# Q: Pedal pop and pulldown resistors



## devnulljp (Mar 18, 2008)

Quick Q about popping pedals and pulldown resistors. 
I have an old pedal that pops loudly when stomped on. I've read about adding pulldown resistors at the input but I'm not clear on how to actually _do_ that. I've seen schematics and ascii diagrams but I'm not good at reading those. 
Could someone decipher and show me where the resistor should actually go? 
I'm a clown with this kind of thing, but this means essentially nothing to me from a practical standpoint I'm afraid: 










I'm guessing you trace the wires from the inputs on the jacks to the board and/or switch and add a 1M resistor at the switch? 

A new switch might be in order too mind you...


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## Solid_Gold_Soundlabs (Sep 20, 2006)

Hey Alan, 

what pedal are you having problems with?

Greg


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## Guest (Jan 6, 2009)

devnulljp said:


> I'm guessing you trace the wires from the inputs on the jacks to the board and/or switch and add a 1M resistor at the switch?


That'd work. Or at the board, where the wire from the switch connects to the board. You just wire the resistor between that pad on the board and ground -- which can be the sleeve part of the input or output jack, which ever is closer to the pad on the board.

If you show us a picture of the guts I can show you exactly were to solder the resistor if you like.

mhammer will explain it better. :smile:


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## mhammer (Nov 30, 2007)

You generally don't want any DC entering into or escaping a pedal. DC is used intentionally within the pedal, but its a DC voltage deliberately defined by the circuit and not some stray DC. Consequently, you will find "DC-blocking capacitors" at the entry point and exit point of circuits to make sure that no stray DC comes in, and that no stray DC gets out and buggers anything else up.

But consider for a moment what capacitors *do*: they store charge. That charge could be stray DC, but it could also arise from your signal passing through them. The cap doesn't really care. All it knows to do is to store charge that comes its way, and leak it out very slowly. The bigger the capacitance value, the more charge is stored. IF there is a path for that stored charge to leak out, it will; as fast as its little electron legs can carry it. But if there is no path provided, it will sit there for as long as the cap's leakage properties permit, and as long as there is still current stored. If you provide a resistive path for the stored charge to drain out to ground, it will drain out at a rate set by the resistor value.

In pedals that are old-fashioned bypassed with a SPDT switch, the user would experience a little thump when first plugging in, but no popping thereafter. The non-true-bypass switch meant that the guitar was always electronically connected to the circuit input. No pop, but boy was there usually loading, which gave the inevitable tone suck and volume loss. Because the output cap on the circuit would usually come before a volume control, there would be no switch-popping at that end.

Now enter the era of DPDT "true bypass" switching where the guitar is lifted from both the input and output of the circuit board when in bypass mode. When the effect was switched in and out then back in again, there might be some residual charge stored in the input capacitor (C1 in the diagram shown). If there was no way for that cuirrent to drain off to ground when the circuit wasn't in use, then when you turned the pedal on and connected your guitar to the circuit input, the guitar would provide a path for whatever charge was stored to drain off instantly. That draining current is heard as a pop. Because the amount of charge stored in the input cap is not always the same, switching on and off might get you a pop sometimes, but not others. 

The solution to that risk of pop is to provide a resistor to ground at the input of the circuit so that the DC-blocking cap always has a path to ground to drain off stored residual charge when it has a moment's rest. That moment's rest could be when in bypass mode, or simply when you stop playing for a moment before switching it off. Of course, the challenge is to find a resistance that a) provides enough speed of drain that you can turn the thing on and off several times in a hurry and expect all excess charge to be drained all the time (i.e., no pop risk), but b) is a high enough resistance that it does not load down the guitar signal. The compromise is usually a value between 470k and 2.2M, with 1M being very popular.

So, if one wishes to "de-pop" a pedal, you look for the solder lug on the stompswitch that goes to the circuit board, you can usually tell by tracing the common lug it connects to and seeing which jack it goes to: in or out. Once you have the lug identified, you can solder a 1-2.2M resistor from that point to ground (a neighbouring jack might provide a ground point), or else trace the wire back to the board, and solder from where it contacts the board to an identified ground point.

Finally, I will note that popping does not only come from caps with free ends and no viable ground point. De rigeur these days in pedals is a 3PDT stompswitch and a status LED to let you know when the pedal is on (I only find this necessary for clean stuff like compressors; if you can't tell your fuzz is on, then you have bigger issues to deal with). LEDs need current. If rushing current is what produces the pop, then anything which poses sudden demands on current can provide a pop too, and poorly planned status LEDs can do that. Essentially, when the switch connects the LED to the voltage source, it may draw so much instantaneous current that you hear a pop.

We've had a ton of discussions about it at the DIY forum. One solution is to replace the existing LED with a "superbright" variety. Just FYI, "garden variety" red LEDs found in the typical pedal often produce a maximum illumination of 300 or less millicandles. Superbrights can often produce in excess of 3000 millicandles with the same current draw. What that lets you do is use much less current to power up the replacement LED to the level of illumination you're used to. So, if there is a garden variety in there now with a 2.2k current limiting resistor, you could easily replace the 2.2k resistor with 15-18k and drop the amount of current drawn when the LED is switched on by a factor of 7-8 or more...if the regular LED is replaced with a superbright.

I'm not sure if that makes it clearer to you what to do, but certainly let me know if it doesn't.


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## devnulljp (Mar 18, 2008)

Wow, great info guys thanks...and the endless patience of mhammer. 
Greg, it's the Boomerang, a bit of a crack when engaged, otherwise clean sweep. 

Some pics below (click for hi-res). Of course it's made more complicated by the fact that the previous owner rewired it using all orange wire! (input and output jacks are reversed too...this pedal originally has the jacks the "wrong" way round like a fuzzface). It's wired TB. Wondering where to put a resistor to ground it out.


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