Cable-to-cable crosstalk (Capacitative Coupling)

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Discussion Overview

The discussion revolves around calculating cable-to-cable crosstalk due to capacitive coupling in a harness involving two cable pairs. Participants explore the implications of various parameters such as cable separation, impedance, and voltage levels, while addressing specific questions related to the calculations and concepts involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how to derive the CCC dB value of -26.4 dB, indicating a need to find the capacitance per unit length of the victim wire.
  • Another participant suggests that the formula for CCC is straightforward and implies that the calculated values are close to expected results.
  • There is a discussion about the meaning of "source logic," with clarification that it refers to the source voltage from a logic integrated circuit.
  • Participants debate the significance of noise levels being below 0.4V, with one explaining that voltages below this threshold are interpreted as a logical "0."
  • Concerns are raised about the calculation of combined resistance ZV, with one participant providing a specific formula that does not yield the expected CCC value.
  • Another participant points out a potential error in adding impedances to admittances, questioning the validity of the voltage divider used in calculations.
  • Disagreement exists regarding the CCC value, with one participant obtaining a more negative result than -26.4 dB.

Areas of Agreement / Disagreement

Participants express differing views on the calculations related to CCC and ZV, indicating that multiple competing interpretations and methods exist without a consensus on the correct approach or values.

Contextual Notes

Participants mention specific parameters such as capacitance per unit length and resistance calculations, but there are unresolved assumptions regarding the values used and the methods applied in the calculations.

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Homework Statement


Compute the cable-to-cable crosstalk due to capacitive coupling in a harness between two cable pairs having an average separation distance of 3 mm and a 10 m in a cable tray. The cable diameters are 1 mm and both cables are operating at a 100 ohm impedance level. Assume h = 5 mm, source logic = 3.5 V with rise/fall times of 500 ns.

The lecturer showed the following diagram when he was talking about the capacitative coupling, and the equivalent (part of) circuit representation.

2818w5.png


Homework Equations


The Capacitive Cross-talk Coupling (CCC) is defined as

CCC(dB) = 20 log (VV/VC)

where VC is the culprit source line voltage and VV is the victim load line voltage.

The Attempt at a Solution



The solution given is as follows.

The second corner frequency, fc, in the spectrum of a pulse (or > pulses) corresponding to rise/fall time of 500 ns is: fc = 1/π τ = 637 kHz

CCV = 12 pF/m (read from a graph), and [ωCCVl]-1 = 2 kΩ
Thus CCC dB = -26.4 dB
Source logic = 3.5 V = 11 dBV, then
Noise = 11 dB – 26.4 dB = -15.4 dBV = 0.17 V < 0.4 V
Thus, switching OK.

I have several questions.

1) I don't know how to get CCC dB = -26.4 dB. I know I can use the voltage divider on the circuit representation to find VV/VC, which means I need to find ZV. But to do that, I need to find CV, which is the capacitance per unit length of the victim wire, and I don't know how to get this value.

2) What is "source logic"?

3) Why does noise < 0.4V imply "switching OK", and why "0.4V"?

Thank you.
 
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wu_weidong said:
I have several questions.
1) I don't know how to get CCC dB = -26.4 dB. I know I can use the voltage divider on the circuit representation to find VV/VC, which means I need to find ZV. But to do that, I need to find CV, which is the capacitance per unit length of the victim wire, and I don't know how to get this value.
You just calculated fc and you know CCV and l, so what's the problem? The given formula is kind of obvious and the answer is indeed close to 2K ohms.
2) What is "source logic"?
The source voltage. By "logic" we mean the source from a logic integrated circuit.
3) Why does noise < 0.4V imply "switching OK", and why "0.4V"?
Because if the "low" voltage limit is 0.4V then anything below that looks like a "0" and that's what you want here.
 
rude man said:
You just calculated fc and you know CCV and l, so what's the problem? The given formula is kind of obvious and the answer is indeed close to 2K ohms.

ZV is close to 2K ohms? Why is that?

I thought the combined resistance ZV is calculated as ZV = [1/100 + 2000 + 1/100]-1 = 0.0005 if CV = CCV. This gives me the voltage divider ZV / (ZV + [jwCCV]-1) = 0.005/(0.005 + 2000) = 2.5 * 10-7, which doesn't give me the CCC value of -26.4dB.
 
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wu_weidong said:
ZV is close to 2K ohms? Why is that?
[ωCCV]l-1 = 2 kΩ.

wu_weidong said:
ZV is close to 2K ohms? Why is that?

I thought the combined resistance ZV is calculated as ZV = [1/100 + 2000 + 1/100]-1 =

if CV = CCV. This gives me the voltage divider ZV / (ZV + [jwCCV]-1) = 0.005/(0.005 + 2000) = 2.5 * 10-7, which doesn't give me the CCC value of -26.4dB.
You're adding impedances to admittances! What is really your voltage divider?
BTW I don't get -26.4 dB. I get a somewhat more negative number than that.