Hi all,
Assuming we have the circuit attached, with Vin amplitude = 240mV, Fin = 6Khz.
What I don't understand is the behavior of C3 on AC signals.
Everywhere I searched for an explanation I saw that C3 is calculated to have a very small reactance at the lowest frequency point so that to be a short circuit for the AC signals with the frequency above the lowest point.
But, what happen if this capacitor is not calculated like this? I want to generalize this problem for all capacitors in this circuit:
0,01uF means Xc= 1/(2*pi*F*C) = 1/(6.28 * 6000 * 10nF) = ~ 2,7k ohm.
I didn't find anywhere an explanation for this use case. What I want is to be able to do the math taking into account all capacitor reactances.
After a spice simulation it is obvious that the output is modifying according to these capacitors values, but I cannot "see" the equations (?!?)
Thanks for any help.
Assuming we have the circuit attached, with Vin amplitude = 240mV, Fin = 6Khz.
What I don't understand is the behavior of C3 on AC signals.
Everywhere I searched for an explanation I saw that C3 is calculated to have a very small reactance at the lowest frequency point so that to be a short circuit for the AC signals with the frequency above the lowest point.
But, what happen if this capacitor is not calculated like this? I want to generalize this problem for all capacitors in this circuit:
0,01uF means Xc= 1/(2*pi*F*C) = 1/(6.28 * 6000 * 10nF) = ~ 2,7k ohm.
I didn't find anywhere an explanation for this use case. What I want is to be able to do the math taking into account all capacitor reactances.
After a spice simulation it is obvious that the output is modifying according to these capacitors values, but I cannot "see" the equations (?!?)
Thanks for any help.
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