# Valvecaster Pedal...home made with problems



## greco (Jul 15, 2007)

My son-in-law built this pedal for me as a gift.
It "pops" loudly when the switch is engaged.
I have looked on the internet and found information that this popping can be reduced by using a 100K to 4.7M resistor on the "outboard" of both the input and output caps.
Does this seem like a reasonable solution to try?
...and where would the resistors be placed (given the following layout diagram)?
This the diagram he used to build the pedal.

Are there any other obvious concerns with this circuit?

Thanks for your comments and help.

Cheers

Dave


----------



## keto (May 23, 2006)

Ignore this post do not do what is quoted below no no no lol


----------



## greco (Jul 15, 2007)

keto said:


> 1M, right between C1 and S1 I think, though I'm not familiar with vero board so much. Mr. Hammer should be along shortly to confirm or refute my hypothesis.


Thanks Keto..this should be the circuit without the 3PDT switch and the LED + dropping resistor.










However, this circuit could easily have been modified (later) as the thread on DIYstomboxes is 139 pages in length.
Here is a link to the thread.
http://www.diystompboxes.com/smfforum/index.php?topic=63479.0 
I tried to read the thread but got lost fairly quickly.


I'd just like to get it working a bit better, as I like the pedal.

Cheers

Dave


----------



## starjag (Jan 30, 2008)

greco... click here, I think this is the solution you're looking for.


----------



## mhammer (Nov 30, 2007)

D'ja ever have to "hold it in" for a long time, fiddle frantically with the house keys to let yourself in, and when you finally got to the john there was this moment of both release AND pain, as you voided the remains of a large assortment of beverages? Switch popping is a lot like that, and occurs for some very analogous reasons.

What do capacitors do? They store charge. The charge accumulates, if it has no where to go. The large/higher the cap value, the more charge it can store. Benjamin Franklin, in the early days of electrical experimentation, before there were digital multimeters, would store charge in capacitor jars, and determine how much was stored by grounding out the capacitor through a live chicken, apparently. The chicken's reaction (or non-reaction after being a grounding point) indexed the amount of stored charge. Sheesh, the things you learhn and remember from college!

Anyhow, when a low-impedance path to ground is provided to a capacitor, the stored charge drains out...quickly. If a "hanging" cap like C*1* is connected to another device feeding it, chances are good that device will provide the aforementioned path to ground. But if it is not connected to that path to ground until the switch is stomped, then that sudden connection makes for a rapid discharge, and the cap will produce an audible "pop", not unlike what you feel when your bladder has its sudden "moment of moksha" ( http://en.wikipedia.org/wiki/Moksha ). Note that since it is an effect occuring with the _accumulation_ of charge, stepping on the on/off switch a few times will appear to result in an absence of popping, by viortue of draining the charge off before it has a chance to accumulate. Make no mistake, though. Play for a bit, then turn the effect off, and next time you turn it on the pop will be right back again....because you allowed charge to accumulate.

So what's the deal with the "terminating resistor" from the free end of C*1* to ground? Such a resistor can provide a path to ground for the cap to discharge when you're not engaging the circuit, such that when you next hit the stompswitch, there will not be any stored charge just waiting to leap to ground. The trick is to identify a resistance value that allows for enough charge to drain off at a reasonably fast clip, without impacting on the input impedance of the circuit.

Just for argument's sake, let's say we had two of these circuits in series, such that the second one was being fed by the wiper of the 100k volume pot in the first one. Okay, what's a good terminating resistor value to drain off C*7*? We could use, say 10k, which would drain off C*7* pretty dang quick. But wait a sec, that 10k resistor would be placed in parallel with the leg of VR3 between the wiper and ground. What if we turned that volume control all the way up so that the wiper was effectively tied directly to the junction of C3 and C4? Well, with a 10k resistor in parallel, that would make the combination of the pot and the terminating resistor behave as if it was a 9.1k resistor to ground. What we want and need is a terminating resistor that is a high enough value that VR3 will behave as if it IS a 100k pot. Can't make it too high a value (e.g., 22meg), or else it'll take forever for the cap to drain, and you'll still get popping. Can't make it too low for reasons already outlined. Values between 470k and 2.7meg are usually pretty good i serving both purposes, with 1M probably the most common.

Make sense?

Incidentally this is true of both input and output caps, and any other place in the audio path where a cap might be in a position to gather charge and have no way of dumping it. Let's say for argument's sake that you wanted a mod to this circuit where you selected between two values of C2: one for full bass, and a smaller value for a thinner tone. If you wired it up to select between C2a and C2b, say 6.8nf and 47nf, you'd hear a pop when switching between them because neither has a path to drain off until you connect it. If, let's say, we left the smaller value in place as the "default", and used a toggle to switch in a larger value in parallel (say, 39nf), you'd hear a pop when the larger one was engaged, but not when you lift the larger one and revert back to the default one _because the default one already has a path to drain off_.

Are you forever saddled with that pop? Not necessarily. Let's say that C2b (39nf) is wired i series with a 1M fixed resistor, such that the C2b/1M network runs in parallel with the "default" cap. With that 1M series resistor, it would have very little tonal impact, but it would have a path to continuously drain off. If a toggle was used to short out that 1M resistor, the effect would be the same as suddenly connecting C2b in parallel, but without the audible pop!

Moral of the story, if you take your capaictor dogs out for a pee regularly, they won't come in the house and pee all over your carpet.


----------



## greco (Jul 15, 2007)

WOW...that is a lot of information. Thank you very much for the tutorial.

I followed everything quite well...including the urological analogies, the history of testing capacitance (chicken must have been a lot cheaper/disposable back then) and Moksha...

I became bogged down at the point "*Just for argument's sake, let's say we had two of these circuits in series"......*I'll need to go over the last 3 paragraphs a few more times. 

In looking over this circuit quickly, do you see anything else that makes it an inherently a poor design in any way....or do you have any further comments? I know it is a very basic circuit from what you are typically involved with, but I actually think the pedal does a good job of adding a bit of grit and fattening the tone. 

I'd like to make this pedal function as best as possible, as my son-in-law spent hours making it and it is a shame not to use it. BTW..He is nearing the completion of his PhD at Carleton.

Again, many thanks.

Cheers

Dave


----------



## keto (May 23, 2006)

I mis spoke, been a while since I worked on pedals. What I should have said is 1 end of a 1M resistor between C1 and S1, with the other end going to ground. BUT R1 in the schematic, which I can read easier than the vero layout, IS ALREADY DOING THAT.

Mark's explanation greatly confused me, especially the repeated references to C7.kkjq


----------



## jcayer (Mar 25, 2007)

keto said:


> Mark's explanation greatly confused me, especially the repeated references to C7.kkjq


Ditto !!! The explanation is very interesting but I can't find any C7 ...


----------



## greco (Jul 15, 2007)

jcayer said:


> Ditto !!! The explanation is very interesting but I can't find any C7 ...


Same here..I printed off Mark's post and read it several times. I ended up assuming that he meant C1 instead of C7.

I am reasonably close to understanding the concept involved...I'm just not sure what to do.

Cheers

Dave


----------



## mhammer (Nov 30, 2007)

Well that's what happens when you're working with a screen that is too small for your eyes!
kqoct
Now that I'm at work, and using a 24" monitor, I went over the post and corrected it. Corrections are in *red*. I think you will find it a *lot* more coherent now.

As for using two of the circuits in series, I just suggested that because you'd have the schematic in front of you and there'd be no need to refer you to a different schematic or draw anything out myself. You can actually insert pretty much anything you want in front of it.

If a guitar is plugged directly into it, and the input is never switched, then you can feel free to omit any terminating resistor, because *C1 (!!) *will have a path to drain through the guitar volume pot. It will also provide optimum impedance matching.

You will note that pretty much ALL Boss pedals have a "hanging cap" at their inputs. This is because their electronic switching takes place after the input buffer and before the output buffer. If you were to plug the output of a Boss pedal into your amp, and then plug into the input jack, unplug, and plug in again, you would hear a pop, for all the reasons outlined earlier. But you would not hear any afterwards because the hanging input cap remains connected to a path to ground for as long as you're plugged in.


----------

