Understanding HP/LP Filters

[akmenpi]

I was taught that high frequency sees capacitors as a short, and DC sees capacitors as an open. This makes sense to me. However I struggle to apply this concept to filters.

Why are low pass filters wired like A and not like B? (see photo)

"B" makes more intuitive sense to me. If I were a hi-frequency wave and I had the option of going through a resistor OR a capacitor, I would go through the capacitor (and vise-versa for DC).

"A" doesn't make intuitive sense. If I were a hi frequency wave and I already went through a resistor, what would then compel me to go through a capacitor?

Does this first-person wave-humanization just simply not work? Or is there a concept I'm missing out on? Or does B also work just as well as A?

I appreciate any insight ya'll have

[Rob Strand]

"B" makes more intuitive sense to me. If I were a hi-frequency wave and I had the option of going through a resistor OR a capacitor, I would go through the capacitor (and vise-versa for DC).

"A" doesn't make intuitive sense. If I were a hi frequency wave and I already went through a resistor, what would then compel me to go through a capacitor?

If you put a 10A load on a car battery the voltage hardly drops but if you put a 10A load on a 12V stack of AA batteries the voltage will drop a lot.   So what does that? On a schematic a battery is just a battery.    The reason the voltage doesn't drop on the car battery is because the internal resistance is low whereas the internal resistance of the AA battery is much higher.     My point here is in order for a voltage source to drop the terminal voltage you need series resistance.   In textbooks voltage sources have no resistance and there is no voltage drop, even with a 1000000A load.

So the reason you might think B is more of a low pass filter is the because there is more current going down the capacitor.   The normal assumption is the circuit feeding the capacitor has no resistance so while the *current* is high the voltage does not change so there is no filtering.   A reason to see B isn't correct is to ask what is the resistor doing?   In order for a resistor to have an effect you need a current going through it.  In case B the output side is open, no current flows, so the resistor is not doing anything.

If we added a resistance in series with a car battery then took the load off the other side of the resistor we could get a voltage drop just like the AA batteries.     You see this effect in your car when the leads have a bad contact.      This is in fact case A in your diagrams.  We are adding a resistance *in order that* the voltage can drop when a load is placed on the output side.

The difference between the battery case and the filter case is we are dealing with AC signals.    Also the capacitor impedance get lower as the frequency increases.    That's how the low-pass filter works the cap impedance gets lower and lower as frequency increases.   The amount of effect of cap not only depends on the cap value but it depends on the cap value relative to the resistance R.      You might see the filter cut-off frequency written as f = 1/(2*pi*R*C)  if R is made smaller by a factor of 10,  C needs to be made 10 times larger to compensate.    You need a larger cap to "short more" to counteract the fact lower resistance value will have less tendency to drop the voltage. 

A guitar tone control is a bad example to understand this concept.   The guitar pickup is not a low resistance source (we normally say low impedance source).  The impedance is quite high and that's why placing a cap across the output drops the highs.   The pickup has the series resistance built-in.   A little better is to think of it as a series resistor and series inductor.

So the concepts you need to understand are voltage vs current, voltage drops occur because of current through a series resistor.   Also ideal voltage sources don't have any resistance/impedance.

FYI: you might see filters from Radio handbooks which have caps or inductors directly across the input terminal.  It looks like your case B.  However, radio signals have a 50 or 75 ohm impedance in the source.   That takes the role of the resistor in case A, it's just not drawn as an added resistor.  It's hidden in the source like a guitar pickup.

[antonis]

It only depends on where you consider the OUT..
(A is a voltage divider where B is "nothing"..)

[Bunkey]

Think how you would wire a volume pot - taking the output off the middle wiper (A)

(B) would be wiring your input to the wiper and taking your output from the top instead.

[akmenpi]

Thanks for the replies. I'm gunna need to think about this for a while.

[ashcat_lt]

Everything useful is a voltage divider (quote from PRR).  A capacitor looks like a big resistor to low frequencies and gets smaller at higher frequencies.  If that is the "top resistor" in a divider, then low frequencies are divided down more than higher.  If it is the "bottom resistor", the high frequencies are divided down more.

B will "work" in most real world situations because the source will act as the top resistor.

[DrAlx]

The problem with "wave humanisation" as you describe it that it puts in mind the idea that something starts off at the input and then ***propagates*** outward from there much like the ripples on a pond when you throw something in. That is the wrong concept for current flow in a circuit.
The reason that thinking is not appropriate is that signal currents are not things that propagate.  They ***flow*** (in a circuit).  The flow between two points can be calculated from the voltages at the two points and the impedance between them (Ohm's law). So you need to have information on two voltage levels to know the flow.
In example A, you have the input signal voltage at one point, and ground at another, and that causes a signal current to flow from one point to the other through both parts. There is no output current as such at the point marked Out. It just sits there, and will see a voltage level that depends on the impedance of the two parts (voltage divider, as pointed out above).
In example B, the signal current flows from input to ground through the capacitor. It does not flow through the resistor at all because there is no voltage applied at Out. Remember the current flow depends on the voltage levels between two points. You don't have an applied voltage level at Out, so both ends of the resistor are at the same voltage level. So Vout is Vin.

[ashcat_lt]

...so both ends of the resistor are at the same voltage level. So Vout is Vin.

In theory, on paper, with an ideal source and load...

IRL anything you put across the free end of that resistor will have some finite impedance, and there must be some division.  Whether it's enough to care about is up to you, but the two ends of the resistor will read different voltages.  Likewise at the other end, a real source will not have infinite current capacity and therefore will present some impedance greater than zero, so that there will be attenuation at some high frequencies.

If there's no source and no load, then what are even doing? 

What we would like to do is build like A and choose the resistor to be enough bigger than the source so that we can ignore the source AND enough smaller than the load that we can ignore the load.  Then we can figure the C against that R to get the cutoff we want and know that it will be close enough and we won't have enough broadband loss to worry about.  If real precision is necessary, you control both the load and the source (buffer on each side), so that you know what you're working with and include them into your calculations.

This is pretty much why a passive magnetic guitar pickup loses so much more treble than an active source like active pickups or pedal outputs through the same cable and input.  It's also at least part of the reason we end up wanting makeup gain after passive tone stacks like in amps and big muffs and things.  You just can't get the source to be enough smaller than the load to be able to pick resistors that "fit" in between.  For audio, we generally figure 1:10 ratio source to load is acceptable, but if we're thinking 1:10 from source to resistor (so we can ignore the source) and then 1:10 R to load (so we can ignore the load), that's 1:100 all the way through, which can be tough to achieve even with opamps sometimes.

[ElectricDruid]

Before I understood RC filters, I'd seen a capacitor charging up through a resistor. A capacitor is like a little battery, and the resistor changes how fast you charge it up. Smaller resistor equals faster charge. Pretty straightforward, especially if you can watch it on a scope or a voltmeter. Do a few experiments, try a few different caps and resistors, get a feel for it. Discharging a capacitor is similar. Short it with a wire and it discharges pretty much instantly. Use a big resistor and it takes a while. Watch that on the meter too.

The next step is feeding frequencies through it. Imagine we feed a sine wave in where we previous used DC. Obviously if the resistor is large, the capacitor won't have had time to charge up fully before the sine wave starts going back down again. So the output tries to follow the sine wave, but the amplitude of the sine wave is reduced. If the resistor is bigger, the capacitor charges less, and the output is quieter. If the frequency goes up, the capacitor doesn't have as much time to charge, so the output is quieter. By now you get the idea.

That was how I got it at first. Highpass took me a bit longer, but once I'd seen the lowpass, it wasn't hard to see it the same way, but with the voltage divider the other way up. But for high pass, you need the voltage divider concept too, which I didn't have initially.

I was probably twelve years old or so, and didn't have (or need) any mathematical understanding of it.

HTH,
Tom

[antonis]

I was probably twelve years old or so,

Barely ten years ago or so..

[ElectricDruid]

I was probably twelve years old or so,

Barely ten years ago or so..

Absolutely right! I'm a fresh-faced youth, and you better believe it!

[akmenpi]

I appreciate the replies, and I've spent the weekend trying to absorb the information.

I've drawn another picture to see if I understand what you guys are saying. Suppose the pot is the volume control of the last pedal before the amp.

If the volume is at 12 o'clock, that means signal can go to ground via 50K (assuming linear pot for simplicity) or 10Meg right? So then 50k/(10Meg+50k) of the current goes to the amp, and 10Meg/(10Meg+50k) of the current goes to ground?

Does that mean that the front of the amp actually receives very little current? Is this the point of high input Z?

I'm sorry that I'm just a ball of questions. I just can't seem to find this information elsewhere.

[ElectricDruid]

It's not 50K / (10Meg +50K) goes to the amp, because the 50K and 10Meg are in parallel, not in series. So you have 50K / (50K || 10M).

The point of high input impedance is that 50K || 10M is roughly equal to 50K (49.75K in this example, 0.5% different). The higher the input impedance, the more true that is.

The other point of high impedance which you don't have on your diagram is the effect of capacitance. Imagine you have a capacitor between the wiper of the volume control and the 10M resistor (most pedals do). That capacitance forms a highpass filter, so either cap needs to be large or the resistor needs to be large to ensure that bass frequencies can pass unhindered.

[antonis]

[ashcat_lt]

For most of the things we do, current is basically irrelevant.  Voltage = Volume and Current just does what it does to get us there.  In that picture the voltage across the amp input will be just less than half of what is across the whole pot, and just enough current will flow through the 1M to accomplish that, and that's about all you need to know until/unless you're trying to specify a power rating for the components.  Then still, I think we'd usually figure the maximum voltage it's going to see, then how much current that's going to flow, then multiply.

I find most of the metaphors whether it's hydrodynamic or whatever to be mostly meaningless.  It just really doesn't matter which electrons are going where or why.  It's a voltage divider.  Learn to see it as such, and it'll tell you about all you need to know.

[Rob Strand]

For most of the things we do, current is basically irrelevant.

For a lot of audio type stuff current is only used as a "bridge" during analysis to get to the voltages you want.   Ultimately we are interested in voltages.

If you example was different where the input resistor  was 100k and the output load was 100k  the two dividers interact.  If you try to do calculations ignoring the loading effects you will get very wrong answers.

There is a way to handle this using simple dividers but you need one extra tool to get the right answer.   It's called Thevenin's theorem.   (You don't really need to use Thevenin here if you break it down the right way.)

https://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem

The general idea of breaking down a complicated circuit into simpler sections which can be analysed using simple ideas is called network reduction.    Typically your tool kit would include,
- parallel resistor formulas
- delta to Y and Y to delta formulas
- voltage dividers
- Thevenin's theorem
- Ohms law

I find most of the metaphors whether it's hydrodynamic or whatever to be mostly meaningless.  It just really doesn't matter which electrons are going where or why.  It's a voltage divider.  Learn to see it as such, and it'll tell you about all you need to know.

Analogies are only good if you already understand one topic and you are tying to understand another.   Electrical engineers don't need to use water analogies they already understand electricity in their terms.     In fact there's more of a tendency to represent other fields in terms of electrical analogies because the analysis methods are very well known.

EDIT:
Another tool is superpositon theorem.