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CP208 Digital Electronics Class Lecture 11 May 13, 2009

CP208 Digital Electronics Class Lecture 11 May 13, 2009. In This Class. We Will Discuss : Chapter 11: Memory and Advanced Digital Circuits. Memory and Advanced Digital Circuits. 3. Introduction.

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CP208 Digital Electronics Class Lecture 11 May 13, 2009

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  1. CP208Digital ElectronicsClass Lecture 11May 13, 2009

  2. In This Class We Will Discuss: Chapter 11: Memory and Advanced Digital Circuits

  3. Memory and Advanced Digital Circuits 3

  4. Introduction • Chapter 10 was about Combinational logic circuits – output of which depends on the present value of the input, that is NO MEMORY • Memory is very important part of digital systems • Memory in digital computers allows for storing programs and data • Also, it is important for temporary storage of the output produced by combinational circuit for use at later time in the operation of a digital system • Logic circuits that incorporate memory are called sequential circuits

  5. Introduction • Sequential circuits have output that depends not only on the present value of input but also on the input’s previous value. Such circuits require a timing generator for their operation

  6. Introduction • How Does It Work … • In first approach application of positive feedback provides a circuit with two stable states. Bistable circuit then stores a BIT of information: One state correspond to stored 0 and the other to stored 1 • Bistable circuit can remain in either state indefinitely and hence named as Static Sequential Circuit

  7. Introduction • In second approachcharge on capacitor is used to realize memory. • Charged capacitor corresponds to stored 1 and • Discharged capacitor corresponds to stored 0. • Because leakage effects cause capacitor to discharge, this form requires periodic recharging of capacitor or Refresh. • These circuit are named as Dynamic Sequential Circuit

  8. 11.1 Latches and Flip-Flops • Latch is the Basic Memory Element, we shall consider a sampling of its application 11.1.1 The Latch • Latch consists of two Cross-coupled Logic Inverters • Inverters form a Positive-feedback Loop

  9. Three Operating Points A, B, and C. Two Stable Operating Points, A and C. At C: vW = vZis High, vX = vYis Low. At A: Reverse is True.

  10. Now, X and Z are Latch Out Puts. In State A: vXis High (VOH) and vZis Low (VOL). In State C: vXis Low (VOL) and vZis High (VOH). Thus, Latch is Bistable Ckt having Two Complementary Outputs

  11. The Stable State in which Latch Operates depends on the External Excitation that Forces it into. The latch then Memorizes this External Action by Staying in the Acquired State Indefinitely (Static Memory). • Latch is capable of Storing One Bit of Information, e.g., we can designate State A corresponding to Logic 1. The other complementary State C would be Logic 0. • What is the mechanism by which Latch can be Triggered to Change State?

  12. Latch with a Triggering Mechanism/Circuitry forms a … … Flip – Flop Simplest Flip-Flop is Set/Reset (SR) Flip-Flop… …NEXT

  13. 11.1.2 The SR Flip-Flop • Simple type of Flip-Flop is Set/Reset (SR) • It is formed by cross coupling two NOR gates and thus Incorporates Latch

  14. Second Input of each NOR Gate Labeled as S and R serve as Trigger Outputs emphasize their Complementarity. Set State (Storing 1): When Reset State (Storing 0): Opposite of Set Rest or Memory State (when we do not wish to change state of Flip-Flop) : Both S and R inputs Low

  15. Consider the Case When Flip-Flop is Storing Logic 0 Since Q is low, Both inputs to G2 will be Low and its Output will therefore be High. High Output of G2 is applied to Input of G1, causing its Output Q to be Low (since R is Low), satisfying the Original Assumption.

  16. To SetFlip-Flop [To Store Logic 1] Raise S to Logic-1 Level While Leave R at 0. Output of G2 to 0 or Low. Two Inputs of G1 will be 0 or Low, causing its Output Q to be 1 or High. Now, if S returns to 0 Flip-Flop remains Set. Raising S to 1 again will make No Change.

  17. To ResetFlip-Flop [To Store Logic 0] Raise R to Logic-1 Level While Leave S at 0. SHOW… The Flip-Flop will be Forced to Reset State and will remain in This State even after R returns to 0. Raising R to 1 again will make No Change. Note that the Trigger Signal Merely Starts Regenerative Action of the Positive-Feedback Loop of the Latch

  18. What Happens When S=R=1 ??? Both NOR Gates will cause to become 0 making Complementary labeling Incorrect. However, if R and S return to Rest State (R=S=0) precisely at the same time, the state of Flip-Flop will be undefined. For this reason this input Combination is Not Allowed. In Practice one of the R and S returns to 0 first, and the Final State is determined by the input that Remains High Longest.

  19. Truth Table: Qn is Value of Q at Time tn just before application of R and S Signals. Qn+1 is Value of Q at time tn+1 after application of input Signals. SR Flip-Flop can also be implemented using two NAND gates. In this case Set Reset Functions are active when Low, therefore labeled as

  20. 11.1.3 CMOS Implementation of SR Flip-Flop • SR Flip-Flop can be directly implemented in CMOS by just replacing each NOR gate by its CMOS circuit realization

  21. Clocked version of SR Flip-Flop using CMOS. Except Clock Inputs, It operates exactly same as the Logic Circuit scheme.

  22. Clock inputs form AND Function with S and R, therefore, Flip-flop can only be Set and Resent when Clock is High.

  23. To Set from Reset State: High (VDD) Signal on S while R is Held at Low (0V). Now, When Clock Goes High both Q5 and Q6 will conduct, Pulling Voltage at Down. When V goes below threshold of (Q3,Q4) Inverter, It will begin switching states and V at Q node will Rise. The Increase at Q is Fed Back to the Input of (Q1, Q2) Inverter, causing its Output at to go down further. The regeneration process is in progress.

  24. Flip-Flop switching (for Set) is predicted on following two assumptions: 1. Q5 and Q6 supply sufficient current to pull node Q at least slightly below threshold of (Q3 and Q4) inverter. This is very important for the regenerative process to begin. Example 11.1 investigates minimum W/L ratios so that Q5 and Q6 must meet this requirement.

  25. 2. The Set signal remains high for an interval long enough to cause regeneration to take over the switching process. An estimate of minimum width required for the set pulse can be obtained as the Sum of Interval during which voltage at is reduced from VDD to VDD/2 and the interval for the voltage at Q to respond and rise to VDD/2 Finally, due to symmetry all remarks apply equally well to Reset process. Example 11.1

  26. 11.1.4 Simpler CMOS Implement-ation of clocked SR Flip-Flop • In simple implementation, Pass-Transistor Logic is employed to implement the Clocked SR Flip-Flop • This ckt is very popular in the design of Static Random-Access Memory (SRAM) chips, where it is used as the Basic Memory Cell

  27. In Next Class We Will Discuss: RAM and ROM Memories

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