Tilt pot (for ramp up/down) with duty cycle independant of rate pot?

[varialbender]

I've been trying to figure out how to add a tilt pot, like the diode trick, that isn't affected by the rate knob.
In other words, if I've got the tilt pot maxed, and that gives me something like a ratio of 50:1 for rampup:rampdown time, I want it to be 50:1 no matter what the rate pot is.
I tried using a dual gang pot for rate, with one part for the normal rate pot, and the other in parallel with the tilt section. Both rate and tilt were 1M. I added a 10k resistor in parallel with the rate portion of the rate pot, and 2 k in series with the tilt portion of the rate pot. Also, in series with the whole thing is a 200 ohm. The cap in the oscillator is 10uF. These are just ideal values. Anyway, I wanted a max ratio of 50:1, and that seemed to give me 50:1 at both ends (Rforward/Rreverse). The only problem is that I figured out the function for the ratio, or duty, if that's the right word, and it's very non-linear. The function I'm getting, using kiloohms instead of ohms, with 1000k for tilt in one direction and 0 in the other direction, is Duty = 1 + (1000Rr + 2000)(Rr + 10) / [(Rr + 1002)(10.04Rr + 0.4)] , where Rr is the resistance set by the rate pot. That gave me lovely, perfect values at all the endpoints, nice range of rates, etc... but in between, it behaved horribly.
Does anyone have any ideas as to how to add a tilt pot without the duty set by this tilt pot being affected by the rate pot?
Do boxes with tilt controls, like the pulsar and catalinbread semaphore, behave this way, or do they just let you deal with the difference?
Thanks a lot!

[gez]

Don't really understand your post.  If you use the pot and 2 diodes trick then 'tilt' is independent of rate.

[varialbender]

Hmmm.... let me try to think this through again.
The formula for rate in the oscillator I'm using is F=100/62(4*R*C)
My understanding is that the Duty cycle is Rforward/Rreverse.
For a 1M tilt pot, and 1M rate pot, with both maxed, you get Duty = (1M+1M)/(1M) = 2
Whereas when the rate pot is at half (0.5M) you get Duty = (1M+0.5M)/(0.5M) = 3
I'm hoping I'm wrong and you're right, and that you can point out my error(s).
Am I right in thinking that the Duty cycle is Rforward/Rreverse?
Anyway, I've breadboarded it, and I noticed that the Duty cycle varies with the rate pot, so unfortunately, I'm thinking I'm right here

[gez]

I haven't really read thru either of your posts in any detail as there's no point without a schematic, or at least an written outline of what the LFO is.  If you do that, I'm sure you'll get some help.

[Paul Perry (Frostwave)]

I can't lay my hands on it, but isn't there a way with a couple of diodes and a pot? You have a triangle generator with a pot having the wiper divide it into a charge and a discharge resistor. The idea being, that charge time & discharge time are proportional to the relevant resistors, so total charge-discharge cycle is constant.

[R.G.]

50:1 is not simple, because the imperfections of simple circuitry become perceptible unless you go to more complex circuits. 10:1 is easily doable.

To do this, one makes a standard integrator/schmitt trigger oscillator, but separates the up down integration voltage by feeding the output of the schmitt to the wiper of the duty cycle pot and taking the output of the duty cycle pot off the end lugs with two diodes, one feeding positive and the other negative to the speed pot. The speed pot can then be a series variable resistor or a voltage divider pot into a fixed single resistor into the integrator.

The + and - excursions out of the schmitt trigger are equal voltages and opposite polaraties around the schmitt reference, so the duty cycle is always the sum of the + and - integration times. In a stock setup, these are equal. But with diodes separating the polarities, you integrate faster one way than the other.

The imperfections of the pots and opamps are likely to show up if you want a real 50:1, so it takes some careful design to get to an accuracy which is imperceptable compared to 2% on the ends. Not impossible, as synths do this, but not trivial.

[varialbender]

Sorry, I tried to describe the circuit a bit, and would post a drawing, but I'm not on my home computer.
I'll post a picture of the circuit soon, hopefully.

R.G. and Paul, I'm not sure that I see the difference between the circuit you describe and the one I'm using.
Might you happen to have a picture of the circuit you're describing?
If not, you can tell me if they're the same when I post mine.
I'm not sure I understand why you're saying 50:1 would be difficult.
What would the error sound like?
Would it be close enough to sound good?
Would it only really be noticeable when it's oscillating in audio frequencies?
I'll try to search for some synth schematics, but can you think of any in particular that are worth checking out?

Thanks a lot

[TELEFUNKON]

synth-diy...

[varialbender]

I messaged Nick from Catalinbread, and he suggested I try adding a unity gain follower between the rate and tilt pot.
I'm not sure how that works, theory-wise. Anyone seen this before? Mind explaining it to me?
Thanks

[Paul Perry (Frostwave)]

thanks RG, that's exactly what I dimly remember! but, buggered if I can find an example on the web..

[gez]

I messaged Nick from Catalinbread, and he suggested I try adding a unity gain follower between the rate and tilt pot.
I'm not sure how that works, theory-wise. Anyone seen this before? Mind explaining it to me?
Thanks

If you have a series resistance between the schmidtt trigger and integrator controlling rate, it will make the falling slope of a tooth more gentle at lower rates.  Divide down the square with a smaller value pot than used for the 'tilt' control and it not only helps overcome this but can give a more 'linear' feel to the rate pot.

I tend to use a 10k pot( plus stop resistor to set min rate).  Wire the outer lugs between the Schmidtt output and half supply - opamp wired as follower is ideal as V ref - then the wiper connects to the wiper of your 'tilt' pot, its outer lugs each connecting to a diode (back to back) which join, connect to a small value stop resistor and then this connects to - input of integrator.  If you buffer the wiper of the pot which divides down the square, in which case it can have a larger value, then even better.  Use the other half of a dual opamp (if you've used it for Vref), just wire it as a follower.  Careful selection of an amp will be required as inputs will be pulled to rails, or close to rails (depending on output of amp used for schmidtt), at max rate.  A series resistor with the pot can prevent this happening.

[varialbender]

I've been thinking that if I varied the capacitance instead, that might solve the problem.
Obviously, I don't want to use a variable capacitor.
I came across this picture, posted by Skreddy, and it was exactly what I had in mind:

Now, I'm not sure about the math that would be involved in calculating the capacitance in that.
Another thing I'm worried about is that the resistor used will still make it non-linear.
Also not too sure on the range I can get out of this.
Can anyone educate me?
Thanks