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So I've been trying to understand how an RC lowpass filter works, and I think I got it, but I just need some clarification.

enter image description here

My understanding: So an RC Lowpass filter in a nutshell is just a voltage divider made to filter out higher frequency signals. The cutoff frequency is dependent on the R and C values of the filter.

The equation for the impedance of a capacitor is 1/jwC. So with low frequency signals (i.e. 20 Hz), the impedance of the capacitor will be high. Now looking at our voltage divider, since we have a large impedance connected to ground, Vout should be close to Vin.

Now let say we have a high frequency signal (i.e. 1GHz). Here, the impedance of the capacitor will be low. And because of that, Vout will instead be a small value, close to 0V.

Can anyone clarify if my thought process is right?

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  • \$\begingroup\$ what does low signals mean? ... low voltage? ... low current? ... what does high signal mean? ... please edit your question with correct terms \$\endgroup\$
    – jsotola
    Jan 2, 2020 at 4:21
  • \$\begingroup\$ What do you mean with high signal? High magnitude or high frequency? If it is "high" frequency, it would make the impedance of the capacitor (1/jwc) tend to zero, thus creating a short between Vout and ground, making Vout ~ 0. \$\endgroup\$
    – MPA95
    Jan 2, 2020 at 4:23
  • \$\begingroup\$ The important thing to remember is when Zc(f) = R then in theory the attenuation is predictable at -6dB/octave or -20dB/decade. In practise the length is inductive ~ 0.5 to 1 nH/mm which limits the upper useful freq. \$\endgroup\$ Jan 2, 2020 at 4:30
  • \$\begingroup\$ Your understanding appears to be correct. The filter is set up like a frequency dependent voltage divider using the resistor and capacitor impedance. As the frequency increases the circuit divides the input by a larger amount due to the reducing capacitor impedance. \$\endgroup\$
    – user4574
    Jun 13, 2021 at 3:35

2 Answers 2

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A simpler and perhaps more intuitive explanation...

As the input moves up and down, the capacitor will charge and discharge via the resistor.

If the input to the resistor moves up and down quickly (high frequency) then the capacitor will not have chance to charge very far and the output will be small.

If the input to the resistor moves up and down slowly (low frequency) then the capacitor will get a better chance of charging to a higher peak voltage and the output will be larger in amplitude.

How far the capacitor is able to charge on each input cycle is determined by the input frequency and the RC time constant.

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You can look at this type of a filter as voltage dividers.

enter image description here

$$ V_{OUT} = I*X_{C1} = V_{IN} * \frac{X_{C1}}{R_1 + X_{C1}}= $$

This circuit should be treated as a voltage divider where the role of the variable resistance is taken by the capacitor. At low frequencies the capacitor has a high reactance (\$X_C\$), so almost the entire input voltage is transferred to the output. But when input frequency increases the reactance of a capacitor decreases and thus causes the output voltage to decrease.

At frequency at which reactance of a capacitor is equal to the value of the resistor, the output voltage is equal:

\$ V_{OUT} = 0.707*V_{IN}\$

And this happens at frequency equal:

\$F = \frac{1}{2 \pi RC} \approx \frac{0.16}{RC}\$

The voltage divider voltage ratio (\$\frac{V_{OUT}}{V_{IN}}\$ gain) of a voltage divider that contains two resistors doesn't depend on frequency, because the resistance of resistors does not change with frequency. But in our case, the divider's voltage ratio changes with frequency. Because \$R\$ is unchanged but capacitor reactance \$X_C\$ is changing with signal frequency.

And the voltage gain is equal to:

$$ A_V = \frac{V_{OUT}}{V_{IN}} = \frac{X_C}{Z} = \frac{\frac{1}{\omega*C}}{\sqrt{ R^2 + (\frac{1}{\omega *C})^2}}=\frac{1}{\sqrt{R^2*\omega^2 *C^2 +1)}} $$

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  • \$\begingroup\$ In your first equation do you mean -jXc? \$\endgroup\$
    – Chu
    Jan 2, 2020 at 8:27

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