bypass, debounce, hot dogs, mechanical, pop, switch, true, unicorn -

What really causes switch pop

 I've been there before.  You know the situation.

The band is groovin.  Shit is in the pocket.  The song is about to hit and the guitar drops out for a half measure.  In preparation, you stomp your favorite pedal and its deafening:  TTTTHHHHUUUMMMPPPP!!


In today's edition of Straight Jive, we'll be discussing "switch pop" and the causes, sources, and solutions to the hassle of mechanical-bypass, often mis-marketed as true-bypass.

Lets get things started by clarifying some of the mis-information and bullshit floating around out there regarding mechanical bypass systems.  I don't use the word "systems" lightly either.  Every bypass scheme is a system, not just a switch.  We'll start with a few.  Pop a brew, pour a scotch, light a doobie, cuz its about to get real in the field.  By the way, LED indicators are omitted from all of the drawings.  Its just easier to explain like that.

A little background on some common bypass schemes...

Below we see the "old-school" method of mechanical bypass, employing an SPDT (that's Single-Pole Dual-Throw) switch.

Pretty simple system.  Here's how it works:

With the switch in the position shown, the Output Jack is directly connected to the input jack.

If the switch were toggled (i.e. you stomp on the pedal), the output jack would connect to the circuit block's output and whatever is supposed to come out of the circuit comes out of the circuit.  We got effect tones.  Yeah!!

Nice n clean eh?  Umm... yeah... about that...

While this system is simple and only requires a SPDT switch, there is an inherent nuance (or weakness as many people consider it) to this system.  Draw your attention to the left side of the diagram:

The Input jack is always connected to the circuit block.  bummer.  That means some signal (however small and "insignificant") is going into our circuit's input and we're presenting a load onto our signal.  Now, it may be a small load, but its a load nonetheless, and frankly, its unnecessary in today's modern day and age.  This scheme is so dated we're not even going to talk about some of the sources of noise in it.  Its just, as they say, a waste of time.

Lets focus our efforts on something useful and worthwhile, jump ahead a few years and start looking at True-Bypass systems!

This is the simple way to accomplish true-bypass, using a dual-pole dual-throw (DPDT switch):

Luckily, this is real easy to see.

When the switch is in the position shown above, the input jack directly connects to the output jack and nothing else, while the input and output to the circuit block just kick it and wait for action (think LOITER).  TRUE-BYPASS!  Yay!!

When you stomp on the switch, the switches change position and the input jack connects to the circuit's input only, and the output jack connects to the circuit't output only giving us our toan.  Totally rad.

So we have true-bypass!  Totally bitchin and now we're done.  Nope.  There's a lot more to think about, and some really vital information we need to look at before we can call ourselves "expert witnesses" or whatever you need to say to know what you're talking about these days.  The diagram above is functional, but it is lacking some critical components that allow everything to work, and also to reduce any switching noise; and this is where shit starts to get real.

Dig the updated diagram below:

Looks about the same, except we've added four parts:  input and output coupling capacitors (99.999% of the time these are necessary, and if you know when they are not, you probably shouldn't be reading this article because you cut people's brains open and replace parts of them), and two resistors, which in this diagram, are functioning as:

"Pull-down" resistors:

You've probably heard that term before, no?  Pull-down resistors.  Guys say:  "Your switch is popping?  Yeah, some moron didn't put pull-down resistors in.  Just put a 2M2 or 10M or *some other arbitrary value* in and that'll fix everything!!  Trust me, I'm a builder."  Yeah.  I've heard that one too.  And I've seen guys do all kinds of crazy shit with these so-called pull-down resistors, often times destroying the operation of the circuit, or simply including a moot part.  (See what I did there?!)

So why would someone say the pedal needs the fabled "pull-down resistor?"  Simple really.  To "pull" the node it is connected to to low potential.  You notice I didn't say ground?  There's a reason for that, but that's an entirely other discussion.  Therein lies the crux of the "pull-down" resistor.  It PULLS SHIT DOWN.  If it were a "pull-up" resistor, what do you think it would do? (And don't get smart with me and say that there are no such thing as pull-up resistors; start reading about digital circuits and then you can buy me a beer.)  Back on track here, Jack.  Back on track.

Why would we need this resistor?

Because nothing in this world is perfect.  Just like we like it.  Capacitors leak & bleed.  That's just how it is.  We probably have some potential (voltage) other than low potential at the input of the circuit block that we want to isolate from our guitar signal, and the capacitor does just that.  Trouble is, that capacitor bleeds a little of that potential through and we end up with an unknown potential right where our switch connects.  Epic bummer.

Because our guitar doesn't have any DC offset (if this is getting overly tech-y, stick with me), that unknown potential from our shitty, leaky, does-even-buy-me-dinner-first capacitor will hit our guitar or whatever is feeding our circuit, and similarly, our amplifier or whatever our circuit is feeding!  That little offset can create quite a pop.  Trust me.

How about some numbers and metrics?

Proof or FACT as I like to call it.  From my research and studies, I have found that any DC bleed in excess of 5mV (that's 0.005 VDC) will cause a small POP, and it gets much worse as voltages increase.  By 20mV, shit is intolerable and you will dread hitting your pedal, especially if it is upstream from a delay pedal.

And that's where our unicorn pull-down resistor comes in.  It sits there biding its time, drinking tea, eating crumpets and politely gives the potential bleeding through that dirty, rotten, good-for-nothing, know-it-all, selfish capacitor somewhere to go, aside from our amplifiers, speakers and ultimately ears.

Yes!  Finally!  20,000,000 words and we have an actual source of switch pop!  DC offset, or DC bleed!  Took you long enough, Jack.

Wait!  There's more!  More ways to wire switches for true-bypass, and also, more sources of switch pop.

I'd like to steer our discussion toward another method of accomplishing true-bypass with a mechanical switch, and this way will cut our parts count!  -- at least, until we look at another source of switch pop.

Ladies and gentlemen, and you too, Johnny, I'd like to call your attention to the grounded input wiring scheme for accomplishing true-bypass with a mechanical switch:

Dig it!  By simply changing the way our switches are connected, we can eliminate the need for that pull-down resistor on the input coupling capacitor!  Hot damn!  We just saved us a part, Cletus!  Lets have a look at how this scheme works, shall we?

When the system is in the bypass state (shown above), the circuit's input is only connected low potential (which should be a stable 0VDC in most effects pedals) through a coupling capacitor and thus the circuit doesn't really process any signal, except whatever may be lurking on low potential.  The output jack is directly connected to the input jack, and the circuit's output just loiters.  DC offset is reduced (not nulled!  reduced) by the pull-down resistor tied to the output coupling capacitor's switch-facing node.  Simple, elegant and effective. AND we nulled out that first pull-down resistor.  Fuckin double-whammy!

When the system is in the effect state, low potential is removed from the circuit's input and replaced with whatever is on the input jack.  The output jack is no longer connected to the input jack, but instead connected to the output of the effect circuit, and BOOYAH!  We have our effect circuit.  Slick, eh?

So we've saved one of our unicorns, and we're doing a great job reducing DC bleed by giving any DC offset on the input coupling capacitor somewhere to go (low potential in this example) and by employing a unicorn pull-down resistor on our output coupling capacitor.  Awesome!  Now we've got a nice, pretty quiet bypass scheme.  This one is so popular, you'll find it on the majority of "boutique" pedals, and many independent manufacturers using true-bypass as their bypass scheme!

But, it could still be better.  After all, we've only talked about one source of switch pop.  The one everyone thinks about:  DC offset, or DC bleed, and its kryptonite:  the pull-down resistor.  Would you believe me if I told you that part of that pop you hear is coming from somewhere other than DC offset?  You probably would, and if you did say yes, you should really be proud.  Introducing:

Switch de-bounce and the eternal exchange of noise and energy

Although it seems instantaneous, when you step on that foot-switch and the contacts move from one position into the next, they bounce around for a period of time.  As the contacts are bouncing, they are actually not connected to anything, which is a terrible byproduct of such a system but is ultimately what we must give to take mechanical switching.  This might not seem like a big deal, but whenever a conductor is left "floating" or un-connected to any intentional potential (hey, it rhymes too!), it acts like an antenna.  That's because it effectively is.  That means that conductor picks up all kinds of stray energy and an alternate potential is established between the conductor (switch rocker in this case) and its contact.

What does this sound like?  DC Offset.  Again.  Shit.

But this is a little different for a number of reasons.  The first being we simply cannot include a unicorn pull-down resistor that connects and disconnects when needed (that just doesn't exist, even in the realm of unicorns).  The second being the bouncing action that happens when the switch contacts settle into position.  Because we have no idea how much voltage could be present on our floating contact (rocker), we have no measure of the capacitance of our leads to/from our contacts, we have no idea what the instantaneous and ever changing resistance between the rocker and contacts will be, and we have no idea how many times our contacts will bounce before settling, we really can't make any type of concrete determination about this recently in(tro)duced potential or the time required for it to discharge through our system and for the entire system to reach beautiful, quiet equilibrium.  The zen of switching, if you will.  And shit, even if you won't.

Sounds like a toughie, huh?  You're right.  Its a pretty tough egg to crack, but when the shell gets thick, you just get a bigger hammer and swing harder.

Dig the last diagram for this edition of Straight Jive and perhaps it'll begin to come into focus:

What's different?  We've got two more resistors.

But I though we just got rid of one resistor!?  We did.  And now we're adding two more.  But these two new resistors aren't pull-down resistors.  And they're not pull-up resistors either.  No, these are current limiting resistors.

Why a current limiting resistor?

Elementary my dear Watson.  We need to put a limit on how much current will be entering and exiting our effect circuit during the transition.  Lets play a fun little example using Ohm's Law (if you don't remember or don't know Ohm's Law, its no big deal.  Its just some simple algebra that explains the relationship between current, voltage, resistance and power.  Click on one of the links above for a quick primer).

For an extreme example, lets say we have 1 Volt headed into our circuit, towards our input coupling capacitor.  If we had 1Ω of resistance, we could have 1 A current hitting that capacitor, and that's just unnecessary.  What if that 1Ω resistance was actually 1MΩ?  We would have a maximum of 1µA (that's 0.000001 A) of current hitting that capacitor.  That's a little more reasonable, but it really doesn't mean anything, because our capacitor doesn't care about current in this case.  It cares about voltage.  But what happens is pretty neat.

That current, be it 1A or 1µA is translated back into a voltage and affects our capacitor by causing a little disruption on the other side of the capacitor.  The translation will be similar to the translation that happened on the way in, so that 1A of current may become a 1V change (BIG FUCKING PROBLEM), and that 1µA of current may become 1µV (not really much of a problem at all).  See where we're headed?

By including a series current limiting resistor with our input and output coupling capacitors, we limit the amount of current they can absorb and discharge and that ultimately will be translated back into a voltage which our amplifiers will amplify and our ears will ears.  I mean hear.

Where does that current come from and why is it there?  Mechanical switches, Watson!  Its just the nature of the beast.  As those contacts bounce around and settle into place, they generate a small amount of current and dump it wherever they can, whenever they can.  In this case, into our input and output coupling capacitors causing "switch pop."  And there it is.  Easy-peezey, lemon-breezey, cheese.

That is some shit they don't teach you at effects pedal college or wherever they get their degrees from, but now you know!  And it was free too.  And frankly, I like free.  Especially when the words Hot Dogs follow the word free.

Does it really get any better?


PLEASE NOTE:  WE DO NOT HAVE THE RESOURCES TO OFFER ANY SUPPORT/TUTORIALS/ADVICE.  There is enough information above to solve just about any mechanical switching concern you could possibly have.  Just re-read it.  :)