Basics: Digital Logic. F(x,y) = x.y F(x,y) = x.y + x'.y

Basics:Digital Logic

Battery

A B

A AND B

Light Bulb

PushButtons

Battery

A B

A AND B

Light Bulb

PushButtons

OR

p

q

rOROR

p

q

r

AND

OR

NAND

NOR

XOR

NOT

AND

OR

NAND

NOR

XOR

NOT

x y

F(x,y)

F(x,y) = x.y

F(x,y) = x.y + x'.y'

x y

F(x,y)

x y

F(x,y)

x y

F(x,y)

x y

F(x,y)

G(x,y,z)=x'yz + x'yz' + xy'z' + x'y'z'

x y z

G(x,y,z)

x y z

G(x,y,z)

Canonical SOP Form

a b c

P(a,b,c)

a b c

P(a,b,c)

Canonical POS Form

(1) real logic gates need to be connected to a power source

(2) real logic gates do not respond instantly to changes in their input values

(3) the outputs of two logic gates cannot be connected directly to each other

(4) there is a limit to the number of inputs that can be connected to a logic gate

(5) integrated circuits stop working if you let out the magic smoke.

Real Logic Gates

Impedance - Impedance is a measure of the resistance of a component to alternating current (AC) signals, and is a function of the frequency of the AC signal for most components. A logic gate has a very high input impedance, which means that they only need a small current to set their voltage levels to high or low. Logic gates have a very low output impedance, which means that they can provide a relatively high current to drive a circuit high or low. Due to these differences in impedances, logic gate inputs can be connected to each other while logic gate outputs cannot.

Power - Power, P, is the rate at which energy is used (dissipated). Measured in Watts, the power used by an electronic component is related to the voltage drop, V, across the component, the resistance, R, of the component, and the current, I, in the component by,

P = VxI or P = V2/R or P = I2xR

Electrical Characteristics of Digial Logic

Power/Impedance Issues

Depth of Circuit

TTL Gate Delay

Fan-In Fan-Out

Common TTL Integrated Circuits

Typical Laboratory Setup

Implementing Order of Precedence in a Circuit

The Relationship Between Venn Diagrams and Karnaugh Maps

Examples of Simplifying Groups of Terms in K-Maps

Karnaugh Maps (K-Maps)

We will use a K-map to simplify the follwing expression.

F(x,y,z) = xy’z’ + x’y’z + xz’ + y’z + x’

First we note that there are three variables in the expression F( ), therefore our K-map will have 23=8 cells.

Next we decide where each variable will be represented.

x

yz

Now we arrange the literal values for x,y and z so that binary vectors that are next to each other differ by only one bit. In other words we make sure that logically adjacent vectors are also physically adjacent.

x

yz

0

1

00 01 11 10

Now we place ones (true values) into the K-map to show which binary vectors satisfiy the expression F( ).

F(x,y,z) = xy’z’ + x’y’z + xz’ + y’z + x’

1 1 1 1

1 1 1

x

yz

0

1

00 01 11 10100 001 1-0

100110

-01

001101

0--

000001010011

Now we find the smallest number of the largest rectangular patterns of 1’s whose sizes are integer powers of two. (Weird) That is the length of a selected rectangle of ones must be of length 1,2,4,. . .,2n. Of course in this case 4 is the limit. We must choose a set of these rectangles such that every 1 is in at least one of the rectangles. Here are a few candidates.

1 1 1 1

1 1 1

x

yz

0

1

00 01 11 10

1 1 1 1

1 1 1

x

yz

0

1

00 01 11 10

1 1 1 1

1 1 1

x

yz

0

1

00 01 11 10

In each case we have contained all the 1’s in three rectangles of the required dimensions. The middle case is better because it uses larger patterns.

Finally we convert these pattern back into terms in the simplified expression for F( ).

F(x,y,z) = x’ + y’ + z’

Comparing K-Map and Algebraic Simplification

There are Many Equivalent K-Maps Configurations

Examples of 4-Variable K-Maps

A K-Map "Tiles the Plane"

These terms are logically adjacent even though they appear on opposite corners of the K-Map

Max-Terms (Zeros) Can Be Used to Populate a K-Map

0

0

The Importance of "Don't Care" States

Sometimes the value of a Boolean function for certain bit patterns is not important. For example, when the input is a binary-coded decimal (BCD) the values for input greater than 9 do not matter.

In these cases we can use the don't care states either as 0's or 1's to simplify the circuit.

Note: For those of you who are unfamiliar with the conversions between SOP, POS, minterms and maxterms please work the following exercises as practice. If you have any questions please feel free to ask in class or contact me directly.

Sum-of-Products Product-of-Sums minterms Maxterms

ABC+AB’C+A’B’C’

(P+Q’+R)(P’+Q+R’)

f(x,y,z)=m(0,1,3,4,6)

g(a,b,c,d)=M(0,11,15)

For each row, use the representation of the Boolean expression given to derive the other representations. Some representations will be too large to fit conveniently in the table.

Sample Problems for Transforming Boolean Expressions to Other Equivalent Forms

Step 1: Express the Boolean function to be simplified in canonical SOP form.

Step 2: Using minterms or by direct conversion, generate the binary vector form for each product term.

Step 3: Arrange the binary vectors into groups according to the number of 1's in the vectors.

Step 4: Compare vectors in consecutive groups replacing each adjacent pair with their corresponding implicant. (Note: When comparing vectors with i 1's to vectors with i+1 1's, two vectors are adjacent when the positions of i 1's match. The implicant is generated by replacing the unmatched 1 with a dash.) Build a new list comprised of the newly generated implicants and any binary vectors not part of an adjacent pair.

Step 5: Repeat Steps 3 and 4 on the newly generated list until no new implicants are found. (Note: A pair of implicants containing i and i+1 1's are adjacent when the positions of all dashes match and the positions of i 1's match. As with binary vectors the new implicant is generated by replacing the unmatched 1 with a dash.) The final list is comprised of the prime implicants.

Step 6: Place list of prime implicants into a Prime Implicant table that compares them to the original list of minterms. Find the essential prime implicants and determine which minterms are covered by this subset of implicants. (Note: The essential prime implicants are those that are the only implicants to cover a particular minterm.)

Step 7: Find the smallest subset of remaining implicants needed to cover the remainder of the minterms of the original Boolean function. (Note: This step can be computationally expensive.)

Quine-McKluskey Method

Quine-McKluskey - An Example Problem

Finding Logically Adjacent Terms

Reduced Terms Must Represent all the Original Terms

Finding Logically Adjacent Reduced Terms

Listing the Prime Implicants

Minimal Set of Essential Prime Implicants