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I have this simple circuit. And from what I calculated the output impedance seen from the OUT node comes to about \$138.765 \Omega\$.

$$ X_c = \frac{1}{2\pi * 890*10^3 * 10^{-9}} = 178.8257\Omega $$ $$ Z_{out} = R2||X_c = \frac{220 * 178.8257}{\sqrt{220^2 + 178.8257^2}} = 138.765\Omega$$ enter image description here

With an input of 890K Hz sine wave of 2v pk-pk, the voltage at OUT node comes to is about 1.2615v pk-pk.

If I connect a resister of the same value of \$138.765 \Omega\$, the voltage at OUT node should be half ed.

enter image description here

But I see a voltage of 697.843mV instead of 630mV. What is the problem, am I calculating the output impedance wrongly?

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    \$\begingroup\$ Since you have a reactive component (the capacitor) in there, the output impedance is a complex number. In order to test this, you need a similar load impedance -- one that also includes a reactive component. \$\endgroup\$ – Dave Tweed Nov 12 '14 at 16:23
  • \$\begingroup\$ How did you arrive at \$\approx 138\Omega\$? \$\endgroup\$ – Null Nov 12 '14 at 16:26
  • \$\begingroup\$ @DaveTweed Thanks, I see. But why is the output 697mv? \$\endgroup\$ – Arjob Mukherjee Nov 12 '14 at 16:49
  • \$\begingroup\$ Because you have failed to account for the phase shifts, which are different with/without the load resistor. \$\endgroup\$ – Dave Tweed Nov 12 '14 at 17:02
  • \$\begingroup\$ Z= R+Jwc wc= capacitive reactance complex conjugate of impedance \$\endgroup\$ – user58210 Nov 13 '14 at 14:04
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The output impedance can be calculated from 1/(1/Xc + 1/R)

Xc = \$1 \over j\omega C \$

R = 220\$\Omega\$

The admittances add in quadrature since one is imaginary and the other is real- It is easy to show(tm) that the magnitude of the impedance Z is as follows:

|Z| = \$ R \cdot (\frac {1}{\omega C}) \over \sqrt{R^2 + \frac {1}{\omega^2 C^2}}\$

That is the reason you are seeing a higher voltage than you expect. The situation will be different again with the added resistor, so you have to consider the above equation with the two resistors effectively in parallel.

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  • \$\begingroup\$ I thought |Z| = (Xc*R)/sqrt(R^2 + Xc^2). \$\endgroup\$ – Arjob Mukherjee Nov 12 '14 at 17:16
  • \$\begingroup\$ @ArjobMukherjee Thanks, fixed. I'd prefer not to write Xc here, because the problem is coming up from ignoring that Xc is imaginary, and Xc is used both ways. \$\endgroup\$ – Spehro Pefhany Nov 12 '14 at 17:47
  • \$\begingroup\$ That is how I have calculated the impedance. How do I account for the phase shift? \$\endgroup\$ – Arjob Mukherjee Nov 12 '14 at 18:33
  • \$\begingroup\$ If you're just measuring the voltage, the phase is of no importance, but as I said you have to add the admittances in quadrature, so use the capacitive reactance and the Thevenin equivalent resistance (the two in parallel). \$\endgroup\$ – Spehro Pefhany Nov 12 '14 at 19:20
  • \$\begingroup\$ I am sorry, but I do not understand what In-phase and quadrature components are. Can you suggest a book or something? \$\endgroup\$ – Arjob Mukherjee Nov 13 '14 at 14:20
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I finally understood what was wrong.

Without the load resistor, the impedance seen at the OUT node is a complex number. $$\dfrac{1}{Z_{out}} = \dfrac{1}{R2} + \dfrac{j}{X_c} $$ Thus, $$ Z_{out} = 87.527 - j107.678\Omega$$

With that Load resistor of 138.765 \$\Omega\$, the total impedance of the circuit becomes, $$ Z_{total} = 138.765 + 87.527 - j107.678$$ $$= 226.292 - j107.678\Omega$$ $$|Z_{total}| = 250.604 \Omega $$

If the unloaded voltage at the OUT node is 1.26v, then, the voltage at the OUT node with the LOAD resistor is a simple voltage divider, and

$$V_{out} = \dfrac{1.26 \cdot 138.765}{|Z_{total}|} $$ $$V_{out} = \dfrac{1.26 \cdot 138.765}{ 250.604} = 697.68mV. $$

The answer checks out, with acceptable error.

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