Contemporary Logic Design Multi-Level Logic © R.H. Katz Transparency No. 11-1 Chapter # 3:...

Contemporary Logic DesignMulti-Level Logic

© R.H. Katz Transparency No. 11-1

Chapter # 3: Multi-Level Combinational Logic3.3 and 3.4 -- Time Response and

Hazards / Glitches

Timing Issues

Chapter # 5: Arithmetic Circuits5.2 -- Networks for Binary Addition



Networks for Binary Addition1. Half Adder -- Review

With twos complement numbers, addition is sufficient

Ai 0 0 1 1

Bi 0 1 0 1

Sum 0 1 1 0

Carry 0 0 0 1

AiBi

0 1

0

1

0 1

1 0

Sum = Ai Bi + Ai Bi

= Ai + Bi

AiBi

0 1

0

1

0 0

10

Carry = Ai Bi

Half-adder Schematic

Carry

Sum A i

B i



Networks for Binary Addition

2. Full Adder -- Review

+

A3 B3

S3

+

A2 B2

S2

+

A1 B1

S1

+

A0 B0

S0C1C2C3

Cascaded Multi-bit Adder

usually interested in adding more than two bits

this motivates the need for the full adder




Full Adder

A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

CI 0 1 0 1 0 1 0 1

S 0 1 1 0 1 0 0 1

CO 0 0 0 1 0 1 1 1

A BCI

0

1

00 01 11 10

0

1

1

0

1

0

0

1

A BCI

0

1

00 01 11 10

0

0

0

1

0

1

1

1

S

CO

S = CI A B

CO = B CI + A CI + A B = CI (A + B) + A B




Full Adder/Half Adder

A

AA

B

BB CI

CIS CO

Alternative Implementation: 5 Gates -- cascading half adders

Half Adder

A

B

Half Adder

A + B

CI

A + B + CIS S

COCOCI (A + B)A B

S

CO

A B + CI (A xor B) = A B + B CI + A CI

Standard Approach: 6 Gates

+




3. Adder/Subtractor

A - B = A + (-B) = A + B + 1

A B

CO

S

+ CI

A B

CO

S

+ CI

A B

CO

S

+ CI

A B

CO

S

+ CI

0 1

Add/Subtract

A 3 B 3 B 3

0 1

A 2 B 2 B 2

0 1

A 1 B 1 B 1

0 1

A 0 B 0 B 0

Sel Sel Sel Sel

S 3 S 2 S 1 S 0

Overflow

Control = 0 ==> AddControl = 1 ==> Subtract



• For the adder circuit, Ai and Bi are summed in parallel. Bu for Si to be correct, the adder requires serial propagation of the carry outputs from the right most lowest-order stage to the left-most highest order stage. The rippling of the carry from one stage to another determines the adder’s ultimate delay.




Networks for Binary Addition4. Carry Lookahead Circuits

A

A

B

B

CI CO

@0@0

@0@0

@N

@1

@1

@N+1

@N+2

Critical delay: the propagation of carry from low to high order stages

latearrivingsignal

two gate delaysto compute CO

Assume the adder stages are implemented as the ‘S’ and ‘CO’ circuits onprevious slide and all gates have the same delay. For the ‘S’ circuit,

At time 0, Ai, Bi, and CI are presented to the adder. Within 2 gatedelays, one delay for each XOR gate, S0 will be valid

For ‘C0’ circuit,Ai=Bi=1, the carryout is computed in two gate delay and independentof the carry-in ==> C0 always = 1Ai=Bi=0, the carryout is computed in two gate delay and independentof the carry-in ==> C0 always = 0AiBi, the calculation of the carry out takes three gate delays (C0depends on the CI)




Carry Lookahead Circuits


4 stageadder

final sum andcarry

A 0

B 0

C 0

S 0 @2

A 1

B 1

C 1 @2

S 1 @3

A 2

B 2

C 2 @4

S 2 @5

A 3

B 3

C 3 @6

S 3 @7

C 4 @8

0

1

2

3

Addition for worst case:

1111 + 00012 2



Networks for Binary AdditionCarry Lookahead Circuits


1111 + 0001worst case

addition

T0: Inputs to the adder are valid

T2: Stage 0 carry out (C1)




T0 T2 T4 T6 T8

S0, C1 Valid S1, C2 Valid S2, C3 Valid S3, C4 Valid



• 8 gate delay seems acceptable for a 4-bit adder

• Cascaded delays become intolerable for adders of greater widths, such as 32 or 64 bits

• A 32 stage adder (32 bits) take 64 gate delays to final stage carryout

• Led to carry look ahead schemes

• Ways to calculate the carry inputs in parallel rather than in series. So that it will never take more than 2 gate delays to compute any of the carry outputs.

Networks for Binary AdditionCarry Lookahead Circuits




Carry Lookahead Logic -- General

Carry Generate Gi = Ai Bi must generate carry when A = B = 1

Carry Propagate Pi = Ai xor Bi carry in will equal carry out here. This[asses propagates the carry across thestage

Si = Ai xor Bi xor Ci = Pi xor Ci

Ci+1 = Ai Bi + Ai Ci + Bi Ci

= Ai Bi + Ci (Ai + Bi)

= Ai Bi + Ci (Ai xor Bi)

= Gi + Ci Pi

Sum and Carry can be reexpressed in terms of generate/propagate:

We introduce 2 new functions,




Carry Lookahead Logic

Reexpress the carry logic as follows:

C1 = G0 + P0 C0

C2 = G1 + P1 C1 = G1 + P1 G0 + P1 P0 C0

C3 = G2 + P2 C2 = G2 + P2 G1 + P2 P1 G0 + P2 P1 P0 C0

C4 = G3 + P3 C3 = G3 + P3 G2 + P3 P2 G1 + P3 P2 P1 G0 + P3 P2 P1 P0 C0

Each of the carry equations can be implemented in a two-level logic network

Variables are the adder inputs and carry in to stage 0 (C0)!




Carry Lookahead Implementation

Adder with Propagate and Generate Outputs

Pi @ 1 gate delay

Ci Si @ 2 gate delays

BiAi

Gi @ 1 gate delay




Carry Lookahead Implementation

Increasingly complex logic

C0C0

C0

C0P0P0

P0

P0

G0G0

G0

G0

C1

P1

P1

P1

P1

P1

P1 G1

G1

G1

C2P2

P2

P2

P2

P2

P2

G2

G2

C3

P3

P3

P3

P3

G3

C4

When Pi and Gi are available, the subsequent C1, C2, C3, C4 are computedafter two more gate delays (3 gate delays total).The sum bits can be computed in just one more gate delay.





5. Cascaded Carry Lookahead

Carry lookaheadlogic generatesindividual carries

sums computedmuch faster

A 0

B 0

C 0

S 0 @2

A 1

B 1

C 1 @3

S 1 @4

A 2

B 2

C 2 @3

S 2 @4

A 3

B 3

C 3 @3

S 3 @4

C 4 @3





Cascaded Carry Lookahead

4 bit adders with internal carry lookahead

second level carry lookahead unit, extends lookahead to 16 bits

4-bit Adder

4 4

4

A [15-12] B [15-12] C 12 C 16

S [15-12]

P G 4-bit Adder

4 4

4

A [1 1-8] B [1 1-8] C 8

S [1 1-8]

P G 4-bit Adder

4 4

4

A [7-4] B [7-4] C 4

S [7-4]

P G 4-bit Adder

4 4

4

A [3-0] B [3-0] C 0

S [3-0]

P G

Lookahead Carry Unit C 0

P 0 G 0 P 1 G 1 P 2 G 2 P 3 G 3 C 3 C 2 C 1

C 0

P 3-0 G 3-0

C 4

@3 @2

@0

@4

@4 @3 @2 @5

@7

@3 @2 @5

@8 @8

@3 @2

@5

@5 @3

@0

C 16



Networks for Binary Addition6. Carry Select Adder

Redundant hardware to make carry calculation go faster

compute the high order sums in parallel

one addition assumes carry in = 0

the other assumes carry in = 1

4-Bit Adder [3:0]

C 0 C 4

4-Bit Adder [7:4]

0 C 8

1 C 8

4 ¥ 2:1 Mux

0 1 0 1 0 1

4-Bit Adder [7:4]

C 8 S 7 S 6 S 5 S 4 S 3 S 2 S 1 S 0

0 1

C 4

Adder High

Adder Low

C8 is selectedfrom adder highwhen C4=1and Is theAND of the carries of adder lowand high when C4=0.



3.3 Time Response in Combinational Networks• emphasis on timing behavior of circuits

• waveforms to visualize what is happening

• simulation to create these waveforms

• momentary change of signals at the outputs: hazards can be useful— pulse shaping circuits can be a problem — glitches: incorrect circuit operation

Terms:

gate delay— time for change at input to cause change at output minimum delay vs. typical/nominal delay vs. maximum delay careful designers design for the worst case!

rise time— time for output to transition from low to high voltage

fall time— time for output to transition from high to low voltage



Time Response in Combinational NetworksGate Delays for Typical TTL Families

Various families of TTL exhibit tradeoffs between delay and power.Faster a component, the more power it consumes.

Delays depend on whether a gate is going from low to high or high to low.

TTLcomponent

Maximum Typical

tpHL(ns)

tbLH(ns)

tpHL(ns)

tpLH(ns)

7400 15 22 7 11

74H00 10 10 6.2 5.9

74L00 60 60 31 35

74LS00 15 15 10 0

74S00 5 4.5 3 3

74LS02 15 15 10 10

74LS86A 22 30 13 20



Time Response in Combinational NetworksPulse Shaping Circuit -- It generates a periodic sequence of pulses.

A' • A = 0

F is not always 0!

3 gate delays (corresponding to the 3 inverters)

D remains high forthree gate delays after

A changes from low to high

A changeat its inputcauses a shortduration pulse atthe output

Each gate has propagation delayof 10 time units



Time Response in Combinational Networks

Hazards/Glitches and How to Avoid Them

Glitch -- unwanted pulse at the output of a combinational logic network.

Hazard -- a circuit with the potential for a glitch is said to have a hazard.Occur because delay paths through the circuit experience different propagation delays

Danger if logic "makes a decision" while output is unstable OR hazard output controls an asynchronous input (these respond immediately to changes rather than waiting for a synchronizing signal called a clock)

Usual solutions: wait until signals are stable (by using a clock)

never, never, never use circuits with asynchronous inputs

design hazard-free circuits

Suggest that first two approaches be used, but we'll tell you about hazard-free design since some components unavoidably have asynchronous inputs.



Time Response in Combinational NetworksHazards/Glitches and How to Avoid Them

Kinds of Hazards

Input change causes output to go from 1 to 0 to 1when it should have remained at 1.

Input change causes output to go from 0 to 1 to 0when it should have remained at 0.

Output signal has the potential to change more thanonce when it is expected to make a single transitionfrom 0 to 1 OR from 1 to 0 .

Static 0-hazard

Dynamic hazards

Static 1-hazard

1 1

0

1

0 0

1

0 0

1

1 1

0 0



Time Response in Combinational CircuitsGlitch Example

F = A' D + A C'

input change within product term-- G1

G1

G2

G3

A

\C

\A

D

F

G1

G2

G3

A\C

\AD

F

1

1

1

1

1

0

0

1

1

1

1

0

0

0

ABCD = 1100 ABCD = 1101

A AB

00 01 11 10

0 0 1 1

1 1 1 1

1 1 0 0

0 0 0 0

00

01

11

10 C

CD

D

B

Lets look at what happens when input changes from 1100 -> 1101



G1

G2

G3

A\C

\AD

F

G1

G2

G3

A\C

\AD

F

0

1

0

0

1

0

0

1

1

1

1

1

0

0

ABCD = 1101 ABCD = 0101 (A is still 0)

G1

G2

G3

A\C

\AD

F

0

1

0

1

1

1

1

ABCD = 0101 (A is 1)

Time Response in Combinational Circuits

Glitch Example

input change that spans product termsoutput changes from 1 to 0 to 1

Examine K-Map:

When the intial and final inputs are covered by the sameprime implicant, no glitch is possible.

When the input change spans prime implicants, a glitch can happen.

gate delay

F = A glitch happened



Time Response in Combinational NetworksGlitch Example -- Eliminate static hazards

General Strategy: add redundant prime implicant terms

F = A' D + A C' becomes A' D + A C' + C' D

This eliminates 1-hazard. How about 0-hazard?

Re-express F in PoS form:

F = (A' + C')(A + D)

Glitch present!

Add term: (C' + D)

This expression is equivalentto the hazard-free SoP form of F

A AB

00 01 11 10

0 0 1 1

1 1 1 1

1 1 0 0

0 0 0 0

00

01

11

10 C

CD

D

B



Time Response in Combinational NetworksGlitch Example -- Eliminate static hazards

Start with expression that is free of static 1-hazards

F = A C' + A' D + C' D

F' = (A C' + A' D + C' D)'

= (A' + C) (A + D') (C + D')

= A C + A C D' + C D' + A' C D' + A' D'

= A C + C D' + A' D'

covers all the adjacent 0's in the K-map

free of static-1 and static-0 hazards!

F = (A’ + C’) (C’ + D) (A + D)

Work with complement:

Alternative approach to eliminate 0-hazards




General Strategy for Static Hazard Elimination

Static 1-Hazards

Examine the K-map to make sure that all adjacent elements of theon-set are covered by a prime implicant.

If not, add redundant prime implicants until all elements of the on-set are covered

Static 0-Hazards

Reexpress function in POS form

Verify that adjacent elements of the off-set are covered by a common prime implicant in the PoS form.

If necessary, add more prime implicants to cover any uncoveredadjacencies.



Time Response in Combinational NetworksDetecting Static Hazards in Multi-Level Circuits

Map the multi-level function into a two-level form called transient output function

variables and complements are treated as independent variables

cannot use X + X' = 1 or X • X' = 0 for simplifications

F = A B C + (A + D) (A' + C')

F1 = A B C + A A' + A C' + A' D + C' D

Example:

2-level form

ABCD: 1111 to 1110, covered by term ABC, so no 1-hazard present

some static hazards are present!

ABCD: 1110 to 1100, term ABC goes low while term AC' goes high

1111 to 01111111 to 1101

A AB

00 01 11 10

0 0 1 1

1 1 1 1

1 1 1 0

0 0 1 0

00

01

11

10 C

CD

D

B

Examine K-map for static 1-hazards



100

A B C D F F 2


Static 1-hazards

Solution:

Add redundant terms to insure all adjacent transitions are covered by terms

F2 = A C' + A' D + C' D + A B + B D

1's hazards in Fcorrected in F2

1111 to 0111, 1111 to 1101



Time Response in Combinational NetworksStatic 0-Hazards

Similar to previous case, but work with the complement of F

If terms of the transient output function cover all 0 transitions, then no 0-hazards are present

F = [A B C + (A + D) (A' + C')]'

= (A' + B' + C') (A' D' + A C)

= A' D' + A' B’ D' + A' C’ D' + A B' C

= A' D' + A B' C

0-hazard on transition from1010 to 0010

+ B' C D'

F3 = (A + D) (A' + B + C') (B + C' + D)

0-hazard free

equivalent to F2 on last slide

A AB

00 01 11 10

0 0 1 1

1 1 1 1

1 1 1 0

0 0 1 0

00

01

11

10 C

CD

D

B

Add redundantterm



100

A B C D F F 3

Time Response in Combinational NetworksStatic 0-Hazards

0-HazardCorrected in F3



Time Response in Combinational NetworksDesigning Networks for Hazard-free operation

Simply place transient output function in a form that guarantees that all adjacent ones are covered by a term

no term of the transient output function contains both a variable and its complement

F(A,B,C,D) = m(1,3,5,7,8,9,12,13,14,15)

F = A B + A' D + B D + A C' + C' D

= (A' + B + C') D + A (B + C')

(factored by distributive law, which does notintroduce hazards since it does not depend onthe complementarity laws for its validity)

A AB

00 01 11 10

0 0 1 1

1 1 1 1

1 1 1 0

0 0 1 0

00

01

11

10 C

CD

D

B



Time Response in Combinational NetworksDynamic Hazards -- happens because of multiple paths in the underlying

multilevel network, each with its own assymetric delayExample with Dynamic Hazard

Three different paths from B or B' to output

ABC = 000, F = 1 to ABC = 010, F = 0 different delays along the paths: G1 slow, G4 very slow

Handling dynamic hazards very complex. Beyond our scope

G1

G2

G3

G5

G4

\A B

\B

\B \C

F

A

0 1

1

1 0

1

0 1

1 0

1 0 1

1 0 0

1 0

1 0 1 0

Slow

V ery slow

fast

fast

fast

A multilevel network free of static hazards may still have dynamic hazards.So if a hazard-free network is needed, it is best to design a two-level statichazard-free network.



HW #11 -- Sections 5.2, 3.3, 3.4

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