Low-pass filter with C but no R

[carbonzit]

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.

 

[Nigel Goodwin]

OK, I guess I'm outta here, or at least moving to the sidelines with my big bag of popcorn while all you long-bearded Pharisees argue heatedly about how many joules can dance on the head of a capacitor lead ...

Have at it! Meanwhile I'll be safely comfortable with my close-enough assumptions here ...

You mean your completely wrong assumptions, about LPF's

 

[ZipZapOuch]

You mean your completely wrong assumptions, about LPF's

You still haven't answered how much resistance a short must have. Tick, tick, tick...

 

[MrAl]

You still haven't answered how much resistance a short must have. Tick, tick, tick...

A short in theory means 0 resistance. That's 0.00000 with infinite number of zeros after the decimal point.
It does not matter if there is 1 picoohm, that's not a short when we consider pure theory. Sure, in practice we will most likely never see that nor use it, but that's theory for ya

Note that when we read about circuit analysis we might see "To create the supernode we short point A and point B together, then do the analysis."
It does not say, "To create the supernode place a 1 picoohm resistor across point A and point B, then do the analysis".

How is this different? It's different in a variety of ways, but of course it's relative just like with the RC low pass filter question.
Let's say we have a voltage divider with R1 and R2 and input voltage 10 volts. The formula would be:
Vout=Vin*R2/(R1+R2)
Now say we had a gap between the battery (Vin) and R1. That would mean no current flow yet. Next we would say "Short out the gap and do the analysis". When we do that, we get the formula:
Vout=Vin*R2/(R1+R2)
as before.
What happens if we use some small resistor value (like a 1picoohm resistor) instead of what we just called a 'short'?
Here's what we get:
Vout=Vin*R2/(R1+R2+R3)
We had to introduce R3 because we don't yet know if it's 1 picohom or 2 picoohms etc.
Of course this is pure theory and probably not practical except in a lab where it matters.

So sometimes we either have a resistor or we don't but this is not the only case where this may come up. If you recall, there is such a thing as a "zero ohm resistor" in an SMD package, and also I think in a 1/4 watt through hole component (maybe others). In that case it's a very practical thing.

So again we run into the casual conversation vs the pure theoretical conversation. In some cases we call it a short and mean a direct short with no exceptions, and other times it does not matter if we use a 1mOhm resistor even.

If you don't feel comfortable with this that's perfectly fine, but sometimes when someone says "a short" they mean an actual short with no resistance whatsoever, and no inductance, no capacitance, no external fields, etc.

If we turn to pure physics on the other hand, we may be able to prove that there is no such thing as a short. That would be a practical result if we could, and yet circuit theory would still go on to say that a short is a perfect zero ohms. The difference here is physics theory vs circuit theory. They both have associated assumptions. If someone says a short is a perfect zero Ohms, then they are talking within the context of circuit theory in its purest form. There is no reason to reject it within that theoretical scope.

 

[MrAl]

OK, I guess I'm outta here, or at least moving to the sidelines with my big bag of popcorn while all you long-bearded Pharisees argue heatedly about how many joules can dance on the head of a capacitor lead ...

Have at it! Meanwhile I'll be safely comfortable with my close-enough assumptions here ...

I guess I do not understand why you asked the question if you intended to go on assuming that your original definition was never going to be questioned. In short, you asked, we all told you

Here's a famous quote from Feynman:
"I would rather have questions that can't be answered than answers that cannot be questioned".

 

[danadak]

Nigel

I presume you meant 100KHz? (not 100Hz), as that's the standard test frequency for ESR measurements.

I looked at several 10K uF datasheets, all ESR values were done at 100 Hz.
Or tables at 120 Hz, 20 Khz, seems like 100 Khz is not THE standard test
frequency.....

 

[carbonzit]

Here's a famous quote from Feynman:
"I would rather have questions that can't be answered than answers that cannot be questioned".

Just keep in mind that this applies as well to your answers.