What is the output impedence of this RC Circuit

Question

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

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

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.

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

Answer 1

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.

Answer 2

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.