Resistors in series with the AC coupling cap -- what do they do?

[midwayfair]

I'm not sure where to look to find more information about this, but it's so common in circuits that I figure a lot of other people might know.

Part of my ongoing quest to understand tone circuits.

Most builds have a resistor after the AC coupling cap. Sometimes it's the master volume, though that also connects to ground, so it's not *exactly* what I'm taking about. The Madbean Fatpants has a 10K before the master volume and a 470K to ground, which I guess is actually like having a second volume knob hardwired into the pedal. In the Klon/Sunking there's a 1K following the "first" circuit, just before the diodes, and following an AC coupling cap. Lots and lots of circuits have a resistor there. Is it doing anything except regulating volume? Does it do anything specific tonally based on its value?

And related ...
I was flipping some stuff around on a breadboard in a simple one-transistor boost circuit (basically a Rangemaster-like test circuit with no boost knob) and put a 1K resistor before the AC coupling cap, so the 9v input went 2.2K > Collector > 1k > .1uF cap. It reduced output, but also may have affected the treble and available distortion some (it's was hard to tell exactly what was going on because of the volume change). I realized that this was kind of like hardwiring the Rangemaster's boost knob, even though it looked a little funny, where the pot divides resistance between the collector, the power source, and the output cap. Fuzz Faces have a kind of similar positioning of resistors in that area. Again, though, I'm wondering if this does something specific tonally.

Anyway, here's my question: When you have an output cap and a resistor in series without a connection to ground (that is, not sending the highs to ground like in a regular treble cut knob), are they still simple low- and high-pass networks, or does their position change what they're doing? And how would I go about calculating the effect, beyond just using my ears and plugging in different values, if I wanted to, for example, cut or boost a specific amount of treble or bass in a given circuit?

[R.G.]

A couple of principles will guide you on this one.

First, the nominal rolloff frequency of a single R and single C is always 1 / (2*pi*R*C).  This only makes sense where the circuits feeding the RC network have internal impedances which make them not affect the R and C, but in isolation, a single RC's frequency is always 1 / (2*pi*R*C).  It's sometimes handy to remember the frequency of a 1K and 1uF - this is 1/ (2*3.14*1000*1E-6) = 159Hz. You can then do in-your-head arithmetic to get 10K and 1uF = 15.9Hz, 1K and 0.1uF = 1.59kHz, etc.

Second: The order of things in series does not matter as long as you don't look at the internal nodes. You get the same voltages and currents at the two ends of a resistor followed by a cap as you do a cap followed by a resistor. This is actually the answer to one of your question: R before C is the same as C before R.

Third: An R and a C can make either a high pass or low pass filter, depending on how they're arranged. You can always figure this out by imagining the C as an open circuit below the R-C frequency and a short circuit above it (although that's a vast oversimplification - it's just useful for getting an intuitive feel for it).

There are four combinations of one R and one C that happen.
1. Signal goes through a series resistor to the output, C goes from output to ground. Using the third principle, C is open up to some frequency, then shorted. So this means frequencies up to the RC frequency pass, and above that are "shorted out". This is a lowpass filter.
2. Signal goes through a series capacitor to the output, and has a shunt resistor from output to ground. This means nothing goes through up to the RC frequency, then all of it goes through. This is a high pass filter at the RC frequency.
3. Resistor and capacitor in parallel to ground. The cap is an open up to the R-C frequency, then is open. This is a lowpass filter, but the impedance to ground never gets bigger than the R value.
4. Series R and C. The signal doesn't go through at all up to the R-C frequency, then goes through when the cap "shorts". This is a highpass filter at the RC frequency.

The last one is the one you asked about. What it different about it from just a series capacitor is that at frequencies where the capacitor's impedance is trivially small (that is, "shorts out" in our simplistic rule) the impedance through the R and C does not go to zero, but goes down only to the value of R.  In some circuits you need to keep the input signal from supplying too much current at AC only. The series R-C does exactly this.

And this last is the answer to your question about why there is a resistor before or after a coupling cap in the simple case. In some instances, a series R after a coupling cap is also a part of the gain-setting of the circuit after the coupling cap. In this case, it's still limiting current after the C is above the R-C frequency, but it may serve other purposes, too.

Always be aware that there may be unseen "resistors" that are inherently part of the circuit driving the RC, and loading it down.

As to tone issues from series resistors - sure, they affect tone. Everything conceptually at least affects tone, including the bass player's bad breath. But it's the value of the R and the C that sets the frequency where the RC combination starts changing what frequencies get through. That's the biggie.

[midwayfair]

A couple of principles will guide you on this one.

First, the nominal rolloff frequency of a single R and single C is always 1 / (2*pi*R*C).  This only makes sense where the circuits feeding the RC network have internal impedances which make them not affect the R and C, but in isolation, a single RC's frequency is always 1 / (2*pi*R*C).  It's sometimes handy to remember the frequency of a 1K and 1uF - this is 1/ (2*3.14*1000*1E-6) = 159Hz. You can then do in-your-head arithmetic to get 10K and 1uF = 15.9Hz, 1K and 0.1uF = 1.59kHz, etc.

Second: The order of things in series does not matter as long as you don't look at the internal nodes. You get the same voltages and currents at the two ends of a resistor followed by a cap as you do a cap followed by a resistor. This is actually the answer to one of your question: R before C is the same as C before R.

Third: An R and a C can make either a high pass or low pass filter, depending on how they're arranged. You can always figure this out by imagining the C as an open circuit below the R-C frequency and a short circuit above it (although that's a vast oversimplification - it's just useful for getting an intuitive feel for it).

There are four combinations of one R and one C that happen.
1. Signal goes through a series resistor to the output, C goes from output to ground. Using the third principle, C is open up to some frequency, then shorted. So this means frequencies up to the RC frequency pass, and above that are "shorted out". This is a lowpass filter.
2. Signal goes through a series capacitor to the output, and has a shunt resistor from output to ground. This means nothing goes through up to the RC frequency, then all of it goes through. This is a high pass filter at the RC frequency.
3. Resistor and capacitor in parallel to ground. The cap is an open up to the R-C frequency, then is open. This is a lowpass filter, but the impedance to ground never gets bigger than the R value.
4. Series R and C. The signal doesn't go through at all up to the R-C frequency, then goes through when the cap "shorts". This is a highpass filter at the RC frequency.

The last one is the one you asked about. What it different about it from just a series capacitor is that at frequencies where the capacitor's impedance is trivially small (that is, "shorts out" in our simplistic rule) the impedance through the R and C does not go to zero, but goes down only to the value of R.  In some circuits you need to keep the input signal from supplying too much current at AC only. The series R-C does exactly this.

And this last is the answer to your question about why there is a resistor before or after a coupling cap in the simple case. In some instances, a series R after a coupling cap is also a part of the gain-setting of the circuit after the coupling cap. In this case, it's still limiting current after the C is above the R-C frequency, but it may serve other purposes, too.

Always be aware that there may be unseen "resistors" that are inherently part of the circuit driving the RC, and loading it down.

As to tone issues from series resistors - sure, they affect tone. Everything conceptually at least affects tone, including the bass player's bad breath. But it's the value of the R and the C that sets the frequency where the RC combination starts changing what frequencies get through. That's the biggie.

Holy crap, this had math and I understood it.

Thanks, R.G., for always having a thorough and understandable answer. This will really help with the experimentation.

[swinginguitar]

3. Resistor and capacitor in parallel to ground. The cap is an open up to the R-C frequency, then is open. This is a lowpass filter, but the impedance to ground never gets bigger than the R value.

Hijack Alert!

I've always been a little confused about this configuration (such as the resistor and bypass cap to ground on the typical triode tube stage)....if the cap offers an effective short to ground to certain  (high)freq, the resistor is ignored and the cap shorts them? OK, so that makes a LPF. As freq decreases, the reactance of the cap opposes the low freq...so how does the resistor come into play then?

RG - can you give a little background on this - RC in parallel?

[R.G.]

I've always been a little confused about this configuration (such as the resistor and bypass cap to ground on the typical triode tube stage)....if the cap offers an effective short to ground to certain  (high)freq, the resistor is ignored and the cap shorts them? OK, so that makes a LPF. As freq decreases, the reactance of the cap opposes the low freq...so how does the resistor come into play then?

RG - can you give a little background on this - RC in parallel?

Sure. First, this is one where the unseen impedance of whatever drives the parallel RC to ground. When frequency is low, the capacitor's impedance is very large, much larger than the R in parallel with it, so the voltage is unaffected by the cap. However, it *is* affected by the impedance of the signal source.

The output impedance of the signal source and the R to ground make a voltage divider. If that signal source is, for instance, about equal to R, then for all signal frequencies where you can ignore the cap as an open circuit, the output voltage is 1/2 of whatever it would be if you opened both the resistor and cap. And in general, if the signal source impedance is Rs, then the output voltage is always V = Vopencircuit * R/(R+Rs).

However, when the frequency rises to where the capacitive impedance is equal to R (that is, F = 1/(2*pi*R*C) and Xc = 1/(2*pi*F*C)  ) then the capacitor is 'eating' as much signal current as the resistor is, and this drops the output voltage, still taking into account Rs.

At very high frequencies, C acts like a short circuit to ground and the output voltage is zero.

... with the caveat that if Rs is small compared to the capacitive impedance, the signal source may be able to drive enough current into just the capacitor to hold up the output voltage far beyond F = 1/(2*pi*R*C) . The signal impedance has an effect on the result.

Actually, this problem is better considered by the Norton (current source and parallel impedance) method of modelling the signal source, at least to my mind. The model which has a signal voltage in series with an impedance is called the Thevenin equivalent. Norton and Thevenin signal sources are equivalent, and always give the same results  if you do the math right, so you can use either one you like better and find easier to solve.

Not a hijack - just an elaboration of the basic question.

Where did I muddy it up more than clearing it up?