate delays for all carry
Cost: more gates

Sums can be calculated from the following equations, where carryout is taken from the carry calculated in the above circuit.
sum_out0 = X 0 Y0 carry_out0
sum_out1 = X 1 Y1 carry_out1
sum_out2 = X 2 Y2 carry_out2
sum_out3 = X 3 Y3 carry_out3

MSI Adder
Adders are available in Medium Scale Integration (MSI) devices
Both TTL and CMOS are available, e.g.
74183: TTL 1-bit Full Adder
7482: TTL 4-bit Carry-Look-Ahead Adder
4008: CMOS 4-bit Carry-Look-Ahead Adder
74182: 4-bit Look-Ahead Carry Generator
4-bit Addition

To add 2  4-bit numbers: Z = X + Y

8-bit Addition

To add 2  8-bit numbers: Z = X + Y

Subtractor
Subtractor circuits take two binary numbers as input and subtract one binary number input from the other binary number input. Similar to adders, it gives out two outputs, difference and borrow (carry-in the case of Adder). There are two types of subtractors.
• Half Subtractor
• Full Subtractor

Half Subtractor

The half-subtractor is a combinational circuit which is used to perform subtraction of two bits. It has two inputs, X (minuend) and Y (subtrahend) and two outputs D (difference) and B (borrow). The logic symbol and truth table are shown below.
Symbol

Truth Table
 X Y D B 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0

From the above table we can draw the Kmap as shown below for "difference" and "          borrow". The boolean expression for the difference and Borrow can be written.
From the equation we can draw the half-subtractor as shown in the figure below.

Full Subtractor

A full subtractor is a combinational circuit that performs subtraction involving three bits, namely minuend, subtrahend, and borrow-in. The logic symbol and truth table are shown below.

Symbol

Truth Table

 X Y Bin D Bout 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1

From above table we can draw the Kmap as shown below for "difference" and "borrow".

The boolean expression for difference and borrow can be written as
D = X'Y'Bin + X'YBin' + XY'Bin' + XYBin
= (X'Y' + XY)Bin + (X'Y + XY')Bin'
= (X Y)'Bin + (X Y)Bin'
= X Y Bin
Bout = X'.Y + X'.Bin + Y.Bin
From the equation we can draw the full-subtractor as shown in figure below.
Full-subtractor circuit is more or less same as a full-adder with slight modification.

Parallel Binary Subtractor
Parallel binary subtractor can be implemented by cascading several full-subtractors. Implementation and associated problems are those of a parallel binary adder, seen before in parallel binary adder section.
Below is the block level representation of a 4-bit parallel binary subtractor, which subtracts 4-bit Y3Y2Y1Y0 from 4-bit X3X2X1X0. It has 4-bit difference output D3D2D1D0 with borrow output Bout.

A serial subtractor can be obtained by converting the serial adder using the 2's complement system. The subtrahend is stored in the Y register and must be 2's complemented before it is added to the minuend stored in the X register.
The circuit for a 4-bit serial subtractor using full-adder is shown in the figure below.

Comparator
Comparator compares binary numbers.

Logic comparing 2 bits: a and b

Magnitude Comparator
Comparator compares binary numbers
4-bit Magnitude Comparator:
Inputs: A3A2A1A0 & B3B2B1B0
Outputs: Y A>B, Y A, Y A=B
For each bit, let:
Si = AiBi + Ai’Bi’ = (AiBi’ + Ai’Bi)’
Si is true when Ai = Bi
For A = B, we must have:
A3=B3 and A2=B2 and A1=B1 and A0=B0
Hence, Y A=B = S3•S2•S1•S0 136
Logic For A > B
For A > B, there are 4 cases:
1.      A3B3 is 10 and A2A1A0 & B2B1B0 can be anything:
A=1xxx, B=0xxx
2.      A3=B3 and A2B2 is 10 and A1A0 & B1B0 can be
anything: A=11xx, B=10xx or A=01xx, B=00xx
3.      A3=B3 and A2=B2 and A1B1=10 and A0B0 is xx: e.g.
A=011x, B=010x
4.      A3=B3 and A2=B2 and A1=B1 and A0B0 is 10: e.g.
A=1011, B=1010

Y A>B=A3B3’+S3A2B2’+S3S2A1B1’+S3S2S1A0B0

Logic For A < B

For A < B, there are also 4 cases:
1)      A3B3 is 01 and A2A1A0 & B2B1B0 can be anything:
1.      A=0xxx, B=1xxx
2)      A3=B3 and A2B2 is 01 and A1A0 & B1B0 can be
1.      anything: A=10xx, B=11xx or A=00xx, B=01xx
3)      A3=B3 and A2=B2 and A1B1=01 and A0B0 is xx: e.g.
1.      A=110x, B=111x
4)      A3=B3 and A2=B2 and A1=B1 and A0B0 is 01: e.g.
1.      A=1000, B=1001

Y A=A3 ’B3+S3A2 ’B2+S3S2A1 ’ B1+S3S2S1A0 ’ B0

4-bit Comparator Logic Circuit

MSI: 7485 4-bit Magnitude Comparator

Comparison of 4-bit Numbers

Comparison of 8 - bit Numbers

Decoder
A decoder is a multiple-input, multiple-output logic circuit that converts coded inputs into coded outputs, where the input and output codes are different.Binary Decoder has n inputs and 2noutputs  also called as n-to-2n decoder.          Inputs have all the 2n combinations and the corresponding output will be activated for each input combinations.Decoding is necessary in applications such as data multiplexing, 7 segment display and memory address decoding. Enable inputs must be on for the decoder to function, otherwise its outputs assume a single "disabled" output code word. Figure below shows the pseudo block of a decoder.
A binary decoder has n inputs and 2n outputs. Only one output is active at any one time, corresponding to the input value. Figure below shows a representation of Binary n-to-2n decoder

                     e.g. 3-to-8 decoder has 3 inputs and 8 outputs

3-to-8 Decoder
Function Table

3-to-8 Decoder Logic Circuit

2-to-4 Decoder with Output Enable

Implement Logic Function with Decoder
• Any n-variable logic function, in canonical sum-of-minterms form can be implemented using a single n-to-2n decoder to generate the minterms, and an OR gate to form the sum.
• The output lines of the decoder corresponding to the minterms of the function are used as inputs to the or gate.
• Any combinational circuit with n inputs and m outputs can be implemented with an n-to-2n decoder with m OR gates.
Suitable when a circuit has many outputs, and each output function is expressed with few minterms.

(Ex) Full adder using decoder
S(x, y, z) = (1,2,4,7)
C(x, y, z) = (3,5,6,7)

Truth Table
 X Y Z C S 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1

From the truth table we know the values for which the sum (s) is active and also the carry (c) is active. Thus we have the equation as shown above and a circuit can be drawn as shown below from the equation derived.

Use a 3-to-8 decoder to implement:
f = x’y’z + xy’z + xyz
(m1   +   m5  + m7)

MSI Decoders

1. 2-to-4 Decoder
2. 3-to-8 Decoder
3. 4-to-16 Decoder
4. BCD-to-Decimal Decoder
5. BCD-to-Seven-Segment Decoder
e.g. Low Power Schottky TTL:
74LS138 3-to-8 Decoder where G1, G2A and G2B are enable pins

Logic Symbol

74LS138 3-to-8 Decoder

Implement Logic Function with74LS138

Use a 3-to-8 decoder to implement:
f = x’y’z + xy’z + xyz
(m1  +   m5  +  m7)

4-to-16 Decoder

Use 2 3-to-8 decoders
Inputs: D, C, B, A
Outputs: Y0 – Y15
When D = 0, top decoder is enabled
When D = 1,bottom decoderis enabled
En’ is enable

Binary Encoders

An encoder is a combinational circuit that performs the inverse operation of a decoder. If a device output code has fewer bits than the input code has, the device is usually called an encoder. e.g. 2n-to-n, priority encoders.
The simplest encoder is a 2n-to-n binary encoder, where it has only one of 2n inputs = 1 and the output is the n-bit binary number corresponding to the active input. It can be built from OR gates

         e.g. 4-to-2 Encoder

Octal-to-Binary Encoder
Octal-to-Binary take 8 inputs and provides 3 outputs, thus doing the opposite of what the 3-to-8 decoder does. At any one time, only one input line has a value of 1. The figure below shows the truth table of an Octal-to-binary encoder.
Truth Table
 I0 I1 I2 I3 I4 I5 I6 I7 Y2 Y1 Y0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 1 1 1 1

For an 8-to-3 binary encoder with inputs I0-I7 the logic expressions of the outputs Y0-Y2 are:
Y0 = I1 + I3 + I5 + I7
Y1= I2 + I3 + I6 + I7
Y2 = I4 + I5 + I6 +I7
Based on the above equations, we can draw the circuit as shown below

Priority Encoder

If more then two inputs are active simultaneously, the output is unpredictable or rather it is not what we expect it to be.This ambiguity is resolved if priority is established so that only one input is encoded, no matter how many inputs are active at a given point of time. The priority encoder includes a priority function. The operation of the priority encoder is such that if two or more inputs are active at the same time, the input having the highest priority will take precedence.
         e.g. 4-to-2 PriorityEncoder
                     A3 has the highest priority
                     A0 has the lower priority

74148 8-to-3 Priority Encoder

16-to-4 Priority Encoder

Cascade two 74148 8-to-3 priority encoders. The Input 15 has highest priority

Multiplexer

A multiplexer (MUX) is a digital switch which connects data from one of n sources to the output. A number of select inputs determine which data source is connected to the output. The block diagram of MUX with n data sources of b bits wide and s bits wide select line is shown in below figure.

MUX acts like a digitally controlled multi-position switch where the binary code applied to the select inputs controls the input source that will be switched on to the output as shown in the figure below. At any given point of time only one input gets selected and is connected to output, based on the select input signal.

The operation of a multiplexer can be better explained using a mechanical switch as shown in the figure below. This rotary switch can touch any of the inputs, which is connected to the output. As you can see at any given point of time only one input gets transferred to output.

2x1 MUX

A 2 to 1 line multiplexer is shown in figure below, each 2 input lines A to B is applied to one input of an AND gate. Selection lines S are decoded to select a particular AND gate. The truth table for the 2:1 mux is given in the table below.

Design of a 2:1 Mux
To derive the gate level implementation of 2:1 mux we need to have truth table as s           hown in figure. And once we have the truth table, we can draw the K-map as shown in figure for all the cases when Y is equal to '1'.
Combining the two 1' as shown in figure, we can drive the output y as shown below
Y = A.S’ + B.S
Truth Table
 B A S Y 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 1 1
Kmap
Circuit

MSI MUX
74150: 16-to-1
74153: Dual 4-to-1
74157: Quad 2-to-1
74151: 8-to-1

16-to-1 MUX

Use two 74151
D = 0 enables top MUX
D = 1 enables bottom MUX
W = Y’
= (Y1+Y2)’
= (W1’+W2’)’
= W1W2

Larger Multiplexers
Larger multiplexers can be constructed from smaller ones. An 8-to-1 multiplexer can be constructed from smaller multiplexers as shown below.
8-to-1 multiplexer from Smaller MUX

16-to-1 multiplexer from 4:1 mux

Quadruple 2-to-1 MUX

It is 2-to-1 MUX with 4 bits for each input
There is 1 output of 4 bits
There is 1 select signal
When 1 input is selected, the whole group of 4 bits goes to the output

Quad 2-to-1 MUX

Implementing Functions Multiplexers
Any n-variable logic function can be implemented using a smaller 2n-1-to-1 multiplexer and a single inverter (e.g 4-to-1 mux to implement 3 variable functions) as follows.
Express function in canonical sum-of-minterms form. Choose n-1 variables as inputs to mux select lines. Construct the truth table for the function, but grouping inputs by selection line values (i.e select lines as most significant inputs).
Determine multiplexer input line i values by comparing the remaining input variable and the function F for the corresponding selection lines value i.
We have four possible mux input line i values:
·         Connect to 0 if the function is 0 for both values of remaining variable.
·         Connect to 1 if the function is 1 for both values of remaining variable.
·         Connect to remaining variable if function is equal to the remaining variable.
·         Connect to the inverted remaining variable if the function is equal to the remaining variable inverted
3-variable Function Using 8-to-1 mux
Implement the function F(X,Y,Z) = S(1,3,5,6) using an 8-to-1 mux. Connect the input variables X, Y, Z to mux select lines. Mux data input lines 1, 3, 5, 6 that correspond to the function minterms are connected to 1. The remaining mux data input lines 0, 2, 4, 7 are connected to 0.

3-variable Function Using 4-to-1 mux

Implement the function F(X,Y,Z) = S(0,1,3,6) using a single 4-to-1 mux and an inverter. We choose the two most significant inputs X, Y as mux select lines.

Truth Table

 Select i X Y Z F Mux Input i 0 0 0 0 1 1 0 0 0 1 1 1 1 0 1 0 0 Z 1 0 1 1 1 Z 2 1 0 0 0 0 2 1 0 1 0 0 3 1 1 0 1 Z' 3 1 1 1 0 Z'

We determine multiplexer input line i values by comparing the remaining    input variable Z and the function F for the corresponding selection lines value i
·         when XY=00 the function F is 1 (for both Z=0, Z=1) thus mux input0 = 1
·         when XY=01 the function F is Z thus mux input1 = Z
·         when XY=10 the function F is 0 (for both Z=0, Z=1) thus mux input2 = 0
·         when XY=11 the function F is Z' thus mux input3 = Z'

Example for logic function implementation using MUX

De-multiplexers
They are digital switches which connect data from one input source to one of n outputs.Usually implemented by using n-to-2n binary decoders where the decoder enable line is used for data input of the de-multiplexer.The figure below shows a de-multiplexer block diagram which has got s-bits-wide select input, one b-bits-wide data input and n b-bits-wide outputs.

The operation of a de-multiplexer can be better explained using a mechanical switch as shown in the figure below. This rotary switch can touch any of the outputs, which is connected to the input. As you can see at any given point of time only one output gets connected to input.

1-to-4 De-multiplexer
Truth Table

 S1 S0 F0 F1 F2 F3 0 0 D 0 0 0 0 1 0 D 0 0 1 0 0 0 D 0 1 1 0 0 0 D

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