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Effect of high frequency current on resistance 
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#1
Aug2710, 12:53 PM

P: 799

Hi,
I found something quite interesting about resistance in my book. It says that at high frequency (MHz scale), resistance R decreases due to "inductance L" and "capacitance C" characteristics of resistor. So the resistor can be modeled as (Ro in series with L) // C (Ro denote the resistance in the model as to distinguish from R, which is the real effective resistance). The model can explain mathematically why R decreases when frequency f increases. However, what I'm concerned about is the nature of the phenomenon. In any geometry of the resistor, there should be magnetic field, and that accounts for L. But if the resistor is just a straight wire, not a coil, how should we explain the existence of C? Besides, is there any paper or text analyzing this effect theoretically? Thank you very much. 


#2
Aug2710, 01:11 PM

PF Gold
P: 2,011

Hello again hikaru.You may find an answer by googling self capacitance and parasitic capacitance.



#3
Aug2710, 01:56 PM

P: 799

Thank you, Dadface
I think if it has something to do with self capacitance then there should be charge buildup somewhere in the circuit (and that also means there is charge depletion somewhere else). How can that happen? And about parasitic capacitance, I guess we have to consider the geometry of the whole circuit. If so, then this effect isn't just on the resistor only, i.e. the same resistor put into different circuits will lead to different results. Am I correct? 


#4
Aug2710, 05:02 PM

PF Gold
P: 2,011

Effect of high frequency current on resistance



#5
Aug2710, 11:32 PM

P: 799

Thanks.
It seems to me that in continuous conducting medium, there shouldn't be charge buildup. Maybe self capacitance is just a minor factor. I was just wondering, if this has something to do with skin effect. Because at frequency in MHz range, for common conductor like copper, skin depth is just about 10^2 mm. That means, there is nonuniform charge distribution along the conductor, so inside the conductor, there exists Efield. 


#6
Aug2810, 02:25 AM

PF Gold
P: 2,011

Consider the resultant B field at any point in the circuit.The field will interact with the charge carriers and deflect them at 90 degrees to the current as in accordance with the left hand rule.An equilibrium will be reached where the electric force on each electron is balanced by the magnetic force.Basically the circuit is inducing a Hall Voltage the value of which at each point depends,amongst other things,on the value of B at that point.Since there can be differing charges at points of different potentials then there is in,in effect,a capacitive effect.Anyway,I have only just started thinking about this.I'm away for several days and shall get back to it.



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