Uh, actually I was right in the first place you were wrong.

My first equation was the basic formula.

My second equation was the same equation brought to the lowest common denominator.

Your equation is wrong. It is not the inverse of the total but rather the actual total resistance.

Otherwise I would have never reached the correct value in the first place.

Now, me tired. Me go sleepy now.

Gary

You were really too tired

1/Rtot = 1/R1 + 1/R2 (it's the actual relation for two resistors in parallel and the one I wrote)
so (common denominator)
1/Rtot = (R1 + R2) / R1*R2
and (reciprocal)
Rtot = R1*R2 / (R1 + R2)

The latter is what you wrote in your previous post and is correct, but cannot be obtained from the one you posted in first place.

You're right. It was 4 AM when I posted that.

Thanks for cutting me some slack

Anyway, a 12Kohm resistor in parallel would effectively turn any 50Kohm pedal into a 10Kohm pedal.

Something to consider when you're cutting compromises on buying a new pedal.

Gary

In fact I wrote the 12K value was fine, and the whole idea is really good
Regards
C.

Actually on second thought I'm not sure if it would actually work.

I'm not certain what sort of curve the resistance would track as the wiper moved along intermediate positions on the pot.

In practice the pedal might act erratically as it approached either end of the pot.

I'd have to plot the actual curve on a graphing calculator or a program like mathcad to see if it actually would give a linear curve with pedal travel.

Unfortunately I have neither at my disposal.

Gary

Yeah it's a no go.

The function 1/x=(1/y)+(1/12) plots out as a hyperbola not a linear function.

https://www.desmos.com/calculator

To make it work linearly you'd have to attach the 12K resistor as a second pot with the wiper paralleled as well.

Yet another failure to die prematurely before it makes it to the shark tank

Gary

Post deleted

Lololololol well you guys have proved a point! As far as the Kurzweil's pedal input is concerned.. You can't make a 50k linear pot seem like a 10k linear with one additional resistor! Or even two!

As I see it the big issue is the input impedance of the Kurzweil, causing more current to flow through the pot via the wiper and so corrupting the linear relationship between the wiper position and the voltage present at the wiper (on the FC7''s 50k pot).

A solution that doesn't involve replacing the pot would be desirable, because, as I've said previously the FC7's pot is quite a tough one to source at 10k being, an Alps, quarter turn full range and physically smaller than most standard pots, with a 5mm D type shaft and thin shank.

I've been mulling over an alternative approach. Really apart from having my head read for wasting so much time on this, I accept it's become a bit of a challenge!

What's needed is a circuit that changes the FC7's output impedance down to a level that doesn't get corrupted by the Forte. Using a low power CMOS op amp as a simple voltage follower/buffer, one that will be happy to be driven from a single rail ( the voltage between ring and screen) and swing to within mVolts of the rails, should do the trick. Something like a TLV2371. These are available in small enough packages to actually be built into a jack plug. So you could incorporate the impedance buffer into the ring/tip switch over lead.

The circuit details need some careful thought as I don't want to blow up a cc pedal input! I'm thinking op amp will need decoupling and the output may need a small value series resistor to limit the current. I will see if it works when I get a minute!

Should I expect smoke to start rising from my PC3? I have used a couple of Yamaha FC7s for a few years, with just the T&R of the TRS jacks reversed and they work fine. Maybe I can't hear the linear or log rate of the pedals as I work them for volume, but they do what I want. Should I be concerned about 50K or 10K pots? Not an engineer here.

Rockinredeye >>> As far as feeding 50K to a PC3 input goes, if it's working for ya, don't worry about it. Reality is that the pedal travel/volume is probably a bit iffy but if you're not hearing any problems and it's acceptable for your performances then you have nothing to worry about. It won't release any magic smoke. It'll just function properly over a narrow range before it maxes out.

hectorspace >>> so what you're basically proposing is some sort of linear DI box for the pedal input. While it sounds good in theory I don't think any sort of passive circuit would work properly. I think you'd have to drive the circuit with a 9 volt battery to avoid overloading the Kurzweil circuit.

Personally on my own PC3K I took the coward's way out and bought a couple of Moogerfooger EP-2 expression pedals which are the correct impedance and they work well enough for my needs. There have been reports of the EP-2's breaking under heavy use but I don't stomp my pedals like godzilla on a good day. Also, Moog has supposedly replaced them with the EP-3 now. I have no idea on how well built the EP-3 is or not.

Gary

You know your idea of a DI sounds doable. Read 50k sweep and send 10k.
Kind of like the Midi Solutions remapper. But then you might as well get the Midi Solutions box.
It can read any pedal, has an adjustable curve, and has very fine control. . $160 USD is a bit pricey.

Oh well. I'm happy with the Boss pedal and the FC7 kinda works if you use an adapter or resolder the leads.
The Moog one works and there are a few others.
The Roland EV-5 is what it is. I've got 2 that I just can't use. Too floppy.

I wonder if anyone has tried the Hammond ones.
They are probably very smooth but expensive.

John

Actually, the Moog EP-3 is pretty cheap at $40 USD and it''s got 4-1/2 *'s rating on Amazon.

http://www.amazon.com/Moog-ACCEP003-EP3 ... =moog+ep-3

100% compatible with Kurzweil.

I'd have to say, why screw around with kludges since this seems to be the best, most economical solution?

Gary

Cool Gary,

It's a lot smaller than the Boss pedal I have.
I might be able to fit two on my pedal board in place of the one.
I really wanted two but pedal board space is tight.

Is the clutch on that adjustable or just right for not flopping around if nudged?
Or maybe anybody here have a quick review?

Thanks,

John

I can't speak as to the EP-3 as I have the EP-2 and the clutch was redesigned on the newest model.

But it has 32 reviews on Amazon and you can post questions to the reviewers on Amazon regarding the hardware quite easily.

You should get answers to all of your questions within a few hours.

Gary

Actually, the Moog EP-3 is pretty cheap at $40 USD and it''s got 4-1/2 *'s rating on Amazon.

http://www.amazon.com/Moog-ACCEP003-EP3 ... =moog+ep-3

100% compatible with Kurzweil.

I'd have to say, why screw around with kludges since this seems to be the best, most economical solution?

Gary

I own an EP-3 pedal, works great with the Forte! And build quality is quite good for the price..