MXR 100 phaser build - theory and trim

Started by alextdel, January 11, 2006, 07:30:54 PM

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alextdel

I am about to build the GGG MXR 100 Phaser. Does anyone have any links to the theory of operation of this opamp / LDR based phaser - expecially what the trim is for and how to set it up.
http://www.generalguitargadgets.com/diagrams/mxr_p100_sc.gif

varialbender

Check this site out, it gives you some good phaser theory:

http://www.geofex.com/Article_Folders/phasers/phase.html

A lot of that stuff is applicable to this diagram.
Also, I'm pretty sure you can set that trim pot by ear. Whatever sounds best.

Luck

Mark Hammer

Quote from: alextdel on January 11, 2006, 07:30:54 PM
I am about to build the GGG MXR 100 Phaser. Does anyone have any links to the theory of operation of this opamp / LDR based phaser - especially what the trim is for and how to set it up.
http://www.generalguitargadgets.com/diagrams/mxr_p100_sc.gif

You will note that R47 is tied to R48.  R48, in turn, sets the current on a voltage taken from the voltage divider (pot) formed by R49 (tied to V+) and R50 (tied to ground).  So R47 adjusts the base of Q3, producing a voltage offset, controlling where the sweep starts from.  This is *roughly* equivalent in function to the Manual or Initial control found in some phasers and flangers.  You may not be able to use all of its rotation, but it may be worth panel mounting a 20k-25k pot for R47 instead of using a trimpot.

Ideally, R47 is set up such that the sweep does not flatten out (i.e., not change even though the LFO is still going) at one end or the other of the sweep cycle.  Within those constraints, though, it should be something you can adjust to taste.  I have an old Rocktek phaser (schem at Experimentalist Anonymous's site), which also has a bias/offset/centre-trim trimpot on the board.  I chassis mounted it and can adjust the flavour of phasing a fair amount, although admittedly the outside 20 degrees of rotation at either end are not useful.

varialbender

In the case where, say, you're using a 100k pot, and it's only useful between 25k and 75k, could you put a 25k resistor in series and 50k in parallel to the pot? Or, of course, a 25k in series with a 50k pot?

Mark Hammer

What you do, ideally, is twiddle the trimpot and identify the range where twiddling does something useful.  Then you measure the outside lug to wiper resistance at each end of that range.

For example, say a 100k trimpot did something useful between the point where the lug-1-to-wiper resistance was 17k, and the point where the wiper-to-lug-3 resistance was 38k.  100k minus 55k gives us 45k.  Obviously, we won't find a 45k pot, but we CAN find nominally 50k pots and 47k pots.  If we stuck a 36k fixed resistor and an 18k resistor on either side of a 47k pot, we would have the functional equivalent of a 18+36+47 = 101k pot (close enough for rock and roll!) that could never be turned beyond the 18k and 36k points.

Those of you with Dr Q clones and derivatives (incl. Dr Quack and Nurse Quacky and the E-H Bassballs) might consider replacing the trimpots between the envelope follower and transistor that tunes the filter with a smaller value pot and selected series resistors that will make the entire rotation of the pot usable (currently only part of the rotation is).

varialbender

Cool, thanks.

I went over that schematic, and there are a couple things I found intersting. It looks like there are 10 stages, but only 6 of them are oscillating. Also, it looks like the phased signal is fed back at the end of the 9th stage. I'm having a bit of a hard time reading the oscillator circuit, but I'm getting there.
Any info would be much appreciated.

Thanks

Mark Hammer

The phase shifter doesn't particularly care how many stages it takes to produce the phase shift or how many of them are swept.  All it "knows" is that when you add up all the phase shift over stages that some frequencies are shifted by enough degrees that cancellation results when combined with a nonshifted copy of the signal.

The fixed phase-shift stages, each contribute up to a maximum of 90 degrees above a certain frequency, and less than that below that frequency.  That amount of phase shift is added to thenumber of degrees produced for each frequency by the swept stages. 

Think of it like money you got from your aunt for Christmas.  It's not *quite* enough to buy you the Xbox 360 you lust after, but if you can save a little more you are soooooooo there.  Because you have a head start from the money you didn't have to earn, you don't have to work quite as long or hard to get the money that puts you over the top.

Similarly, the phase shift provided by the fixed stages provides a kind of reserve bank that doesn't do an enormous amount on its own, but when added to by the swept stages allows you to easily create more notches in more places.

Because the allpass/phase-shift stage used in the P100 is inverting, you need to feed it back an *odd* number of stages in order for it to accentuate the cancellation and create resonances on each side of the notches (because an even number of negatives is always positive, right?).  To do that, many phasers feedback the last stage to the second one.  IN this instance, because the last two stages are not swept, they feed back the ninth to the first.

alextdel

Despite starting this thread I am not replying as I am unworthy :icon_mrgreen:
However I am monitoring the thread and learning heaps.
Mark and Varial, which components comprise stage 1 and 9 as I can't see a feedback path?
My thoughts....IC4a is a buffer, IC5b is stage1 this would make IC2a stage 9 but isn't R18 the feedback path to stage 1, this comes from IC3b (stage 10??)

I'm a bit lost ???

varialbender

I think schematics are made a little easier for building than for learning. It can be hard to look at something the way that it's layed out sometimes. I can't seem to get the image loaded, so unfortunately I can't redraw it right now, but you have to try to think about the circuit a little different. Those 10k resistors at the top are in a nice pretty line, but it would probably make more sense to you if they were reorganized. Try and redraw tham so that it looks like a phaser you understand. Take the 10k resistors going around the opamps and draw them in closer to the opamps, like a nice little loop around it. Then, from the output of the opamps, draw a single line that splits up to a 10k resistor at the top side of the opamp, and a cap on the bottom side (with the extra junk around there). Hopefully this'll help you see that you can take the output from the 9th phase stage and feed it back up while making a bit more sense.
Make some sense?
It sounds like you're understanding alright, keep going over the geofex article, it's a really good one.

Makes sense that it's feeding back like that. You can take the ouput at the 10th stage and feed it to another opamp with a pot between the positive and negative input and take the output from that and feed it to the 1st stage right? This'll let you do positive and negative feedback?

Also, not varying 4 of the stages, does that spread the notches out as it's swept more than if the 4 stages were also tied to an LFO? Sorta like, the notches are a bit... lazier at keeping up with the others? One end will move a lot less than the other?

Thanks a lot

Transmogrifox

Upon staring at the schematic for a bit I have determined it is unlikely that you could keep the circuit stable by implementing positive feedback as suggested.  For very low frequencies (particularly DC), this circuit looks like a bunch of cascaded inverting op amps.  If they are somehow arranged to a positive feedback, they'll drift to one of the supply rails and stay pinned there indefinitely.  You would probably get 0 sound out of the phaser.

For high frequencies, the capacitor looks like a short and in in fact contributes little or no phase shift of its own, so it also makes the circuit look like a long string of noninverting op amps; and again, to rearrange this circuit to get positive feedback through the noninverting inputs will make it go unstable, though for this config it would make a high-pitched squeal as opposed to just being pinned to a rail.  The feedback implemented cleverly uses an op-amp inverting input to set the inverted phase right off for the high frequencies, which experience no more appreaciable phase shift in the following stages.  For the DC, it goes through the right number of inversions to where it is negative feedback back by the time it makes it around the loop.  Everything in between the extreme low and extreme high frequency is where the interesting resonance stuff happens-- for at these frequencies the capacitors are actually generating some real delay, not just this "flip upside-down" type phase relationship.  Over the transition band there are some frequencies that get positive feedback, and others that get negative feedback.  This works to create some resonance.

If you put a bandpass filter in the feedback loop, you might be able to push the envelope a bit, but bear in mind you're teetering on the edge of putting some of the poles in the right half plane  (in non-engineering terms: teetering on the edge of a loud squeal or not workingness).

As stated before, you can adjust the trimpot by ear.  It will be obvious what it does for you.  This is actually a very simple type of circuit to build because there's a lot of repetition and the components lay out nicely around the op amps.  Furthermore, there are few external controls (pots and switches) to mess you up.  Pretty much plug and play.  Just keep your solder joints done well and make your circuit tidy.
trans·mog·ri·fy
tr.v. trans·mog·ri·fied, trans·mog·ri·fy·ing, trans·mog·ri·fies To change into a different shape or form, especially one that is fantastic or bizarre.