How does the Orange Squeezer work? Mods?

[rockgardenlove]

Hmmm...
So in the OS, when the current is flowing through the channel, are we talking about the current flowing through Q2, and then to ground?  This is what the AC from the signal is doing, right?
If the drain is more negative though, the DC would have to be flowing up through the channel from source to drain, huh?\
I know I sound dumb.   

[R.G.]

Ah. OK, I think I know where the confusion is. You're not dumb - I confused you.

What I was blathering on about is that JFETs have two different ways to work. They can work as a variable resistor in the low-voltage/low-current triode region, or as a voltage variable current source in the pinchoff region.

Q1 is working in the pinchoff region. That is, it's set up as a constant current source. All the stuff about channel voltage drop and pinchoff apply to Q1, not Q2.

Q2 is working in the triode region, as a voltage variable resistor. The drain of Q2 is at the same DC voltage as its source, since it's blocked in all directions by capcitors. But it swings more positive and more negative than the source voltage. The source of Q2 is fixed at some voltage above ground. The gate of Q2 is pulled lower than the source by the resistors R6 and R12. For the JFETs we commonly work with, A voltage of maybe 4V on the source of Q2 and 0V on the gate will turn Q2 off completely. That's why the bias voltage on R7 is important. It raises Q2's source above ground so that with R6 and R12 pulling it down, it can go to ground and turn Q2 fully off.

[rockgardenlove]

Yes!  Haha, hurrah!  Things make WAY more sense all of a sudden. 
It made no sense that it would be flowing from source to drain because of those capacitors, I feel smart now.  Woohoo!
I'm gonna try and clear my head and then go through it all again.
Thanks! 

[Jay Doyle]

http://www.vishay.com/docs/70598/70598.pdf

[markm]

http://www.vishay.com/docs/70598/70598.pdf

Very useful info there Jay!
Thank you Sir!! 

[Mark Hammer]

Not mentioned much so far is that the "virtual pot" formed by the FET and fixed resistor has two legs, one of which IS the fixed resistor.  The amount of signal attenuation resulting from dropping the drain-source resistance of Q2 will certainly depend on the FET and the envelope signal applied to Q2's gate, but it will also depend on the value of the "other leg of the pot", namely the 82k resistor.  As that resistance gets smaller/lower, small to medium changes in the FET resistance have less and less impact.

Consider the case of a passive divider made from a fixed resistor and a pot in variable resistor (rheostat) mode.  If we have 500k from input to "wiper" (midpoint), and 500k of variable/adjustable resistance from the pot, then what we see at the junction/midpoint can never be greater than half of what is at the input, right?  Any reduction in the pot resistance immediately starts to take away from signal amplitude by dividing down the input more and more.  Drop that pot resistance to 100k and you've chopped quite a bit of the input signal.

Okay, now make the input leg of our imaginary pot a 1k fixed resistor.  You would have to drop the resistance of the 500k pot a great deal to start to divide down the input signal appreciably.

The long the short of this is that the OS, and indeed ANY sort of compressor that uses a virtual resistive divider, whether fixed resistor + FET or fixed resistor + LDR, can have its response altered by changing the fixed resistance of the divider.

Although I've never seen one or heard of one, in principle one could create a resistive divider using two variable elements to create a transfer function which is nonlinear and adapts to overall signal level and transients.  For instance, if the 82k resistor were, say, a 220k resistor in parallel with a slightly sluggish LDR, and the LDR was set for slightly less response to the envelope signal (one assumes here that the envelope drives a light source), then the state of the LDR half of the virtual pot could be made to operate differently than the FET half.  The FET retains its lightening fast response time in all conditions, whereas the LDR may take some time to recover from sharp transients.  So, if a hard struck note sends our LDR to low resistance, then subsequent FET-resistance changes occurring before the LDR fully recovers will/should result in less relative attenuation.  If it does what I think it does, it should permit the unit to compress nicely to the first note in a string, but not make subsequent notes a slave to whatever the first note dictates.

Make sense?

[big bustle]

Thanks to all of you who participated in this discussion.

I think i'll learn more from what has been said here than I did in junior high.

I'm printing this to devour when I build my Orange Squeezer next week.

[welcomb]

Thanks R.G. for your write up. I was googling trying to understand how compressors like the OSQ works but couldn't find anything. I even looked up your "Technology of the..." section in Geofex but it wasn't there. So I gave up...

It's been wat, say 2 years(?) since you wrote this, why not just stick it into Geofex? I sort of found this by fluke trying to look for help debugging my OS build. I really enjoy all your articles there; they help me appreciate how all these effects work.

[Havaden]

Thank you guys, i finally understand how this works
But as a novice DIY pedal noob, i can't help but to wonder: won't the rectifier make the signal clip like any other diode clipping distortion pedal?

[Thecomedian]

Which rectifier is that? What schematic are you using?

ah nvm, I didnt even consider the diode stage that feeds back as a rectifying circuit..

[duck_arse]

the rectifier section is driven by the output of the opamp, but is isolated from it, and the rectifier feeds its "clipped, smoothed" output to the gate circuit of the fet, Q2, which does the compressing. the output of the rectifier does not appear in the signal output.

[Mark Hammer]

Thank you guys, i finally understand how this works
But as a novice DIY pedal noob, i can't help but to wonder: won't the rectifier make the signal clip like any other diode clipping distortion pedal?

There are two "versions" of the signal leaving the op-amp.  One becomes the audio output.  The other passes through a few components, most importantly a diode which removes half of the signal.  By virtue of that version of the signal becoming something that only varies from "zero" in one direction, rather than somethng that goes both positive and negative from zero, it is magically transformed into "DC", or a sloppy version of it.  And it is DC we need to make the JFETs behave as they do.

That version of the signal may have been derived from the audio, but is being used as a "control" voltage, not as audio.  As such, the "clipping" you refer to is simply never heard.

[Thecomedian]

Mark, would the "DC sine wave" also increase and decrease the depletion region, thus changing the amount of current flow?

[Mark Hammer]

Beats me.  I don't understand any of that stuff....really.