How to invert an LFO (in an LED/LDR combo)

[mrsage]

I actually posted about this in my other thread (about using optocouplers as variable resistors), but since this isn't really about making a variable resistor I thought I would start a new thread.

I'm trying to put an optocoupler in place of a potentiometer in the Kay Fuzztone I'm building. I found out from my other thread that I need the optocoupler to act as a potential divider instead of a variable resistor. And I need to make two LFOs...and one of them needs to be inverted.

It all makes sense conceptually, but I'm having trouble finding any information on how to invert one of the LFOs. A lot of the posts in the archives assume the builder already knows about inverting and non-inverting things...I'm just trying to figure out how to do that.

I already built one working LFO out of this (the auxiliary LFO for the Phaseur Fleur):
http://www.commonsound.com/phaseur/auxlfo_modulate.pdf

Now I need a second one, only inverted. My question is how to do that? Do I simply swap the + and - legs on each side of the opamp? Using a 4558 chip, that would mean legs 2 & 3 and legs 5 & 6:
http://www.semiconductors.philips.com/acrobat/datasheets/NE_SA4558.pdf

That seems a bit simplistic and I'm sure I'm missing something...but I can't find a good explanation for how to invert one of the LFOs. I'm hoping someone can help me out.

Once that's ready, I can wire the two up together like a potentiometer.

I'm told that the Tremulus Panneur has two LFOs, one inverted and one non-inverted, but I can't tell the difference between the two...they look identical:

http://www.commonsound.org/panneur/panneurscheme.gif

Can anyone help out? Thanks for the help thus far, Colin...I just thought I'd put all this in a new thread since it's a slightly different topic from optocouplers as variable resistors...

[loscha]

You pose an interesting question: If you just wanted one LED on at a time, you'd just make a Sine LFO, have two diodes back to back, and have the other ends of the diodes biased so one half of the wave one is on, and then the other if off, but, this would only give you one LED on at a time, and I'm assuming you want a "crossfade" between the two.

I'll have to think about this one, but I might have a circuit in a book here, I'll have a forage and get back to y'all

[Paul Perry (Frostwave)]

I'm told that the Tremulus Panneur has two LFOs, one inverted and one non-inverted, but I can't tell the difference between the two...they look identical:
http://www.commonsound.org/panneur/panneurscheme.gif
...

See that trem/pan switch?
When it is in the trem position, then the two circuits do the same.
When the switch is in the pan position, then the output is inverted, compared to the other one.

And when the other switch is in sync position, they both connect to the same LFO, and when not in the sync position, the two sections are controlled by different LFOs.

[ExpAnonColin]

To elaborate on Paul, all you need to do is find the output of your LFO on the aux LFO thing, and attach everything after the sync/unsync switch, with it in the upper position.

-Colin

[puretube]

http://diystompboxes.com/sboxforum/viewtopic.php?t=29374

[ExpAnonColin]

http://diystompboxes.com/sboxforum/viewtopic.php?t=29374

Aw come on, that is way too easy.

-Colin

[mrsage]

http://diystompboxes.com/sboxforum/viewtopic.php?t=29374

:shock:

Wow...okay...

That seems like almost exactly what I need. Actually, it probably is exactly what I need...


So at the risk of sounding like a complete newbie, can someone walk me through how to apply that to my LFO? :oops:


I have built this already (exactly as shown):
http://tinypic.com/5361py

That works perfectly as a varying resistor. The two leads on the far left (marked "To pin 7" and "Main speed 3") are just like the two leads of a standard resistor...the only difference being that its value changes (speed and depth controlled by the respective knobs).


So now do I just add an additional resistor and LED to my existing circuit?

If so, where?

[darren]

Hey mrsage, happy to walk you through it  :-)

first up, the amount of light output by a diode is dependant on the current through it.  Since the diode has a voltage drop but no resistance, the current is controlled by an external limiting resistor and the voltage across it according to ohms' law.

If we have an LED, a 1k resistor, and a 9V battery, and the spec sheet for the diode says that it drops 1.5V, we can work out that the voltage across the resistor is 9V-1.5V=7.5V and that by ohms' law the current through both the resistor and diode is 7.5V/1000ohms = 7.5mA.

Say we vary the voltage as a sine wave between 0V and 9V.  Below 1.5V there will be no current as the diode won't conduct until the 1.5V threshold is reached.  Above this, we can see that at 2V we get 0.5V/1000ohms=0.5mA current, and at 5V we get 3.5V/1000ohms=3.5mA current.  So, as we change the voltage across the diode and resistor string, the current goes up and down.  So far so good, you probably had that bit figured.  :-)

Looking at puretube's diagram, let's assume that the output of the oscillator (triangle thingy) is a sine wave varying between 0V and 9V, and that the ground is at 0V and that the B+ is at 9V.  It's intuitive to see that the voltage across the bottom resistor and diode is whatever the voltage at the output of the oscillator is.

Less intuitive, and key to understanding this circuit, is that the voltage across the TOP resistor/diode is the difference in voltage between 9V and the output of the oscillator.  So if the oscillator's output is at 0V, the voltage across the top diode/resistor string is 9V.  If the output of the oscillator is at 3V, the voltage across the top diode/resistor is 9V-3V=6V.

Cool huh.  :-)

[mrsage]

[sigh]

Yeah, that's pretty cool, but I think I'm getting in over my head with all this...

I understand all of what you said from a conceptual standpoint, and it makes a lot of sense...but I just can't wrap my head around it enough to figure out how to apply it to the circuit I have here in front of me.

I have a nice little LFO controlling an LED/LDR combo, and I can adjust the depth and rate nicely with two pots...I just can't figure out how to take that circuit and make a second one that sends the LED/resistor to 9V+ instead of to ground. And I'm also realizing that even if I did, I'm not sure how I would keep the depth and rate in sync with the first LFO...(since it doesn't seem like I could send wires from two different boards to the same depth and rate pots)

So I'm admitting defeat. I'll step away and maybe attempt something like this when I'm a bit further along in my stompbox life.

Thanks anyway guys...I'm sure I'll come back to this thread in the future and shake my head in dismay at how easy this should have been.

[darren]

is that what you were after?

[mrsage]

is that what you were after?

Maybe...

Thanks for all the help. I'm going out of town for work, but I'll pick all this up when I return...hopefully I'll be able to make something of all this....

[Joe Kramer]

Hi!

The circuits offered by puretube and Darren will give inverted and noninverted, or biphase LFO, only under one condition that hasn't been mentioned here: the LFO circuit has to be bipolar supply so that its output swings both positive and negative in reference to ground.  

I may be missing something in the circuit you posted, but I think it's single-supply, which means its output is swinging only positive in reference to ground.   If you run the LFO circuit with +/- 9V, say, then the biphase trick will work fine.

Joe

[puretube]

are you so sure?

[Joe Kramer]

are you so sure?

Well, no I'm not.  But I'm eager to learn!  I said what I did because I tried that trick with the standard Boss-type sigle supply LFO.  The output swings between about +2V to about +7V, so the LED only worked in one direction for me.  I was using a little untested hypothesizing to arrive at the bipolar conclusion.  Wouldn't be shocked to find out that I was wrong--happens all the time!  That said, is what I said true or not?  If not--oops. . . .
Joe

[puretube]

...that makes 2.5V below to 2.5V above "Vbias" (=4.5V in a single 9V supply circuit);
now if you don`t look at the output of the opamp as a "voltage provider" referring output to ground,
but rather as a "source/drain" referring to the rails, you`re getting close...

[Paul Perry (Frostwave)]

Maybe a VERY LARGE cap would overcome the offset problem?
Or something coulod be done wiht a zener?

[Joe Kramer]

...that makes 2.5V below to 2.5V above "Vbias" (=4.5V in a single 9V supply circuit);
now if you don`t look at the output of the opamp as a "voltage provider" referring output to ground,
but rather as a "source/drain" referring to the rails, you`re getting close...

Are you saying that if the positively connected LED is biased at 1/2V+ like the opamp, it will work?  In other words, one LED goes opamp to ground, the other goes opamp to +4.5V?  If so, I've never tried it that way.  Then again, it seems like this wouldn't function symmetrically: the 4.5V LED would have half the range as the grounded one, and there would be overlap between the two.   I may be wrong about that too though!  

Regards,
Joe

PS: I just went and tried the 4.5V bias trick and it didn't work at all.  It occured to me that maybe BOTH LEDs had to be connected for it to work (I previously only tried them singly), but I tried this and still nothing.  So, at this point, I have to say I'm pretty sure the biphase LED trick will not work with the Boss-type single supply LFO similar to CH-2, here:

http://home.hetnet.nl/~chrisdus/download/ce2.gif

My "hypothesis" is, it will only work with a bipolar-supply LFO, but I'm still willing to be shown otherwise.  --Joe

[puretube]

http://diystompboxes.com/sboxforum/viewtopic.php?t=29374

Aw come on, that is way too easy.

-Colin

well, err, I just looked into a very popular pedal: they use that too!