Hi all,
I am making a ton of Kosmo modules in quarantine, and I have been looking to build a nice design to create a simple expression pedal interface - I’m sure it couldn’t be terribly hard since expression pedals are basically just a potentiometer rigged to send and return on a TRS cable. In doing a little digging, I found this schematic from the late, great Ray Wilson of Music from Outer Space:
http://musicfromouterspace.com/analogsynth_new/HOT_TIPS/joystick_ideas.html
It’s for a joystick, but one could essentially replace the connections to the joystick potentiometer with the tip and sleeve connections on the expression pedal jack. Simple enough, and this schematic is for two axis, meaning you could have two expression pedal interfaces coming off the same TL072 op amp. Noice.
But here’s where my question comes in - I addition to modulating a DC voltage (which this module should do in it’s “normaled” configuration), I would also like the module to serve as an attenuator for an incoming control voltage signal, if one is plugged in. We a switching jack for the CV input, and when plugged in, the jack would need to break the connections between the + and - 12v rails and the pot, and instead instead send the incoming CV through the pot, to be attenuated and sent through the TL072. I haven’t fully worked out how this would work, but if anyone has ideas, I’d love to bounce something around, and come up with a stripboard layout to share. I think expression pedals are somewhat overlooked as a way to control parameters on a modular synth, but I bet we could come up with one that has a very minimal part count!
Take care,
-Wesley
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Hi, welcome!
Don’t know if you’ve checked out this video or if it’s useful to you, but it seems worthwhile if you haven’t:
You could have the pedal entirely passive, and if you want it to “generate” a CV, just send it +12V for input.
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Separate jacks and circuits for 0-12v and -12:+12 would do it.
Apparently the pedal he’s using in this video is built around a 100k pot with a 2k fixed resistor at the upper end. Looks like that’s high compared to a lot of pedals: The M-Audio EX-P uses a 12k pot with a 50k low limit pot on the bottom end and a series 1k resistor on the wiper. Similarly the Roland EV-5 has a 10k pot plus a 50k low limit pot and, per the unofficial schematic I’ve seen, no fixed resistor anywhere.
So for an interface for such a pedal presumably its pot and fixed resistor should be lowered correspondingly, right? Which would mean it would draw more current. With fixed 1k above the pot, pot at minimum and nothing plugged in it’d be 12 mV which seems a lot for a module doing nothing.
Or should I not be trying to think before breakfast?
Edit: Still haven’t finished breakfast, but would it make more sense to put the interface pot after the pedal? At least it then wouldn’t draw current with nothing plugged in.
If you’re talking about this one:
the maximum current use there is with the range pot in max position, which puts it fully in parallel with the pedal pot, so 12 / (10k + 100k||102k) = ~0.2 mA (ignoring the LEDs and whatever the output is connected to).
I’m talking about using a pedal with only a 12k pot (in series with a 50k low limit pot as variable resistor, which however could be set near zero):
in which case you’d presumably want the 10k fixed and 100k pot at the front to be more like 1k and 10k. But yeah, I was thinking wrong about the max current. In that case 12 / (1k+10k||12k).
The only reason to lower the values would be to lower the output impedance (i.e. make your pedal able to source more current into a low impedance input without dropping too much voltage, cf the discussion here). I’d consider adding an output buffer instead so you only use more current when you actually need it, alternatively making sure to always route it via a high-impedance input (haven’t watched the entire video, maybe he does that?).
In case I wasn’t clear, I was talking about modifying the interface to use with a different, lower impedence pedal. My pre-breakfast thinking was that as drawn, with a 1k+12k pedal (and the low limit pot at zero), then the highest voltage you can get at the pedal is 12((100k||13k)/(10k+100k||13k) ~ 6V if, again, my brain is even working. To get closer to 12V you would need to change the voltage divider to something like 1k+10k (or at least change the fixed resistor) and then it’s 12(10k||13k)/(1k+10k||13k).
But wouldn’t it be better to have a 10k+100k interface after the pedal, like this?
Edit And I guess a buffer before the output, as you mention.
Edit Removed hastily drawn schematic before someone thinks it’s a good idea to build it
Sorry, I lost track of what you’re planning to build. I thought you were considering building the one in the video but with lower resistor values, since some commercial pedals have that, but now I’m not sure 
All right, let’s try it again.
He built his for a pedal with a 100k pot.
Most pedals out there seem to have 10k to 20k pots (see https://www.harmonycentral.com/forums/topic/1501978-expression-pedals-and-their-pot-values/). Unless I’m even more clueless than I thought I was, putting a 10k resistor before such a pedal, as in the video, would reduce the voltage range significantly. And the response to the interface pot would be nonlinear wouldn’t it?
So if using such a pedal:
- Would it be better to reduce the 10k fixed resistor in the interface to 1k?
- In that case, would it be better to reduce the 100k pot in the interface to 10k?
- Would it instead be better to build the interface such that the 10k fixed + 100k pot voltage divider comes after the pedal rather than before?
- In that case, would it be best to add a buffer on the output?
Or more succinctly and generally, what would be a good design for an interface to a ~10k expression pedal?
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I’m loving this discussion, thanks so much for giving so much attention to this idea. Some of your suggestions are a bit over my noobie head, but I think I’d like to go with a buffered output, since I have op amps handy and it seems that there is an advantage to that method, in that it can drive all kinds of inputs without wasting current. I also saw a useful suggestion that maybe it’s not necessary to have a negative output voltage, since it’s counter intuitive for a foot pedal to have a neutral position at it’s center - rather you want the foot pedal to be neutral at the bottom and be able to drive the voltage high when pushed forward (or at least that’s how I would find it intuitive).
I think that would simplify the schematic a bit, and make it easier to use a switching jack to turn the pedal into CV attenuator when a CV cable is plugged in.
I am currently doing a big revamp on my case so the whole synth is dis-assembled, but I’m hoping I can test out some designs and maybe crank out a DIYLC stripboard layout. I also 3D print my cosmo panels and I can post the .STL’s if anyone likes to do the same.
EDIT: One potential design concern to keep in mind (I haven’t fully thought this through yet) but we should also try to make sure that if +12v is going into the pedal, we don’t have a situation where power is connecting to ground momentarily as the jack slides in (assumign the rig is under power when you plug the pedal in). My brain is a bit addled at the moment, I’m going to try to read through this discussion in more detail tomorrow. Thanks again all!
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The +12V is (in the above video) going into a voltage divider, so it’s not going to connect to ground. The hasty schematic I posted above has +12V going to the jack and it needs a limiting resistor in series. (But there’s other stuff wrong with it too.)
Quincas’s design works for him and it has the advantage of simplicity, basically a pot and a resistor and a couple jacks (and the LED and resistor which are optional). Same thing with reduced values for the pot and resistor ought to work for a ~10k pedal, though buffering might improve it at the cost of making it a good deal less simple, needing the op amps and -12V bus.
I agree it doesn’t make sense to think of the pedal as a joystick with ± outputs.
As you say, you could use a switching jack to provide a CV input normalled to +12V.
There are others here better qualified to comment, but I’ll take a shot. I have a hard time understanding circuits from stripboard layouts so I attempted to convert yours to a schematic.
I assume the black wire to the jacks on the bottom is supposed to connect to ground, but it doesn’t appear to. It looks as though it connects to the second strip from the bottom which connects to the third strip from the bottom which connects to nothing else.
I assume the orange wire from CV Input 2 is supposed to go to the sixth strip from the top, not the fifth.
Given those assumptions I come up with this:
Edit: Added ground connection to pedal jacks
My comments:
- I’d use 0.1uF, not 0.01uF, for the bypass caps.
- I think your output stage is a non inverting amp with gain 2. Is that the intention? Why the gain of 2?
- R1 (20k) and input attenuator 1 (5k), when nothing is plugged in on CV In 1, form a voltage divider; then the voltage going to the pedal (for large pedal resistance) can vary from 0V to 12(5k/25k) = 2.4V. But if you have a control voltage plugged in, R1 is absent and the voltage going to the pedal varies from 0V to the full value of the input CV. This doesn’t seem likely to be what you want.
Compare with Moreira’s video, in which (with no CV input jack) he has 12V going to a voltage divider with a 10k resistor on the top and a 100k potentiometer on the bottom, for a voltage range from 0V to 12(100k/110k) ~ 10.9V, and no amplifier on the output.
If your intention is to do similarly, but with either CV or 12V as input and with buffering on the output, I’d think (depending on the value of the pot in your pedal, I’ll assume ~10k) you’d want 12V going through say a 1k resistor to the jack switch terminal, the tip going to a 10k pot. Then I’d think you’d just want a unity gain voltage follower on the output (pedal tip to pin 3, pin 1 to pin 2 and to output.) And maybe a 1k resistor in series on the output to protect the op amp.
If I’ve gotten this all wrong I trust someone else will speak up!
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Hey, thanks for the well considered reply. I’ll need to re-read it when I’m really awake but regarding the ground wires - in my layout I assumed a Eurorack style power connector with three rows dedicated to ground, so in my mind, everything black is grounded. I can totally see how that wasn’t clear though.
As for the switching jack tips - you are correct, they are each intended to have their own row, connected to 12V through a 20k resistor.
It seems like I got a lot of the values wrong, but hopefully we are off to a good start with the general architecture. I think I want the output to be either 0-5 volts DC, or if the CV jack is used, to go from full attenuation to unity passthrough. I think anything past 5 Volts might be enough to damage the inputs on some of my other semi-modular gear - better to make it as idiot proof as possible around me. More head scratching to come. Thanks again for weighing in on this.
-Wes
Assuming 1) @analogoutput’s schematic is correct (I too prefer looking at a schematic rather than a stripboard layout) and 2) that an expression pedal is essentially a variable resistor going from 0 to a maximum value under 1M, then because of the very high input impedance of the opamp, the pedal in this circuit would have no discernible effect i.e. the output will be the same regardless of the position of the pedal.
If the pedal is indeed just a two pin variable resistor, you have to use it in a voltage divider for it to have the desired effect. For example, ground the T of the pedal’s jack and connect the + input of the opamp to the R of the jack. (That would risk shorting the CV to ground, but at least the pedal would have an effect. Another resistor should be place in series with RV1/RV2 to avoid a possible short to ground.)
EDIT: Never mind, turns out both my assumptions were wrong. (Good thing I stated them clearly.)
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See Prototyping DIY Expression Pedal interface for Kosmo for typical pedal schematic. Unless I’m misunderstanding something this is a voltage divider, not a variable resistor.
I omitted the pedal jack’s connection from sleeve to ground in the schematic from last night.
Ah, I see about the power. Yes, I took the incoming wires literally.
For unity gain with CV input you need either to change the resistors in the output stage, or omit them and connect the op amp as a voltage follower. Then if you want to reduce the upper limit in DC mode adjust that fixed resistor: for 5V you need R_pot/(R1+R_pot) = 5/12, so for instance a 10k pot with a 12k or 15k resistor for 0 to 5.5 or 4.8 V. Though that doesn’t take the pedal resistance into account, which would lower the voltage.
Edit: With the M-Audio EX-P, for instance, the minimum pedal resistance to ground is 12k. Then with a 10k pot, a fixed 6.8k or 8.2k resistor would give a maximum DC output (at the op amp input) of 5.3V or 4.8V.
Most modules I know of are robustly designed enough to take greater than 5V input but you know what you need. If that’s an issue make sure you don’t plug in a CV source that can go higher than 5V since that won’t be attenuated at full pedal. Or add a Zener to ground to limit the voltage if you think that’s a possibility.
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I breadboarded this
Video:
With the M-Audio EX-P pedal, low range set to zero and RV1 to full range, R1 knocks the 12V input down to about 11V and R2/LED take about 10% off that, for output of 0 to ~10V.
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This looks amazing, can’t wait to breadboard it myself. I got distracted trying to coax a midimuso to life. So many projects, so little time!
Thanks for the great contributions on this thread.
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