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I configured a simple series RC circuit. Across the circuit I have placed the terminals of a sinusoidal AC voltage source (I'm using my phone as a function generator). My goal has been to measure the resistor/voltage gain and compare my results with theory.

Data was measured using the RMS voltage setting on a voltmeter across the source $(V_{\rm in})$ and across the resistor $(V_{\rm out})$. I calculated the gain as $V_{\rm out}/V_{\rm in}$, and plotted it against the frequency being generated. I have attached the plot (Figure 1) along with a logarithmic trend line.

Note: $C=10\,\rm \mu F$ and $R=50\,\Omega$.

enter image description here

The trend in Figure 1 is expected; lower frequencies get cut more than high ones. It is to my belief that my issue then resides in my attempts to generate the same plot using theory.

I am looking for a sort of "ground up" explanation of this circuit. So far this is all I have:

If I represent by input signal as $V(t)=A\sin(\omega t),$ where $A$ is the maximum voltage, $\omega$ is the $2\pi$ of the frequency, and $t$ is time, then the current through the circuit as a function of time should be

$$I=\frac{A}{Z}\sin(\omega t+P)$$

where $P$ is the complementary angle to the phase angle given by $\arctan[1/(\omega RC)],$ and $Z$, the impedance, is given by

$$Z=\left(R^2 + \left(\frac{1}{\omega C}\right)^2\right)^{1/2}.$$

After this point, I'm not sure how to get functions of time that describe the voltage across the resistor and capacitor (assuming my equation for current is correct).

Where do I go from here?

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  • $\begingroup$ On the experimental side of things, 50 ohms is a rather small value, and might require more current than your phone is happy putting out. I would suggest 5k ohms, with a C of .1 microfarads. $\endgroup$ Mar 1 at 14:53

1 Answer 1

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When it comes to calculating the response of a circuit to a sinusoidal signal, it is often useful to use the complex representation of signals. We represent a given signal $S = S_0 \sin(\omega t + \phi)$ with $\underline{S} = S_0 \exp(j(\omega t + \phi)) = \underline{S_0} \exp(j \omega t)$, where $j^2 = -1$ and $\underline{S_0} = S_0 \exp(j\phi)$. Then, $S = \Re(\underline{S})$, $\phi = \arg(\underline{S_0})$ and $S = |\underline{S_0}| = |\underline{S}|$.

In your case, you have

$$\begin{cases} \underline{V_{in}} = \underline{V_0} \exp(j\omega t) \\ \underline{V_{out}} = \frac{R}{R+\frac{1}{C\omega j}} \times \underline{V_{in}} = \frac{1}{1+\frac{1}{RC\omega j}} \times \underline{V_{in}}\\ \end{cases}$$

I've found the relation between $\underline{V_{out}}$ and $\underline{V_{in}}$ using a voltage divisor, knowing that the impedance of a capacitor is $\frac{1}{C\omega j}$ (you can try to find it on yourself, it is a good exercise).

Now, we have

$$ \underline{G} = \frac{\underline{V_{out}}}{\underline{V_{in}}} = \frac{1}{1+\frac{1}{RC\omega j}}$$

which gives

$$ G = |\underline{G}| = \frac{1}{\sqrt{1+(\frac{1}{RC\omega})^2}}$$

and $$ \arg{G} = \arctan(\frac{1}{RC\omega})$$

so $$ V_{out} = \Re(\underline{G}\times \underline{V_{in}})= \Re(\underline{V_0}G\exp{j(\omega t + \arg{G})})$$

Finally, $$V_{out} = \frac{V_0}{\sqrt{1+(\frac{1}{RC\omega})^2}}\sin(\omega t + \arctan(\frac{1}{RC\omega}))$$

where $V_0$ is the peak input voltage value.

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  • $\begingroup$ Where A is the peak input voltage? $\endgroup$
    – Ben
    Apr 29, 2017 at 6:13

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