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Timing and Hazards. Logical levels Positive Logic Negative Logic 5v = 1 0v = 0 0v = 1 5v = 0.

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Presentation on theme: "Timing and Hazards. Logical levels Positive Logic Negative Logic 5v = 1 0v = 0 0v = 1 5v = 0."— Presentation transcript:

1 Timing and Hazards

2 Logical levels Positive Logic Negative Logic 5v = 1 0v = 0 0v = 1 5v = 0

3 Noise – basic concepts 0 1 Ideal World

4 Noise – basic concepts 0=0v 1=5v Ideal World 0 1 Real World Positive Logic Voltage

5 Noise – basic concepts 0 1 Real World Positive Logic

6 Noise – basic concepts 5v 0v 1v 1.5v Output Range Input Range Output Range 4v 3.5v High Value Low Value

7 Timing – basic concepts 0 1 Ideal World “zero” time to change

8 Timing 0 1 Ideal World “zero” time to change 0 1 Real World measurable time to change

9 Timing 0 1 Ideal WorldInput 0 1 Output

10 Timing 0 1 Ideal WorldInput 0 1 Output 0 1 Real World 0 1

11 Timing – Delay Time 0 5V Input Output Time 50% t HL t LH Propagation delay = t PD = max(t HL,t LH )

12 Timing – Contamination time 0 5V Input Output Time 50% t CD The minimum time in which the input changed (50%) and the output didn’t change.

13 Timing Rise and Fall time (measured on the output) 0 1 10% 90% trtr Rise Time 10% 90% tftf Fall Time Output

14 The t pd of a circuit The function : X 1 *X 2 + X 3 ’ A C B X3X3 X2X2 X1X1 Data in nsABC t HL 1009080 t LH 11070100 t CD 12810 trtr 201214 TfTf 171319 Y

15 The t pd of a circuit a single assignment The function : X 1 *X 2 + X 3 ’ A C B X 3 =1 X 2 =0  1 X 1 =1 Data in nsABC t HL 1009080 t LH 11070100 t CD 12810 trtr 201214 TfTf 171319 Y

16 The t pd of a circuit a single assignment The function : X 1 *X 2 + X 3 ’ A C B X 3 =1 X 2 =0  1 X 1 =1 Data in nsABC t HL 1009080 t LH 11070100 t CD 12810 trtr 201214 TfTf 171319 0101

17 The t pd of a circuit a single assignment The function : X 1 *X 2 + X 3 ’ A C B X 3 =1 X 2 =0  1 X 1 =1 Data in nsABC t HL 1009080 t LH 11070100 t CD 12810 trtr 201214 TfTf 171319 0101 Y 0  1

18 The t pd of a circuit a single assignment X 2 =0  1 Output of A Output of C t LH (A) t LH (B)

19 The t pd of a circuit a single assignment X 2 =0  1 Output of A Output of C t LH (A) t LH (B) t LH (A) + t LH (B)

20 The t pd of a circuit The t pd of a circuit is the maximal value Over all the possible single assignments

21 The t cd of a circuit a single assignment X 2 =0  1 Output of A Output of C Real t CD (Circuit)

22 The t cd of a circuit a single assignment X 2 =0  1 Output of A Output of C t CD (A) t CD (B) R

23 The t cd of a circuit a single assignment X 2 =0  1 Output of A Output of C t CD (A) t CD (B) R Real t CD (Circuit) = t CD (A) + t CD (B) + R

24 The t cd of a circuit a single assignment t CD (Circuit) = t CD (A) + t CD (B) < t CD (A) + t CD (B) +R = Real t CD (circuit) Since t CD is the minimum

25 The t CD of a circuit The t CD of a circuit is the minimal value Over all the possible single assignments

26 Timing The minimum time for the output to change What is the minimal time that the output will change over a single assignment?

27 The minimum time for the output to change a single assignment X 2 =0  1 Output of A Output of C t LH (A) t LH (B) t LH (A) + t LH (B)

28 The minimum time for the output to change a single assignment X 2 =0  1 Output of A Output of C t LH (A) t LH (B) t LH (A) + t LH (B)

29 The minimum time for the output to change a single assignment Real Minimum Time = t LH (A) + t HL (A) + T r (X2) + T r (C) 2

30 The minimum time for the output to change a single assignment Real Minimum Time = t LH (A) + t HL (A) + T r (X2) + T r (C) 2 < Max(Tr(X2), Tr(C))

31 The minimum time for the output to change a single assignment Estimated Minimum Time = t LH (A) + t HL (A) +Max(T r (X2),T r (C)) > Real Minimum time

32 Problems in timing Y = X2*X3 + X1*X2’ B D A C X1 X2 X3 Y Assume t HL =t LH > 0 And equal for all gates

33 Problems in timing Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1 X3=1 Y=1

34 Problems in timing Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1 X3=1 Y=1 0 0 1

35 Change in input – final state Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1  0 X3=1 Y=1 We changed the input But the result remained the same 1 1 0

36 Change in input – Time slice 1 Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1  0 X3=1 Y=1

37 Change in input – Time slice 2 Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1  0 X3=1 Y=1 0101 0 1010 1

38 Change in input – Time slice 3 Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1  0 X3=1 Y=0 1 0101 0 1010

39 Change in input – Time slice 4 Y = X2*X3 + X1*X2’ B D A C X1=1 X2=1  0 X3=1 Y=1 1 1 0 0101

40 Input Change Time 0 X2 A B C Y 1 0 0 1 1 0 1 1 0 0

41 Input Change Time 1 X2 A B C Y 1 0 0 1 1 0 1 1 0 0

42 Input Change Time 2 X2 A B C Y 1 0 0 1 1 0 1 1 0 0

43 Input Change Time 3 X2 A B C Y 1 0 0 1 1 0 1 1 0 0

44 Input Change Time 4 X2 A B C Y 1 0 0 1 1 0 1 1 0 0

45 Static Hazard X2 A B C Y 1 0 0 1 1 0 1 1 0 0 Static Hazard – A change in a single input That should not change the output Leads to a momentary change.

46 Static hazard – another point of view 00011110 01 1111 X2X3X2X3 X1X1

47 Static hazard – another point of view 00011110 01 1111 X2X3X2X3 X1X1

48 Static hazard – another point of view 00011110 01 1111 X2X3X2X3 X1X1

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