Input buffer question

[rankot]

Hi guys, I've been away from electronics for some time (too much job), but now I'm back! I have a question regarding this input buffer:

What's the purpose of those 100k resistor and 1u capacitor in series, connected to pin 2 of input buffer opamp? I suppose it's some kind of filter, but is it really necessary?

[GibsonGM]

Hmm, it's an active high pass....the cutoff freq. appears to be about 1.6 Hz, tho!  Attenuates frequencies less than that....so really, how functional IS it??   No clue why it's there, to be honest.  If the values were different, and Fc were higher, it would make a lot of sense, but does not (to me) with those values.  Example, if the R were 1k, the cutoff would be 159Hz, a much nicer cutoff point for guitar!     So, good question!
Component values might not be correct?

[ElectricDruid]

Seems pointless to me too, Ranko. The solid connection from pin 1 to pin 2 means there's no point having any kind of divider to Ground or Vbias. The cap would make sense if you were trying to take the lower end of a feedback divider to Ground.

To be honest, it looks like "left overs". You start with a circuit that has gain. You discover that by reducing the feedback resistor you can reduce the gain. Eventually, you replace the feedback with a wire and get unity gain. Ok, great! We've just designed a unity gain buffer. What does this other stuff do? Dunno! Ok, better leave it in, just in case...

I'm happy to learn why I'm wrong though, if anyone would like to step in.

There are other things about that circuit that don't look well considered. 1u/510K for the input HPF is roughly 0.3Hz. Now, I don't think even bass guitars go that low. Ten times higher (3Hz!) would have been plenty low enough, and 100n is a *much* cheaper and easier value to find than 1u. But who knows? Maybe Maxon got a job lot for next to nothing. Or maybe that schematic is not quite right (I agree with Gibson).

Gibson, how do you see that as an active highpass? I've never seen that arrangement before. Have you got a link to something similar please?

[antonis]

edit..
(wrong post,,)

[FSFX]

. 1u/510K for the input HPF is roughly 0.3Hz. Now, I don't think even bass guitars go that low.

Quite often it seems that people use far higher values of coupling capacitors than necessary and most likely never actually consider the frequency response of their circuit.
I usually design for low end cutoff frequency of around 70Hz to 80Hz. That way it is adequate for a typical guitar signal.   

[Rob Strand]

What's the purpose of those 100k resistor and 1u capacitor in series, connected to pin 2 of input buffer opamp? I suppose it's some kind of filter, but is it really necessary?

Do you have an example schematic?

[FiveseveN]

Yeah, I was confused by that as well. There's a schematic embedded in the original post but the host has some issues with SSL that prevents it from displaying. Try this:

[Rob Strand]

Yeah, I was confused by that as well. There's a schematic embedded in the original post but the host has some issues with SSL that prevents it from displaying. Try this:

That one I can explain.   The 1uF and 100k serve no real purpose in the context of that schematic.   What they are is left over parts which were used for the buffered clean signal in the "full schematic" of the pedal.

[antonis]

I wouldn't be surprised in case of original effect PCB had a wire jumper across a printed resistor between pins 1 & 2..

[GibsonGM]

Gibson, how do you see that as an active highpass? I've never seen that arrangement before. Have you got a link to something similar please?

I think you may have caught me at 'active', Tom, but I'm not actually sure...since this is a buffer, perhaps just 'high pass' or 'low cut' is the proper technical term?  Since it's a buffer maybe the term 'active' was a bit overly ambitious on my part?   

[Dormammu]

Yeah, I was confused by that as well. There's a schematic embedded in the original post but the host has some issues with SSL that prevents it from displaying.

I'll call it — a meaningless buffers (2 pieces).   

[antonis]

The whole thing is a TubeScreamner replica with dual op-amp substitution for IN/OUT BJT buffers..
(which buffers, of course, served for by-pass purpose only 'cause they were useless for the original circuit.)

Probably, the designer had the idea to add a little extra Input gain but finally realised that is no bueno..

[rankot]

Thanks to everyone for answers! This also looked completely pointless to me, but I just needed confirmation from people with more knowledge.

[antonis]

since this is a buffer, perhaps just 'high pass' or 'low cut' is the proper technical term?  Since it's a buffer maybe the term 'active' was a bit overly ambitious on my part?

Since it's a buffer (unity gain non-inverting configuration with no feedback resistor) 1μF cap and 100k resistor could be considered as an output "extra" load, forming a LPF with IC1a output impedance, of ridiculously high corner frequency due to Zout/100k ratio..

[GibsonGM]

Ok, yes, but since it's in the inverting loop, does that not make it a HPF? As viewed from the output, as a circuit block, it is passing frequencies higher than cutoff.   
That is how the same network is described in a Tube Screamer - which is in a loop that does have gain, yes, but the filtering action is quite similar.

[FSFX]

That's what always seems to happen when people produce Frankenstein type of circuits.
They cut and paste from other designs without thinking about whether some things they are including are needed or actually do anything.

[GibsonGM]

It's a 'legacy chunk' LOL

[antonis]

Ok, yes, but since it's in the inverting loop, does that not make it a HPF? As viewed from the output, as a circuit block, it is passing frequencies higher than cutoff.   
That is how the same network is described in a Tube Screamer - which is in a loop that does have gain, yes, but the filtering action is quite similar.

From terminology point of view, you might be *marginally* right, Sir..

But here, there isn't any feedback resistor to "isolate" it from op-amp output and practically there isn't any inverting loop..

On the other hand, considering pins 1 & 2 overlapped, you're free to formulate the "inverting point" term..

[GibsonGM]

OK    

[rankot]

Guys, you're so nice and precious! Love you all! <3