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Resistanc and capacitance extraction

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Junus2012

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Dear friends and teachers

in this image , a block with internal capacitance and resistance. Kindly, what is the simulation and measurements method to extract the value of both

Note: consider please the only two ports from the circuit

Thank you

 

You can excite with an AC voltage source and measure the current or use a current source and measure the voltage.
then you know that:
V/I = Z = R/ (1+ jwRC) that is the impedance of the RC circuit.
w = 2* pi * frequency
Take into account that V, I and Z are complex numbers. So you have to measure magnitude and phase.
Calculate V/I from your measurements. You also know at what frequency the measurement took place.
Rearrange the expression of Z and convert V/I so both have the form a+jb.
Re(V/I)= Re(Z) -- gives R
Im(V/I)=Im(Z) -- knowing w allows you to calculate C.

Hope that this is not very confusing
 
dear albert

thank you for your explanation

But how you assured that the real part gives you only the resistance, i think it cant be coz

real = R/root(1+sqr(WRC)), so you see from the real part that both variables (R and C) are still together


albert22 said:
You can excite with an AC voltage source and measure the current or use a current source and measure the voltage.
then you know that:
V/I = Z = R/ (1+ jwRC) that is the impedance of the RC circuit.
w = 2* pi * frequency
Take into account that V, I and Z are complex numbers. So you have to measure magnitude and phase.
Calculate V/I from your measurements. You also know at what frequency the measurement took place.
Rearrange the expression of Z and convert V/I so both have the form a+jb.
Re(V/I)= Re(Z) -- gives R
Im(V/I)=Im(Z) -- knowing w allows you to calculate C.

Hope that this is not very confusing
Click to expand...
 

I see the point. The eq. gets complicated.
Try using the admittance:
I/V= 1/Z=Y= (1/R)+( JwC)
Z should give the same results I will to try to solve it.
 

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