# What is the output impedence of this RC Circuit

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?

• 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. – Dave Tweed Nov 12 '14 at 16:23
• How did you arrive at $\approx 138\Omega$? – Null Nov 12 '14 at 16:26
• @DaveTweed Thanks, I see. But why is the output 697mv? – Arjob Mukherjee Nov 12 '14 at 16:49
• Because you have failed to account for the phase shifts, which are different with/without the load resistor. – Dave Tweed Nov 12 '14 at 17:02
• Z= R+Jwc wc= capacitive reactance complex conjugate of impedance – user58210 Nov 13 '14 at 14:04

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.

• I thought |Z| = (Xc*R)/sqrt(R^2 + Xc^2). – Arjob Mukherjee Nov 12 '14 at 17:16
• @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. – Spehro Pefhany Nov 12 '14 at 17:47
• That is how I have calculated the impedance. How do I account for the phase shift? – Arjob Mukherjee Nov 12 '14 at 18:33
• 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). – Spehro Pefhany Nov 12 '14 at 19:20
• I am sorry, but I do not understand what In-phase and quadrature components are. Can you suggest a book or something? – Arjob Mukherjee Nov 13 '14 at 14:20

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.