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baremountain wrote:It's a type B/linear pot.

Have you *measured* it or are you going off printed code?

A drawing might help me visualize why the taper response flips in your application.

Yeah I second the drawing. Are you using the pot to control one of the feedback resistors or are you varying the input level with constant gain? I've usually stuck to the latter personally, but both can work.

crochambeau wrote:

baremountain wrote:It's a type B/linear pot.

Have you *measured* it or are you going off printed code?

A drawing might help me visualize why the taper response flips in your application.

Fair point, I have only read the written code. I can check that when I get home.

As for the drawing, I've got this quick sketch and then the whole circuit diagrammed, but that's a little messier. I don't mind sharing at all though. Let me know if you'd need to see more. The tricky thing is in one configuration (CV amplifier) it is at the input jack, but that jack becomes the output jack when it's in attenuator mode. So when the switches are to the right it's at the input jack before the op amp input, and when the switch is to the left it is between the op amp output and the output jack.

So, your voltage input to be attenuated/adjusted is occurring at the 1/4" jack?

Your switch drawing broke my brain, remind me - what are you doing here? Switch in one position and the block diagram would look a little like:
And flipping the switch would result in:


What is ground reference doing there? Is the switch nested in the opamp circuit?

Need to see the active circuitry and all supporting components... but before going down that rabbit hole, double check the pot. That might be the root of the problem.

Haha yeah for sure, I will do that first thing when I get back home.
But basically it's a 3PDT switch. One position (amplifier) configures the 1/4" jack to be the input (as 1/4" is the standard for guitar fx), takes that input CV and amplifies it to +/- 10V (euro level). It does so by configuring the poles as such: (1) the 1/4" jack routes through the pot to an input resistor and the op amp, (2) the 1/8" output resistor to feed into the 1/8" jack, and (3) grounding the unused side of the dual op-amp I have (TL072) [w/o this grounding switch, the amplifier channel wouldn't work for whatever reason]. In the other position (attenuator), the 1/4" jack is treated as the output. (1) 1/4" jack is placed after the output resistor of the second side of the op-amp, (2) the 1/8" jack is placed before an input resistor on the second side's non-inverting input, and (3) the first side is sent to ground.

Here's the diagram before I realized I needed to add a third pole to the ground (sketched in quickly if you look close) and before I included the B100K.

I hope you can read it haha

VR1 & VR2, the cap that's blocking 12VDC on the input of that regulator is a typo, right?

crochambeau wrote:VR1 & VR2, the cap that's blocking 12VDC on the input of that regulator is a typo, right?

IT;s not a blocking cap, its a DC supply drawn backwards.

Also none of the ideas presented will work that well.

Why linear pot gives non linear response at input:
its in llel with R8 in your drawing.

Why linear pot gives non linear resonse at output
it is in llel wiht the load of the circuit it is driving. its a tl072. who knows many reasons why it might not work.

Why even this won;t work for a circuit that draws any appreciable current and why voltage sag is still a dark art that is different with every pedal:

crochambeau wrote:

JonnyAngle wrote:I'll take up your offer on a diagram RMA

Evan, it's just a variable resistor from 9v to 0v

Okay, Fig 1 depicts what you've got with the limited useful window of rotation. The upper point and lower point of rotation will have to be direct measured for the voltages that apply to your unit, I've just tossed a couple numbers there that are meaningless beyond the example.

So, first thing to do is determine the percentage of important voltages (in this case 6.8 is JUST BEFORE you start to observe the starved effect, and 3 is when the circuit falls flat), and know your supply voltage.

Supply voltage is 9.0, so 0.09 is one percent.
6.8 divided by 0.09 is roughly 75.56%.
3 volts is 33.33%

the circuit will make a llel load between the wiper and ground. the measured voltage with nothing plugged in and the measured voltage with something connected will be different. if the load is low in resistance it will be drastically different. by low i mean even close to teh scale of teh resistance of teh pot, so a load of even 5mA or qqc is high.

use a current buffer ?
not trying to come off as negative, just it is not hard to see why a "linear" response is not observed. think about the whole circuit when the load is present.

s

My earlier drawing has no relevance to the circuit baremountain speaks of. I realize the load down stream will be in parallel with the lower resistor, hopefully the value derived would have come about in the circuit under test and would be a non issue. Agree a buffered approach would be more solid, but cavemen like to play as well.

multi_s wrote:

crochambeau wrote:VR1 & VR2, the cap that's blocking 12VDC on the input of that regulator is a typo, right?

IT;s not a blocking cap, its a DC supply drawn backwards.

Ah cool, I would not have interpreted it in that manner if I had picked the schematic up off the sidewalk.

Next morning edit, complete with caffeine:

multi_s wrote:Why even this won;t work for a circuit that draws any appreciable current and why voltage sag is still a dark art that is different with every pedal:

True, I'm still mired in the land of old school electronics that are happy enough to run off a dry cell. In this incarnation the voltage divider sees negligible loading through the gate of the mosfet and the BJT can be somewhat scaled up to support heavier loading. The pot section could just be exchanged for the opamp scaling (and since the opamps have high input impedance and are happy to drive the BE current of the BJT the mosfet buffer can also be ousted).



Those small caps are not going to clean much, so for filtering one would want to plant caps either after R3 to ground, or if some lag after adjust is your thing you can filter the emitter of the BJT (which is supplying the rest of circuit)

Yeah getting a current amp in there is what I was trying to get at.

Baremountain, I wish I could help more but I really don't know what's going on in your drawing haha. One suggestion I could make though, since you're going for a DC amp here (right? the CV is at most super low frequency witha DC offset?) you could go for a rail-to-rail op amp. Then you'll have access to the full range of voltage you're powering them with and won't need to handle any reference shenanigans.

I'm thinking along the lines of giving each op amp +12/-12, reference the + pin to ground, and use a rotary switch to swap in different values for R2 for different gains and use an input pot to trim if need be. I have no idea if I'm even remotely on base with what you're trying to do though

multi_s wrote:

crochambeau wrote:VR1 & VR2, the cap that's blocking 12VDC on the input of that regulator is a typo, right?

IT;s not a blocking cap, its a DC supply drawn backwards.

Also none of the ideas presented will work that well.

Why linear pot gives non linear response at input:
its in llel with R8 in your drawing.

Why linear pot gives non linear resonse at output
it is in llel wiht the load of the circuit it is driving. its a tl072. who knows many reasons why it might not work.

Thanks for the help, this makes a lot of sense. I'm still learning so there's always more to understand. And yeah that's just a poorly drawn DC supply
I'm gonna go ahead and can the trimming idea for now. It works fine for my purposes now & I can maybe revisit it when I'm a little more well-learned.

cherler wrote: Baremountain, I wish I could help more but I really don't know what's going on in your drawing haha. One suggestion I could make though, since you're going for a DC amp here (right? the CV is at most super low frequency witha DC offset?) you could go for a rail-to-rail op amp. Then you'll have access to the full range of voltage you're powering them with and won't need to handle any reference shenanigans.

I'm thinking along the lines of giving each op amp +12/-12, reference the + pin to ground, and use a rotary switch to swap in different values for R2 for different gains and use an input pot to trim if need be. I have no idea if I'm even remotely on base with what you're trying to do though

Alright so my terminology may be a little shaky. I use a second board that's essentailly converting to 12V bipolar power rails inside - that's AC, yeah?
The TL072 works well for the 3 operations I had in mind (0-5V -> +/-12V; +/- 12V -> 0-5V; +/-12V -> 0-3.3 V), and I think losing a little bit at the upper and lower limits is good because Euro stuff really works with +/- 10V which is within 0.5V of what I measure at full and minimum gate.
I use a SPDT to swap gains right now, but a 3-position rotary might be a pretty sexy alternative.

baremountain wrote:Alright so my terminology may be a little shaky. I use a second board that's essentailly converting to 12V bipolar power rails inside - that's AC, yeah?

Shaky terminology happens to all of us. Bipolar DC is not AC, it's only AC if the current source is alternating between a positive and a negative excursion.

Yeah hopefully your power is DC, I was referring to the CV signal. If that's changing less than like 10Hz might as well call it DC. Then you don't need op amps catered towards high frequency signals and more catered towards this application, like having rail-to-rail output and stuff. If the TL072 works though then great!

This thread is where dick jokes go to die.
Fuck you guys.

Iommic Pope wrote:Fuck you guys.

TWSS

Damn you even gave us a whole month and we still blew it.