Eh lpb-1

[nopainkiller]

I found two version of this pedal

Here's the one I built:

http://www.beavisaudio.com/techpages/HIW/hiw1.gif

I've seen other versions but this one was on the same site.

http://www.beavisaudio.com/bboard/projects/bbp_LPB1.pdf

I was wondering what the idfferences in the sounds would be. I know I could just build them, but I'd like to understand what the changes actually DO without having to strain my ears for something only a dog can hear. Specifically the one meg input resistor. What does this do? What does leaving it out do? Also there's a 43k resister replaced with a 100k resistor in the second schematic. What's going on?

[nopainkiller]

also the 430k resistor is changed to 1 meg!!

Could someone explain to me what bias is, and how it affects sound/tone??

[Mark Hammer]

The 430k/43k network, and the 1M/100k network both act as voltage dividers, that divide down the 9v supply voltage such that a specific voltage is presented at their junction, which also happens to be where the base of the transistor is connected to. You will note that 1M = 100k x 10 and 430k is 43k x 10. So the two networks are doing exactly the same thing....except you stand a reasonably good chance of actually finding a 1M and 100k resistor in your town, in comparison to finding a 430k and 43k unit.

The 1M resistor on the input makes a big difference in sound, and the difference is that if you are going to have a footswitch to switch the LPB-1 in and out, that 1M resistor will make the difference between a unit that has an audible pop when switching it on, and one that switches silently. The 1M resistor is an improvement on the original.

The 2N5133 transistor was a unit that EHX used on a great many circuits at the time, so I'm assuming they bought a lot of them at once and then figured out what to do with them. They can be found on the LPB-1, Screaming Bird and Mole, Muff Fuzz, Big Muff Pi, and many others. Nothing special about them. EHX now uses 2N5088s, which are also decent all-purpose transistors that will have no audible impact on the LPB-1 circuit.

Build either circuit and you'll get the same thing, sound-wise, except that you may probably find the parts and finish the second circuit before ever finding the parts for the first one.

[nopainkiller]

oh thanks! I have the original transistors actually. Those are what I'm using. I've already built the first circuit. I actually don't have any 100k resistors! I thought the 1meg resistor at the input might change the sound. I guess I'll wire it up in there, it's only one part after all. Thanks!

[Mark Hammer]

It wasn't until the practice of using DPDT switches for "true bypass" that resistors like the 1M resistor to ground became more commonplace. Traditionally, the player would plug into the pedal, and the stompswitch would simply select between the output of the circuit board or whatever was at the input jack. But it would not disconnect the input lead from the circuit board at any time.

Why was this an issue, and why did it require something like a 1M resistor? Capacitors store current. That's what they do. When the circuit board is disconnected, there can be stored current sitting in the input cap with nowhere to go. To dissipate, it needs to drain off to ground, and requires connections at both ends to do so. When the unconnected cap becomes reconnected again, that stored current rushes out all at once at the first available opportunity. The user, and unfortunately the audience, hears that as a pop.

Of course, if the cap remains eternally connected to the input jack, you get only one pop when you plug in, but nothing after that, no matter how many times you step on the switch because the cap always has a path to drain off via your guitar or whatever else is plugged into it. That's why no one ever bothered about resistors like the 1M found on the newer LPB-1.

The price one pays for having several pedals connected in this manner is that they are all perpetually tied to the guitar and act in parallel to load the guitar signal down, so that you lose signal even when in "bypass" mode. (Note: pedals which use an input buffer and FET switching do not suffer from this challenge) This is why many manufacturers changed to DPDT and 3PDT "true bypass" switches that would simultaneously connect/disconnect both the input and output.

So, again, why the 1M resistor? This resistor provides a path for the input cap to bleed off stored charge to ground, while in bypass mode. That way, when you switch the booster on, there is no audible pop.

But why 1M and not 100k? The rate at which the stored charge completely drains off depends on how much charge is stored in the first place, and how much the path restricts or resists passing current. At the same time, the quality of the signal being sensed by the pedal will depend on how favourably the guitar can "compete" with other things tied to the input of the circuit. A resistance value in the range of 470k-2M2 or so provides a nice compromise between all those factors. It is low enough that the input cap will drain off fairly quickly, high enough that it won't load down the guitar, and is a commonly available (hence cheap) value. I've used values between 470k and 2M7 without any problem, but 1M has become a standard because pedal makers have them lying around anyways.

[nopainkiller]

thanks alot man. So I'll just put 1 meg resistors in every pedal I make eh!