Timeline for Crosstalk: How is the victim node voltage increasing through the leakage current?
Current License: CC BY-SA 4.0
5 events
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| Oct 10, 2022 at 10:44 | comment | added | Criticizing Israel not allowed | @KEE97 Suppose I have a circuit with a resistor, capacitor, another resistor. Let's say 100\$\mu\$F and 100k\$\Omega\$ (quite big so the time constant is long). Then I tap my magic wand on one side of the capacitor and add some voltage to it. In the next microsecond, what happens to the voltage across the capacitor? Which equation calculates the voltage across a capacitor? | |
| Oct 10, 2022 at 7:58 | comment | added | danmcb | perhaps it might help the OP, in addition to the excellent explanation above, to consider that at the moment when the transistion occurs there is a a short burst of HF signal in the frequency domain which is not there when the signal is in steady state. | |
| Oct 10, 2022 at 3:14 | comment | added | Polynomial | There's no magic with the current. A current flows when there's a potential difference between two points. The magnitude of the current that flows is inversely proportional to the impedance. The impedance of a mutual capacitance is dependent on the frequency of the signals along with the physical properties of the two transmission lines and the environment. Ultimately it's about the behaviour of the electric fields around the transmission line; typically you would use a field solver to figure out what the coupling will look like for a given set of transmission lines on a PCB. | |
| Oct 10, 2022 at 3:05 | comment | added | KEE97 | This makes a lot of sense when I look at it like this and in terms of the formulas and math. I also wanted to think about it in terms of currents and properties of a capacitor and failed to understand, was wondering if you could shed some light! Thank you so much for your response. | |
| Oct 10, 2022 at 2:50 | history | answered | Polynomial | CC BY-SA 4.0 |