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In an earlier post i discussed about the effects of voltage derating on a capacitor. From here i measured the output with DC and AC applied to a highpass filter and found the cutoff frequency with the applied voltage and calculated the capacitance change from these results.

Now i wish to calculate the capacitance from my input and output with a known resistance. I know the formula for the reactance of a capacitor, and if i recall correctly a Highpass filter is a voltage divider, where it attenuate lower frequencies. With this knowledge i was able to solve for the capacitance. enter image description here

What seems to the issue with my calculation is, the capacitance calculated from this seems to be double or even more than its normal value. I am not sure if my measurements are wrong, since the results are the same on different measurement equipments.

Edit: So i tried another approach at this, but the results still seems to be the same as before: enter image description here

Any suggestion on what might could have gone wrong?

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    \$\begingroup\$ You're treating the capacitor as a resistive component. Remember when you calculate the capacitor impedance (and voltage) it's going to be perpendicular to the impedance/voltage across the resistor. You'll have to use Pythagoras. \$\endgroup\$ – Cristobol Polychronopolis Mar 18 at 15:46
  • \$\begingroup\$ Thanks for the response, do you mind showing an example on how i can proceed to do as you suggested? \$\endgroup\$ – Vinh Trung Thai Mar 18 at 15:51
  • \$\begingroup\$ Again, just treat them as perpendicular vectors. If you measure the voltage across the resistor and add it to the voltage across the capacitor, your total won't be Vc + Vr, it'll be sqrt( Vc^2 + Vr^2 ) because they're 90 degrees out of phase. (The current through them is in phase because they're in series.) \$\endgroup\$ – Cristobol Polychronopolis Mar 18 at 15:58
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Your another approach to calculate the capacitance is correct, but you have made an algebraic mistake. I hope this helps.

enter image description here

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    \$\begingroup\$ Thanks!! The Results from this is closer to the results i expected. \$\endgroup\$ – Vinh Trung Thai Mar 20 at 7:35
  • \$\begingroup\$ While my results were very close, i can't seem to get the exact result. I tried running a simulation, and then took the results from the simulation and applied it to this formula. It resultet in a capacity slighty of the original value \$\endgroup\$ – Vinh Trung Thai Mar 20 at 12:06
  • \$\begingroup\$ There is no such thing as "exact" in real world. Your capacitor is an ideal but the capacitor used in simulators are modeled differently and it contains resistance and inductance too. You can read more about this here. Here is the capacitor model used in LTspice circuit simulator. This affects the value of capacitance and this is why you are not getting an exact value. \$\endgroup\$ – Yunus SA Mar 20 at 12:48
  • \$\begingroup\$ You're right about the exact in real world, and i do know that. But i assumed with the simulation results applied to the formula, i would get capacitance used in the simulation, since i expected it to be ideal. \$\endgroup\$ – Vinh Trung Thai Mar 20 at 13:12
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When Xc(f)=R you have the breakpoint f-3dB= amplitude at 70.7% of input where 20log 1/√(1+1)= -3dB. and the phase shift is 50% of 90 deg= 45 deg.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\omega_0{(-3dB)}= 1/RC = 2\pi f_0\$

You can prove this with algebra you learn in univ. I just showed the intuitive graphical way. as the cap represents -j1/ωC at -90 deg impedance (Xc) to voltage.

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