Loading in 5 sec....

CENG 241 Digital Design 1 Lecture 7PowerPoint Presentation

CENG 241 Digital Design 1 Lecture 7

- 111 Views
- Uploaded on
- Presentation posted in: General

CENG 241 Digital Design 1 Lecture 7

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

CENG 241Digital Design 1Lecture 7

Amirali Baniasadi

amirali@ece.uvic.ca

4-bit by 3-bit Binary Multiplier

B3 B2 B1 B0

A2 A1 A0

A0B3 A0B2 A0B1 A0B0

A1B3 A1B2 A1B1 A1B0

A2B3 A2B2 A2B1 A2B0

- When dealing with decimal numbers BCD code is used.
- A decimal adders requires at least 9 inputs and 5 outputs.
- BCD adder: each input does not exceed 9, the output can not exceed 19
- How are decimal numbers presented in BCD?
- Decimal Binary BCD
- 9 1001 1001
- 19 10011 (0001)(1001)
- 1 9

- Decimal numbers should be represented in binary code number.
- Example: BCD adder
- Suppose we apply two BCD numbers to a binary adder then:
- The result will be in binary and ranges from 0 through 19.
- Binary sum: K(carry) Z8 Z4 Z2 Z1
- BCD sum : C(carry) S8 S4 S2 S1
- For numbers equal or less than 1001 binary and BCD are identical.
- For numbers more than 1001, we should add 6(0110) to binary to get BCD.
- example: 10011(binary) = 11001(BCD) =19
- ADD 6 to correct.

BCD adder

Numbers that need correction (add 6) are:

01010 (10)

01011 (11)

01100 (12)

01101 (13)

01110 (14)

01111 (15)

10000 (16)

10001 (17)

10010 (18)

10011 (19)

Decides to add 6?

Adds 6

BCD adder

Numbers that need correction (add 6) are:

K Z8 Z4 Z2 Z1

0 1 0 1 0 (10)

0 1 0 1 1 (11)

0 1 1 0 0 (12)

0 1 1 0 1 (13)

0 1 1 1 0 (14)

0 1 1 1 1 (15)

1 0 0 0 0 (16)

1 0 0 0 1 (17)

1 0 0 1 0 (18)

1 0 0 1 1 (19)

C = K + Z8Z4 +Z8Z2

- Compares two numbers, determines their relative magnitude.
- We look at a 4-bit magnitude comparator;
- A=A3A2A1A0, B=B3B2B1B0
- Two numbers are equal if all bits are equal.
- A=B if A3=B3 AND A2=B2 AND A1=B1 AND A0=B0
- Xi= AiBi + Ai’Bi’ ; Ai=Bi Xi=1 (remember exclusive NOR?)

- How do we know if A>B?
- 1.Compare bits starting from the most significant pair of digits
- 2.If the two are equal, compare the next lower significant bits
- 3.Continue until a pair of unequal digits are reached
- 4.Once the unequal digits are reached, A>B if Ai=1 and Bi=0, A<B if Ai=0 and Bi = 1
- A>B = A3B3’+X3A2B2’+X3X2A1B1’+X3X2X1A0B0’
- A<B = A3’B3+X3A2’B2+X3X2A1’B1+X3X2X1A0’B0
- Xi=1 if Ai=Bi

Magnitude Comparators

A3=B3 ?

X3A2’B2

- A decoder converts binary information from n input lines to a maximum of 2n output lines
- Also known as n-to-m line decoders where m< 2n
- Example 3-to-8 decoders.

- X Y Z D0 D1 D2 D3 D4 D5 D6 D7
- 0 0 0 1 0 0 0 0 0 0 0
- 0 0 1 0 1 0 0 0 0 0 0
- 0 1 0 0 0 1 0 0 0 0 0
- 0 1 1 0 0 0 1 0 0 0 0
- 1 0 0 0 0 0 0 1 0 0 0
- 1 0 1 0 0 0 0 0 1 0 0
- 1 1 0 0 0 0 0 0 0 1 0
- 1 1 1 0 0 0 0 0 0 0 1

Decoders: AND implementation

2-to-4 Decoder: NAND implementation

Decoder is enabled when E=0

How to build bigger decoders?

We can combine two 3-to-8 decoders to build a 4-to-16 decoder.

Generates from

0000 to 0111

Generates from

1000 to 1111

- A decoder provides the 2n minterms of n input variables.
- Any function is can be expressed in sum of minterms.
- Use a decoder to make the minterms and an external OR gate to make the sum.
- Example: consider a full adder.
- S(x,y,z) = Σ(1,2,4,7)
- C(x,y,z) = Σ (3,5,6,7)

Combinational Logic implementation

- Encoders perform the inverse operation of a decoder:
- Encoders have 2n input lines and n output line.
- Output lines generate the binary code corresponding to the input value.

- OutputsInputs
- X Y Z D0 D1 D2 D3 D4 D5 D6 D7
- 0 0 0 1 0 0 0 0 0 0 0
- 0 0 1 0 1 0 0 0 0 0 0
- 0 1 0 0 0 1 0 0 0 0 0
- 0 1 1 0 0 0 1 0 0 0 0
- 1 0 0 0 0 0 0 1 0 0 0
- 1 0 1 0 0 0 0 0 1 0 0
- 1 1 0 0 0 0 0 0 0 1 0
- 1 1 1 0 0 0 0 0 0 0 1
- z=D1+D3+D5+D7 y=D2+D3+D6+D7 x=D4+D5+D6+D7

- Encoder limitations:
- If two inputs are active, the output is undefined.
- Solution: we need to take into account priority.
- What if all inputs are 0?
- Solution: we need a valid bit
- Input Output
- D0 D1 D2 D3 x y v
- 0 0 0 0 X X 0
- 1 0 0 0 0 0 1
- X 1 0 0 0 1 1
- X X 1 0 1 0 1
- X X X 1 1 1 1

Priority Encoders: Map

Priority Encoders: Circuit

- Multiplexer: selects one binary input from many selections
- example: 2-to-1 MUX

Directs 1 of the 4 inputs to the output

- Multiplexers can be combined with common selection inputs to support multi-bit selection logic

- General rules for implementing any Boolean function with n variables:
- Use a multiplexer with n-1 selection inputs and 2 n-1 data inputs
- List the truth tabel
- Apply the first n-1 variables to the selection inputs of multiplexer
- For each combination evaluate the output as a function of the last variable.
- The function can be 0, 1 the variable or the complement of the variable.

Implementing Boolean functions w/ MUX

- Reading up to page 156