Popping in a true bypass box

[MikeH]

I built a true bypass box, and it works ok with one effect in it, but when I put it on my board and put 4 or so effects in the loop it pops when I turn it on.  If I hit the switch a few times in a row it pops loudly the first time, then less each time until about the fifth time, when there's no pop.  Sounds like a drop-down resistor type problem, but how do you fix that in a bypass box?  I've got a boss TR-2, a Visual sound H2O, a boss phaser and a EH holy grail in the loop, btw.

[smnm]

it's all on GEO if you look:

http://www.geofex.com/Article_Folders/box_pop.htm

[MikeH]

Right, right, right.  I've already read that.  But this is a little different, because this is just a bypass box.  It has no input or output capacitors to put a dropdown resistor on.

[Mark Hammer]

The Boss pedals, and anything similar to them that uses FET switching, all have a "floating" input cap that only has a drain path to ground when you plug an instrument in.  Of course, because the switching is all done electronically AFTER that input buffer, the working assumption on the part of the designer/manufacturer is that the only pop that will ever be audible is the first one when you plug in.  When one uses a separate TB box for loop selection, and the Boss-type boxes are first in line, it is as if you are unplugging and plugging in repeatedly.  The solution is to provide the sort of resistive path to ground on the Boss pedal that one would normally find in a pedal that uses mechanical bypass switching as its normal mode.  This would only be true of the first pedal in the loop, and would be unnecessary for subsequent pedals in that loop since one is presumably using their onboard electronic switching for individual pedals, in contrast to the mechanical bypass for the entire loop.  It is the input to the first thing in the loop that causes the popping.

But why does the amount of popping change with use?  Good question.  That input cap, like ANY capacitor, stores charge.  When there has not been any recent opportunity for that charge to drain off, it is no different than those tales you hear about working on tube amps and how you should drain off power supply caps to avoid getting shocked...except on a much smaller scale.  If you are playing, and selecting/deselecting the loop, the floating cap on that first pedal has opportunities to drain off through the guitar electronics once the guitar is connected to the input.  The suggested extra resistor to ground from that input cap simply mimics what the guitar itself would naturally do.

In general, the likelihood of having audible pop is a function of the size/value of that input cap, it's leakage properties, and the amount of time that has passed since it was last offered a path of some sort to ground.  So, a high-quality 0.1uf input cap that was last connected to ground through 2meg very briefly 20 minutes ago will likely pop more audibly than a .001uf cap that was tied to ground through 10k for 10 minutes 4 minutes ago.  Think in terms of how much opportunity you have provided for charge to accumulate, vs how much opportunity there has been for charge to dissipate.  Bigger storage capacity plus less drain opportunity = more pop.

Make sense?

The way to solve the problem is to simply run a 2M2 resistor from the hot lug of the input jack on the individual pedal (first one) to ground on that pedal.  That's it.  No need to figure out anything else because the input jack is normally hardwired to that input cap and the switching stuff is elsewhere.  Easy, huh?

[MikeH]

Awesome Mark Hammer.  You always have the most informative, interesting and well-to-the-point answers.  Thank you.

[Pushtone]

Would this fix the situation?...

Wiring the effect send jack in the loop box so it's tip (hot lug)
gets grounded by the 3PDT switch when the loop is bypassed?

This would give that floating input cap a path to ground when the loop is bypassed.

Like the way JD wires his 3PDT to ground the effect input on so many GGG layouts.

Or is that how you have wired your footswitch already?

[MikeH]

That's not how it's wired now, I should try that...

But-  If the solution is wiring a dropdown across the input jack of the first pedal in the loop, why couldn't I just wire it across the send jack of the looper?  That would create the same path, and it would work for any pedal I put first in the chain.

[Pushtone]

That's not how it's wired now, I should try that...

But-  If the solution is wiring a dropdown across the input jack of the first pedal in the loop, why couldn't I just wire it across the send jack of the looper?  That would create the same path, and it would work for any pedal I put first in the chain.

Since that is inside the loop box its easy enough just to test out.
So is re-wiring the switch so the input is grounded when bypassed.

Please let us know if it works.