Cable-to-cable crosstalk (Capacitative Coupling)

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SUMMARY

The discussion focuses on calculating cable-to-cable crosstalk due to capacitive coupling in a harness with specific parameters: a separation distance of 3 mm, cable diameters of 1 mm, and a 100 ohm impedance level. The Capacitive Cross-talk Coupling (CCC) is calculated using the formula CCC(dB) = 20 log (VV/VC), resulting in a value of -26.4 dB. Key parameters include a second corner frequency (fc) of 637 kHz and a victim load line voltage (VV) derived from the source logic of 3.5 V. The noise level calculated is -15.4 dBV, which is below the threshold of 0.4 V, indicating acceptable switching conditions.

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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.