High- and Low-Pass Filter in Feedback Loop (GGG-Schematic Problems)

[pokus]

Hey, I'm quite new in the pedal universe. I've already built a few clones but now I want to modify and create some schematics myself.
I found this article called "design your own distortion" on generalguitargadgets.com. Mostly everything written there made sense to me. But when it comes to the end, there is a circuit of a simple distortion/overdrive pedal. I tried to breadbord it and it also worked, but it had a very less distorted sound and a big lack of bass.



So i just tried some different values, making R5 390 Ohm solved the distortion problem. As C4 I put in some largely higher values an it got more bass, but it still could have some more. I also put the Big Muff tone stack instead of R6 and C6 and it just worked but could not solve the tone problem.

And now we are slowly getting to me question. The text in the article says that R5 and C4 form a low pass filter and the pot R4 and C3 form a high pass filter. So I decided to figure out the corner frequencies of that design and then it got confusing.
The corner frequency of the low pass filter R5 and C4 was 1592.4 HZ and the CF of high pass with R4 turned all the way up to 100k was 33.8 kHZ.

So I thought when it had been the other way around HP is 1592.4 HZ and LP is 33.8 kHZ it would match exactly with what I can hear. Now my question(s): Is the text right saying R5 and C4 form HP and R4 and C3 form LP? And can the values in the original schematic be right for a common distortion pedal? Or did I any mistakes by trying to understand the schematics?
Here's the complete article: http://www.generalguitargadgets.com/how-to-build-it/technical-help/articles/design-distortion/
Thanks for reading and ensuing answers!

[GibsonGM]

Hi Pokus, welcome!  Good work on going thru what's happening there.

Sounds like you might be having a mind block over the opamp...remember, the input that the fed-back frequencies is going into is the INVERTING one.   So, whatever is going on in the loop...the output will show as "opposite", or inverted with respect to them!    The more signal you send thru that feedback loop, the LOWER those frequencies will be in the output.   

EX: you cut mids in the loop, you will have a mid HUMP in the output.  If you send highs back in the loop, you will get less highs in the output...see?
Hope that helps.

[EBK]

Some useful reading material with the answer:
http://www.geofex.com/article_folders/tstech/tsxtech.htm (scroll down to the Clipping Stage section). 

[ElectricDruid]

This is a question of terminology really. I'd always describe C3 as providing a lowpass action, since that's what it does overall, but that is achieved as Gibson says by feeding back more highs - e.g. it lets more highs pass.

Here's a reference:

http://www.physics.unlv.edu/~bill/PHYS483/op_amp_filt.pdf

It could just be a typo in the article, to be honest. He seems to use the terms the other way around earlier on.

If it's lacking bass, I'd whack C4 up to something much larger, especially if you've lowered R5 to get more gain. Try 4.7uF. That puts the bottom end down at 86Hz. Calculated here:

https://electricdruid.net/rc-filter-calc/?f=100&r=390r&r_series=3&r_errors=1&c=&c_series=1&c_error=10

There's one thing that's pretty odd about that circuit, which is if you close switch 1, you're clipping the signal at a level *below* what's required to make the other diodes clip. So you can have one set or the other set, but not both. Changing D1/D2 for something with higher forward voltage (like an LED or a pair of series diodes) would give you enough level to get the output diodes clipping too. I mean, if you close both switches, it's because you want crunch plus more crunch, right?!

HTH,
Tom

[pokus]

Hey thanks a lot you all really helped me out. I guess I just ignored the fact that there is a inverting and non-inverting Input.
So am I right assuming that the high-pass filter in the inverting feedback loop acts like a low-pass filter in general and cuts off the amplified highs from my signal?
According to my tone problem it should get better when setting the corner frequency of LP R5 and C4 in the loop to something lower like ElectricDruid said. But I already put in a higher values for C4 and it didn't make me completely satisfied. Or do I still get something wrong?
By the way the CF of low-pass R5 and C4 is 3184.7 HZ, no idea how I got to 1592.4 HZ

The problem with the diodes is absolutely right, I put in some LED's for both clipping stages and when both are in the circuit the hard clipping ones don't light up, I guess the switch should be a SPST so only one stage is connected.

As C5 and the Volume pot R7 form a high pass filter the CF when the pot is at low Volume is also cutting some lows, isn't it?

[antonis]

This is a question of terminology really.

It doesn't mind at all as far as we shake hands on the actual function of that cap..

[GibsonGM]

It doesn't mind at all as far as we shake hands on the actual function of that cap..

Exactly!  That sort of thing can make someone crazy!     

If you keep in mind, Pokus, that the opamp's inverting pin will make what you put into it "backwards", that might make it easier.   A LPF is still a LPF - it is passing lows to the inverting pin, which then gives the opposite phase on the output!  It's counter-intuitive.   

You will get the hang of it, no worries.

[Kipper4]

Nice one Tom.
I like those. Saved for future reference.

[GibsonGM]

That IS a good one, and gives a better explanation of the 'phenomenon' by describing this as a decrease in GAIN as frequency increases, and so on...adding reactance to the FB loop...nice!  Better than my overly simplistic statement 

[pokus]

Ok, so summed up: A resistor and cap directly shunt to ground form always a "real" low-pass filter. But if you want to cut off the highs of the out-signal in a circuit using an non-inv-opamp you need a "real" HP filter in the feedback loop with the corner frequency somewhere around 20kHZ.
And in the example from ElectricDruid's link Page 3. R1 and C1 of the bandpass filter in the feedback loop of an non-inv opamp always set the low CF and R2 and C2 the high CF coming out at the output.
Is that all right except the fact that I'm attached too much to the terms high-pass and low-pass? 

[ElectricDruid]

R1 and C1 set the lower limit (the highpass cutoff) and R2 and C2 set the upper limit (the lowpass cutoff). I'm attached to those terms too.

What you're calling a "real" lowpass filter is more properly called a "passive" lowpass filter. These ones we're looking at and discussing are all lowpass (or highpass) filters too and no less real, but they're "active" filters (they include an op-amp or a transistor or something to provide gain) and those can have more different forms. The paper I posted shows quite a few, but these are all active single-pole filters. Active filters with two poles come in several flavours too.

HTH,
Tom

[ashcat_lt]

I always call this negative feedback loop "Bizarro World" because up is down and everything is inside out.  Attenuation in the loop leads to gain at the output and vice versa.  In fact, I never bothered to learn the formula for NI opamp gain because it is literally exactly the inverse of the attenuation in the feedback loop, and I already know that formula pretty intimately.  When one considers that a passive filter is really just a voltage divider where one of the "resistors" gets bigger at lower frequencies...

So I do look at C4/R5 AND R4 as a lowpass filter on the feedback loop which just happens to cause gain at the output proportional to the attenuation at a given frequency.  Likewise C3/R4 AND R5 act as a highpass filter in the loop.  Together they act as a very lossy band cut. 

But it's important to remember that these are shelving filters.  Neither one can actually get to infinite attenuation at any frequency because of the way that the resistors are around them.  At high frequencies, C4 looks like a short, and R5 vs R4 sets the maximum amount of attenuation.  At low frequencies, C5 looks like an open circuit, and if R4 was removed, the opamp would (try to) add infinite gain to the DC bias voltage and slam the whole thing against the positive rail so hard that nothing would come out.

So yeah, it's kind of complex and definitely weird, but not that hard to get your head around once you understand the upside-down nature of the feedback loop.

[PRR]

> instead of R6 and C6

As drawn, R6 C6 do nothing; at least nothing good.

Look at the extreme, R6 at zero resistance. The R-C product is 1nF against.... well, the output resistance of the op-amp. Which is ideally "zero", and in practice maybe 1 Ohm. The answer is way up in the TV band. (Not that simple because the opamp output is inductive, rising to hundreds of Ohms above the audio band, and in fact many opamps just go screwy driving caps with no resistance in series.)

So the ideal opamp you can hang caps on the end all day and nothing happens; real opamps nothing happens until they go bonkers.

There should be some series resistance for R6 C6 to work against.

Stealing known-good plans is also a good bet.

[ElectricDruid]

But it's important to remember that these are shelving filters.

Yes, this is a good point. That's an important difference between this and some other arrangements of LP/HP.

[antonis]

@pokus: Perhaps the integrator/diferentiator point of view could be more confusing so you might prefer to see C3 as a R4 by-passing device..
(it sets highs out of NFB loop hence reducing their gain - the oposite action of C4 which passes highs through R5 to GND..)

[pokus]

Alright, at first thank you guys all of your posts brought me forward.

To use my new knowledge i was looking at the circuit of a marshall shredmaster, which for my ears sounds pretty well. Besides I've already built a clone of one of these.




So starting at the Input the 2.2M resistor and the 10nF Cap let 7,2 Hz and the frequencies below that pass the signal path. But as the signal, regarded isolated from all the other R's and C's, now goes to the first IC it keeps the same in reference to the freq. So it stays + in my mind. Then it hits the inverting pin of IC2, so it became -, or inverted. Same with IC3 (+) and IC4 (-). And what you will get out is now the (-) inverted signal, in which the frequencies higher than 7,2 Hz have passed through. Is that correct in terms of the function and logic and not of electronic correctness?

I also wonder why quite many pedals use a frequency about 1kHz set by the 47nF cap and the 3,3k resistor in the feedback of IC1. I saw that in many gain-stages. Isn't that cutting to much useful bass at the end?

Sorry for the sometimes might unclear language in this one, but as you might noticed I'm not an native english speaker and it's quite hard for me to find the exact words for what's going on in my head. 

[GibsonGM]

Hi Pokus,

You are close, but....the 2.2M is a pulldown resistor, to 'reference' the input to ground.  The input filter consists of the 10nF cap with the 1M resistor to Vref in parallel with the opamp's input impedance, which we could take to be HUGE, so let's just assume 1Meg for simplicity, ok?!
     
10nF, 1 Meg is about 16Hz cutoff at the input.  Since input Z is probably more, that will be a lower frequency, in reality...so, this lets in just about EVERYTHING.

There is nothing wrong with your logic - the way the signal then passes is correct, but the components in the feedback paths of all opamps do affect the signal, of course. 

The 1kHz cutoff you mention...it seems like it is cutting very much, but these first order filters have a slope rather than truly CUTTING the signal.  The amount of bass reduction is gradual, not sudden, so there is still much bass content there.    Many effects, especially distortions, suffer from having too much bass content.   Remember, the feedback loop is controlling the gain applied to the signal....so the bass content is still there from the original signal, but it simply is not amplified as much as other frequencies that the designer thought 'more desirable'  (often, a boost in treble is wanted in overdrives, distortions). 
~~Also remember to take a look and see if the 'cut' is in the negative feedback loop, making it "normal gain" instead of a cut!      

If you wanted to, you could preferentially boost the bass later, if you thought the circuit was too bright, shrill, or 'tinny' sounding - so, you would have achieved good overdrive with bright harmonics, and no 'boomy, farty' sounds, but can still bring some bass back in the tone stack or another gain stage!  Bass often does not sound very good when it is hard-clipped in distortion.

[antonis]

Nothing to add on Sir MIke's analysis..

Except, perhaps, for 47nF/3k3 cut-off frequency (1026Hz) severely attenuated by 68nF/8k2 at 285HZ..(at Drive pot full set)

So, we actually have a "mild" mids attenuation followed by a severe lows one..

[pokus]

Maybe this doesn't fit the topic anymore, but what's about the bias thing. I mean I can mostly understand what the first one connected trough the 1M resistor does. But why all the others? Quite everywhere where I'd expecting gnd is now bias. And another thing is, when I'm "designing" a circuit with couple of opamps in it, do I have to bias the signal in front of every one, because the first IC should give me the necessary voltage amplification, dosen't it? And what's about the frequencies coming in the circuit by biasing, do I have to get rid of these?
I guess it's time now to study some basic literature of amps and pre-amps, any tips? 

[antonis]

Quite everywhere where I'd expecting gnd is now bias.

You should only expect GND in case of symmetrical dual supply..

when I'm "designing" a circuit with couple of opamps in it, do I have to bias the signal in front of every one, because the first IC should give me the necessary voltage amplification, dosen't it?

We bias op-amp inputs and NOT signal..

And what's about the frequencies coming in the circuit by biasing, do I have to get rid of these?

Bias method should have taken care of this..