How can I Thevenize the following circuit looking into the out terminal:

enter image description here

I want to deduce this circuit to a source and a single series impedance.

Here what I tried:

\$\frac{R1\cdot R2}{R1+R2}\$ is in series with C1

This becomes \$\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 }\$

And now \$\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 }\$ is in parallel with R5

I cannot proceed more. How could this be achieved if mine is wrong as an alternative way?

is the following correct?: \$\frac{R1\cdot R2}{R1+R2}\$ is in series with C1

\$Z_{thevenin} = \frac {\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 } \cdot R5} {\sqrt{\frac{R1\cdot R2}{R1+R2}^2 + \frac{1}{\omega C1}^2 } + R5}\$

And the Thevenin voltage needed to be found as well.

Vth = V1 * ((|R5+(1/jwc)| // R2)/ (R1 + (|R5+(1/jwc)| // R2))) * R5/|R5+(1/jwc)|

  • \$\begingroup\$ Why can't you proceed more? Do you know how to use phasors? Do you know the frequency of operation? \$\endgroup\$ – Elliot Alderson Mar 26 '19 at 12:02
  • \$\begingroup\$ @ElliotAlderson Is the stage where I am at correct? Frequency is ω so it is variable. \$\endgroup\$ – user1999 Mar 26 '19 at 12:03
  • \$\begingroup\$ You have lost the phase information. Does that matter to you? I would use impedances and phasors. Do you know how to use phasors? \$\endgroup\$ – Elliot Alderson Mar 26 '19 at 12:06
  • \$\begingroup\$ Yes I know how to use phasors, phase does not matter just the amplitude \$\endgroup\$ – user1999 Mar 26 '19 at 12:06
  • 1
    \$\begingroup\$ Looks to me like you're overcomplicating things. What if C1 was a resistor, could you then solve this? If yes, replace C1 with an impedance \$Z_{C1}\$. Then treat \$Z_{C1}\$ as if it is a resistor. Apply Thevenin. Then fill in \$Z_{C1} = 1/j\omega C_1\$ and you have your answer. \$\endgroup\$ – Bimpelrekkie Mar 26 '19 at 12:31

If I do not make any mistake in the math the correct equation for \$Z_{th}\$ will look like this:

$$Z_{th} = \frac {\sqrt{R5^2 + (\omega\ C_1\ R_T\ R_5)^2}}{\sqrt{1+(\omega\ C_1 (R_T+R_5))^2}}$$


$$R_T = \frac{R_1 \cdot R_2}{R_1+R_2}$$

And the Thevenin voltage

$$Vth = V_{IN}\cdot\frac{R_2||R_5}{R_1 + R_2||R_5}\cdot \frac{\omega\ (R_T +R_5)C_1}{\sqrt{1 +(\omega\ (R_T + R_5)C_1)^2} }$$

  • \$\begingroup\$ Can you also add Vth? Then it would be complete and to reduced one source and a single impedance. \$\endgroup\$ – user1999 Mar 26 '19 at 17:08
  • \$\begingroup\$ I believe it should be handled down a canonical ratio of polynomials to have a better understanding of what this impedance does over frequency \$\endgroup\$ – carloc Mar 26 '19 at 17:51
  • \$\begingroup\$ I add Vth voltage. \$\endgroup\$ – G36 Mar 26 '19 at 18:11

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