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Switching Theory / Schakeltechniek 5A050

Lab session Sequential circuits

Highlights previous session

• From binary to decimal:

11010 = 1 x 2⁴+ 1 x 2³+ 0 x 2²+ 1 x 2¹+ 0 x 2⁰= 26

• From decimal to binary:

successive division

47 in binary ? 47 / 2 = 23 23 / 2 = 11 11 / 2 = 5 5 / 2 = 2 2 / 2 = 1 1/2 = 0

remainder 1

“ 1

“ 1

“ 1

“ 0“ 1

1 0 1 1 1 1 check: 32 + 8 + 4 + 2 + 1 = 47 !

Number systems • 2’s complement:

2’s c: 0100 dec: -4 2’s c: 1100 complement

1011 ^{add 1}

Highlights previous session

Arithmetic circuits: ripple carry adder

FA A3 B3

C S3

FA A2 B2

S2

FA A1 B1

S1

HA A0 B0

S0

A3 A2 A1 A0 + B3 B2 B1 B0 C S3 S2 S1 S0

Addition: A

B S

C

Circuits with memory

Memory:

behavior of a circuit is determined by the current inputs plus inputs from the past

0 10

20 40 30 50 60

70

• Lock only opens when a designated sequence of inputs is received

• Combinational circuits do not have memory

A simple memory element

0 / 1 1 / 0

• values are stable: it remembers a 0 or 1 value

• difficult to change a value

Idea:

• reset Q (Q becomes 0) if R is 1

• set Q (Q becomes 1) if S is 1

S R

Q

x 0 0 1 1

c + x 0 1 0 0 0 c

0 1 1

invert x

reset output c: control

The R-S latch

S R

Q

S

R Q

Q

Truth table:

R 0 1 0 1

Q Q 1 0 0 S

0 0 1 1

hold reset / set

Q Q 0

0 1

Gated R-S latch

Clock:

• independent periodic reference signal

• allows control over changes in memory values S

R Q

Q

clk Changes only possible if clk is high

But … ?

D flip-flop

clk

D Q Q copies the value on input D at each rising clock edge

→(positive/leading) edge triggered ff

clk D Q

states

Parity checker: state diagram

Design a circuit that provides a high output if and only if the number of 1s received on its input is odd

Idea: 2 states (even and odd)

even

odd

xyz state names [0]

[1]

[uv] output values 0

st state transition 1

0 1

initial state

Parity checker: transition table

even

odd [0]

[1]

0

1

0 1

Q even even odd odd

i 0 1 0 1

Q⁺ even odd odd

even Q: current state Q⁺: next state

Q even odd

o 0 1

next state: output:

Q even even odd odd

i 0 1 0 1

Q⁺ even odd odd even

o 0 0 1 1 transition table:

Q even odd

o 0 1 i=0

even odd

i=1 odd even Q⁺ state table:

alternative representations

Parity checker: truth table

even

odd [0]

[1]

0

1

0 1

Q even even odd odd

i 0 1 0 1

Q⁺ even odd odd even

o 0 0 1 1

transition table: choices:

• binary coding of states

• binary coding of in-/outputs

• flip-flop type(s) here:

• even = 0; odd = 1

• ok already

• 1 D flip-flop

i 0 1 0 1

D 0 1 1 0 truth table:

Q 0 0 1 1

o 0 0 1 1

Parity checker: implementation

even

odd [0]

[1]

0

1

0 1

i 0 1 0 1

D 0 1 1 0 truth table:

Q 0 0 1 1

o 0 0 1 1

clk

D Q

i o

expressions:

D = i ⊕Q o = Q

current state next state

Sequential circuits: design trajectory

• state diagram

• transition table

• binary coding of states and i/o, and selection of flip-flops

• truth table

• expressions and optimisation (K-maps, multi-level)

• implementation

Example: vending machine

design a controller for a machine that accepts 5c and 10c coins, and provides coffee if it has received (at least) 15c

[0]0c 00

10 01 [0]5c 00

10 [0]10c 00

10,01 [1]15c xx

01

inputs coded:

x y = 5c 10c 11 is invalid input

Vending machine: transition table

[0]0c 00

10 01 [0]5c 00

10 [0]10c 00

10,01 [1]15c xx

01

5c10c 0 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 x x

Q⁺ 0c 10c

5c x 5c 15c 10c x 10c 15c 15c x 15c Q

0c

5c

10c

15c

o 0 0 0 0 0 0 0 0 0 0 0 0 1

transition table contains exactly the same information as the state diagram

Vending machine: truth table

[0]0c 00

10 01 [0]5c 00

10 [0]10c 00

10,01 [1]15c xx

01

5c10c 0 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 x x

D0D1 0 0 1 0 0 1 x x 0 1 1 1 1 0 x x 1 0 1 1 1 1 x x 1 1 Q0Q1

0 0

0 1

1 0

1 1

o 0 0 0 0 0 0 0 0 0 0 0 0 1 state coding

0c = 00 5c = 01 10c = 10 15c = 11 2 D flip-flops

Vending machine: K-maps

5c

10c x

1 1 1 1

Q0

x 1

1 Q1

1

1 D0:

1

1 x

5c

10c x

1 1 1

1

Q0

x 1

1 Q1

1

1 D1:

1 x

Q0

1 Q1 o:

D0 = Q0 + 10c + 5cQ1

D1 = 10cQ0 + 5cQ1 + 5cQ0 + 5cQ1 o = Q0Q1

don’t cares !?

does it make a difference?

Vending machine: implementation

clk

D0 Q0

5c

clk

D1 Q1

o

5c 5c 10c

10c 5c

Moore machines

combinational logic for next state

combinational logic for outputs flip-flops