0
\$\begingroup\$

I am getting confused with LC-filters (also with emi-filters). Capacitors eliminate noise from signals. It bypass it to ground or noise source. The inductor does not allow high frequency to pass and also reduce current noise and I think it is correct but not sure??

Inductor reactance are linearly promotional to frequency and it is opposite for capacitive reactance. I am also familiar with frequency resonance in RLC ... I studied first order RC-filters one time. I know how they work.

When I think of LC filters first it comes to my mind is that capacitor limit frequency but not sure what the inductor does?? Signals are DC square waves!!

Could you please tell me which one limit frequency and which one eliminate noises ??

:(

\$\endgroup\$
2
  • \$\begingroup\$ Both are frequency dependent and can operate together to eliminate unwanted noise. What is your specific use case? There are a number of basic configurations (L, T and PI). \$\endgroup\$ Sep 5, 2019 at 15:51
  • \$\begingroup\$ This question and answer is a good start: electronics.stackexchange.com/questions/332458/… \$\endgroup\$ Sep 5, 2019 at 16:01

1 Answer 1

3
\$\begingroup\$

Here is a PI filter:

schematic

simulate this circuit – Schematic created using CircuitLab

When the frequency is very low, the capacitors are almost open circuit and the inductor is almost a short circuit.

As the frequency increases, the capacitors become a lower reactance (thus sinking some of the input signal to ground) and the inductor increases its reactance, limiting the flow of current to the output.

The reactance of a capacitor is \$ \frac {1} {2 \pi fC}\$ and therefore decreases with increasing frequency. The reactance of an inductor is \$2 \pi fL\$ and therefore increases with increasing frequency.

So the capacitors and the inductor are achieving the same thing (they form a low pass filter); a series inductor and a parallel capacitor can both form a low pass filter on their own in conjunction with a resistor.

The cutoff frequency assuming C1 and C2 are equal (they normally are but not always) is:

\$Fc = \frac {1} {2\pi \sqrt LC}\$

You can derive this:

The cutoff frequency is when \$ \frac {1} {2 \pi fc} = 2 \pi fl\$ and is fairly simple algebra to simplify.

\$\endgroup\$
2
  • \$\begingroup\$ Hey Peter! I think you gave me the right answer or what I am looking for. I just need to think enough of the links you have shared. By '' low frequency capacitors are almost open '' you are saying this because of high impedance?? \$\endgroup\$
    – PMP
    Sep 5, 2019 at 16:23
  • \$\begingroup\$ The impedance is a result of the applied signal frequency; at DC capacitors are (for a perfect device) an open circuit and a perfect inductor is a short circuit. \$\endgroup\$ Sep 5, 2019 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.