Virtual WalletDesign of BCD Binary Converters

To create a handheld device which will save money and time through budget assistance and improve the shopping experience.

• Gates Winkler • Jordan Samuel Fei• Yin Shen

• October 19 , 2009

StatusFinished• Flow Chart• Behavioral Verilog• Transistor Estimate • Floor Plan• Structure Proposal• Structural Verilog• Schematic

To Do • Layout• Testing

Introduction

• Binary – Representing numbers on a “2” scale. – Ex. 19 = 10011

• BCD – Representing numbers on a “10” scale

using a “2” scale. – Each set of 4 bits stores 1 base 10 digit.

• Called a nibble.

– Ex. 19 = 0001 1001 (2 nibbles)• 0001 = 1; 1001 = 9

When Converting

• It is important to remember that though both are represented with 0 and 1s they will have different lengths when converted.

• BCD bit length will always be a multiple of 4

Number Number of Bits

Binary BCD

0-1 1 4

16-20 5 8

100 7 12

10000 14 20

10e6 20 28

Binary To BCD Algorithm

• Assumes the necessary number of bits to store the nibbles are all initialized to 0.

• Ex. Binary 51 = 110011

• Needs 2 nibbles all set to 0 => 0000 0000

Step 1 – Check For >=5

• Check all nibbles to see if any are >= 5.– 0000 < 5– 0000 < 5

• If none are >= do nothing.

Step 2 – Shift Left

• Shift all bits left shifting the MSB of the binary number into the LSB of the BCD number. Anything can be shifted in.

• Initial: 0000 0000 : 110011

• After Shift: 0000 0001: 10011X

Step 3 – Repeat

• Repeat x times where x = length of binary number.

• 1 Run: 0000 0001: 10011X

• 2 Run: 0000 0011: 0011XX

• 3 Run: 0000 0110: 011XXX

• Wait… After 3 runs there is a nibble that that is >= 5.

Step 1b – Greater or Equal 5

• If any nibbles are >= 5 then add 3 (0011) to that nibble. No nibble can ever be >9 so overflow is impossible.

• 0000 < 5• 0110 >5 thus

– 0110 + 0011 = 1001

• After adding continue with step 2– Initial: 0000 0110: 011XXX– After Increasing Nibbles: 0000 1001: 011XXX– After Shift: 0001 0010: 11XXXX

Step 3 – Repeat (Again)

• Finish repeating.• 4 Run: 0001 0010: 11XXXX• 5 Run: 0010 0101: 1XXXXX• 6 Run After Adding: 0010 1000: 1XXXXX

– 6 After Shifting: 0101 0001: XXXXXX

• Final answer: 51• 0101 = 5• 0001 = 1

BCD To Binary Algorithm

• Practically identical but reversed. Here are the changes. • Step 1 Shift

– Shifts right instead of left– LSB of BCD is shifted into MSB of Binary– Must Shift in 0s

• Step 2 Greater than/equal to 8– Compares each BCD nibble to 8 instead.

• If less than do nothing.• If greater than/equal subtract 3 from the nibble.

• Step 3 Repeat (again)• Repeat x times where x = length of binary number.

Structural

• Binary To BCD– Needs 48 Flip Flops– 7 4-bit adders (28 1-bit adders)– 7 compare blocks.

• BCD To Binary– Needs 48 Flip Flops– 7 4-bit subtracters (28 1-bit subtracters)– 7 compare blocks.

Flip Flops – Output

• The Output Flip Flops can be regular Sequential Flip Flops.

Q

QSET

CLR

D

Q

QSET

CLR

D

Q

QSET

CLR

D

Q

QSET

CLR

D

clock

input…Q9 Q8 Q1 Q0

Flip Flops – Input

• The Input Flip Flops will be parallel Flip-Flops which allows us to chose whether we want to shift or write.

• The nand gates allow for the flip flops to be written to on low write while it acts as a shifter on high write.

Compare to 5

Assumes 4 bit input B[3:0]. Use a Truth table to figure out block configuration

Binary – BCD 1 Nibble

Compare to 8

• Easy, if Cin = 1 then the nibble will become 8 or greater.

• Don’t need a compare.

BCD – Binary 1 Nibble

Note

• In a BCD – Binary converter the BCD is the input. Thus the Flip Flops are actually the ones shown on slide 13.

• Binary Just Shifts without anything fancy. Just Needs respective Flip Flops.

Schematics Binary - BCD

BCD - Binary

Binary – BCD: Binary

Parallel Flip Flop

Flip Flop

Nand Block

Binary – BCD: BCD

Binary – BCD: 1 nibble

Compare to 5

BCD – Binary: Binary

BCD – Binary: BCD

BCD – Binary: BCD 1 nibble