Sag knobs

[JonnyAngle]

I know what a sag knob is and how it works

I know that changing the value and polarity will change how the pot sweeps

99/100 times there's only a 1/4 turn at best of useable turn before the signal completely drops out

On the new Frazz, there's no signal cutout at any pot position. How is that possible and can I adapt other pedals to be like that?

[Chankgeez]

[crochambeau]

The pot is just a voltage divider in this application?

Sometimes a useful range is very obvious across a design, and the pot will be "preset" to the useful window by adding a resistor between the pot and ground (setting lowest range) and/or a resistor in line with the pot (setting highest range). The best values are derived from experiment in the design phase or on the bench. For retrofit into an existing circuit, you can either observe the sweet spot in the rotation and then measure the low & high points across the over all resistance - recreating them with a ratio where your entire pot travel is confined to the sweep of the sweet spot and fixed resistors carry the dead zone (again, with a ratio) - OR - you can just slap some trim pots in there and twiddle until it sounds right.

I can draw that if my block of text is confusing, and if this is something like a variable resistance in a feedback path or a simple rheostat the above approach will not work.

[baremountain]

Just to be sure, a sag knob just runs from the power to the ground,yeah? It just divides how much of the input voltage hits the pedal?

[JonnyAngle]

I'll take up your offer on a diagram RMA

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

[crochambeau]

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%

Now you need to know the size of the pot, again 10K is just a conjured number, and the percentage of range that pot will fill:

10K needs to fill in the blank between 75.6 and 33.3%; therefore 10K = 42.3% (75.6 minus 33.3), and the over all branch resistance needs to be roughly 23640 ohms (10K divided by .423, or 42 some odd percent), sandwiched between 33.3% of nine volts at the low end, and 75.6% of nine volts at the high end.

One percent of 23640 ohms is 236.4 ohms.

Therefore 33.3 x 236.4 will give you the low resistor (7872 ohms) and 24.4% (the difference up top) turns into 5768 ohms.

The eagle eyes amongst you will pick up the discrepancy between what I just typed and what I originally drew, that ten ohm disparity is there because I started rounded my math at a different point, and none of this is critical because all these numbers can be shoved through a 5-10% tolerance (and rounding up/down to the nearest reasonable value, like 8K1 and 5K6).

Conversely, you can just figure it's good in roughly the center third of the turn and sandwich the 10K between two other 10K resistors and be in the ballpark.

For trim pots, just swap out the resistors for a trim pot with the wiper connected to one of the outer legs and forget the math (resistors are neater and cheaper though).

Did I clarify or crash the plane into the mountain?

[crochambeau]

It should also be mentioned that I outlined a worse case scenario, and usually you just need to "shore up" the bottom end where shit falls flat. Makes it much simpler to sort.

[goroth]

viewtopic.php?f=151&t=16743

[JonnyAngle]

Thanks! Research to be done.

[cherler]

I also think it's worth mentioning that for a pedal with any kind of significant current draw, a simple voltage divider won't necessarily work as expected. You only get 4.5V on the output of a voltage divider at 50% if all of the current is going down the resistor to ground, and none of the current is going through the output itself. As soon as your voltage divider starts sourcing current that voltage drops further. If you toss in some kind of buffer on the output, something like an op amp buffer with really high input impedance, then you can more faithfully recreate the voltages off the divider. Then you get a situation where the op amp is actually supplying the current to power your circuit, and the starve knob is essentially just telling it what voltage to use. This won't work for super high current pedals, but even then it's an improvement from a raw voltage divider.

[D Rock]

I love sag knobs as well, I really want to build a Meathead clone soon with a voltage starve. I was thinking it would be cool to starve only one transistor as well and see how that plays. Could be pretty cool.

[Iommic Pope]

I thought this would be a thread about getting older.

[baremountain]

Pardon me if this seems unrelated, but I think we're kinda talking about the same principles here.
I'm doing something somewhat similar to a sag knob with the V3 of my CV amp/attenuator. Basically I had the idea to hook up a pot at one of the jacks - I would hook it up permanently at the input or the output, but that changes with some switching, and I think I would need a 5PDT switch to make that work? But anyways, I put a 100k pot at the jack. In one configuration it sends an input voltage that's then multiplied by an op-amp to go to eurorack levels. (Maybe this is similar to the scenario you listed above, Cherler) When I use the pot in this configuration, I'd say 95% of the voltage drop occurs within the first 1/5 turn, and the rest of the travel is super slow. This suggests maybe using an A100K or something instead? However, when I switch up the order, and make the 1/4" jack the output for an attenuated Euro-level signal, 5% of the travel occurs within the first 4/5ths of the turn, and only at the very end do I see any noticeable voltage drop. Why does this knob (lug 1=jack tip, lug 2=amp/attenuator (input/output) switch, lug 3= ground) behave opposite when the jack is treated as the input vs the output??

[crochambeau]

Are you using a linear or a tapered pot?

Fast and dirty check = rotate knob to half way and measure resistance between wiper and one of the outer legs. In a linear it should reflect ~50% of the pots resistance. I would think linear would be what you want based on your description, but I might be wrong.

[baremountain]

It's a type B/linear pot.