When pots go "out of range"

[motterpaul]

I know that the more expert builders here have the answer I need.  I just built a Mockman and it sounds great - except for one thing. I replaced a 1M fixed resistor on the output with a 1M pot wired 1 to resistor in; 2 to resistor out; and 3 to ground.

In this image it is the pot (1M) closest to the IC on the right connected to its pin 7 and by jumper to pin 6.

http://1.bp.blogspot.com/-wL5RLKMbm6o/UPqTTz8E1eI/AAAAAAAAAzY/PVT6VSX70Qk/s1600/ROG-Mockman.png

The pot works fine, it adds control over the gain, except there is a portion of the sweep towards the top where the sound just cuts out. I know this is common problem, but how do I fix it? Everything else in the circuit is the same as in the picture.

Thank you all for your help and for this great site.

[rocket88]

if i'm not mistaken since you're wiring the pot in place of the resistor, you only need to use 2 of the lugs. think about it, the resistor makes 2 connections, and neither of them are to ground. i would disconnect the lug you have going to ground and try it out. that should take care of the issue.

[IvIark]

Yes, just use lugs 1 and 2 and don't make the ground connection

[motterpaul]

Thank you - I sound "stupid" but I am still learning and I still don't know all the different ways to use pots. I thought I needed the ground connection to have a way to filter away the the part of the signal I don't need. (I know there is usually a third connection).

Obviously, I am new to all this, and some things make sense right away, and some don't.

Thank you!

[rocket88]

no worries man, we all start somewhere. i still ask a tone of questions when i don't know. i wise man always knows when to ask for help. always happy to help

[motterpaul]

Hey guys - one thing. I wanted to make this pot more like a Gain control, but that change just made it more like another volume pot where when it is all the way down it goes off.

What if I wanted to do something like this: http://www.muzique.com/lab/swtc.htm

(the stupidly wonderful tome control) - where in the circuit would that go?

[rocket88]

the SWTC will add a tone control at the end of the circuit, not a gain control. does the level of gain change when you turn the pot now, or does the volume just drop? if it does control the gain, then what you need to do is solder a resistor between the two lugs, so that when the pot is all the way down then the signal can still flow through the circuit, otherwise what's happening is that there is so much resistance that it cuts the signal completely.

[motterpaul]

second answer first - this change made it seem more like a volume control, not a gain control. I still get all of the same amount of gain IOI had before, and not really any less when I turn it down (maybe I get a little less gain, and more treble, but it actually works much more like just another volume pot, which I already have up front.

I can try adding a resistor across the lugs and see if it changes.

Or maybe I should just put the fixed resistor back and add this SWTCinstead. All I really wanted to to do was add some variation to the sound (because I had an extra hole drilled in my box for another pot ;-)

That way I get a tone control instead of what I have. So, the SWTC goes just before the output, correct?

[induction]

The resistor in question is the feedback resistor (Rf) of an inverting amplifier (see the schematic). The gain of such an amplifier stage is -Rf/Ri, where Ri is the input resistor (in this case 10k). When Rf goes to zero, the gain goes to zero and the sound cuts out. If you don't want this to happen, you can add a small resistor in series with the pot. You'll have to experiment with what value gives you the minimum gain you prefer. This can be done easily on the vero by replacing the given resistor with the new one, but move the bottom leg down one row (row 2 from bottom, column 12 from left) and put a cut to the left of it (row 2, column 11). Then put lug 1 of your pot on row 2 all the way to the right, and lug 2 on row 3 all the way to the right. Warning: the maximum gain will increase because Rf is now the pot setting plus the new resistor value. If it oscillates, you might want to play with the pot value, the value of Rf, and/or the value of Ri (that's the 10k just to the right of the 1M, you can increase it to reduce the gain). You can do the math in advance (calculate Rf/Ri for minimum and maximum pot settings before and after the mod), or you can use trial and error. I understand you've already built the vero, so trial and error will be a pain. You could always just recreate the circuit on a breadboard to find the values, and then do the final substitution on the vero.

Tone controls can go almost anywhere in a circuit, but they often get tacked on just before the output. Like many tone controls, the SWTC is lossy. So if this circuit has a nice boost, it'll be okay. Otherwise, you'll need a makeup gain stage after it. That's something for the breadboard, though. I wouldn't want to experiment with tonestacks and makeup stages on a vero that's already built.

[motterpaul]

Excellent advice - thank you very much. All the stuff I wish I knew. But I do understand what you are saying and I do have a breadboard handy - so maybe I can just build the tone control there. as for playing around with a resistor to connect the lugs, the pot I have has tiny holes above the lugs, so it is easy to just drop a resistor in there - not permanently, I know, but a way to test.

[induction]

Terminology: gain refers to how much the voltage signal is amplified by an amplifier (op-amp or transistor) stage. The products of gain can be volume, distortion, or both. This circuit creates distortion by increasing the volume until the op-amp runs out of clean headroom. So your new gain control is both a volume and distortion control because that's how the circuit works. If you decrease the gain you get both less volume and less distortion.

But the first gain stage probably contributes more to the distortion, while the second stage probably contributes more to the volume. (Speaking on general principles here, I haven't built this circuit.) So you might have better luck with the first op-amp stage. That is, leave the right half of the circuit stock, and replace the feedback resistor in the first half of the circuit (the 1M all the way on the left) with a pot. You'll probably get more of a distortion control then. (I haven't tried it though, so you'll have to tell us if it works.) You can do all the same tricks discussed above (like adding a series or parallel resistor to put bounds on the gain) because both op-amp stages are identical (see the schematic).

By the way, adding a resistor in parallel with the pot will limit the maximum gain (parallel resistors reduce the total resistance), but it will still go silent at the minimum gain setting.

Finally, I don't recommend soldering resistors (or anything else) in those holes in the pot. The heat will expand the metal, and the lugs will lose electrical connectivity with the resistive element. The pot will more than likely fail. I speak from experience.

[motterpaul]

Okay - so I need to add a resistor is series with the lugs not parallel - sorry (I had been reading about people changing the values of pots by putting resistors in parallel with the lugs so that is what i thought you meant).

It's funny - pots are easy to understand - except I don't quite yet understand all the ways to use the third lug. I need to read up on it.

I see many circuits where both lugs 2 and 3 will go to ground, and many where lug 2 goes to the output (like this one) but lug one is in and three goes to ground.

As far as this circuit goes - it is a Mockman 2 - which is a very distinct sound (Tom Scholz - Boston) but it does seem just a little dark to me - like you are always on the neck pickup (even when you are not).

[induction]

There are two common ways to use pots:

1. Variable resistor: If you want the resistance to increase as you turn up the knob (like in your case), lugs 1 and 2 are used as the resistor legs. If you want resistance to decrease as you turn up the knob, lugs 2 and 3 are used as the resistor legs. In both cases, the unused lug can be tied to the wiper (lug 2). This is done in case the wiper fails, so you'll get the maximum resistance of the pot instead of an open circuit. At least the circuit will still work, it just won't respond to the pot setting.

2. Voltage divider: Reduces voltage by a percentage that depends on the setting and the taper. For linear pots: Vout = Vin * pot setting (as a percentage). More generally: Vout = Vin * R12/Rpot, where R12 is the resistance between lugs 1 and 2, and Rpot is the total value of the pot.

Variable resistors are common in gain controls and tone stacks. Voltage dividers are common in volume controls.

If you want to make your circuit brighter, try reducing the 470p capacitor (Cx in the schematic). It bleeds high frequencies to ground just before the output. (Technically, it forms a low-pass filter in conjunction with the 100k resistor. Reducing the value increases the corner frequency of the filter.) You could also reduce the 47p's to amplify more treble, or reduce the 10n to let less bass into the circuit, but either of those are more likely to alter the distortion character (probably make it harsher), so if you like the sound of the pedal and just want it less muffled, I'd start by reducing that 470p.

[motterpaul]

I know I need to understand voltage better. resistance is easy, but burning out a few LEDs gets more expensive.

Anyway - that 470p cap is funny because it is part of a "rock/jazz" switch that I believe gives you a choice between that and a 68p - but I can barely tell them apart and I have seen other people write the same.

Anyway - Tried breadboarding the SWTC and it got pretty clumsy. I started getting oscillation (which turned out to be in the original veroboard - something was picking up RF or something, but I didn't know that at first.  Anyway - once I got the SWTC working for a very short term it seemed too lossy and did not give me much tone variation. I seemed to lose more bass and mids than I gained in brightness.

The circuit on its own sounds really good - I'd give it a 9 out of 10, I just want a little more treble.

What would you change that 470p to?

[induction]

You just have to try a few values and see what works best for you. Maybe start with 100p. If it's still too dark go lower, if it's too bright go higher.

Even better, just take it out entirely. That's as bright as it will get without changing something else. If that's not bright enough, you'll have to start looking at other mods.

Also, what op-amp are you using? A TL072 or a NE5532 will be brighter than a 4558 or an LM741.

[ξεναγος νεκροπολης]

yes man! i agree with induction. don't solder anything on that pot.
try this

http://tagboardeffects.blogspot.gr/2012/02/offboard-wiring.html

cut a small 5x2 or 5x3 vero and do all soldering on there first, and then add the pot.
i've burned 2 pots when i first did that "on-pot-soldering", so i've learned something...
plus, when you'll box the effect, you will have serious problems cause that resistor will sort with the enclosure( or you will not even be able to put it in the right place because you'll loose space). i also did that and took me 2 hours to debag it...!!!

[motterpaul]

I started with a TL072 but when I switched to a 4559 it just sounded richer and cleaner, a big improvement, so I believe I will stick with that.

Thanks for the pots tip. Those made for circuit boards definitely need that treatment.

[motterpaul]

Thanks, I have the circuit working in a way where I am happy. Not perfect but I don't want to start from scratch. I went back to the old 1M pot, and changed the 470p to 100.

NOW my question is footswitch wiring, for those of us who think pedals should have batteries - how do you wire the footswitch. I would use the Beavis method but it is different than here, and I already started this footswitch in the manner seen on this site (offboard)

My question is ground. I nthe firsty image he shows it coming from the DC connector right to the board, but not off the footswitch.

any diagrams form this site with battery connectors included?

[induction]

Use a stereo input jack, and wire the battery negative to the ring. Attach all other ground points to the sleeve.

For the most part, all wires that are connected can be considered as a single point, so it doesn't matter too much how the grounds are routed as long as they are all connected to each other. The exception in this case is using the stereo input jack, which keeps the battery ungrounded when there's nothing plugged into the pedal input. To prevent the battery from draining when you aren't using the pedal, just unplug the input.

[rocket88]

take a look at this. it explains how to hook up a battery to a pedal with a stereo jack. i use mark's posted offboard wiring, and if i want to add a battery too i just make the changes here.