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Numerical Cable Equalization Hooman Hashemi4/9/13 1

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Cable Driving & Equalization Good 2 rule of thumb: –Linear increase of dBs with cable length –Number of dBs increases with square root of F(MHz) Solutions we offer –Use amplifiers to boost the higher frequency –Divide extreme equalizing designs into multiple identical cascaded stages –Excel sheet to compute the component values 2

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CAT5e (shielded twisted pair) 100ft response in log log plot Cable Driving & Equalization Here you see a typical log log plot of cable loss vs. frequency The Excel file provided uses this attenuation profile / expression (but it can be modified for a different cable)

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Component Design Numerical Solution –Application note discusses this: Numerical Design Numerical Design –Uses Excel Solver function –It is modular- relatively easy to add more equalization banks –Component values can be changed on the fly with results immediately visible! –Uses Excel complex algebra capabilities to simplify computations 2-banks response on the left compared to 4-banks on the right Equalization Stage with 2 Banks

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Component Design (Numerical Solution) f (Hz) Complex R1+1/sC1R2+1/sC2R3+1/sC3 50,000,000 5- 318.309886 183791i 20- 159.15494 3091895i 420- 10.61032953 94597i 65,000,000 5- 244.853758 602916i 20- 122.42687 9301458i 420- 8.161791953 43053i Equivalent Admittance = G1+G2+G3

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Component Design (Numerical Solution) f (Hz) Cell D11Cell E11Cell F11 R1+1/sC1R2+1/sC2R3+1/sC3 50,000,000 5- 318.3098861 83791i 20- 159.154943 091895i 420- 10.6103295394 597i 65,000,000 5- 244.8537586 02916i 20- 122.426879 301458i 420- 8.16179195343 053i f (Hz) Cell D11Cell E11Cell F11 R1+1/sC1R2+1/sC2R3+1/sC3 50,000,000 Cell H11 =IMSUM(IMDIV(1,D11),IMDIV(1,E11),IMDIV(1,F11)) Cell I11= IMDIV(1,IMSUM(1/RG,H11)) Use Complex function to form the impedance made up of the resistance and the imaginary reactance Use IMDIV to invert the bank impedances (ex: IMDIV(1,D11) ) before summing all the admittances with IMSUM to get the overall bank admittances Add the overall bank admittances to 1/R G and invert in one shot to get Z G_eq Continue using complex functions like IMDIV, IMSUM, and IMABS to get to the 1+R F /Z G _eq which is the circuit gain expression Superimpose plot of gain on cable attenuation to optimize design

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Component Design (Numerical Solution) Limit each stage boost to 25dB or less –More than this may cause secondary effects, parasitic effects and other unexpected phenomenon to get in the way –Use N identical stages, each limited to 25dB max boost, cascaded –Design each stage boost for 1/N boost of the total boost needed (e.g. 50dB total boost with 2 stages would be 25dB / stage) Use the Excel Solve function to minimize the difference between stage gain and cable attenuation with iteration –Define the Delta Function^2: (Difference between gain and cable attenuation)^2 –Select this as Set Objective and use Min –Specify the variables to change (in this case the R, and C values of the banks of one of the N stages) –Assign constraints –Click Solve –Start from low frequency and work your way upwards –May have to do this several times in succession Verify your design using Spice Build the amplifier and test with the cable Excel Solve window and its Settings 1-of-2-stages response on the left vs. 2-cascaded stages on the right

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Component Design (Numerical Solution) From Top LHS to Bottom RHS: –Used nominal starting values for R, and C (10k and 10pF) –Consecutive Solve iterations –Starting from low frequency and working upwards –May need to do several iterations –May decide to reduce # of banks if impact is low

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These are the entries that need to be made in the Excel: –RF –RG –Bank values (R, and C) which is 4 in this case –N stages –Cable attenuation @ 10MHz for the length being used Excel will compute the Gain and Cable attenuation plots and uses the following: –Delta^2= (Gain-Attenuation)^2 @ each frequency –Delta_Sum= Cumulative sum (from low to high frequency) of all Delta^2 terms This is the value which you use Solve, from low to high frequency, to optimize thereby driving down the total error between gain and attenuation Bank values, and RG are the variables available for manipulation by Solve –RF should be kept constant as it is usually not something with a lot of flexibility 9 Component Design (Numerical Solution)

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Amplifier Choice Desirable characteristics –Current feedback type (less effect on response with high frequency boost) –Here is the expression for current feedback closed loop gain: –Rf+Ri(1+Rf/Rg) can be defined as Feedback Transimpedance –Lower Ri (internal buffer output impedance) to lower unwanted response effects –Appropriate slew rate and large signal bandwidth for the signal amplitude and frequency –Capable to supporting the swing and output current needed 10

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Simulation of Excel Results The pole of these RC banks is the frequency beyond which the capacitor is considered a short and the bank impedance is equal to R Looking at the poles for the Excel solution, the most likely pole would be the 1.7GHz one (R1, C1) By reducing this pole, one has the highest chance of eliminating the phase margin issue that simulation has uncovered The blog also investigates the design in TINA 11

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THANK YOU

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