1 Strategies for Coping with Non-linear and Non-time Invariant Behavior for High Speed Serial Buffer Modeling Richard Mellitz Results from DesignCon2008.

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Presentation transcript:

1 Strategies for Coping with Non-linear and Non-time Invariant Behavior for High Speed Serial Buffer Modeling Richard Mellitz Results from DesignCon2008 paper with Steve Pytel, Michael Tsuk, and Toni Donisi

2 Linearity Enables Superposition A linear system possesses the property of superposition, in other words, the system possesses both the additive and homogeneity properties. If Then, by the additive property, And by the homogeneity property where a is a constant.

3 Example of Non-Linearity  A resistance, capacitance, or inductance that changes with voltage creates non linear circuit behavior.  All transistors are non-linear  Many buffers have linear region of operations  IBIS is used to represent non linear characteristics  Full transistor models may include time variant effects. –Not discussed today

4 Simple Superposition Example: “Tales of a lone bit”  The lone pulse can be used to determine the response digital pulse stream.  This is true as long as superposition holds or the system is linear –The interconnect channel is linear.  We will use an example to how a lone pulse with cursor value of 0.75 and post cursor tap of 0.25 results is an bit stream that can be recognized as 6dB pre-emphasis Cursor=0.75 Post cursor =0.25 base = 0 Bit Time or Unit Interval

5 Use superposition to string together a bit pattern out of lone bits with the amplitude of the taps Bits ¾ ½ ½ -¼ 0 0 ¾ ½ ½ -¼ Value 

6 We now have a familiar waveform  Notice the familiar de-emphasized waveform which is a composition of lone bits –Observe that Vshelf is ½ and Vswing is 1. –For 2 tap systems we would call this 6dB de-emphasis 20*log(0.5)  Using this concept simulate or measure one lone bit and with out performing any more simulation we can: –Determine the response of an arbitrary string bits –Determining best or worst case signal distortion. –Determining the eye opening due algorithmically piecing string to that produce aggregated performance Renormalize to 1 peak to peak: Value-1/4 -¼ -¼ -¼ -¼ -¼ -¼ -¼ ½ ¼ ¼ -½ -¼ -¼ ½ ¼ ¼ -½ -¼ -¼ -¼ -¼ renorm Vshelf = ½ Vswing = Bits ¾ ½ ½ -¼ 0 0 ¾ ½ ½ -¼ Value

7 High Speed Signaling tools  Use superposition of edges to create long bits streams  Edge are altered in time to create jitter  Others convolve a channel system function with jitter and data  Adaptive equalization can be determined from bit streams, system function, or pulse response.  Ansoft’s QuickEye™ resembles some of the above  All these type of tools make the assumption that linear superposition is valid

8 Rest of the Agenda  Review CML buffer  Making the CML buffer non linear  Determine effect of % of linearity for different equalization interpretations  Conclusion

9 Simple Current Mode Logic (CML) Differential Buffer Edge filter inverter FIR Filter Level Shifter CMLSOURCE Data Stream Vout diff Routp Routn Coutp Coutn

10 FIR filter data stream UI 2*UI 4*UI 3*UI delay delay delay delay Pre post1 post2 post 

11 Non-Linear Experiment Vout diff Routp Routn Coutp Coutn

12 Non-linear Termination example Rmax Rmin V R V I

13 Experiment setup Buffer channel load Tx Linear Equalizer Math Process Rx Linear Equalizer Set taps at Tx Set taps at Rx Buffer I/V loads Bitwise eye Bit stream Edge Convolved eye ~ QuickEye™

14 Experiments  CML Buffer Loads –50  Rout –30 to 70  Rout variation –10 to 100  Rout variation  Data Pattern PRBS15  Loads –50  both legs  Channels –12” of a 72  differential transmission line (50  SE termination) –2 connector “real” channel  UI=125ps  Simulation time 100ns

15 Routn=Routp=load=50  Rout range: 10 to 90  load=50  Bit stream (1) and edge convolution (2) are equal, if taps are set at the transmitter Rout range: 10 to 90  Rout = 50  “Single resistor equivalent” Rx Mathematical equalization (3): 24.5 mv error Taps = 0.79 and -0.21

16 Channel =72  12” line Rout range: 10 to 90  Bit stream (1) and edge convolution (2) are close, if tap are set in the transmitter Rout range: 10 to 90  Rx Mathematical equalization(3): Taps = 0.79 and -0.21

17 Channel = 72  12” line Rout range: 10 to 90  Bit stream Rx Mathematical equalization w/convolution: Convolution Taps = 0.79 and Zoomed in 1 2 3

18 Channel = 72  12” line Rout range: 10 to 90  Convolution Taps set at Tx Bit Stream ~ 10 mv & 0.1ps difference Taps = 0.79 and

19 Channel = 72  12” line Rout range: 10 to 90  Bit Stream Convolution Taps set at Rx ~ 2 mv & 2ps difference Taps = 0.79 and

20 Channel = 72  12” line Rout range: 10 to 90  ISI Jitter distributions Taps = 0.79 and

21 Channel = 72  12” line Rout range: 30 to 70  Bit stream (1) and edge convolution(2) are very close, if tap are set in the transmitter Rout range: 30 to 70  Rx Mathematical equalization(3): Taps = 0.79 and -0.21

22 Channel = 72  12” line Rout range: 30 to 70  Bit stream Rx Mathematical equalization w/convolution: Convolution Taps = 0.79 and

23 Channel = 72  12” line Rout range: 30 to 70  Bit Stream Convolution Taps set at Tx About the same EO Taps = 0.79 and

24 Channel = 72  12” line Rout range: 30 to 70  Bit Stream Convolution Taps set at Rx 1.5mV & 6 ps E0 Difference Taps = 0.79 and

25 Channel =2 connector real system Rout range: 10 to 90  Bit Stream Convolution Taps set at Tx 30 min Simulation Time 2 min Simulation Time Taps = 0.79 and mV & 1.5 ps E0 Difference

26 Channel =2 connector real system Rout range: 10 to 90  Bit Stream Convolution Taps set at Rx Bit Stream 30 min Simulation Time 2 min Simulation Time Taps = 0.79 and mV & 1.1 ps E0 Difference 3

27 Channel =2 connector real system Rout range: 10 to 90  Bit stream (1) and edge convolution(2) are very close, if tap are set in the transmitter Rout range: 30 to 70  Rx Mathematical equalization(3): Taps = 0.79 and -0.21

28 Channel = 72  12” line Rout range: 0.5pf to 1.5pF Bit Stream Taps = 0.79 and mV & 0.5 ps E0 Difference Convolution Taps set at Tx 2

29 Channel = 72  12” line Rout range: 0.5pf to 1.5pF Bit Stream Taps = 0.79 and mV & 7.5 ps E0 Difference Convolution Taps set at Rx

30 Conclusion  Single resistor equivalent models are insufficient  Convolution Eye is OK if: –Equalize taps are set at the Tx or if buffer impedance range is < 40% from nominal … or –Predictive algorithms for solution space don’t require more than a few ps or mv of resolution  Adaptive equalization methods may be impacted by non-linearity –Jitter distribution varies with method. More work is needed on impact of ISI jitter distributions on adaptive algorithms –More work is required to determine if algorithms will hunt out correct Rx equalization  Result may be worse for higher data rates –More data needed here