《数字电》英文版 chapter3-2 Combinational logic Circuits

Combinational logic circuits Chapter 3 Combinational logic hazards

Combinational logic circuits Chapter 3 Combinational logic hazards

Circuits parameters a Vcc, power supply voltage 口Icc, power supply current cch CcL OL,VoH,士oL,oH d VIl, VIH, fIL, fIH 日tPLH,tpHL a Fan-out

Circuits Parameters  Vcc, power supply voltage  Icc, power supply current ◼ ICCH, ICCL  VOL, VOH , IOL, IOH  VIL, VIH, IIL, IIH  tPLH, tPHL  Fan-out

Hazard a 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 a hazard exists any time the potential for itches is present a Glitches may not be a problem or they may create havoc. It depends on the specif ic application

 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 D a static-1 hazard exists when an output variable should be a logic 1, but goes to o momentarily as a result of input variable changing a A static-o 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  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 口P=Xy"+yZ ■=1.z=1 G4 y+y- Xy a Assuming all the G2 G3 gate have the same delay time

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 At time to, y went from 1 to O to G41 +1 At time t1, y went from 0 to 1 Xy Xy +2 At time t2, xy went from 0 to 1 G2 G3 Zy t1l At time t1, zy went from 0 to 1 Between time t1 to t2 neither inputs of G4 was a 1 P glitch Accounting for the delay t through G4 the momentary time is from t2(t1+t)to t3(t2+t)

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 hazare 口 Boolean algebra aA input variable x and its complement xare both resented in the function equation a The equation would be simplified to(x+x)or xin some conditions when x changes, the circuit correspond to the equation would produce a static hazard because of the different propagation delay a Solution Add redundant terms

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 口 Boolean algebra a Solution ■ Add redundant terms. 口P=(Z+y)(X+Y) ■zZ=X=0.P=y"y=0 a static hazard o P=(Z+)(X+》)(X+Z)

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 口 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 a Solution Generate new pi that bridge two ad jacent epi

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 口P(X,Y,Z)=2m(3457)=X+yz 口 TWo EPI 2000111-10 026 ■EPI(4,5) 0 EPI2 (3, 7 a they are tangent a Static hazard 1 口PT(75) 口P=Xy+yZ+XZ

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 1 1 1  Two EPI ◼ EPI1(4,5) ◼ EPI2(3,7) ◼ They are tangent.  Static hazard 1.  PI(7,5)  P=XY’+YZ+XZ