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
