MODIFICATION Teaser: I've got a Circulator mod in the works

[Raspymcnasty]

4. Inserted 10K resistors (R103-R106) in series with pins 1 & 16 on IC1 & IC2. Pins 1 & 16 program the operating current and gain of each amplifier. The way EQD connected them all in parallel, there is no guarantee that they will share current equally. Putting those resistors in forces them to share. Should make for smoother phase sweep. Optional, but highly recommended.

Hey Chuck,

I have a question regarding mod #4.
So am I supposed to clip the 1 & 16 leg on the IC1 and IC2 and solder on a 10k resistor like the paint image below?

 

[Chuck D. Bones]

Yes. Don't clip the lead too short, just enough so that it doesn't contact the pad on the board. I found it easier to bend the top lead of the resistor in an upside-down "U" and solder that to the stub IC lead.

 

[Raspymcnasty]

I finally got working on your mods, tbh I was dreading it but here are some pictures of what i did in case any other newbie like me wants some pictures. Not sure if this is bad practice or not but i didn’t have 1/8w resistors to fit nicely.

I’ll report back with how it sounds

 

[Chuck D. Bones]

Looks good to me!

Correction: you have the 3.3M soldered to IC1-4, should be IC1-3. It will still work the way you built it, but it's not 100% correct.

 

[Preverb]

I was considering ordering this PCB and came across this thread. Was there still a plan for a boneyard edition?

 

[giovanni]

This may have been answered already, but how do the EQD production units avoid the LFO stalling issue? Carefully selected components?!?

 

[Chuck D. Bones]

LoL, good question. I don't know if EQD does this, but some electronics manufacturers build a bunch, test 'em, sell the ones that work and throw the rest on the scrap heap. No joke.

 

[jeffwhitfield]

LoL, good question. I don't know if EQD does this, but some electronics manufacturers build a bunch, test 'em, sell the ones that work and throw the rest on the scrap heap. No joke.

That’s exactly what I do! I pray that I get at least a 90% success rate though.

 

[Chuck D. Bones]

I worked with a hi-rel computer supplier who was satisfied with a 67% yield. Made me wonder about the long-term reliability of the "good" ones.

 

[giovanni]

I worked with a hi-rel computer supplier who was satisfied with a 67% yield. Made me wonder about the long-term reliability of the "good" ones.

Yikes!

 

[jeffwhitfield]

I worked with a hi-rel computer supplier who was satisfied with a 67% yield. Made me wonder about the long-term reliability of the "good" ones.

Yeah, that's a supplier than needs to be slapped senseless.

 

[Mibonti]

Hi everybody,

I've been trying to get this build to work for some time but still no luck.

All the pots are affecting the sound and if I manually turn the sweep knob, I do get a phaser. The PV switch also affects the sound but the rate switch obviously doesn't do anything. I did test that switch for continuity and it does work.
I've already checked all the components in the right spot and polarity (I might still have missed something?) and went through all the solder joints to make sure they were good. I added mods 8 and 9 as described in this thread, it should prevent the LFO from stalling but it still appears to do so in my case.

Somebody posted their voltages somewhere on the forum and they all were quite close except for every pin 2 on IC1, 2 and 3, where I'm only getting 0.1V. Perhaps this could also be a hint to the problem, I'm not experienced enough to figure that out.

I've added some pictures for reference, yes those mod resistors are not pretty but I just put them on quickly for testing.

Any help would be very much appreciated!

 

[zgrav]

One suggestion is to move your request to the troubleshooting thread where more people might see it and offer suggestions.

I also suggest trying to take better pictures with good lighting and closer focus on both sides of the board. If you look at the schematic in the build document, the part that makes the automatic sweep is in the parts around IC3. You will want to verify the connections between those parts and part values.

What are your measured voltages at all of the pins on IC3?

 

[Chuck D. Bones]

What he said.

 

[Bricksnbeatles]

Just came across this thread. Was never much interested in the circulator because I’m generally not that interested by EQD stuff outside of their FV-1 stuff, but these mods make it seem like a pretty worthwhile compact phaser.

Since the sweep control is set up as a voltage divider, what effect would be had from reducing the pot value down to 10k? Curious because I’d be interested to put an expression jack on the control, and most expression pedals are 10k.

 

[Chuck D. Bones]

10K is probably too low and will cause the LFO to stall. There is phaser code for the FV-1, so that may be your best solution if you want to add an expression pedal.

 

[FleeRemi]

DEPTH is a blend control. If you want to expand the top end of rotation, try a C-taper pot. I probably should have used that because like you say, the bottom end of a blend is not all that interesting.

Apologies for resurrecting an old thread, but I was wondering if another C100K would be good for this one (DEPTH)? Still very new at this and not entirely sure how the differences in pot-resistance value factors in along everything else, and the schematic calls for a B25K there originally. Thanks for any help!

 

[oliverg]

Apologies for resurrecting an old thread, but I was wondering if another C100K would be good for this one (DEPTH)? Still very new at this and not entirely sure how the differences in pot-resistance value factors in along everything else, and the schematic calls for a B25K there originally. Thanks for any help!

I've just built a clone of this circuit, albeit on my own PCB. As a disclaimer, I haven't built this exact circuit so can't give a direct comparison of the sounds. However, in this circuit, it is important to understand how the effect is "blended" with the clean signal in order to pick the right components.

TL;DR a C100K will be different to a C25K (which is only changing the taper), although C100K is also fine - just different to the effect of simply changing from a linear taper to a reverse log taper.

Amplifier U5A is simply a unity gain buffer for the input, the output of which is fed to both the circulator and an inverting mixing amplifier, U5B. The gain of a simple inverting amplifier is -Rf/Rin - for the clean signal, Rin (R1 in this circuit) is 10K, as is Rf (R4) so the gain for the clean signal is just -1. For the effected signal, the value of "Rf" is determined by a combination of the "blend" pot and R24 (because the output impedance of the final circulator stage is low enough to be ignored.

Considering the blend pot at the extremes, when it is fully clockwise then the 25K of the pot contributes nothing the input impedance seen by U5B and so the maximum gain of the mixer with respect to the effected signal is simply -10K/8.2K = -1.22. What this means is that the maximum amount of "effect" you can get from U5B is -(Clean + 1.2 * Effect). When it is fully counter-clockwise the signal presented to R24 is necessarily zero so all you get out is the unadulterated clean signal. Note that U4B is simply another unity gain inverting stage which just restores the phase of the original signal.

Note that in the analysis above, the conditions at the extremes of the pot rotation are not dependent on the value of the pot. HOWEVER, what happens in between is a very different matter. For a voltage divider (which is usually how a pot is configured) the output impedance of the divider is equal to the resistance of each "leg" of the divider in parallel. If we consider the scenario where the pot is linear and it is set in the middle, we can consider the output of the divider to be 0.5 the input voltage and the output impedance will be 0.5 * half the value of the pot (we have split it into two equal legs which combine in series to make the full value of the pot), i.e. 0.25 * the pot resistance. This resistance is then in series with R24 and so significantly affects the gain of U5B with respect to the effected signal. For a 25K pot, the effective resistance will be 14.45K and the overall effect gain (including the reduction in effective driving voltage) will be -0.5 * 10/14.45 = -0.346. Overall the output signal would then be Clean + 0.346 Effect. If we change the pot to 100K then the same pot position will yield a divider output impedance of 25K, thus changing the effect gain to -0.5 * 10/33.2 = -0.151, which is clearly less than half what we would get with the value of 25K.

Because the "blend" pot (it's not strictly a blend control as it simply changes the amount of wet signal that is added to the dry signal) is sitting in front of an inverting amplifier, the effect of the control is significantly non-linear, even with a linear pot. Using a C pot will certainly change the curve and may well yield a better "feel" to the control but upping it to 100K will probably have a far greater effect. Using a C25K pot is clearly what was intended for the suggested mod!

FWIW, in my version I use a TLE2426 to "split the rails" - in this circuit that is achieved by amplifier U4A, This device is in a TO92 package so takes up very little space. I then use a single amplifier to buffer the output that is, like the input buffer, a simple non-inverting unity gain stage. In front of this, I put a true blend control which is simply a pot (of arbitrary value - I used B100K) with the clean signal connected to the CCW end, the effected signal connected to the CW end, and the wiper connected to the output buffer. Because the output buffer has very high input impedance the output is simply a linear mix of the wet and dry signals (from 100% dry to 100% wet). Again, you don't have to use a linear pot - that just determines the point in the pot rotation where you get a 50/50 mix. With a C pot (reverse log taper) it will be before 12 o'clock, with an A pot (log or "audio" taper) it will be after 12 o'clock, and with a B pot (linear taper) it will be bang on 12 o'clock.

 

[oliverg]

I've just built a clone of this circuit, albeit on my own PCB. As a disclaimer, I haven't built this exact circuit so can't give a direct comparison of the sounds. However, in this circuit, it is important to understand how the effect is "blended" with the clean signal in order to pick the right components.

TL;DR a C100K will be different to a C25K (which is only changing the taper), although C100K is also fine -, just different to the effect of changing from a linear taper to a reverse log taper.

Amplifier U5A is simply a unity gain buffer for the input, the output of which is fed to both the circulator and an inverting mixing amplifier, U5B. The gain of a simple inverting amplifier is -Rf/Rin - for the clean signal, Rin (R1 in this circuit) is 10K, as is Rf (R4) so the gain for the clean signal is just -1. For the effected signal, the value of "Rf" is determined by a combination of the "blend" pot and R24 (because the output impedance of the final circulator stage is low enough to be ignored.

Considering the blend pot at the extremes, when it is fully clockwise then the 25K of the pot contributes nothing the input impedance seen by U5B and so the maximum gain of the mixer with respect to the effected signal is simply -10K/8.2K = -1.22. What this means is that the maximum amount of "effect" you can get from U5B is -(Clean + 1.2 * Effect). When it is fully counter-clockwise the signal presented to R24 is necessarily zero so all you get out is the unadulterated clean signal. Note that U4B is simply another unity gain inverting stage which just restores the phase of the original signal.

Note that in the analysis above, the conditions at the extremes of the pot rotation are not dependent on the value of the pot. HOWEVER, what happens in between is a very different matter. For a voltage divider (which is usually how a pot is configured) the output impedance of the divider is equal to the resistance of each "leg" of the divider in parallel. If we consider the scenario where the pot is linear and it is set in the middle, we can consider the output of the divider to be 0.5 the input voltage and the output impedance will be 0.5 * half the value of the pot (we have split it into two equal legs which combine in series to make the full value of the pot), i.e. 0.25 * the pot resistance. This resistance is then in series with R24 and so significantly affects the gain of U5B with respect to the effected signal. For a 25K pot, the effective resistance will be 14.45K and the overall effect gain (including the reduction in effective driving voltage) will be -0.5 * 10/14.45 = -0.346. Overall the output signal would then be Clean + 0.346 Effect. If we change the pot to 100K then the same pot position will yield a divider output impedance of 25K, thus changing the effect gain to -0.5 * 10/33.2 = -0.151, which is clearly less than half what we would get with the value of 25K.

Because the "blend" pot (it's not strictly a blend control as it simply changes the amount of wet signal that is added to the dry signal) is sitting in front of an inverting amplifier, the effect of the control is significantly non-linear, even with a linear pot. Using a C pot will certainly change the curve and may well yield a better "feel" to the control but upping it to 100K will probably have a far greater effect. Using a C25K pot is clearly what was intended for the suggested mod!

FWIW, in my version I use a TLE2426 to "split the rails" - in this circuit that is achieved by amplifier U4A, This device is in a TO92 package so takes up very little space. I then use a single amplifier to buffer the output that is, like the input buffer, a simple non-inverting unity gain stage. In front of this, I put a true blend control which is simply a pot (of arbitrary value - I used B100K) with the clean signal connected to the CCW end, the effected signal connected to the CW end, and the wiper connected to the output buffer. Because the output buffer has very high input impedance the output is simply a linear mix of the wet and dry signals (from 100% dry to 100% wet). Again, you don't have to use a linear pot - that just determines the point in the pot rotation where you get a 50/50 mix. With a C pot (reverse log taper) it will be before 12 o'clock, with an A pot (log or "audio" taper) it will be after 12 o'clock, and with a B pot (linear taper) it will be bang on 12 o'clock.

With the assistance of GnuPlot, the attached plots will give you an idea of how different pots will change the control (each plot shows the amount of effected signal that will be added to the clean signal by U5B against pot rotation). You will see that the compound effect of the C25K and C100K pots is broadly similar. As you should expect, using a really high value like C1M means that high output impedance of the pot dominates the curve.

 

[FleeRemi]

I've just built a clone of this circuit, albeit on my own PCB. As a disclaimer, I haven't built this exact circuit so can't give a direct comparison of the sounds. However, in this circuit, it is important to understand how the effect is "blended" with the clean signal in order to pick the right components.

TL;DR a C100K will be different to a C25K (which is only changing the taper), although C100K is also fine - just different to the effect of simply changing from a linear taper to a reverse log taper.

Amplifier U5A is simply a unity gain buffer for the input, the output of which is fed to both the circulator and an inverting mixing amplifier, U5B. The gain of a simple inverting amplifier is -Rf/Rin - for the clean signal, Rin (R1 in this circuit) is 10K, as is Rf (R4) so the gain for the clean signal is just -1. For the effected signal, the value of "Rf" is determined by a combination of the "blend" pot and R24 (because the output impedance of the final circulator stage is low enough to be ignored.

Considering the blend pot at the extremes, when it is fully clockwise then the 25K of the pot contributes nothing the input impedance seen by U5B and so the maximum gain of the mixer with respect to the effected signal is simply -10K/8.2K = -1.22. What this means is that the maximum amount of "effect" you can get from U5B is -(Clean + 1.2 * Effect). When it is fully counter-clockwise the signal presented to R24 is necessarily zero so all you get out is the unadulterated clean signal. Note that U4B is simply another unity gain inverting stage which just restores the phase of the original signal.

Note that in the analysis above, the conditions at the extremes of the pot rotation are not dependent on the value of the pot. HOWEVER, what happens in between is a very different matter. For a voltage divider (which is usually how a pot is configured) the output impedance of the divider is equal to the resistance of each "leg" of the divider in parallel. If we consider the scenario where the pot is linear and it is set in the middle, we can consider the output of the divider to be 0.5 the input voltage and the output impedance will be 0.5 * half the value of the pot (we have split it into two equal legs which combine in series to make the full value of the pot), i.e. 0.25 * the pot resistance. This resistance is then in series with R24 and so significantly affects the gain of U5B with respect to the effected signal. For a 25K pot, the effective resistance will be 14.45K and the overall effect gain (including the reduction in effective driving voltage) will be -0.5 * 10/14.45 = -0.346. Overall the output signal would then be Clean + 0.346 Effect. If we change the pot to 100K then the same pot position will yield a divider output impedance of 25K, thus changing the effect gain to -0.5 * 10/33.2 = -0.151, which is clearly less than half what we would get with the value of 25K.

Because the "blend" pot (it's not strictly a blend control as it simply changes the amount of wet signal that is added to the dry signal) is sitting in front of an inverting amplifier, the effect of the control is significantly non-linear, even with a linear pot. Using a C pot will certainly change the curve and may well yield a better "feel" to the control but upping it to 100K will probably have a far greater effect. Using a C25K pot is clearly what was intended for the suggested mod!

FWIW, in my version I use a TLE2426 to "split the rails" - in this circuit that is achieved by amplifier U4A, This device is in a TO92 package so takes up very little space. I then use a single amplifier to buffer the output that is, like the input buffer, a simple non-inverting unity gain stage. In front of this, I put a true blend control which is simply a pot (of arbitrary value - I used B100K) with the clean signal connected to the CCW end, the effected signal connected to the CW end, and the wiper connected to the output buffer. Because the output buffer has very high input impedance the output is simply a linear mix of the wet and dry signals (from 100% dry to 100% wet). Again, you don't have to use a linear pot - that just determines the point in the pot rotation where you get a 50/50 mix. With a C pot (reverse log taper) it will be before 12 o'clock, with an A pot (log or "audio" taper) it will be after 12 o'clock, and with a B pot (linear taper) it will be bang on 12 o'clock.

Thank you for all of this, very interesting/helpful. Weirdly couldn't find any C25k pots on tayda or the other site I usually order from, which is why I initally asked this because I was unsure it existed, and guessed C100k just because that's what 2 of the others had been changed to. I did however just look a little harder and found them on stompboxparts. Very much appreciate your help! The graphs actually helped my understanding too (i'm definitely a visual kind of learner).