1 / 19

CSE 140L Lecture 6 Interface and State Assignment

CSE 140L Lecture 6 Interface and State Assignment. Professor CK Cheng CSE Dept. UC San Diego. Interface: Quality of Input Signal. Debouncers Double-Throw Switch Single-Throw Switch Synchronizers Metastability Synchronization. Debouncers: Double-Throw Switch. SR debouncer.

giddensm
Download Presentation

CSE 140L Lecture 6 Interface and State Assignment

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CSE 140L Lecture 6Interface and State Assignment Professor CK Cheng CSE Dept. UC San Diego

  2. Interface: Quality of Input Signal • Debouncers • Double-Throw Switch • Single-Throw Switch • Synchronizers • Metastability • Synchronization

  3. Debouncers: Double-Throw Switch SR debouncer

  4. Debouncers: Single-Throw Switch Pull down Vcap/Vinitial = e –t/RC R2C= -T/ln(Vth/Vdd) Push up Vcap/Vfinal= 1-e-t/RC (R1+R2)C= -T/ln(1-Vth/Vdd) Bounce time < T=20msec

  5. Synchronizer: Input may occur any time • Asynchronous (for example, user) inputs might violate the dynamic discipline

  6. Synchronizer: Metastability • Metastability: A synchronizer failure that the voltage falls between 0 and 1.

  7. Metastability • Any bistable device has two stable states and a metastable state between them • A flip-flop has two stable states (1 and 0) and one metastable state • If a flip-flop lands in the metastable state, it could stay there for an undetermined amount of time

  8. Flip-flop Internals • Because the flip-flop has feedback, if Q is somewhere between 1 and 0, the cross-coupled gates will eventually drive the output to either rail (1 or 0, depending on which one it is closer to). • A signal is considered metastable if it hasn’t resolved to 1 or 0 • If a flip-flop input changes at a random time, the probability that the output Q is metastable after waiting some time, t, is: • P(tres > t) = (T0/Tc ) e-t/τ • tres : time to resolve to 1 or 0 • T0, τ : properties of the circuit

  9. Metastability • Intuitively: • T0/Tc describes the probability that the input changes at a bad time, i.e., during the aperture time • P(tres > t) = (T0/Tc ) e-t/τ • τ is a time constant indicating how fast the flip-flop moves away from the metastable state; it is related to the delay through the cross-coupled gates in the flip-flop • P(tres > t) = (T0/Tc ) e-t/τ • In short, if a flip-flop samples a metastable input, if you wait long enough (t), the output will have resolved to 1 or 0 with high probability.

  10. Synchronizers • Asynchronous inputs (D) are inevitable (user interfaces, systems with different clocks interacting, etc.). • The goal of a synchronizer is to make the probability of failure (the output Q still being metastable) low. • A synchronizer cannot make the probability of failure 0.

  11. Synchronizer Internals • A synchronizer can be built with two back-to-back flip-flops. • Suppose the input D is transitioning when it is sampled by flip-flop 1, F1. • The amount of time the internal signal D2 can resolve to a 1 or 0 is (Tc - tsetup).

  12. Synchronizer: Shifter Allocate one clock cycle for logic to settle at 0 or 1. For each sample, the probability of failure of this synchronizer is: P(failure) = (T0/Tc ) e-(Tc - tsetup)/τ

  13. Synchronizer Mean Time Before Failure • If the asynchronous input changes once per second, the probability of failure per second of the synchronizer is simply P(failure). • In general, if the input changes N times per second, the probability of failure per second of the synchronizer is: • P(failure)/second = (NT0/Tc) e-(Tc - tsetup)/τ • Thus, the synchronizer fails, on average, 1/[P(failure)/second] • This is called the mean time between failures, MTBF: • MTBF = 1/[P(failure)/second] = (Tc/NT0) e(Tc - tsetup)/τ

  14. Example Synchronizer • Suppose: Tc = 1/500 MHz = 2 ns τ = 200 ps • T0 = 150 ps tsetup = 100 ps • N = 1 events per second • What is the probability of failure? MTBF? • P(failure) = (150 ps/2 ns) e-(1.9 ns)/200 ps • = 5.6 × 10-6 • P(failure)/second = 10 × (5.6 × 10-6 ) • = 5.6 × 10-5 / second • MTBF = 1/[P(failure)/second] ≈ 5 hours

  15. State Assignment • Binary Code: log2n flip-flops for n states • Gray Code: log2n flip-flops for n states • Johnson Code: n/2 flip-flops for n states • “Don’t Care” State Encoding • One Hot Code: n flip-flops for n states

  16. “Don’t Care” State Encoding • State i  Qj= 0 for all j > i, Qi= 1 Qj= x for all j < i.

  17. State Assignment: One Hot Code • For n states, we use n flip-flops. • Each state corresponds to a flip-flop. • Qi= 1 iff present state = state i. Thus, one and only one flip-flop contains a “1”. • We use OR gate to collect input edges. • We use Mux gate to distribute the output edges.

  18. State Assignment: One Hot 0 0 s0s 1 1 S2 S0 S1 0 1

  19. 3. Transformation from Mealy to Moore Machine Moore Machine: y(t) = f(x(t), s(t)) Mealy Machine: y(t) = f(s(t)) s(t+1) = g(x(t), s(t)) x(t) x(t) C1 C2 y(t) C1 C2 y(t) CLK CLK s(t) s(t) Moore Machine Mealy Machine

More Related