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Post by mr_sooty on Oct 12, 2009 23:25:28 GMT -5
Hey guys, I've made a pedal using the General Guitar Gadgets ITS808 kit. Here's the stock schematic: www.generalguitargadgets.com/pdf/ggg_its8_sc.pdfI added this 'AMZ fat' switch, which bypasses C3 (see page 2, here: www.generalguitargadgets.com/pdf/ggg_its8_mods.pdf ) C3 is originally 0.047uf, but I have also done the 'Keeley mod', which changes the value of that cap to 0.1 uf. The pedal sounds great, but the 'fat' switch sounds a bit farty when engaged (bypassing C3). It's a little too much, the bass response is a bit over the top, making it really muddy sounding, even with a Strat. What I'm wondering is, can I add a different value cap between the switch and the circuit board to make this fat switch slightly less intense? I thought something that was in between the 0.1uf I have in C3 and 'no cap at all'. Do I go to a higher value or a lower value? And if there's another cap in the bypass route, will it still bypass C3? Thanks. (PS, here's the pedal....)  |
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Post by mr_sooty on Oct 13, 2009 0:50:19 GMT -5
Not sure why this was moved, I'm not 'discussing my favourite stompbox', I'm asking a question regarding electronics and wiring. If this was in the wrong section, why is there an electronics and wiring section at all? If pedal electronics and wiring questions should be under pedals, surely all guitar electronics and wiring questions should be under guitars and amp ones under amps, leaving - nothing.
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Post by ashcatlt on Oct 13, 2009 1:08:17 GMT -5
Find the value of the cap that you want for your fat switch. Easiest way is probably to remove C3 and the switch, but leave the wires for the switch connected to the board. Swap caps in between those two wires till you find what you want.
When you find something, put it in place of C3
Now, figure out the difference between that value and your 0.1µ. Find a cap close to that value and put it between one leg of the switch and the board.
The switch now flips "backwards", so you might what to turn it 180°.
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Post by ashcatlt on Oct 13, 2009 1:10:42 GMT -5
Do I go to a higher value or a lower value? 0 < 0.1µ. Nope...err...maybe sort of...err...They'll be in parallel, and their values will add...but the way I described it above, this is now the "normal" mode. |
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Post by mr_sooty on Oct 13, 2009 2:24:08 GMT -5
Do I go to a higher value or a lower value? 0 < 0.1µ. Nope...err...maybe sort of...err...They'll be in parallel, and their values will add...but the way I described it above, this is now the "normal" mode. Ahh right, so the idea is that with the switch 'off', we're running both caps, and they'll add up to 1 uf, right? Then when we switch it to 'fat', we're taking out the one connected to the switch? I think I got ya. So how it is currently, the bypass switch works because of the 'path of least resistance' theory, right? The current goes through the switch because the cap is more to get through than the switch. But if I'm switching in another cap, it'll still go through both, because the caps offer the same amount of resistance as each other, because they're caps, not resistors. Am I getting this right? |
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Post by newey on Oct 13, 2009 6:05:35 GMT -5
Sooty-
SG moved you here because you're discussing wiring of a stompbox; we try to limit E&W to guitar wiring questions, leaving amp wiring for "amps" and effects wiring for "effects".
The labels on the categories were set up originally by RandomHero, and things have evolved over time. I agree it is a little confusing, but if you look around the effects section, you'll see at least as many wiring questions and schematics as discussion of "favorite stompboxes".
Many of these things could go either place, but SG and I try to maintain some order to the proceedings insofar as we can. So, please don't take offense that your post was moved. It is more likely to be seen here anyway.
At the rate things go in the "effects" section, maybe I should just rename it as "flateric's corner" ;D ;D
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Post by ashcatlt on Oct 13, 2009 11:21:55 GMT -5
Ahh right, so the idea is that with the switch 'off', we're running both caps, and they'll add up to 1 uf, right? Then when we switch it to 'fat', we're taking out the one connected to the switch? I think I got ya. Yep, that's what I was shooting for. This is a dangerous way of looking at it. Electricity will take every path it can. It will try to move from a place of high potential to somewhere lower. In the original mod, the switch (when closed) shorts both ends of the cap together, so there's no potential difference between the two ends, so nothing flows. In my proposed mod the caps are in parallel (with the switch closed). "If you put two capacitors in parallel - the total capacitance is bigger... in fact it's the addition of the two capacitances (because you're effectively making the plates bigger)."The link takes you to a discussion of the basics of capacitance, but I'm afraid doesn't really explain the parallel thing particularly well. Maybe sumbody around here can expound a bit better than I. |
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Post by sumgai on Oct 13, 2009 13:30:33 GMT -5
OK, a quick "jump" and I'll have to get back to what I do best, making the HausMaus happy! ;D First: Mr. Sooty, I apologize for not "warning" you about the move, but I left the "moved message" so I figured you'd still be able to find it in this new location. And as newey points out, this is the proper location for all things related to stompboxes, I'm sure you'll agree. Again, sorry.  Second: in the realm of capacitors... sooty, you've understood ash correctly, and I think you're going to get everything lined up the way you want it to act. As ash notes, it's best to experiement first with the two capacitor values you wish - one for "main cap all by itself" and the other is whatever cap you place in parallel with the main cap, thanks to the switch. I dare suggest that if you found 0.1µf to be a bit too much, then you'll probably want to back off to 0.068µf. There is the possibility that you'll like 0.082µf, but that value is getting harder to find in a single cap. It's possible, but not as easy. Still and all, you aren't locked in to any one value. Think - if you're going to use two caps in parallel anyway (thanks to the switch) then you can just as easily used two caps in parallel for the "main cap", right? This lets you reach all kinds of intermediate values that aren't for sale at all. And I'm sure you'll appreciate the value of experimentation here, you can dial in exactly what you want, no compromise necessary. I need to run, so I've not made any diagrams for this post, but there are some already on this forum. ash, can you find a link or two for mr. sooty, please? Look for Unklmickey, which means of course that it'll be at least 2 years ago and more. I'll try to be more helpful next time out.  HTH sumgai |
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Post by mr_sooty on Oct 13, 2009 14:02:58 GMT -5
OK, a quick "jump" and I'll have to get back to what I do best, making the HausMaus happy! ;D First: Mr. Sooty, I apologize for not "warning" you about the move, but I left the "moved message" so I figured you'd still be able to find it in this new location. And as newey points out, this is the proper location for all things related to stompboxes, I'm sure you'll agree. Again, sorry. It's OK, I'm over it. No worries. I understand now. No, it's not the 0.1mf that's too much, it's the effect of bypassing that 0.1uf that's too much (as I understand it). So instead of bypassing it completely, I want some kind of middle ground, say switching to 0.047uf (which incidently is the stock cap in this spot before the Keeley mod). So Ash's idea suggests if I added something like a 0.053 to the switch and had this hypothetical 0.047 value as the main cap on the board, I would be getting the 0.1 uf value when both were engaged, but bypassing the 0.053 with the switch to get the 0.047 value on it's own. I don't think I need a diagram here - I've come a long way in the last couple of years! |
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Post by ashcatlt on Oct 13, 2009 15:10:40 GMT -5
I didn't figure it took a diagram, either.
To be honest, I'm not quite sure how this mod accomplishes what it does. I'm just going based on the fact that you said bypassing the cap makes it go the direction you want, but just a little too far. So I figured a value somewhere between the one you've got and zero is what you would want.
You may find that there is some popping when you flip the switch. This is one of the problems that come from switching one leg of a capacitor. It stores up energy while it's connected, and holds onto it for quite a while when disconnected. When you switch it back in it sometimes will discharge very quickly.
This is usually fixed with a "pulldown" resistor connected to ground on the switch side of the cap. Now the stored energy has someplace to go. Can't tell you how this will affect the rest of this particular circuit, though.
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Post by mr_sooty on Oct 13, 2009 15:38:04 GMT -5
To be honest, I'm not quite sure how this mod accomplishes what it does... Me neither, except that it bypasses the cap, apparently. Have you looked at the links I posted? Cool, ballpark resistor value? I'm guessing too high would be bad, because this 'stored energy' wouldn't get through. What I'm not really getting is that the Keeley mod involves changing C3 from 0.047 to 0.1, which is a HIGHER value. If we're 'fat switching' to 0.047, we're just going back to the original stock 808 cap. |
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Post by ashcatlt on Oct 13, 2009 21:30:08 GMT -5
Cool, ballpark resistor value? I'm guessing too high would be bad, because this 'stored energy' wouldn't get through. They actually are usually pretty large to avoid loading down whatever they're connected to. That's a good point, and this just confuses things further!  |
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Post by mr_sooty on Oct 13, 2009 22:10:53 GMT -5
OK, in someways that actually clarifies things, because all the suggested mod values get progressively higher, to a max of 1.0uf (my one being an intermediate 0.1 uf). Is says that bypassing the cap (listed here as C2, but I'm assuming it's the same cap that is called C3 on the GGG schematic) will only work properly if you follow this:
Note: The capacitor and resistor coming from the inverting input of the opamp must be connected to Vr (+4.5v) and not to ground for the switch mod to be implemented - you did listen to me on this, didn't you? If they are connected to ground, bypassing the C2 capacitor will upset the dc bias on the opamp and make it function improperly (and sound bad too).
I wonder if that's an issue? Can you possibly check the links in my OP to make sure the bypass points are in some way connected to this 'Vr' point? (I don't know what Vr is).
Otherwise, what I probably need is a way to switch in a bigger cap. Maybe leave the current one where it is and add someting bigger into the switch. Or switch the C3 position back to 0.47 (as per original 808) and add a bigger one to the switch.
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Post by mr_sooty on Oct 13, 2009 22:42:12 GMT -5
OK, I still don't know what Vr is, but it does seem that the switch is connected to Vr, so that doesn't really explain anything.
I could still just try to have the switch go between the stock value and a larger value.
Hmm.
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Post by ashcatlt on Oct 13, 2009 22:51:55 GMT -5
Yeah, I think I figured it out. The answer comes from this: "If the switch is closed, the capacitor is bypassed and the low end frequency response extends down as far as allowed by the Vr bias network..." ...which includes the capacitor noted as C7 in the schematic you posted in your OP. This is in series with C3, and is very large in comparison. Since caps in series combine the way resistors in parallel do, the total capacitance of C3 and C7 is very close to the value of C3. The "10 x rule of thumb" says you can basically ignore C7 in most calculations. On the other hand, shorting across C3 leaves only C7, and drops the cutoff significantly. So, I was wrong, and you are right. Sorry about that. Maybe try a 47µf, this (in parallel with the .1, and series with the other 47) will get you pretty close to half of C7 by itself.Edit - while the math is right, it's going to leave your cutoff down in the region of 14Hz (if we can trust mr orman's table). You probably want a much smaller change. Try the .047 you replaced, or another .1. |
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Post by sumgai on Oct 14, 2009 1:27:45 GMT -5
Guys, V r is shorthand for Voltage reference. It's derived from a voltage divider network between the two poles of the power supply. Usually, these IC's are powered by a single voltage supply (as versus a dual supply, where you give it perhaps plus and minus 9 volts or plus/minus 12 volts or whatever). However, dual-supply IC's came about because of the nature of AC - it swings above and below the zero voltage point (commonly called ground). Operational amplifiers were designed to take advantage of this, and with dual power supplies, they can amplify both halves of the signal equally. But it's much more common "in the wild" (that's you and me, the unwashed public) to have only one source of power, not a dual source. Hence, we must substitute a new value into the op-amp for reference purposes. This is denoted as V r. It comes from connecting two resistors of the same value in series, and between the incoming positive voltage and ground. At the "half-way" point, the voltage measured in comparison to one side or the other (meaning, compared to the positive voltage, or to ground) is supposed to be exactly half of the total voltage supplied to the circuit. Of course, what with resistor tolerances and such, that's a tough criterion to meet. Suffice it to say that if it's within about 1/10 th of a volt, most folks aren't gonna care. To get really long winded, what's happening in this circuit is that the two diodes and their associated caps and resistors are part of the feedback loop. This op-amp is configured to run at maximum gain, all things considered. The feedback loop keeps it from going to max gain by feeding some of the output signal back to the input in inverted phase. That translates to, all of the fedback signal is out of phase with the op-amp's input, but it's not all at the same signal strength! Mark that down in your notes, there'll be a test at the end of the period. ;D The feedback loop is sending some of it's signal content to V r (remember, that's the equivalent of ground, meaning that's it's "zero volts"), and the rest of it at full strength to the op-amp. What's going where? Glad you asked. A capacitor is in the circuit, and it's going to V r. This looks suspiciously like our standard guitar tone controls (aka, treble cut controls), doesn't it? Yeppers, that's what's going on here - the higher frequencies are being shunted to ground, and the lower frequencies are getting through at full strenght to the op-amp. Net result? The amp sees more negation of the signal at low frequencies, and less negation at higher frequencies. Does that make sense? Remember, I said that the feedback loop is designed to knock out or cancel nearly all of the gain. It does so by pitting a negative going signal against a positive going signal. If the two voltages are the exact same level, then we have full cancellation - no output would occur! If the feedback signal is not as strong as the input signal, then only partial cancellation will occur. Since the maximum strength signals coming in from the feedback loop are low frequencies, these are strong enough to nearly cancel all of the low frequency portion of the input signal. Now does this all make sense? I sure hope so, I'm starting to run out of words here.  Long story short, bypassing the cap means that nearly all the frequencies are being shunted to V r, which translates to not very much feedback signal at all, thus not much negation of anything, thus a lot of overdrive coming out of the box. I will say that for most people, musicians and non-musicians alike, too much of a good thing is not good at all!  That's it, stick a fork in me, I'm done!  HTH sumgai |
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Post by mr_sooty on Oct 14, 2009 2:06:55 GMT -5
Mark that down in your notes, there'll be a test at the end of the period. ;D Hmm, I don't see myself passing this test somehow. I got the jist of some of it, in theory at least. I read and tried to comprehend, but it was a bit beyond me. Anyway, the experiments are about to begin. |
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Post by mr_sooty on Oct 14, 2009 4:41:01 GMT -5
Success! I installed a cap from the middle terminal of the switch to one of the other terminals, so that it either connects to the circuit or shorts itself (I hope this is OK practice). No popping, which is good.
As far as the value was concerned, I had a 1 x 0.047uf and 2 x 0.22uf caps left over from the kit. 0.047 wasn't enough to be worth it. I liked 0.22, and almost went with it, but in the end dcided it was borderline muddy (we're testing with a Strat here). So close.
Anyway, I happened to have a 0.1uf lying around, and liked that, although it is definitely more subtle. I ummed and ahhed, going between 0.1 and 0.22. Really wanted 0.22 to be the one, because there are two spare with the kit, and the only 0.1 I have is a big fat old greencap. But in the end I had to admit that the 0.1 was probably the go. I did try the 0.047 across the 0.1 for a compromise, but that increment was too subtle a change.
The downer is that every time I make this pedal for someone I'll need an extra 0.1 uf cap. Of well, I already have to get a few extra bits anyway.
Thanks for all the help as usual!
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Post by ashcatlt on Oct 14, 2009 11:58:11 GMT -5
Remember, I said that the feedback loop is designed to knock out or cancel nearly all of the gain. It does so by pitting a negative going signal against a positive going signal. If the two voltages are the exact same level, then we have full cancellation - no output would occur! If the feedback signal is not as strong as the input signal, then only partial cancellation will occur. Since the maximum strength signals coming in from the feedback loop are low frequencies, these are strong enough to nearly cancel all of the low frequency portion of the input signal. This part is a little confusing. I dig what you're saying, but I also know that if we have nothing in the feedback loop but a straight wire, so that all of the output is connected back to the negative input, we end up with a unity gain buffer. That is, the output is not silent, but rather it is exactly the same as the input. I've always read that the opamp tries to equalize its positive and negative signals, by way of multiplication. Might be just a different way of saying what you're saying, though. Anyway, I'm glad you got it working there sooty. Great job using the switch (you never identified it) to avoid the popping, as well! |
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Post by sumgai on Oct 14, 2009 13:11:15 GMT -5
ash, Whenever you hear/see the word "multiplication" in this context, just remember that you're dealing with a mathematician who likes vectors. (Remind you of anyone in particular?  ) This statement is basically true because the negation action is mathematically known as a differentiator, which is derived some functions specific to calculus. I've forgotten most of my training and exercises in this discipline, so I'm no longer much help..... but I do seem to remember the lingo, does that help?  Yes, unity gain in an op-amp is the application of all the output back to the inverting input. But at unity gain, if all we have is the input, without modification (in theory), then we can't have any overdrive, can we? Thus, to make the circuit work for us, we need to modify something. In the old days (prior to op-amps), we usually messed with the input signal. But op-amps made it obvious that another path to achieve the same results was to modify the feedback loop, and things took off from there. Does that clear it up for you? If not, don't hesitate to ask, this is one area where I have not forgotten much of whatever got beaten into my widdle pointy head. ;D I'm more than willing to sit and discuss this until cynical1 has finished mucking out the horse stalls! And this particular Forum is an excellent place to go ever-so-deep into electronics theory. For stompboxes, we don't need to know molecular theory, but we do need to know more than Ohm's Law. It's not off-topic to ask, so anyone - fire away!  Oops, almost forgot to mention that the caps probably won't pop when switched - there's not enough current involved to build up a significant charge of electrons. Remember, we're working with no more than half of the power supply voltage here (good ol' V r). If you work out the formulas, you'll see that the reactance of a medium value cap at 9 vAC (the theoretical maximum that can appear at the op-amp's output) is quite large. When we plug that into Ohm's Law, we get a very low current, nearly in the micro-amp range. Not quite, but nearly. HTH sumgai |
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Post by mr_sooty on Oct 14, 2009 15:29:43 GMT -5
Oops, almost forgot to mention that the caps probably won't pop when switched - there's not enough current involved to build up a significant charge of electrons. Remember, we're working with no more than half of the power supply voltage here (good ol' V r). If you work out the formulas, you'll see that the reactance of a medium value cap at 9 vAC (the theoretical maximum that can appear at the op-amp's output) is quite large. When we plug that into Ohm's Law, we get a very low current, nearly in the micro-amp range. Not quite, but nearly. It's all good, in all honesty, I wired it up to the switch that way cos it looked neater than going between the switch and the board! |
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Post by ashcatlt on Oct 14, 2009 21:44:29 GMT -5
Oops, almost forgot to mention that the caps probably won't pop when switched - there's not enough current involved to build up a significant charge of electrons. Remember, we're working with no more than half of the power supply voltage here (good ol' V r). If you work out the formulas, you'll see that the reactance of a medium value cap at 9 vAC (the theoretical maximum that can appear at the op-amp's output) is quite large. When we plug that into Ohm's Law, we get a very low current, nearly in the micro-amp range. Not quite, but nearly. Better safe than sorry, though. There is quite a lot of amplification going on here, so a small click at the input could turn into a pretty big snap at the output. There's usually a pull-down resistor at the input of the effect, partly to help set the input impedance, partly to help set the roll-off of the HP and LPFs you find at the front, and partly to keep the AC coupling cap from popping when switching to and from bypass. Here we're only talking about the minuscule current generated by the pickups. In this circuit, you'll find a similar pull-down resistor after the AC coupler on the output, but before the switch, and for about the same reason. That little pop there could eventually be amplified up to the >500V rail level. |
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HTH
) This statement is basically true because the negation action is mathematically known as a differentiator, which is derived some functions specific to calculus. I've forgotten most of my training and exercises in this discipline, so I'm no longer much help..... but I do seem to remember the lingo, does that help? 
