Low-pass filter with C but no R

carbonzit

Well-Known Member
This is something that's bugged me for some time now.

I hasten to point out that IANAEE (I am not an electronics expert); I know enough to get into trouble, and maybe a little more. Hence my puzzlement on this point.

In another thread there's a discussion of why, when a guy added a series resistor to the input of an amplifier, it degraded the sound quality by what sounded like attenuating high frequencies, the classic effect of a low-pass filter. The theory advanced was that by adding the series resistor, it formed a low-pass filter with an already-existing capacitor at the input of the amp.

This does not make sense to me. Let me explain why.

This shows the classic low-pass filter with both a C and R, but also simply a C:



We know the C-R network forms a low-pass filter. But doesn't a C by itself also form one? It's simply the degenerate case where R=0, right? It will still function to attenuate high frequencies. Look at power supplies that have filter capacitors but no series resistors (some have them, some don't). There's your R-less LPF.

Just to be clear, I am not suggesting that one doesn't ever need an R to make a LPF; it's certainly needed to create the proper response curve, rolloff and cutoff frequency. But if that amplifier already had a C across its input, I don't see how adding an R would suddenly create a LPF where none existed before. It would certainly change the response of the filter, but my guess is that it would have less effect, overall, than the C already has.

So what am I missing here? I'll let the experts here respond.
 
Solution

No, but if a circuit adds a capacitor like that to act as an LPF, then it's because it's relying on a suitable source impedance - in your case the circuit is meaningless, because it only shows part of the circuit.

It's quite simple, for such a crude LPF as this - simply calculate the impedance of the capacitor at the frequency you require, then the series resistor (or source impedance) needs to be that resistance.

For example, a 1000Hz LPF with a...
Because your answers were so theoretical and/or useless that it made the rest of what you said hard to believe or take seriously.

You mean a zero Ohm resistance is too theoretical for you to understand? How on earth did you learn about a supernode in circuit analysis then?

Note I am not arguing with anyone here just trying to understand how they came up with their point of view, that's all. I also like to understand things like that in addition to electrical theory, or maybe I better say electrical stuff

My main question was why would someone come to ask a question that they do not understand yet and then reject the answers of several people here with many years of experience. And that includes Nigel who has tons of practical experience with consumer devices!
 
Oh, please.
Saying a short caused by a poorly etched board is not a short because the copper has a resistance is comical. As per the rest of what you said, you are simply trying to reverse the process and tell me I need to know the theory to start. The comical part, you've ever designed a product and brought it to a manufacturable process - the first 1 to 10% of the effort is about the theoretical first pass (done in a few days or hours). The reality takes the time - that's what I call designing a PRODUCT. When you claim I use theory already. That's designing a concept. I'm sorry you have trouble understanding the difference.
 
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