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Parasitic capacitance due to pulse of voltage.

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electricity86

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If two traces are close to each other, and in one trace there is suddenly a pulse of voltage, then a capacitance is created between the two traces.

I dont understand why, and what parameters that capacitance's size depends on.

I know that C = I / (dV/dt), C = Aε/d, and that Xc = 1 / (jωc).

I jst dont understand that, and would love some help.

Thank you.
 

MikeMl

Well-Known Member
Most Helpful Member
If two traces are close to each other, and in one trace there is suddenly a pulse of voltage, then a capacitance is created between the two traces.

I dont understand why, and what parameters that capacitance's size depends on.

I know that ... C = Aε/d ...
What is A in the formula above?

What is d in the formula above?

hint: it has to do with the physical layout and dimensions of the two traces, like length, spacing, thickness :D

It also has something to do with Maxwells equations and electric fields:(
 

dknguyen

Well-Known Member
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The capacitance is always there whether or not there is a pulse. THe pulse does not create the capacitance- that's a bit like saying "If I don't see it, it doesn't exist!" THe capacitance increases as the traces are moved closer together and the longer the distance that the traces run beside each other.
 
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electricity86

New Member
A is the area of the capacitor's plates - the area of the traces - and d is the distance between the plates - the traces.

Could you please explain what you said about Maxwell?

Because if C = Aε/d, than how does it relate to Maxwell?
 

Sceadwian

Banned
The capacitance isn't created by the voltage pulse, the capacitance is always there. The voltage pulse on one side allows the natural capacitance to cause a voltage transfer between the traces. Under normal conditions (lower voltage slower rise/fall times) the capacitance can be disregarded. High frequency or surge situations are very different.
 
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electricity86

New Member
The capacitance is always there whether or not there is a pulse. THe pulse does not create the capacitance- that's a bit like saying "If I don't see it, it doesn't exist!" THe capacitance increases as the traces are moved closer together and the longer the distance that the traces run beside each other.
So we will notice the capacitance's affect only when the frequency changes, and then
Xc = 1/(jωc) gets smaller?

Meaning that its not that the voltage's frequency that only matters, but in the same importance, the current frequency matters, because ω is the frequency of both of them.
Is it correct?
 
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MikeMl

Well-Known Member
Most Helpful Member
A is the area of the capacitor's plates - the area of the traces - and d is the distance between the plates - the traces.

Could you please explain what you said about Maxwell?

Because if C = Aε/d, than how does it relate to Maxwell?
Maxwell's theory explains capacitance based on electric fields in the general sense, and requires understanding of Calculus, and generally requires a finite element computer modelling to solve. C = Aε/d is a highly simplified equation (derived from Maxwell) that sort of works for a very simple case of a parallel plate capacitor. Here is a picture for you to think about:
 

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