Reading Schematics - Tone control question

[Jopn]

So I'm reading, and building, and reading some more, and tweaking, and breadboarding, and doing more reading to learn what I can about this wonderful world of DIY stompboxes.

One of the things I like to try to do is look over a popular pedal schematic and try to talk my way through what's going on.

Over my lunch break today I pulled up the ever popular Zendrive:
http://4.bp.blogspot.com/_y4AYtND8Hz8/SDNHYwkDnhI/AAAAAAAAALA/WHIn40nhzu4/s1600/Z_drive01.gif

I'm fairly comfortable with most of the stages there, except for that tone pot.  How does increasing the resistance between the clipping and gain recovery stage change the tone?  Or am I completely off on this one?

Thanks!

John

[bluebunny]

OK, I'm going in!!   

This is just my mental regurgitation of what I've learned here.  I hope I get this right...  The resistance (50K pot and 10K resistor), together with the cap to ground (3.3nF) form a filter.  As the resistance changes, so do the characteristics of the filter.  Voilà - tone control!

@learned folks: How'd I do?? 

[Bill Mountain]

http://www.muzique.com/schem/filter.htm

That link is for a filter calculator.  What you have in this schematic is a Low pass filter.  As you adjust the pot it changes the resistance value which alters the filter's cut off frequency.  When the pot is turned all the way towards R4 the filter is 10k and 3.3n which has a cut off at 4.8kHz.  When the pot is turned all the way in the other direction the filter is 60k and 3.3n which has a cut off at 800Hz.

[Jopn]

Got it!

Back to reading I go...

Thanks guys!

[induction]

@learned folks: How'd I do?? 

That's not me but I'll chip in anyway.  bluebunny is correct.  Specifically, it's a low-pass filter, which has a corner frequency of fc=1/(2*pi*R*C).  It attenuates all signals above fc in proportion to their frequency (technically in proportion to log(f-fc)).  As you turn up the knob, you decrease the resistance and increase the corner frequency, allowing more highs through.

ninja'd by Bill Mountain, but now you have the formula, anyway.

[ch1naski]

Awesome simple description. Even I get it.

[Hemmel]

Sorry to hijack the thread, but I noticed something I've never seen in the schematic : the 2N7000.
I've seen PNPs and NPNs, but I just want to make sure I understand how to connect the MOSFET here.

For example Q1, do I connect D3 to pin 1 (S) and short pins 2 and 3 (G and D) together ?

[Jopn]

@learned folks: How'd I do?? 

That's not me but I'll chip in anyway.  bluebunny is correct.  Specifically, it's a low-pass filter, which has a corner frequency of fc=1/(2*pi*R*C).  It attenuates all signals above fc in proportion to their frequency (technically in proportion to log(f-fc)).  As you turn up the knob, you decrease the resistance and increase the corner frequency, allowing more highs through.

ninja'd by Bill Mountain, but now you have the formula, anyway.

Awesome.  I started with Bill's link and have already found myself here: http://en.wikipedia.org/wiki/Low-pass_filter#Passive_electronic_realization

[Jopn]

Sorry to hijack the thread, but I noticed something I've never seen in the schematic : the 2N7000.
I've seen PNPs and NPNs, but I just want to make sure I understand how to connect the MOSFET here.

For example Q1, do I connect D3 to pin 1 (S) and short pins 2 and 3 (G and D) together ?

Sounds right based on what I'm seeing in the schematic.  And looks confirmed based on what I'm seeing in an example layout:
http://1.bp.blogspot.com/_y4AYtND8Hz8/SDNHEkkN8uI/AAAAAAAAAKo/ubR7cyIUjbw/s1600/revisiona.jpg

[bluebunny]

Awesome simple description. Even I get it.

Me too!   

"Low pass" was the bit that eluded my tired ol' grey cells this evening, as many things do these days.  Thanks for quoting the formula - I wasn't going to dare try remember that one.  One minor miracle is enough for one day...   

Sorry to hijack the thread, but I noticed something I've never seen in the schematic : the 2N7000.
I've seen PNPs and NPNs, but I just want to make sure I understand how to connect the MOSFET here.

For example Q1, do I connect D3 to pin 1 (S) and short pins 2 and 3 (G and D) together ?

Yep - you're using the MOSFET as a diode.

[Hemmel]

Yep - you're using the MOSFET as a diode.

Oh ! I did not get that. Thanks !
Thanks to all  

[ch1naski]

The 2n7000 appears to be wired as a  diode


Edit: oops, someone beat me to it....guess I should read the full thread before coming in.....

[midwayfair]

Might as well link to some excellent discussion of how those MOSFETs are working:

http://www.diystompboxes.com/smfforum/index.php?topic=90474.0

And I'll just get in a little jab here: Note how the Zen Drive actually does the MOSFET active mode clippers right, whereas in the Fulltone Fulldrive the second MOSFET with the diode in series does nothing thanks to body diode conduction from the second diode.

You can save a few parts by omitting the second MOSFET and all the Schottky diodes as well. You'll get a lot more distortion that way, though.

[ashcat_lt]

Filters sometimes seem kind of mystical.  There's "time constants" and Greek w's and imaginary numbers and that formula with the pi...

But, it really works out to a frequency-dependent voltage divider.  

A "normal" voltage divider is two resistors in series to ground (technically parallel to the load...) , with the output taken from their junction.  In the schematic provided in the OP, R1 and R2 create a voltage divider to get the 4.5V bias voltage.  The volume control (along with R3) is another voltage divider.  "Sure, we know what a voltage divider is!!!"

The filter is essentially the same thing.  In this case, the "top resistor" is made up of R4+the Tone pot.  It would also inclulde the output impedance of the previous opamap stage, but that is going to be extremely small compared even to the 10K, and can be ignored.  Anyway, add that all up and you get the "top resistor".

The "bottom resistor" is a little tougher to figure.  It is made up of the capacitor in parallel with the in-Z of the opamp following.  That opamp input can generally be assumed to be too large to affect things much.  The resistance of the cap is frequency dependent, exhibiting higher resistance for lower frequencies and vice versa.  So...

For low frequencies, the "bottom resistor" looks a lot bigger than the "top resistor", and most of the input voltage drops across the cap itself.  At some frequency, the resistance of the cap equals the "top resistor", and you have 6db attenuation for that frequency.  For extremely high frequencies the cap starts to look like a short.

So, the frequency-dependent resistance of that cap stays constant, and by varying the size of the "top resistor", you are changing the proportional relationship in the voltage divider.

Don't know if that helps, but I personally had a pretty big revelation when I read PRR's statement that "everything useful is a voltage divider".

[PRR]

> How does increasing the resistance between the clipping and gain recovery stage change the tone?

It's not a gain recovery stage. It has no gain. IC1a's output could easily drive any pedal load.

As Marc says: C4 is the other shoe-drop. It leaks highs to ground. The first shoe is R4+TONE resistance. If R4 is small it can handle any leak-off in C4. If R4 is large, C4's leak-off weakens the signal.

Bill points to an R-C calculator. IMHO it is good exercise to work-up a hand-writ table with various R across the top, various C down the side, and frequencies at the intersection. It's enough to go 1K 10K 100K, because that lets you absorb the general trend, and also lets you figure that 3K or 5K will be between 1K and 10K and give you a good-enuff answer without always going to the calculator.

ashcat_lt's concept of "voltage divider" is VERY useful.

So why is IC2b there? Second, because we get two opamps per blob at the same price as one. First, because any load on the R-C network changes both gain and roll-off. IC2b's input is "no load" (>100Meg), so the R-C network has consistent action no matter what stuff it is plugged into.

[ashcat_lt]

ashcat_lt's concept of "voltage divider" is VERY useful.

I'm not taking credit for that.  It's all you!   

[Hemmel]

I'm attempting to perfboard this right now (just realised I don't have any 2N7000's...) but one thing caught my eye :
The V2 (or 4.5V) on pin 3 of IC1. This is also where the audio input goes. Shouldn't there be a diode somewhere, maybe between C6 and R5, to prevent voltage from going towards audio input ?
Maybe the voltage isn't that high to matter ?

[ashcat_lt]

C6 blocks the DC bias voltage from affecting, or being affected by, the source.