Combinational logic
circuits
Chapter 3
Combinational logic hazards
Circuits Parameters
Vcc,power supply voltage
Icc,power supply current
ICCH,ICCL
VOL,VOH,IOL,IOH
VIL,VIH,IIL,IIH
tPLH,tPHL
Fan-out
Boolean algebra does not account for
propagation delays through signal paths of
actually circuits.
The delay can cause glitches to occur,A glitch
is an unwanted signal,usually a short pulse
caused by the transient behavior of signal
path that have different delays
A hazard exists any time the potential for
glitches is present.
Glitches may not be a problem or they may
create havoc,It depends on the specific
application.
Hazard
Static hazard
A static-1 hazard exists when an output
variable should be a logic 1,but goes to 0
momentarily as a result of input variable
changing.
A static-0 hazard exists when an output
variable should be a logic 0,but goes to 1
momentarily as a result of input variable
changing.
Static hazard
P=XY’+YZ
x=1,z=1
P=y’+y=1
Assuming all the
gate have the
same delay time,
t Z
P
X
Y
XY’
ZY
G1
G2 G3
G4
Y’
Static hazard
Z
P
X
Y
XY’
ZY
G1
G2 G3
G4
Y
Y’
XY’
ZY
P
t0
t1
t2
t1
t3 glitch
At time t0,y went from 1 to 0
At time t1,y’ went from 0 to 1;
At time t1,zy went from 0 to 1;
At time t2,xy’ went from 0 to 1;
Between time t1 to t2,
neither inputs of G4 was a 1,
Accounting for the delay t through G4
the momentary 0 time is from t2(t1+t) to t3(t2+t).
Y’
Find the potential for static hazard
Boolean algebra
A input variable x and its complement x’ are both
presented in the function equation.
The equation would be simplified to (x+x’) or xx’ in
some conditions.
When x changes,the circuit correspond to the
equation would produce a static hazard because of
the different propagation delay
Solution
Add redundant terms.
Find the potential for static hazard
Boolean algebra
Solution
Add redundant terms.
P=(Z+Y’)(X+Y)
Z=X=0,P=Y’Y=0
Static hazard 0
P=(Z+Y’)(X+Y)(X+Z)
Find the potential for static hazard
Karnaugh map
There are all the EPI in the k-map when we map
the expression and they are tangent and not
overlapping.
Any signal input variable change that would
result in a change in produce-term coverage may
cause a static hazard
Solution
Generate new PI that bridge two adjacent EPI
Find the potential for static hazard
P(X,Y,Z)=∑m(3,4,5,7) =XY’+YZ
00 01 11 10
0
1
0
1
2
3
6
7
4
5
XY
Z
MSB
LSB
1
11 1
Two EPI
EPI1(4,5)
EPI2(3,7)
They are tangent.
Static hazard 1.
PI(7,5)
P=XY’+YZ+XZ
Find the potential for static hazard
P(A,B,C,D)=∏M(2,6,9,11,13,15)
Two EPI
EPI1(2,6)
EPI2(9,11,13,15)
They are tangent.
Static hazard 0?
00 01 11 10ABCD
00
01
11
10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
MSB=A ; LSB=D
0
0
0 0
0
0
No static hazard
A PI can be generated to bridge the two EPIs.
Dynamic hazard
The output change more when once
as a result of a signal input variable
change
0-1-0-1
1-0-1-0
HOMEWORK
HOMEWORK
Chapter2
21,24i,27b,29a
Chapter3
10c,14a
Chapter4
1,25(1),29,16/34(any one),51(b)
Exercise
There is a digital augur can foresee blood type for
baby according to parent’s blood type.
If parents have same blood type,the baby is
either O type or to be inherit his parents..
If one parent is A type,the other is O type,the
baby is either A type or O type.
If one parent is B type,the other is O type,the
baby is either B type or O type.
If one parent is AB type,the other is O type,the
baby is either A type or B type.
If one parent is AB type,the other is either A type
or B type,the baby would not be O type.
If one parent is A type,the other is B type,the
baby is neither O type nor AB type.