LFO parts substitution.

[11-90-an]

I'm planning on making a "Sea Device" pedal (clone of E*D Sea Mac*ine)
https://pcbguitarmania.com/wp-content/uploads/2018/10/Sea-Device-Building-Docs.pdf

What I'm curious about is the LFO section here:


I'm planning to buy pots locally for now, since the ones here are cheaper than Tayda, but the local store doesn't have any 1M pots, 500k is the biggest value.
So in other words, I'm planning to replace the 1M pot with a 500k pot.

Now I know that the pot value is essential to the entire LFO, and that swapping it out would take out most of the range, etc.
So I'm wanting to ask, how do I calculate the minimum and maximum frequencies? And which components should I also change the value of to get the same amount of LFO speed range as the stock pedal?

Now, a beginner's question: could swapping out the LM324 with a TL074 work out well? Or would it just start misbehaving due to being near negative power rails?

[Rob Strand]

The voltages on pin 3 of your IC3_A determines the range of voltages in the circuit.   They seem to be in range of the TL072.   The LM324 output swings lower than the TL072 so it will have a small effect.

The way pot works is a bit deceptive in that circuit.   It's wired like a divider but the 10k on the integrator input is so low the pot almost looks like a series connected pot like Boss BF2.

You could scale down the 10k's and increase the 470nF cap to 1uF to compensate for the 500k to 1M pot change.

The pot is actually a C taper so a B taper isn't going to be as nice to adjust.   You might be able to use add a resistor across the pot to change the taper (as mentioned on RG's geofx pages) but you would have to change the pot to work as a divider.

[11-90-an]

https://www.diystompboxes.com/smfforum/index.php?topic=66471.0
http://www.geofex.com/article_folders/potsecrets/potscret.htm

So from all these useful links...

To convert a Linear taper to Reverse Log, I would have to do something like this?


When I try to calculate it, the calculator gives me this:



I'm guessing the negative kilo-ohm value is counted as no connection?

[duck_arse]

you can't get a 499 pot from a 500 pot. you need much more pot to start with, and the tapering resistor value will be a fraction of that. but not 499/500. try actual = 50, desired =5 and see what results.

[composition4]

Honestly just do what Rob said and impedance scale, it's going to be way easier and better results than messing about with tapering resistors. 500k pot, change R25 & R26 to 4.7k or 5.1k, and change C20 to 1uF

[11-90-an]

Honestly just do what Rob said and impedance scale, it's going to be way easier and better results than messing about with tapering resistors. 500k pot, change R25 & R26 to 4.7k or 5.1k, and change C20 to 1uF

Yup. I'm gonna do that in the pedal... Rob just said that the LFO pot is *better* to fiddle around as a reverse log pot, but then I have only linear pots at hand....

you can't get a 499 pot from a 500 pot. you need much more pot to start with, and the tapering resistor value will be a fraction of that. but not 499/500. try actual = 50, desired =5 and see what results.

500k is the biggest value i can get...maybe if I scale the other resistors down and convert it to say, a 250k pot?



The values make more sense now...

[Kipper4]

You can't just scale the r without scaling the caps too.

Two series 500k pots if no 1M?

[Rob Strand]

If you want to keep the impedance high as possible for a pot taper tweak it's best not to use any lower resistor.

Suppose you want make 15% reverse log from a linear.    When the linear pot is in the middle
the top half will be 15% and bottom half will be 100-15 = 85%.

If you start with a 500k linear pot that means the middle is going to be 250k + 250k.

With no lower "tweak" resistor that means the 85% part is 250k.
If the lower part is 85% and 250k that means the top part must be (15/85)*250 = 44.1k.

The top part of the linear pot is 250k  so you need to add a parallel resistor to the top that makes that 44.1k,
ie you need to add a 53.5k resistor.

Now, when you look at this as a whole pot the end-to-end resistance is 44.1k + 250k = 294.1k.

So that process converts a 500k linear pot into a 294.1k 15% revere log pot. (but only when used as a divider).  When maxed out it will look like 500k to  ground.  At minimum it will look like 53.5k in parallel with 500k  = 48k.     The variable impedance is why the circuit needs to work like the pot is a voltage divider and not a series resistance.

In the original circuit the resistor in the bottom leg of the pot will need to be reduced quite a bit to get the same behaviour as the original circuit, since on the minimum setting it's like the pot is 48k.

[Rob Strand]

One thing, you might need to increase the value of the 10k resistor on the input of integrator.  If that is a high-ish value then the pot can operate as a voltage divider.

[11-90-an]

After reading your post over 5 times over, i think I'm starting to get it... 

Are there any formulas for this LFO?
and also... will I have to change R27, R28, and the Shape pot?

[Rob Strand]

Are there any formulas for this LFO?

There are formulas but they are for min/max settings.

If you want the same frequencies you want the same current going down the 10k integrator resistor.
So you would take the original design and work out the 10k resistor current when the pot was at max, midway and min.

If you kept the timing cap the same, then to match the speed when the pot is on full you would
have to keep the 10k integrator resistor the same.

At different pot settings you have to convert the pot into an Thevenin equivalent voltage source and Thevenin equivalent source Resistance.   Then that Thevenin equivalent circuit drives into the 10k integrator resistor.  From that you can work out the current into the integrator resistor and match that current for the old design and the new design.

If you used a cap two times the size on the new design then instead of matching you would match twice the current down the 10k integrator resistor in the old design.

It's straight forward but quite very messy to do.

A rough attempt in matching the old design with a 15%/85%  reverse log 1M pot, and using a 500k pot on the new design I got

C = same as before
R integrator = 10k (same as before)
Rpot = 500k linear
R to Vref off the pot 1.8k (was 10k in old design)
Tweak resistor from pot wiper to cw lug  680k.

Much different to what you would expect from pot taper tweaking.   The reason for that might be the 10k integrator resistor is so low.

The other reason is I might have made a mistake     it's very messy to work out.

After all that the LFO frequency matches the original design at min, midway and max positions.

and also... will I have to change R27, R28, and the Shape pot?

The idea would be to keep the LFO output swing the same, so no change there.

[Rob Strand]

Find attached a rough spreadsheet.  The spreadsheet is a .ods spreadsheet from libre office.

- Save attached the .txt file with a .ods extension
- If you have Excel drag the .ods file into Excel.   IIRC you have to do something weird like that.
- If you have open office or libre office it should just open.

Might not make sense, since it has the bare minimum info to do the job.

[11-90-an]

and so it seems that there are C500k pots after all...    (Sorry Rob... )

So I'm guessing i still have to change R to Vref off the pot to 1.8k... and probably change the integrator input resistor, and the capacitor, right? it seems wrong not to do so...

[Rob Strand]

and so it seems that there are C500k pots after all...    (Sorry Rob... )

C taper is much nicer to use.

So I'm guessing i still have to change R to Vref off the pot to 1.8k.

In that case the best solution is to half the two 10ks and double the integrator cap.

That's just scaling the circuit down by a factor of 2.  It should will work *exactly* the same as the original.

FYI, the 1.8k change is required because of the 500k pot change but also the added taper-tweak resistor.

[11-90-an]

So now that you're saying that everything will be scaling the circuit down, and that there would be no need for a taper p-tweak resistor as the store sells C500k... shouldn't the R to vref be 5k ohms?

Does the 47n cap have to be changed too? Maybe to 100n?

[Rob Strand]

So now that you're saying that everything will be scaling the circuit down, and that there would be no need for a taper p-tweak resistor as the store sells C500k... shouldn't the R to vref be 5k ohms?

That's right.

was: pot = 1MC, R25 10k, R26 10k, C20 470n
now: pot = 500kC, R25 4k7, R26 4k7, C20 1u

Does the 47n cap have to be changed too? Maybe to 100n?

No, you shouldn't change the 47nF cap.

In the original the signal comes out of IC3A pin 1, goes into the pot, then comes out on IC3D pin 14.

We have scaled the part of the circuit between IC3A pin 1, starting at the pot, and all the way upto IC3D pin 14.
The idea is the scaling doesn't affect the *behaviour* of the circuit at all, only the values.
The rest of the circuit can't tell we have changed anything, so we don't need to change it.

[11-90-an]

Thanks a lot, Rob... 

One last thing... the 2k Animation pot...
I'm assuming I can use a B5k pot and tack a 3.3k across both outer lugs and it would be fine?

[deadastronaut]

just my 2p.... i often use 1MA pots wired in reverse, when i cant be arsed to order a fancy pants fussy 'C' pot...

works for me....

[Rob Strand]

I'm assuming I can use a B5k pot and tack a 3.3k across both outer lugs and it would be fine?

Should work fine.     The middle position will be off, 1.4k vs 1k, but you can't change or fix that.

[11-90-an]

I'm making a perf layout right now, and with having some space issues...

Why is there a separate voltage divider made with R23 and R22? can't they just be fed with VB? (a 100k - 100k voltage divider that supplies for the rest of the LFO...)