3
\$\begingroup\$

I am trying to construct a simple low-pass filter for the following scenario: A high-voltage DC signal with lots of noise. I looked at Wikipedia's article on a simple passive low-pass filter. I understand how the capacitor "looks like a short" to high-frequency signals and open for DC. What I don't understand is the role of the resistor. I read what Wikipedia has to say, but am not quite grasping how the resistor influences the "time for the capacitor to charge up". I vaguely grasp the idea of resonant RC circuits, but don't really understand the relevance of that principle in this context. Any insight would be greatly appreciated.

Thanks in advance, Brian

\$\endgroup\$

3 Answers 3

Reset to default
3
\$\begingroup\$

Some Basics

One way to think about it is the resistor is setting a current limit on the capacitor. In an ideal situation with no resistance, you could potentially have very very large amounts of current being drawn in this "short" situation. Because of this you need to add a resistor. The resistor limits how much current is able to flow which results in the frequency characteristics changing.

Non-ideal Case

Now in the not so ideal case, everything has some resistance. This will include the wires/traces and the internal resistance of the capacitor. This allows us to use capacitors between power and ground to filter noise while not having huge spikes of current draw.

Mathematical Point of View

From a mathematical point of view, you can look at the transfer function of the system. The basic ideas of a transfer function comes from the simple electrical science concepts. If you had 2 resistors, you could solve for the voltage between the 2 resistors. In the case of a capacitor or inductor your "resistance" is no longer called a resistance, instead it is called an impedance. Essentially they are the same thing, except impedances typically have imaginary parts to them. In the case of a transfer function, imaginary parts of circuits are frequency dependent. If you solve the math you can find out what you expect the circuit to act like across many frequencies, and then plot it.

\$\endgroup\$
1
  • \$\begingroup\$ I see. This is assuming the system's power supply is voltage controlled I guess? Thanks for the explanation! \$\endgroup\$
    – BASnappl
    Mar 16, 2011 at 20:14
4
\$\begingroup\$

If you only use the capacitor you would have a load, like a resistor, only this time frequency dependent. The thing is that this load doesn't change the signal.
Enter the resistor. Now you have a voltage divider. For low frequencies the capacitor's "resistance" will be high compared to the resistor, so most of the signal will arrive at the output. For high frequencies the capacitor's "resistance" is low compared to the resistor, and the divider will attenuate the input signal: only a small portion of it will appear at the output.
I put the capacitor's "resistance" in quotes because it's not just a real resistance, it's more complex, in every meaning of the word..

\$\endgroup\$
2
  • \$\begingroup\$ I had the same concept which stevenvh explained here. Resistor and capacitor together acts like a voltage divider with the output taken across capacitor. Higher the frequency lower will be the capacitor's impedance and less will be the output voltage which is the drop across the capacitor. But doesn't the whole design breaks when we connect the output to another load? \$\endgroup\$
    – 0xakhil
    May 29, 2011 at 18:35
  • \$\begingroup\$ @oxakhil - If you connect a load to the divider you're indeed changing the divider's ratio. Usually you'll have some impedance parallel to the capacitor. If that's unwanted you'll have to place a high impedance buffer after the filter. \$\endgroup\$
    – stevenvh
    May 30, 2011 at 5:59
0
\$\begingroup\$

Here is a good explanation, using the RC filter transfer function.

\$\endgroup\$
4
  • \$\begingroup\$ Is there any nice way to figure the transfer function obtained by cascading resistors and capacitors? I know the normal style of filter uses one amp for every two orders, but I would expect that cascading N simple RC filters would yield a response somewhere between a first-order filter and an N-order filter. If R1C1 is the first stage, and R2C2 is the second, and R2 is much greater than R1 (C2 much smaller than C1), I would think the response would be the product of the R1C1 response and the R2C2 response (basically, a second-order filter). The general difficulty.... \$\endgroup\$
    – supercat
    Apr 11, 2011 at 14:56
  • \$\begingroup\$ ...of such a circuit would be that if R2 is not much greater than R1, the behavior of C2 will affect the signal at C1, and making each stage's resistance much larger than the previous stage's would be difficult if there were very many RC stages between amplifier stages. I designed a music box some years ago with an ad-hoc filter using four resistors and four caps; I wonder how good or bad the filter actually was (it sounded fine aesthetically). \$\endgroup\$
    – supercat
    Apr 11, 2011 at 14:59
  • 1
    \$\begingroup\$ this should be a comment. You are not explaining anything. \$\endgroup\$
    – Federico Russo
    May 30, 2011 at 5:35
  • \$\begingroup\$ the link is dead... \$\endgroup\$
    – PetPaulsen
    Jul 27, 2011 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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