flip flops summary

8/7/2019 flip flops summary

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ECS 465: Digital Systems

Lecture Notes # 7

(A) Introduction to Sequential Circuits

(B) Latches and Flip-Flops

(C) Counter Design

SHANTANU DUTT

Department of Electrical and Computer Engineering

University of Illinois, Chicago

Phone: (312) 355-1314: e-mail: [email protected]

URL: http://www.eecs.uic.edu/~dutt


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(A) Introduction to Sequential Circuits

Current o/p depends on the current i/p and past history of all i/ps

seen by the circuit.

Where the relevant past history should be representable by a finite

number of classes or states

No RedRed Light

State

Reset

Encode as state=0

Light = Red

O/P = 0

Light = Green

O/P = 1

Light = Not green

O/P = 0

Light = Not red

O/P = 0

Encode as

state = 1

State Transition Diagram

Design Problem: Output of the circuit is 1 only if it has seen a red

light in the past and currently light is green.


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Circuit-Level Model of a Sequential Circuit.

Combinational

Circuit

Memory

Unit

Z0

Zm-1

Going to

external

world

x0

xn-1

I/p from

external

point

Statebits of

seq. ckt.

Current

State

Next

State

yk-1

y0

yk-1

y0


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Components to store bits ( latches or flip flops )

1)

1

1

0

1

Problem: cant storenew data

Cascade of inverter

0

I/P 1/0 O/PA

A

LD

LD

LD

LD

Will conduct when

A=1, and open when

A=0

2)

(B) Latches and Flip-Flops


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Another storage element:

3) Cross coupled NOR gates ( R-S latch )

R=0

0

Q

1

S=1

0

Q

Q

R

Q

S

Q

(Set)

(Reset)

NOR gates ( R-S latch )

Property of a NOR gate

A=0

B

B

BBA !

When one I/P of NOR is 0, it acts like an inverter.

When one I/P is 1, then O/P=0.

Different I/P conditions for R-S latch:

i) R=S=0, current I/P is stored indefinitely

( becomes cascade of inverters)Hold


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ii) R=1, S=0, when we want to store a 0 in the R-S latch.

Q=0, 1!Q

iii) R=0, S=1, when we want to store a 1 in the latch.Q=1, 0!Q

iv) R=1, S=1; Forbidden inputs!

Both Q = 0, : Q and its complement have the same value !

Will play havoc in the rest of the logic circuit.

Transit to: R=0, S=0.

R=1, S=1, both Q and and 0.Q

0 0 O/P oscillates.

Q

Q

R=1p0

S=1p0

0

1

0p1

0p1

0pppp

pppp

Oscillates

between 1 and 0

when we transit from

R=S=1 to R=S=0.

0!Q


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Cross-coupled NORR

Q

S

Q

Cross-coupled NAND

R

S

Q

Q

Hold State R=S=0 Hold State R=S=1

Q

QR=1, S=1

R=0, S=0

Forbidden I/Ps

From R, S = 1, 1 transit to R=0, S=1

then Q, transit to 1, 0 ( correctly )

From R, S = 1, 1 transit to R=1, S=0

then Q, correctly transit to 0, 1

Q

Q

Two implementations for R-S latch:


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4) The D-Latch

R-SLatch

DS

RR1

S1

Q

enbQ

Clocked Latch

(level-sensitive clock latch)

see terminology defined

later.

D=1, S=1, R=0

Q=1,

D=0, S=0, R=1

Q=0,

enb=0 R1,S1=0 (hold state)

0!Q

1!Q

enb=1

enb=1


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5) The J-K Latch:

Proposed to get rid of the forbidden I/P problem of R-S

i) J=1, K=0: (a) Let Q=1, p R=0,S=0

p Hold state of R-S p Q=1,

1!Q 0!Q

0!Q

0!Q

ii) J=0, K=1 Q=0, using a similar analysis

iii) J=K=0 | Hold state

iv) J=K=1, suppose Q=1, =0 R=1, S=0 Q=0, =1

S=1, R=0 Q=1, =0

This type of toggling continues as long as J=K=1, and the latch

is enabled ( CLK=1 below )

Q

Q

Q

(b) Let Q=0, , R=0, S=1 p Q=1,

1!Q

R

R-S

Q

QS

RQ

1

Q 0p1 0p

1

1p0 1p0

CLK

K

J


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Latch classification with respect to response to control signal

Terminology:Note that the terminology below applies to all types of latches:R-S, D, J-K, T, etc., though the examples are given for the R-S latch.

i) Transparent Latch: O/P responds to latch I/Ps without any enable or clock signal.

R Q

S

Q

Clock:

Fixed frequency alternating 1 and 0 signal

ii) Clocked or Level-Sensitive Latch:

Q

Q

S

Clock

or

enb

R

O/P responds to I/Ps only when enb orclock is at a pre-determined level (high

or low In this example, it is High)

R

S

Q

Q

Symbol:

R

S

Q

Q

CLK(high enable)

Symbol: or R

S

Q

Q

CLK (low enable)


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iii) Edge-Triggered Flip-Flop (FF) or simply Flip-Flop

O/P will respond to I/Ps only at either:

(a) the positive or rising edge of the enb/clock signal (positive

edge-triggered FF), or

(b) the negative or falling edge of the enb/clock signal (negative

edge-triggered FF).

Clock:

O/P response

period for a

positive

edge-triggered

FF.

O/P response

period for a

negative

edge-triggeredFF

O/P response

period for a

HIGH-enable/clock

level-sensitive

latch

O/P resp. period for

a low-enable/clock

level sensitive latch

Symbol:

R

S

Q

Q

CLK

Symbol:

R

SQQ

CLK


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Setup Times and Hold Time of FFs and Latches

Assume, positive edge-triggered D-FFTHoldn relates to propagation delay

of another part of circuit.

D

CLK

TSetupn relates to propagation delays of

various gates in the FF.The high point ofthe CLK determines the positive edges arrival.

If negative edge-triggered

CLK

D

THoldTSetup

Negative edge arrival

If D-Latch is high-level sensitive: Tsetup and Thold have to be around the negative edge

of clock (more specifically, when the clock begins to go low), similar to negative edge-triggered.

If D-Latch is low-level sensitive: Tsetup and Thold have to be around the positive

edge of clock, similar to positive edge-triggered.


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Solutions to Race Condition Problem with Level Sensitive Latches

Solution 1: Master-Slave FF:

Master-Slave J-K is a solution to race-condition problem: Any change in Q,

during CLK=0 is not propagated to P, and hence back to Q, during the

same CLK=0. Any change to Q, will occur in next CLK=0 period.

J

K

Q

Q

(O/P responds when CLK goes

From 1 to 0)

J-K

M-S

Master-Slave J-K works similarto a J-K latch: E.g. Let

J=1, K=1, CLK=1

Q=1, =0 =1 , P=0

When CLK=0

Q=0, =1

Q

Q

Q

R-S

Latch

CLK

K

J

Q

R

S

R-SQm

mQ R

S

sQ

Qs

1

0

1

0P

P1

0

1

0

Master R-S is level sensitive. Slave R-S is level sensitive.

Q

Q

P Q

Q

P


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Solution 2: Edge-Triggered FF:

Q

Q

R

S

D

Clk=1

D

D

0

0

Holds D whenclock goes low

Holds when

clock goes low

D

Q =D

DQ !

R

S

D

Clk=0

D

D

D

Assume D=1

D=1=S

Q=1,

RD !! 0

0!Q

CLK

Q

Q

R

S

D

Clk=0

D

D

0

0

Q responds to

internal S signal;

responds to

internal R signal.

Q

When CLK is 1 D I/P is

internally sampled but

does not appear at the O/P.O/P is held (changing D does not

cause any change in internal

signals in the FF or in its output)

O/P appears (Q=D)

D

D

D

D


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Characteristic Equations of Latches/FFs

The next O/P Q+ defined in terms of the current O/P Q and the I/P.

(FF/Latch is the simplest possible sequential ckt.)

1) R-S Latch Truth Table:

S(t) R(t) Q(t) Q+ = Q( t+( )

0 0 0 0

0 0 1 10 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 x1 1 1 x

Values at time t

Hold

Reset

Set

Forbidden

Q(t)\SR 00 01 11 10

0 0 0 x 1

1 1 0 x 1

Q+= S+ Q

(Characteristic equation)R


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Similarly: Characteristic Equations of

2) J-K, Q

+

= Q + J.3) D-FF, Q+ = D

4) Toggle FF/Latch Q+ = T + Q

or T-FF / Latch

K Q

Q T

Whenever I/P T is high,the FF will toggle, i.e., Q+ = .

When T=0, Q+=Q.

Q

Of course, these characteristic equations come into play onlywhen the FF/Latch is enabled.

Q

QT

Symbol:


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Excitation Table

Reversed Truth Table

What the inputs to FFs should be for given outputtransitions (Q p Q+)

Q Q

+

R S J K T D0 0 x 0 0 x 0 0

0 1 0 1 1 x 1 1

1 0 1 0 x 1 1 0

1 1 0 x x 0 0 1


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Conversion between FFs

Example: J-K to D

D-FF

D

0 1

x x

0 1

0

1

Q

D

x x

1 0

DQ

D

O/P function = JJ=D

Function = KK= D

Map the D,Q input combination to a QpQ+ transition

and then map this to J-K excitation required.

Thus, when D=1, Q=0, Q+

=1

J,K = 1,xD=0, Q=0, Q+=0 J,K = 0,x

D=1, Q=1, Q+=1 J,K = x,0

D=0, Q=1, Q+=0 J,K = x,1.

This should

behave like a

D-FF.

J

K

QQ

D

CLK

Logic

J

K

Q

QD

CLK


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Example 2: D p J-K

Excitation Table for D

Q Q+ D

0 p 0 0

0 p 1 1

1 p 0 0

1 p 1 1

J K Q Q+

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 0

Q

Q

DJ

K

CLK

TT for J-K

JKQ

0

1

00 01 11 10

0 0 1 1

1 0 0 1

Function is

Q

Q

D

CLK

J

K

Q

Q

J-K FF/LatchQKQJD !

QJ

QK

Logic

should work like a J-K


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(C) Counter Design

A counter is a special case of an FSM that cycles through its states

on receiving triggering clock pluses. It does not have any external data I/Ps.

A

B

C

D

EReset

000

001

010

011

100

FFs

LogicCounter O/P

Next State

bits

nn

CLK

The states need to be encoded by binary bits.

No external I/Ps


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1

Synthesis (3-Bit Up Counter)

000 111

001 110

010 101

011 100

Reset

(a) State Transition

Diagram

C B A C+ B+ A+ TC TB TA

0 0 0 0 0 1 0 0 1

0 0 1 0 1 0 0 1 1

0 1 0 0 1 1 0 0 1

0 1 1 1 0 0 1 1 11 0 0 1 0 1 0 0 1

1 0 1 1 1 0 0 1 1

1 1 0 1 1 1 0 0 1

1 1 1 0 0 0 1 1 1

Input

Present State

Output

Next State

Toggle Flip-Flop

Inputs

(b) State Transition TableFF Excitation Table Revisited

Q Q+ R S J K T D

0 0 x 0 0 x 0 0

0 1 0 1 1 x 1 1

1 0 1 0 x 1 1 0

1 1 0 x x 0 0 1Excitation table for R-S, J-K, T, and D Flip-Flops

(What next statewill be given the

current state.)

State Transition Diagram and Table for a 3-bit Binary Up-Counter


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From excitation table for FF inputs, get K-map for the FF inputs.

1 1 1 1

1 1 1 1

00 01 11 10

0

1

CB

A

TA=1

0 0 0 0

0 1 1 0

00 01 11 10

0

1

A

CB

TC=AB

0 0 0 0

1 1 1 1

00 01 11 100

1

A

CB

TB=A

K-maps for Up-Counter Using ToggleFlip-Flops.

Obtain logic expr. for FF I/Ps (as functions of current state bits A,B, C, --- A=QA, B=QB, C=QC) and realize the counter


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Counters with More Complex Sequencing (Non-Consecutive Binary Outputs)

000 110

010 101

011

State Transition

Diagram

C B A C+ B+ A+

0 0 0 0 1 0

0 0 1 x x x

0 1 0 0 1 1

0 1 1 1 0 1

1 0 0 x x x

1 0 1 1 1 0

1 1 0 0 0 01 1 1 x x x

Present State Next State

State Transition Table

Implementation Using J-K FFs:

C B A C+ B+ A+ JC KC JB KB JA KA

0 0 0 0 1 0 0 x 1 x 0 x

0 0 1 x x x x x x x x x

0 1 0 0 1 1 0 x x 0 1 x

0 1 1 1 0 1 1 x x 1 x 0

1 0 0 x x x x x x x x x

1 0 1 1 1 0 x 0 1 x x 1

1 1 0 0 0 0 x 1 x 1 0 x

1 1 1 x x x x x x x x x

Present

State

Next

State

Remapped Next

State

State Transition Table and Remapped Next-State Functions

Q Q+ J K

0 0 0 x0 1 1 x

1 0 x 1

1 1 x 0

QKQJQ !

J-K Flip-Flop Excitation Table


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Next State Functions

CBJJ

AJ

A

B

C

!

!

!

1CK

CAK

AK

A

B

C

!

!

!

0 0 x x

x 1 x x

00 01 11 10

0

1

CB

A

JC

x x 1 x

x x x 0

CBA 00 01 11 10

0

1KC

1 x x x

x x x 1

CBA 00 01 11 10

0

1JB

x 0 1 x

x 1 x x

CBA 00 01 11 10

0

1KB

0 1 0 x

x x x x

00 01 11 10

0

1

CB

A

JA x x x x

x 0 x 1

00 01 11 10

0

1

CB

A

KA

Remapped K-Maps for J-K Implementation.


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Actual Implementation ( Using J-K)

J Q

CLK

K Q

J Q

CLK

K Q

J Q

CLK

K Q

+

Count

signal

AC

KB

B JA

C

A

JAB

KB

A

C

J-K Flip-Flop Implementation of3 Bit Counter.

A C B A

C

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