Calculating R-C Filters without Resistors?

[thehallofshields]

Hey guys. I'm working on a project guitar and researching the Varitone, Decade Switch and Anderton's Tone-Control and I've uncovered a major gap in my knowledge.

1.) How do we calculate a Cap->Ground treble-filter's cut-off points when there is no series resistor?



2.) What about Bass-Blocking-Capacitor when there is no Resistor->Ground?



3.) Would 1. and 2. follow the -6db/octave roll-off rule?

I'm assuming with 1., we use the output impedance of the preceding stage, and in 2. we use the input impedance of the following stage, but I think I could use some help here.

Thanks.

[amptramp]

In the first case, you use the source impedance in parallel with the input impedance of the following stage.  In most cases, the source impedance is much lower than the input impedance of the next stage, so the input impedance can be ignored.

In the second case, you use the input impedance of the following stage as you suggest, but there will be some effect from series resistance of the source.  If you have a high source resistance, the frequency response will be that of a resistive voltage divider with the source resistance in series with the input impedance of the following stage.  Thus, the impedances add for the purpose of calculating the low-frequency turnover.

[thehallofshields]

Okay, so assume the sans-resistor LPF is followed by a 100k volume pot, which is then on its way to a buffered pedal at 1M Input Impedance.

Would simply plugging in 100,000 as R, give me a close approximation?

[thehallofshields]

I've also been trying to calculate the Q Rolloff for the Jaguar's 'Strangle' Circuit.



Larger Diagram: ('Detail C' in the Image) http://www.offsetguitars.com/personal/sookwinder/various/jaguar_wiring_diagram____nov%2009.jpg

Sorry the switching on this one is so complicated, but basically the source is an inductor with about 7k Dc resistance, sent through a 0.003uf Cap, and leaves though a 1M Voltage Divider.

If I simply try to add the Pickup's Dc->Ground Resistance to the the Volume Pot->Ground Resistance I end up with:
R=1,007,000 , C=0.003, Q=52.7hz  ...Far too low of a corner for Bass roll-off to be noticeable.

What am I missing?

[PRR]

> an inductor

It matters. How big?

> with about 7k Dc resistance, sent through a 0.003uf Cap, and leaves though a 1M Voltage Divider. If I simply try to add the Pickup's Dc->Ground Resistance to the the Volume Pot->Ground Resistance I end up with: R=1,007,000 , C=0.003, Q=52.7hz  ...Far too low of a corner for Bass roll-off to be noticeable.

What would make you happy? 5300Hz, 100X higher? Then maybe it isn't the 1Meg resistor, but something about 100X smaller. Like 10K. Or perhaps that 7K??

> Sorry the switching on this one is so complicated,

You *really* have to re-draw it, to show yourself what is really happening in the switch-position in question.

I am suspecting you will find a series L-C network. If I ass-ume 10H and 0.003uFd, that resonates at like 1KHz (or does it? Check my math!!). Impedance will be low there and higher everywhere else. The lowest impedance is limited by the 7K hidden inside the coil. 7K is "nothing" compared to 1meg pot (*and* ~~1Meg amp input!) so no loss there. L-C impedance is higher everywhere else, so more loss, but maybe not a lot of loss? I suspect there is another resistor in there. I'm looking at a 56K but am too lazy/tired to work out the tangled drawing.

[thehallofshields]

All right PRR. I'm a little unsophisticated and dumped an hour into MS Paint. I hope you print this out and hang it on your refrigerator.



The Jaguar circuit is a little unusual. But I'm really only interested in learning how to calculate the roll-off of the Bass-Blocking Cap.

[thehallofshields]

You know, thinking about this has got me questioning why R-C Filters need to have a certain orientation since we're primarily concerned with AC voltage transfer and much less concerned with the flow of current.



- I guess I'm just lost yet again.

[GibsonGM]

I'll leave the high-test stuff to Paul (PRR) since he's very good at explaining the very complex interactions going on with the Jaguar circuit!  Understanding that can be a lot of work.

But for the above question:   If you move the parts of the filter, you change their fundamental relationship - therefore, they are no longer "low pass", and they can't be "low pass backwards".   It's a good question.

If you get into using LT Spice (free), you can easily simulate what's going on there.

Normal high pass of course will give us the expected first order response.  If you move the R and C the way you did, you effectively made a loop  that consists of your source and the resistor....the cap is now a coupling cap, for all intents and purposes.  So the response of the cap to your signal will be a whole different thing....which is the reactance of the cap in relation to the input impedance of the following stage.    There, it will form ANOTHER HPF, just one that's not working with that initial resistance.   

If you're not prepared for it, the response will be something you'd be very surprised by and probably weren't trying to get, he he.

[R.G.]

A subtlety that doesn't hit most beginners for a while is that there are no isolated signal circuits. A signal processing circuit always has a signal source that drives its input, and a load on its output. The source and load always have internal impedances. The impedances may be very large or very small, but they're always there.

In the case of your low pass backwards, the effective series resistance is the source impedance; and from a guitar pickup, this is not a simple resistance, it's a complex impedance with resistance, inductance, and internal capacitance. In any event, there is a series impedance there - you just didn't draw it in by drawing in the input source. The original seres resistance for the lowpass filter is now simply a resistor in series with the load.

In the case of your high pass backwards, there is still an impedance to ground, it's just the impedance of the load, which again is not drawn in, but is definitely there. The original shunt resistor is now just a parallel load on the source, and does not affect frequency response.

[GibsonGM]

+1  Right, R.G.    What I neglected to really point out is that with the backwards HP, the resistor to ground is in one loop, but there is a missing component for the C to play off of, leaving us in the dark as to what comes next.   

My missing statement would be that as long as that remains undefined, we really can't know too much about how that C will behave...if you load it with 1K, for example, you will have a much different cutoff than if your following impedance/resistance were 1M or more. 

When I took a look in LT Spice at how this behaves, I did have to add a "pot" to ground after the C to get any current to flow...without knowing what comes next, the C is irrelevant, just hanging there, with the R acting as a parallel load as you stated.    So the "backwards HPF" really represents an incomplete circuit...

[thehallofshields]

In the case of your high pass backwards, there is still an impedance to ground, it's just the impedance of the load, which again is not drawn in, but is definitely there. The original shunt resistor is now just a parallel load on the source, and does not affect frequency response.

Gotcha. So in the case of output coupling caps in pedals, our bass roll-off gets determined by the Z of the following stage, usually a 100k Volume control or the ~1M Input Impedance of our next pedal.

It now makes sense to me why, with Fuzz Circuits, changes to the input coupling cap are so much more dramatic than changes to the output cap. The relatively low Input-Z of transistors (~1k-10k for Common Emitter configurations if I understand correctly) makes for 10-100x higher corner frequency compared to the 100k Volume Pot.

So for regular HPF applications can I usually ignore the Source Impedance? (Do you abbreviate that as Zs?)
It doesn't seem to work for the Jaguar circuit.

[thehallofshields]

In the first case, you use the source impedance in parallel with the input impedance of the following stage.  In most cases, the source impedance is much lower than the input impedance of the next stage, so the input impedance can be ignored.

Okay so in a typical stompbox we have something like a Common-Emitter boost or a Common-Collector buffer often followed by a treble filtering-cap and a 100k Volume Pot.

In the case of the Common-Emitter we'll likely have an output Z of 50k, in Parallel with our 100k Pot making our R value = 33k.

In the case of a Common-Collector buffer, our output Z is much lower, something like 300 in parallel with 100k,  giving us an R value = 300.

So to equal the treble roll-off in a Common-Collector Source, the Common-Emitter Source would either need a cap over 100x larger or a 33k series resistor. - Does that sound correct?

[thehallofshields]

Thinking about this some more. I noticed Jack Orman's R-C calculator notes that it assumes a Low Zs. What would a good rule of thumb be for 'Low Impedance'? Under 10k? Under 1k?

Would it be fair to summarize that the standard LPF is dependant on Zs while the HPF is dependant on Zl?

[R.G.]

Gotcha. So in the case of output coupling caps in pedals, our bass roll-off gets determined by the Z of the following stage, usually a 100k Volume control or the ~1M Input Impedance of our next pedal.

... and the cable capacitance, which matters compared to the 1M input resistance of the next pedal.

This is bringing back a post I made long ago, titled something like "What matters, what doesn't". 

I have this theory that there is a C&W song appropriate to every situation if you can just remember it. In this case, it's "The Gambler", and the concept is "You gotta know when to hold 'em, know when to fold 'em, know when to walk away, know when to run."  A professor in my circuits design courses had a really good rule-of-thumb on what matters - if it causes less than a 10% change, you can generally ignore it.

So in this case, if the input impedance of the following stage is more than 10X (i.e. 1/10 the loading) of the output impedance of this stage, it can be ignored for a first, crude evaluation.

It now makes sense to me why, with Fuzz Circuits, changes to the input coupling cap are so much more dramatic than changes to the output cap. The relatively low Input-Z of transistors (~1k-10k for Common Emitter configurations if I understand correctly) makes for 10-100x higher corner frequency compared to the 100k Volume Pot.

Yes. This is also why a lot of Boss and Ibanez pedals have an emitter follower as the first thing the signal hits. An emitter follower stage has an input impedance that's typically hfe times the impedance of a CE stage.

So for regular HPF applications can I usually ignore the Source Impedance? (Do you abbreviate that as Zs?)

It's the old concept - trust, but verify. Yes, you'll design the filter part only ignoring the source impedance, but then you'll go back and at least think once "Ah, yes. This is being driven by an emitter follower (or opamp output, or low-collector resistor CE stage), and that is less than 1/10 the input impedance of this filter." If you notice that this is NOT true, you go make it true with a follower or opamp output buffer, etc.

As for Zs, generalized complex impedance Z as opposed to resistance R includes the ideas of inductive or capacitive impedances added to a resistor. An impedance is of the form R +/-j*Z where the +/-j*Z is the sum of the inductive and/or capacitive impedance **at the frequency of interest**, which obviously varies.  I just lost many readers, which is why I don't often get off into what impedance really is here in this forum.

A lot of the work in making a design "polite" and well behaved for audio consists of removing the funny effects of inductance and capacitance for the audio band.

Tag this concept "more to learn here".

It doesn't seem to work for the Jaguar circuit.

And that's because the Jaguar circuit has a huge, complex impedance driving it - that 68K+j3H pickup impedance in series with the signal voltage.

I noticed Jack Orman's R-C calculator notes that it assumes a Low Zs. What would a good rule of thumb be for 'Low Impedance'? Under 10k? Under 1k?

Welcome "The Fallacy of Calculators". Calculators can inhibit learning the underlying concepts by making getting a sometimes-specious answer so easy.

It assumes a low Zs because that's the only way to write a simple calculator. A more complete calculator would include the source impedance, which would require the simple user to know much more than they do know, and therefore make the calculator much less used.

The only good rule of thumb for "low impedance" or "high impedance" I've ever found is that 10:1 thing - if the driving source impedance is less than 1/10 of the load impedance, that's a "low impedance" compared to the load it's driving. If the load is more than 10 times the source impedance, that's "high impedance" compared to the source, and is not very loading to the source.

Would it be fair to summarize that the standard LPF is dependant on Zs while the HPF is dependant on Zl?

I would put it a different way. I would say that all RC filters are dependent on the source and load impedance until proved otherwise. "Proved otherwise" means you gotta think about it, however briefly.

[GibsonGM]

Nice, R.G.!  I've been following this, even thought briefly of trying to answer the question (a very good question!) last night.  Couldn't come up with a good way to 'explain' what I barely understand, myself, ha ha.

I like that you throw out there the "what doesn't matter" stuff.  We're not generally building microwave receivers here, and some odd effects from NOT considering Zs (among other things) really may not throw us off enough to care.   The best thing I'd suggest is to get LT Spice or another good simulator, and go thru these, see how your freq. response changes with varying Zs.   I did a couple last night and yes, you can see some changes.    If those changes are significant enough for the ear to pick up, I'm not sure.   Get the Z too far from what you THINK it is, and I'm sure you'd get some results you didn't expect...

Great topic.

[thehallofshields]

Thank you R.G.

I actually get it.

I think I'm ready for the Blue-Belt test in Audio Electronics now.

[R.G.]

Watch out for the sneaky foot sweep from the brown belt they have you spar with in the test. 

[PRR]

>> ...output coupling caps in pedals, our bass roll-off gets determined by the Z of the following stage, usually a 100k...or the ~1M...
> ... and the cable capacitance, which matters compared to the 1M input resistance of the next pedal.

...uhh, bass roll-off is rarely affected by cable capacitance.

In a very severe case (cable C similar to output cap C), there is broadband attenuation and the bass -3dB point is an octave lower than with no cable. But even for a puny 0.01uFd output cap, this doesn't happen until you got 300 feet of cable. (And long before that, it will be picking-up all the buzz in the room.)

[thehallofshields]

Watch out for the sneaky foot sweep from the brown belt they have you spar with in the test. 

You know, at my gym, I'm actually that guy with the foot-sweeps beginners never see coming. You'd be surprised how much Judo you can learn from the internet... or maybe not.

But here I am on a hot April afternoon, still pondering how to calculate a freaking guitar tone-control.

I get it... the Pickup creates a highly Q dependent, complex Z, and the layman's formula may not work.

If Spice is needed, could someone give me some idea of how to set up a simulation?

[R.G.]

The particular test from my memory was tai kwon do, but the brown belt still tried to sweep my foot.

... I had wrestled in high school, so I just shifted and picked up the foot. He only caught a little of my big toe, and missed the roundhouse punch coming in the confusion.