Understanding filters

[Gumby212]

I understand that a high pass filter is a capacitor followed by a resistor to ground. And i understand that a resistor followed by a capacitor to ground is a low pass filter. But what does it mean when the second component doesnt go to ground and just stays in the signal path? Or the differance between a capacitor in the signal path, like at the input of a circuit, or when that cap goes to ground? What would the simplest way to explain the relationship between the values of the components, and the frequency they cut or boost?

[Mr.Kite]

Generally speaking, smaller coupling caps cut more bass, while bigger caps to ground cut more highs...

[Rob Strand]

But what does it mean when the second component doesnt go to ground and just stays in the signal path? Or the differance between a capacitor in the signal path, like at the input of a circuit, or when that cap goes to ground? What would the simplest way to explain the relationship between the values of the components, and the frequency they cut or boost?

In order to form a filter the circuit *has* to have some path the ground.   In some cases the trick is to work out what that path to ground is.

An easy case is an inverting opamp.  From the perpective of filtering the the opamp's '-' input looks like it connects to ground.
https://electrosome.com/wp-content/uploads/2016/12/Single-Supply-Inverting-Amplifier-using-Opamp.jpg
The input signal is actually the current flowing through the input cap in series with the input resistor.   At low frequencies the current is low, in fact zero at DC, because the cap looks more like an open circuit.  At high frequencies the cap looks like a short circuit and the current levels off to Vin / R1.     For the two extremes you can deduce this forms a high pass filter (the cut-off is 1/(2*pi*R1*C1)).

[Rob Strand]

Here's another good example that can confuse people.   Look at the top circuit.  This is a high-pass filter and a divider.

https://i.stack.imgur.com/6cgYI.png

The way to break it down is like this:
- Swap the order of C1 and R1.  Changing the order of series components has not effect on the circuit.
- First think of the signal at the point where C1 and R1 join and ignore the real output point.
  In this form R1 and R2 appear in series.
  So the output where C1 and R1 join is normal high pass filter,
  with C=C1=1uF and R = R1 + R2 = 100+100 = 200ohm
  and f3 = 1/(2*pi*C1*R) = 1/(2*pi*C1*(R1+R2))
- The actual output is at R2.  You can see a voltage divider formed by R1 and R2, which divides the
   high-passed signal.   
   The output voltage at R2 is the high pass signal * R2 / (R1+R2) = high-pass signal * 100 / (100+100)
     = high-pass signal * 0.5

So the circuit does high-pass filtering and divider the level by 2.


The key point is the high-pass filter depends on R1 + R2.  There is no filtering related to R1 and C1 ie. no meaning to the frequency 1/(2*pi*R1*C1).

[Gumby212]

So, a capacitor and resistor in series wont create a filter until the path meets ground correct? Does that mean that in that diagram, c1 and r2 is the filter, not r1 and c1? If so, what is the first resistor doing in the signal path? I see a lot of schematics that have a cap and resistor in series after the op amp or transistor, just before where the diodes clips to ground. In that case, do the diodes going to ground help make the filter? Thanks for sny help on this, as im trying to dig deeper into how these circuits really work and this has me bery confused.

[Rob Strand]

So, a capacitor and resistor in series wont create a filter until the path meets ground correct? Does that mean that in that diagram, c1 and r2 is the filter, not r1 and c1?

For the second example the filter is formed by C1 and (R1+R2) both resistors are involved in filtering.   In this case thinking about C1 + R1 or C1 + R2 has no meaning.   Seeing that difference is an important step to understand what determines the filter - it's hard to understand at first.

If so, what is the first resistor doing in the signal path?

The first resistor is part of the divider formed from R1 and R2.  Ignore the cap and you should see the divider.

I see a lot of schematics that have a cap and resistor in series after the op amp or transistor, just before where the diodes clips to ground. In that case, do the diodes going to ground help make the filter?

Like in this case, with C4 and R5,
http://www.muzique.com/schem/shaka5.gif

Yes, the diodes form the path to ground and form part of the filter.    You can think of this somewhat like the second example where the diodes replace R2.     From a filtering perspective the diodes *kind of* makes R2 vary so the filter varies.    Understanding the frequency response of this type of circuit is a little complicated, so you shouldn't think of the diodes being an actual varying resistance too much because it's not entirely true.  [The real answer is the diodes make R1 look bigger when they don't conduct.   How much bigger depends on how much time the diodes are on and off.  How all this works is far too complicated to explain at this point.]

When the signal is large and the diodes clip. It's like R2 is small.  If the diodes clip hard the filter frequency *is* determined by "R1" and "C1"  (C4 and R5 on that schematic).

When the signal is small the diodes have no effect and there is no filtering.
In the real circuit example I gave we don't just have diodes we also have R12 which takes the place of "R2".
In this case the high pass filter is formed by C4 + R5 + R12  which is the same type of circuit as my second example:  C1 <=> C4,  R5 <=> R1, R12 <=> R2.

[Gumby212]

Ok, this helps a lot. So in other terms, whenever there is a cap and resistor in series, or vice versa, the next component that follows going to ground helps create part of the filter? In a simple transistor circuit like an electra distortion, where theres an input and output cap, i take it the input cap creates its filtering with the help of the transistor and the resistor attatched to the emmiter going to ground? I see that the output cap on that circuit is follwed by diodes to ground, but if you remove those diodes, is that cap now not doing anything? Or does it find ground via the output jack? Just so im clear, an input signal followed by any value cap, going straight to the output does nothing to your signal? Ground has to be seen for any tonal changes in signal? It seems that a single resistor in the path chokes the volume the lower the value, but does no filtering to the eq, is that correct?

[Rob Strand]

Ok, this helps a lot. So in other terms, whenever there is a cap and resistor in series, or vice versa, the next component that follows going to ground helps create part of the filter?

Yes, that's correct.   Series components *alone* don't modify the signal.

i take it the input cap creates its filtering with the help of the transistor and the resistor attatched to the emmiter going to ground?

Yes, that's the correct way to think about this type of thing.    You might not know what that resistance is but the important thing is you have some "thing" and it looks like a resistance to ground.   To work out what that resistance is is requires calculations; sometimes easy, sometimes difficult.   (The input impedance of the Electra also includes the effect of the 2.2MEG resistor across the collector and base.)

is that cap now not doing anything? Or does it find ground via the output jack?

In isolation the cap does nothing.  In reality you will connect the Electra output to another effects pedal or and amplifier.  When you do that you are effectively connecting the input impedance of the amplifier, which is a resistance to ground,  to the Electra output cap.   

The input impedance next device is typically 470k to 1MEG but it could be 100k to 10MEG.  You just don't know.   So the cut-off frequency of the high pass filter is indeterminate.   Some effects might add a 100k from output of the effect to ground, doing that means the resistance to ground is at least 100k and that reduces the uncertainty in the cut-off of the high pass filter.

Just so im clear, an input signal followed by any value cap, going straight to the output does nothing to your signal?

Yes, but at some point you have to connect it to something.

Ground has to be seen for any tonal changes in signal?

Yes, otherwise no filtering (or loading) can be created.

It seems that a single resistor in the path chokes the volume the lower the value, but does no filtering to the eq, is that correct?

For a series resistor it won't choke the volume unless there is a path to ground.   Yes no filtering.

You have have to be careful about isolating the effect of a single component too much.   Take my second example above, the series resistor R1 affects both the divider and the filter.    In general all the parts interact.   One of the biggest steps in electronics is knowing just what interactions are import and which ones are not.    The best way to start is to understand things roughly, with some "wrongness", then once you get past that point you can see more of the finer details.

[Gumby212]

Awesome advice guys. Thanks!