Cable-to-cable crosstalk (Capacitative Coupling)

[wu_weidong]

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.

Homework Equations

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

CCC(dB) = 20 log (V_V/V_C)

where V_C is the culprit source line voltage and V_V is the victim load line voltage.

The Attempt at a Solution

The solution given is as follows.

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

C_CV = 12 pF/m (read from a graph), and [ωC_CVl]^-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 V_V/V_C, which means I need to find Z_V. But to do that, I need to find C_V, 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.

 

[rude man]

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 V_V/V_C, which means I need to find Z_V. But to do that, I need to find C_V, which is the capacitance per unit length of the victim wire, and I don't know how to get this value.

You just calculated f_c and you know C_CV 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.

 

[wu_weidong]

You just calculated f_c and you know C_CV and l, so what's the problem? The given formula is kind of obvious and the answer is indeed close to 2K ohms.

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

I thought the combined resistance Z_V is calculated as Z_V = [1/100 + 2000 + 1/100]^-1 = 0.0005 if C_V = C_CV. This gives me the voltage divider Z_V / (Z_V + [jwC_CV]^-1) = 0.005/(0.005 + 2000) = 2.5 * 10^-7, which doesn't give me the CCC value of -26.4dB.

 

[rude man]

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

[ωC_CV]l-1 = 2 kΩ.

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

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

if C_V = C_CV. This gives me the voltage divider Z_V / (Z_V + [jwC_CV]^-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.