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© H. Heck 2008Section 2.41 Module 2:Transmission Lines Topic 4: Parasitic Discontinuities OGI EE564 Howard Heck.

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Presentation on theme: "© H. Heck 2008Section 2.41 Module 2:Transmission Lines Topic 4: Parasitic Discontinuities OGI EE564 Howard Heck."— Presentation transcript:

1 © H. Heck 2008Section 2.41 Module 2:Transmission Lines Topic 4: Parasitic Discontinuities OGI EE564 Howard Heck

2 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.42 Where Are We? 1.Introduction 2.Transmission Line Basics 1.Transmission Line Theory 2.Basic I/O Circuits 3.Reflections 4.Parasitic Discontinuities 5.Modeling, Simulation, & Spice 6.Measurement: Basic Equipment 7.Measurement: Time Domain Reflectometry 3.Analysis Tools 4.Metrics & Methodology 5.Advanced Transmission Lines 6.Multi-Gb/s Signaling 7.Special Topics

3 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.43 Contents Capacitive Load: Qualitative Look Capacitive Load: Step Response Analytical Solution Discrete Capacitive and Inductances: Delays and Reflections Distributed Loads: Loaded Velocity and Impedance Summary References Appendix: Reflections Due to Capacitive Load Appendix: Inductive Load Step Response

4 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.44 Capacitive Loading: Qualitative Look What does the signal at z=l, t=  d  l+ look like?  The driver circuit sends a voltage step at t=0.  The step travels along the line (as a wave) to z=l.  The capacitor at z=l sees a voltage step input. How does the capacitor respond to this input?  When the incoming wave arrives, the capacitor initially acts like a short.  At steady state, the capacitor is an open circuit.  The transmission line looks resistive to the capacitor.  So, perhaps we should expect to get an RC circuit-like (exponential) response. Z 0,  d V S (t)CLCL RSRS z = 0 z = l V S (t) = V S ·u(t) 0 VSVS t = 0

5 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.45 Quantitative Analysis  Start at z =0, t =0+: Z 0,  d V S (t) CLCL RSRS z = 0 z = l V S (t) = V S ·u(t) 0 VSVS t = 0 [2.5.1] [2.5.2]  The wave propagates to z = l, t =  d : [2.5.3] [2.5.4]

6 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.46 Quantitative Analysis #2 [2.5.5] [2.5.6] We have a linear ordinary differential equation. Find the form of the solution by solving the homogeneous portion: Homogenous equation:[2.5.7] Solution form: [2.5.8] [2.5.4] For a capacitor,. Apply it to [2.5.4]: Note the “ Z 0 C ” time constant.

7 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.47 Quantitative Analysis #3 [2.5.6] General solution to O.D.E.:  General solution: Boundary condition:. Apply it to [2.5.6]: [2.5.9] [2.5.10] [2.5.11]

8 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.48 Quantitative Analysis #4 Substitute [2.5.13] into general solution [2.5.11]:  Initial condition: V ( z = l, t =  d l )=0. [2.5.13] [2.5.12] [2.5.14] [2.5.15] The expression for becomes: [2.5.16] [2.5.17]

9 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.49 Quantitative Analysis #5 Equation [2.5.18] is valid for  d l  t <3  d l. If R S = Z 0, [2.5.18] is valid for  d l  t.  Solution: [2.5.17] [2.5.18]

10 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.410 Quantitative Analysis Check Let R S = Z 0. The following boundary conditions must be satisfied: The wave components are: This checks out.

11 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.411 time [ns] 01234 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 voltage [V] WaveformsCalculated Simulated V(z=0) V(z=l) 01234 time [ns] 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 voltage [V] V(z=0) V(z=l) 50 , 1.0 ns 2.0V 5.0 pF 50  z = 0 z = l V S (t) = V S ·u(t) 0 VSVS t = 0 t r = 1 ps

12 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.412 Capacitive Reflections & Delay Z 0,  d V S (t) CLCL RSRS z = 0 z = l V S (t) = V S ·u(t) 0 VSVS t = 0 Refer back to equation [2.5.18] : [2.5.18] Assume that 50% of the full signal swing is required for switching: Solve for t (assume  d l = 0 for simplicity): [2.5.19] [2.5.20] Many sources give the following: [2.5.21]

13 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.413 Capacitive Reflections & Delay #2 With the load at an intermediate point, the capacitor is charged to half as much voltage as for a capacitor at the end of the line. [2.5.22] Z 0,  d V S (t) CLCL RSRS z = 0 z = l Z 0,  d Reflections due to the capacitor (see the appendix):  Mid-line  End [2.5.23] [2.5.24]

14 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.414 Inductive Loads Z 0,  d V S (t) RSRSL Z 0,  d V S (t) RSRSL [2.5.25] [2.5.26] [2.5.27] [2.5.28]

15 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.415 Distributed Capacitance Z0, dZ0, d Z0, dZ0, d Z0, dZ0, d Z0, dZ0, d Z0, dZ0, d CLCL RSRS Z 0,  d CLCL CLCL CLCL CLCL V S (t) Bi-directional network: Each receiver presents a capacitive load: The loads are uniformly distributed, separated by length l. The effective capacitance of each transmission line segment is: [2.5.24] C 0 is the transmission line capacitance per unit length.

16 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.416 Distributed Capacitance #2 Since the distributed capacitance has increased, the propagation velocity and impedance are changed: We can also express the loaded impedance and velocity in terms of the characteristic impedance and non-loaded velocity: Z L is the “loaded” impedance of the transmission line. v L is the “loaded” propagation velocity of the transmission line. [2.5.28] [2.5.25] [2.5.26] [2.5.27]

17 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.417 Summary Capacitive loads slow the rise/fall times and add delay. Capacitive loads generate reflected waves of opposite polarity to the incoming wave. Inductive loads also affect the delay and generate reflections. Distributed capacitive loads increase the propagation delay of the transmission line and reduce the effective impedance.

18 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.418 References S. Hall, G. Hall, and J. McCall, High Speed Digital System Design, John Wiley & Sons, Inc. (Wiley Interscience), 2000, 1 st edition. W. Dally and J. Poulton, Digital Systems Engineering, Cambridge University Press, 1998. Ramo, Whinnery, and Van Duzer, Fields and Waves in Communication Electronics, 1985. R. Poon, Computer Circuits Electrical Design, Prentice Hall, 1 st edition, 1995. Ramo, Whinnery, and Van Duzer, Fields and Waves in Communication Electronics, 1985. R.E. Matick, Transmission Lines for Digital and Communication Networks, IEEE Press, 1995 H.B.Bakoglu, Circuits, Interconnections, and Packaging for VLSI, Addison Wesley, 1990. “Transmission Line Effects in PCB Applications,” Motorola Application Note AN1051, 1990. W.R. Blood, MECL System Design Handbook, Motorola, Inc., 1988.

19 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.419 Appendix: Reflections from Capacitive Load Z 0,  d CLCL ViVi VrVr Recall: Combine [2.5.a1] and [2.5.a2]: Where: [2.5.A1] [2.5.A2] [2.5.A3]

20 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.420 Appendix: Reflections from Capacitive Load #2 Substitute [2.5.a3] into the expression for  : We can rewrite [2.5.a5] in terms of V r and V i : [2.5.A4] Assume. Then [2.5.a4] simplifies to: [2.5.A5] [2.5.A6]

21 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.421 Appendix: Reflections from Capacitive Load #3 We need to make use of the following time domain relationship: For, the time derivative is, which must be satisfied for all . For our case,  0 (t) is V i. Substitute into [2.5.a6] : [2.5.A7] Letting, we get [2.5.a8]: Limitation: We must satisfy Z 0 C L << t r. Why?  Recall  = 2  f and f  0.35/ t r. Apply to our assumption that. [2.5.A8]

22 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.422 Appendix: Mathematical Expression for Inductive Load  Start at z =0, t =0+: [2.5.B1] [2.5.B2]  The wave propagates to z = l, t =  d l : [2.5.B3] [2.5.B4] Z 0,  d V S (t) L RSRS z = 0 z = l V S (t) = V S u(t) 0 VSVS t =0

23 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.423 Inductive Loading #2 Rearrange [2.5.b6]: [2.5.B4] For an inductor,. Apply it to [2.5.b4]: [2.5.B5] [2.5.B6] [2.5.B7] Linear O.D.E.

24 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.424 Inductive Loading #3  Homogeneous solution: [2.5.B7] [2.5.B9] [2.5.B8]  General solution: Use the boundary condition:. Therefore, [2.5.B10] [2.5.B11]

25 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.425 Inductive Loading #4 General solution is the sum of the homogeneous solution and special solution: [2.5.B12]  Initial condition: Apply the initial condition at t=  d l (the inductor acts like an open circuit): [2.5.B13] [2.5.B14] Substitute into general solution: [2.5.B15] [2.5.B16]

26 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.426 Inductive Loading #5 Equation [2.5.18] is valid for  d l  t <3  d l. If R S = Z 0, [2.5.18] is valid for  d l  t. Substitute for A in [2.5.B16]: [2.5.B17]  Solution: [2.5.B18]

27 Parasitics Discontinuities EE 564 © H. Heck 2008 Section 2.427 Check Let R S = Z 0. The following boundary conditions must be satisfied: The wave components are: This checks out.


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