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Numerical Cable Equalization Hooman Hashemi 4/9/13 1
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Numerical Cable Equalization. Hooman Hashemi4/9/13. Cable Driving & Equalization. Good 2 rule of thumb: Linear increase of dB‘s with cable length Number of dB‘s increases with square root of F(MHz) Solutions we offer Use amplifiers to boost the higher frequency - PowerPoint PPT Presentation

### Transcript of Numerical Cable Equalization

Numerical Cable Equalization

Hooman Hashemi 4/9/13

1

2

Cable Driving & Equalization• Good 2 rule of thumb:

– Linear increase of dB‘s with cable length

– Number of dB‘s increases with square root of F(MHz)

• Solutions we offer– Use amplifiers to boost the higher

frequency– Divide extreme equalizing designs

– Excel sheet to compute the component values

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)

Component Design• Numerical Solution

– Application note discusses this: 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

10Hz 100Hz 1KHz 10KHz 100KHz 1MHz 10MHz 100MHz-88dB

-80dB

-72dB

-64dB

-56dB

-48dB

-40dB

-32dB

-24dB

-16dB

-8dB

0dB

8dB

-210°

-180°

-150°

-120°

-90°

-60°

-30°

30°

60°

90°

120°

150° V(cable_loss) V(output)Cable Loss & Overall Gain (2 banks)

Response with 4-banks

10Hz 100Hz 1KHz 10KHz 100KHz 1MHz 10MHz 100MHz-81dB

-72dB

-63dB

-54dB

-45dB

-36dB

-27dB

-18dB

-9dB

0dB

9dB

18dB

27dB

-210°

-180°

-150°

-120°

-90°

-60°

-30°

30°

60°

90°

120°

150° V(cable_loss) V(output)

Response with 2banks

2-banks response on the left compared to 4-banks on the right

Equalization Stage with 2 Banks

Component Design (Numerical Solution)• First need to compute the expression

for gain• Can look difficult – at first• Can be simplified by considering

equalization banks as “admittances” in parallel

• Use Excel’s Complex arithmetic• Compute ” for each bank for all

frequency of interest (in columns) and invert it to get the admittance of each bank

• Sum all admittances, add to 1/RG, and invert to get ZG_eq

• Having obtained ZG_eq, divide RF by it and add 1 to get “Vout/Vin”

• Now you have the circuit’s gain expression plot

• With more banks, copy and paste the cells to expand the spreadsheet

f (Hz)

Complex Complex Complex

R1+1/sC1 R2+1/sC2 R3+1/sC3

50,000,000

5-318.309886183791i

20-159.154943091895i

420-10.6103295394597i

65,000,000

5-244.853758602916i

20-122.426879301458i

420-8.16179195343053i

Component Design (Numerical Solution)

f (Hz)

Cell D11 Cell E11 Cell F11

R1+1/sC1 R2+1/sC2 R3+1/sC3

50,000,000

5-318.309886183791i

20-159.154943091895i

420-10.6103295394597i

65,000,000

5-244.853758602916i

20-122.426879301458i

420-8.16179195343053i

f (Hz)

Cell D11 Cell E11 Cell F11

R1+1/sC1 R2+1/sC2 R3+1/sC3

50,000,000 =COMPLEX =COMPLEX =COMPLEX

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/RG and invert in one shot to get ZG_eq

• Continue using complex functions like “IMDIV”, “IMSUM”, and “IMABS” to get to the “1+RF/ZG_eq” which is the circuit gain expression

• Superimpose plot of gain on cable attenuation to optimize design

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

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

9

• 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

Component Design (Numerical Solution)

10

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

11

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