MXR Phase 45/90 and the function of it's LFO

[YouAre]

I'm trying to understand the function/output of the Phase 45/90's LFO so that I can tinker with it.

After reading this thread:

http://www.diystompboxes.com/smfforum/index.php?topic=82069.0

I breadboarded just the LFO from a phase 90 with a large resistor in place of the rate pot (to make it oscillate super slow), and measured the voltage at the 10uF cap. It seemed to vary between 3.8v and 5.8v, which makes sense to me because the input to the LFO is about 4.5-4.8volts. So it seems like a voltage swing of +/-1v.

I can't seem to get a good reading on current, so I don't know what's going on there.

So it seems that the output of the LFO is a varying voltage, with little current. Does that make sense? But It seems that the output of the LFO for the phase 90 goes to the source and gate of the FETs, which (if I didn't fall asleep too much in my digital electroncis class) can set up as variable resistors depending on the source voltage, right? Is that what the LFO in the phase 90 is doing? Turning those FETs into variable resistors?

My goal here is to replace the phase 90 LFO with an LDR, so that it can be controlled with a tremulus lune's LFO. I'll worry about finding the right LDR, and properly setting up the LDR as a voltage divider later....Or is replacing the LFO with a voltage divider going to spell trouble?

Can anyone offer insight on the function of the LFO in the phase 90? And can anyone offer advice on replacing the LFO with something else?

Thanks for the help again!

[daverdave]

Well if I'm right the phase 90 has a triangle wave lfo, and yes the FETs are ebing used as variable resistors for the sweep of the allpass filters. It'd be pretty easy to replace the FETs with LDRs I'd have thought, you'd probably want a larger voltage swing out of the LFO though to get a nice sweep.

You could easily set a potentiometer as a voltage divider into the led segments of the ldrs, just remember to add current limiting resistors to protect the leds.

I designed a phaser for my first year uni project, it's not the most amazing or genius thing ever but it uses LDRs and I think it works great. I'll try and draw up the schematic later and post it if you're interested in seeing it.

[YouAre]

Well if I'm right the phase 90 has a triangle wave lfo, and yes the FETs are ebing used as variable resistors for the sweep of the allpass filters. It'd be pretty easy to replace the FETs with LDRs I'd have thought, you'd probably want a larger voltage swing out of the LFO though to get a nice sweep.

You could easily set a potentiometer as a voltage divider into the led segments of the ldrs, just remember to add current limiting resistors to protect the leds.

I designed a phaser for my first year uni project, it's not the most amazing or genius thing ever but it uses LDRs and I think it works great. I'll try and draw up the schematic later and post it if you're interested in seeing it.

Awesome, thanks! I'm wondering though, what would the circuit look like if the FETs were replaced by LED's? What components would we take out, and what is the positive input of each stage's opamp going to exactly?

I see what you're saying about replacing the FET's with LDR's, and the need to adjust the LFO for that (that's kinda like the univibe, but with photocells, right?). And I'd love to see that schematic, thanks!

[daverdave]

You're replacing the FETs with LDRs, one half of an optoisolator, generally all you need is the LDR with a resistor in parallel with it to set the max resistance.
I'll have a better look when I finish work.

[frequencycentral]

There are a number of phasers which use LDRs instead of FETs as the VCRs. Here's a couple:

http://guitarwork.ru/electronic/firm/Phaser/MXR%20Phase%20100.gif

http://commonsound.org/phaseur/pcb1.1f/schematic.pdf


...also, Roland implemented a control voltage input on their 172 FET based phaser:

http://dl.dropbox.com/u/967492/M172-B.jpg

[YouAre]

Gentlemen, I've gone cross-eyed. But I think the schematics you sent me cleared up a little bit of what I was confused with

Daverdave, I see that now. I was looking at the wrong schematic, and I thought the FET's looked a little more involved than they actually are. Question is, what resistance range do they vary between?

FreqCent: The MXR 100 and Roland Phaser are a tiiiiiiiny bit too complicated for me at the moment, by the Phaser Fleur helped a lot actually. I'm first trying to understand the output of the LFO, and then what it does to the circuit. Looking at this schematic:

http://www.geofex.com/FX_images/p180plus.gif

I see a few thing that the + input of the LFO opamp is connected to the Source of each FET. Is this like a Vref? Is Vref created by putting the 5.1v zener diode to ground from 9v? And does the 250k bias help set vref? (Phew! I'll mail a cookie to whoever answers all those)



Also, any insight on the output of the LFO?

[frequencycentral]

I see a few thing that the + input of the LFO opamp is connected to the Source of each FET. Is this like a Vref?

Yes, the sources of the FETs together withe the +ve input of the LFO opamp (via a resistor), and a few other things are connected to vref

Is Vref created by putting the 5.1v zener diode to ground from 9v?

The vref starts at the junction of the zener and the 10k resistor to +ve.

And does the 250k bias help set vref? (Phew! I'll mail a cookie to whoever answers all those)

Yes, the bias trims the vref down to a point which is desirable to the particular set of FETs used.

Also, any insight on the output of the LFO?

Triangle, a few volts either side of 4.5v.

(Phew! I'll mail a cookie to whoever answers all those)

Choc chip?

[daverdave]

Most optical phasers use about 220k resistors to cap the ammount of resistance the LDRs can go to, that's what I used and it produces a pretty wide sweep as far as I can tell.
My lfo swings from 1V to 9V, the top of the power supply. The sweep is pretty damn big as far as I can tell. The LEDs are protected from excesive current by 220 ohm resistors.
I personally think using LDRs is the easiest method, that's for ME personally, and the results are pretty sweet. I made the optoisolators out of tellow leds and generic LDRs, they work a treat but at faster speeds the ldrs have a hard time keeping up, so you do lose the bottom end of the sweep.

As I said, I'll post a schematic, which I've been meaning to do for ages, when I get it done.

[R.G.]

So it seems that the output of the LFO is a varying voltage, with little current. Does that make sense?

That is correct.

But It seems that the output of the LFO for the phase 90 goes to the source and gate of the FETs, which (if I didn't fall asleep too much in my digital electroncis class) can set up as variable resistors depending on the source voltage, right? Is that what the LFO in the phase 90 is doing? Turning those FETs into variable resistors?

Yes.

My goal here is to replace the phase 90 LFO with an LDR, so that it can be controlled with a tremulus lune's LFO. I'll worry about finding the right LDR, and properly setting up the LDR as a voltage divider later....Or is replacing the LFO with a voltage divider going to spell trouble?

I'm a little unclear on what you're intending. Do you want to (1) replace the P90 LFO with a different one, (2) control the existing LFO from a second one (e.g. the tremulus lune), or (3) replace the JFETs with LDRs *and* the LFO with a different one? Maybe I haven't had enough coffee yet this morning.   It's early here.

Can anyone offer insight on the function of the LFO in the phase 90?

The P90 LFO is a very specific implementation of the standard integrator-schmitt trigger oscillator. The integrator is replaced by a single cap. The cap ramps up and down in voltage at a rate which depends on the current going into/out of it as controlled by the speed pot and any other series resistance. The speed pot is in turn driven by the output of a schmitt trigger (the opamp).

Sidebar 1: The Schmitt Trigger
The Schmitt trigger is a comparator with hysteresis. An ordinary comparator has a reference voltage and an input. The input is compared to the reference, and if the input is greater than the reference it does one thing, and if the input is less than the reference, it does another. This neat explanation of comparators leaves aside what happens if the input is more or less exactly the reference. In practice, you get ugly, unpredictable results if that happens. The Schmitt circuit sidesteps the question of input equals reference by using a little positive feedback. The output is coupled back to the reference through resistors in a way that lets the output change the reference that the comparator actually sees. This splits the reference into two references, a higher and a lower trip point, centered on the actual reference. This makes a "dead zone" in the middle. When the input is below the actual Vref, it must get higher than Vref+Vdelta before the output flips. When the output flips, it changes the effective reference that the comparator sees to Vref-Vdelta, so that now the input must go below Vref-Vdelta before the output will flip again. Vdelta is set up by the choice of the positive feedback resistors. The designer can put in a few millivolts of hysteresis just to eliminate any oscillatory fuzziness in the middle, or can set the trip points to very wide apart just by the choice of the resistors.

In the P90 LFO, the reference and trip points are set to be the right amount for driving the JFETs in their variable resistor range.

And can anyone offer advice on replacing the LFO with something else?

Yes. Remove the LFO. Replace its output with a variable voltage source (pot, perhaps) to drive the JFET gates a variable amount below the source voltages (Vref). Put an oscilloscope on the dry signal and the signal at the end of the phase line, and drive the input to the line with a sine wave. Vary the JFET gate variable back-voltage while you watch the relative phase of the wet and dry signals. Twiddling the JFET back voltage will cause the phase to change. Note the voltages for where the phase stops changing both forward and backwards. That is the usable voltage range for putting in a control voltage to the JFETs **for those specific JFETs**. Once you know the voltage range which runs the JFETs through their usable range in the effect, you can supply any voltage in that range and get the JFETs to do their stuff.

This will involve both a DC offset from zero and a range of usable voltages. In the P90, this is on the order of +1 to +3V end to end. You will have to supply a matching voltage from any new LFO you put in. Otherwise, the gates will be driven out of their variable resistance range, and usable phasing will stop. It may do interesting things outside the variable resistance range, but it won't be phasing.   Your job in replacing just the LFO is to make the JFETs get the gate voltages they expect, whatever you put in to replace the LFO.

That's all if you wanted to replace the LFO only, leaving the JFETs. If you intended to replace the JFETs with LDRs, then you have to drive the LDRs with the correct range of light intensity. In every modulation scheme, you have to (1) understand the usable range of inputs for the modulator, whether that's voltage, current, light, sound, gamma rays, whatever, and (2) provide a control signal in that range.

Awesome, thanks! I'm wondering though, what would the circuit look like if the FETs were replaced by LED's? What components would we take out, and what is the positive input of each stage's opamp going to exactly?

LEDs or LDRs? LEDs put light out. LDRs are controlled by it. Mostly. In fact, both do mostly one, but a little of the other.

I see what you're saying about replacing the FET's with LDR's, and the need to adjust the LFO for that (that's kinda like the univibe, but with photocells, right?). And I'd love to see that schematic, thanks!

The Univibe *is* photocells. Photocells is the other common name for LDRs.

Question is, what resistance range do they vary between?

All JFETs have a variable resistance range between r_dson from the datasheet to off, effectively open circuit. R

dson

for most JFETs is between 10 ohms and a few hundred ohms. The better question is what's the usable range of resistances that makes for an audible change of phase on the phase shift circuit? In this case, the answer lies in the way the phase shift circuit works. The phase shift is done by varying the "R" in the R*C circuit on the + inputs. The cap on the input has an impedance which varies with frequency, as Xc = 1/(2*pi*f*C). The effect of the resistor in controlling the output phase starts one decade below the RC crossover of the R*C network and extends to one decade above it. So if the crossover frequency of the cap and JFET resistance is 1000 Hz, the lowest frequency that shows any noticeable effect is 100Hz and the highest is 10000Hz. In practice, only about an octave above and below makes any difference, so the noticeable range for RC=>1000Hz is 500 to 2000. As you change the R by varying the JFET resistance, this two-octave phase change region moves around, following the R*C frequency.

All that is a preamble to saying: it depends on the capacitance value. The JFET *can* vary over a hugely bigger range than you need. It's most sensitive in the 1K - 100K region. You pick the frequency range by selecting the cap value to put the sensitive phase change regions in the middle of the human hearing sensitivity region for the biggest audible effect. Call it 500Hz to 1kHz. And put in enough wobble to wave it completely through the region, then season to taste.

I'm first trying to understand the output of the LFO, and then what it does to the circuit. Looking at this schematic:
http://www.geofex.com/FX_images/p180plus.gif
I see a few thing that the + input of the LFO opamp is connected to the Source of each FET. Is this like a Vref?

You intuition is good. It is Vref. The opamps and JFETs happen to share Vref. Their connection to each other is coincidental to that.

Is Vref created by putting the 5.1v zener diode to ground from 9v? And does the 250k bias help set vref? (Phew! I'll mail a cookie to whoever answers all those)

Yes, yes, and I like chocolate chip.  

Also, any insight on the output of the LFO?

See above long winded explanation of LFOs, Schmitts, and stuff.

[frequencycentral]

Does everyone who answers get a cookie? Or just the first to answer? Is there more than one cookie on offer? Can we answer more than once and get more cookies?

[R.G.]

Don't care. I just like chocolate chip. 

[YouAre]

WOW GUYS!

Ok R.G., as usual, your replies/answer never totally satisfy me because...you make me search for more! Haha, your informative posts have driven me to get off my ass and LEARN. You don't get a cookie. Why would I do such a thing, promise you a cookie then deny it you ask? Because you're one of the dudes who inspired me to actually learn how everything works. I used to be one of those lazy d-bags who just comes on this forum to whine "waaaah, I cloned this boutique pedal and it doesn't work. fix it for me!" Now I'm working on custom designs and mods on my own, thanks largely to your posts/emails. So no, you're not getting a cookie. But you WILL be getting royalty/licensing checks from me in the future when I sell my own designs. Hopefully those designs will cause many a headache amongst the future n00bs of this forum when they find an inaccurate schematic of them on some russian/japanese website . Besides, imagine all the cookies you can buy in the future with the licensing!

FreqCent, i have to find that dunkin donuts coupon for a cookie around here somewhere, and I'll mail it to you if you want. If I can't find it, I'll mail you the buck twenty-five it costs to get one 

R.G., I need to read all of your response in more detail before I ask more questions because yes, there WILL be blood....I mean more questions. Thanks again guys! I'm gonna read these responses and hit the books and breadboard to visualize all this stuff! I'll be back!

[YouAre]

.

Question is, what resistance range do they vary between?

All JFETs have a variable resistance range between r_dson from the datasheet to off, effectively open circuit. R

dson

for most JFETs is between 10 ohms and a few hundred ohms. The better question is what's the usable range of resistances that makes for an audible change of phase on the phase shift circuit? In this case, the answer lies in the way the phase shift circuit works. The phase shift is done by varying the "R" in the R*C circuit on the + inputs. The cap on the input has an impedance which varies with frequency, as Xc = 1/(2*pi*f*C). The effect of the resistor in controlling the output phase starts one decade below the RC crossover of the R*C network and extends to one decade above it. So if the crossover frequency of the cap and JFET resistance is 1000 Hz, the lowest frequency that shows any noticeable effect is 100Hz and the highest is 10000Hz. In practice, only about an octave above and below makes any difference, so the noticeable range for RC=>1000Hz is 500 to 2000. As you change the R by varying the JFET resistance, this two-octave phase change region moves around, following the R*C frequency.

All that is a preamble to saying: it depends on the capacitance value. The JFET *can* vary over a hugely bigger range than you need. It's most sensitive in the 1K - 100K region. You pick the frequency range by selecting the cap value to put the sensitive phase change regions in the middle of the human hearing sensitivity region for the biggest audible effect. Call it 500Hz to 1kHz. And put in enough wobble to wave it completely through the region, then season to taste.

OK after having read through this I'd like to clarify. Using the Capacitor feeding the positive input of each opamp and the effective resistance connecting it to Vref, the equation for the frequency is 1/(2*pi*r*c), right? And you're saying to basically vary the frequency between 500-1000Hz? So if i wanted to replace the FET's with LDR's, I'd want to tune the resistances such that the LDR in parallel with the resistor to Vref effectively equal the appropriate resistances that satisfy the 500-1000 = 1/(2*pi*r*c), right?

Throwing those frequencies into that equation yields effective resistances (LDR || the resistor to Vref) ~= to 3.1k to 6.4k. That sound about right?

[askwho69]

bump

[PRR]

....That sound about right?

Yes.

bump

After 5 months of thought, a 2-hour scrolling bump?

[YouAre]

....That sound about right?

Yes.

bump

After 5 months of thought, a 2-hour scrolling bump?

Awesome. thank you!

And i'm sorry about it taking 5 months, haha. Life kind of got in the way of pedal building.

[askwho69]

It a good thread to wake up! sorry for that im planning to build phaser one of this days

[stringsthings]

And can anyone offer advice on replacing the LFO with something else?

Yes. Remove the LFO. Replace its output with a variable voltage source (pot, perhaps) to drive the JFET gates a variable amount below the source voltages (Vref). Put an oscilloscope on the dry signal and the signal at the end of the phase line, and drive the input to the line with a sine wave. Vary the JFET gate variable back-voltage while you watch the relative phase of the wet and dry signals. Twiddling the JFET back voltage will cause the phase to change. Note the voltages for where the phase stops changing both forward and backwards. That is the usable voltage range for putting in a control voltage to the JFETs **for those specific JFETs**. Once you know the voltage range which runs the JFETs through their usable range in the effect, you can supply any voltage in that range and get the JFETs to do their stuff.

This will involve both a DC offset from zero and a range of usable voltages. In the P90, this is on the order of +1 to +3V end to end. You will have to supply a matching voltage from any new LFO you put in. Otherwise, the gates will be driven out of their variable resistance range, and usable phasing will stop. It may do interesting things outside the variable resistance range, but it won't be phasing.  

[savethewhales]

Hey guys!

I've read the whole thread because I am building a Phaser pedal inspired by the MXR P90.
However, I wanted to fully understand the LFO part, which is not what's happening.

I already made simulations on softwares, and what I got was a square wave at the end of the schmitt-trigger and a triangular wave out of the whole LFO part.

But I can't understand how does the whole schematic for the LFO work. Like why are those values of resistances put there in the schmitt-trigger? And why is it not simpler, just like it's shown on the internet? Why is the capacitor of the Schmitt-Trigger around the value of 0.01 uF? What does it implies besides delay on starting to deliver the square wave (as I tested it)?

Also, how are the vaues of the  "Triangular part" chosen? How do I make the calculations/assumptions?

I'm sorry for the big post, but I'm kinda desperate and kinda lost right now.

Greetings, Fred.