I'm confused!

[DWBH]

So, lets focus on the LPB-1.

After the first capacitor we have two resistors that form a voltage divider network.

According to the image above, the output voltage created by these two resistors is 8,1v.
If you insert these values (input voltage and the values of the two resistors) into a potential divider calculator (http://www.electronics2000.co.uk/calc/calcdiv.php, for example), you'll get and output voltage (V2) of 818.182 milivolts, which is what I get on my board (0,82 volts).

Again, here (http://www.beavisaudio.com/techpages/SchematicToReality/), you can read

R2 and R4 form a voltage divider. This simple snippet is in charge of providing half of the 9 volt source voltage as a reference point to the base of the transistor. This reference point helps tell the transistor how much to amplify the signal.

Half of the 9 volt source voltage, would mean two resistors of the same value. However, in that page, in this schem (http://www.beavisaudio.com/techpages/SchematicToReality/am1.gif), the resistors used are 1M and 100k, which, according to the math, also form an ouput voltage of 818.182 milivolts.

Moving on, in this layout here (http://www.aronnelson.com/gallery/main.php/v/DRAGONFLY-LAYOUTS_0/album14/album150/EH_LPB_PERF_LAYOUT.gif.html), 830k and 100k resistors are used, forming, once again, an output voltage < 1 volt (about 970 milivolts).

My question is: what's the voltage I'm supposed to get on the transistor's base? 8,2 volts, 4,5-ish volts, or the 0,82 volts that I'm getting.

PS: I should say that I haven't tried the booster yet, and that the voltages I measured were measured without the transistor in the socket (should this make a difference?)

[oskar]

So, lets focus on the LPB-1.

After the first capacitor we have two resistors that form a voltage divider network.
According to the image above, the output voltage created by these two resistors is 8,1v.
If you insert these values (input voltage and the values of the two resistors) into a potential divider calculator (http://www.electronics2000.co.uk/calc/calcdiv.php, for example), you'll get and output voltage (V2) of 818.182 milivolts, which is what I get on my board (0,82 volts).

You are right.

My question is: what's the voltage I'm supposed to get on the transistor's base? 8,2 volts, 4,5-ish volts, or the 0,82 volts that I'm getting.

Precisely what you're getting...  ~0.8-ish...

[DWBH]

Ok, thanks.

[ayayay!]

Half of the 9 volt source voltage, would mean two resistors of the same value. However, in that page, in this schem (http://www.beavisaudio.com/techpages/SchematicToReality/am1.gif), the resistors used are 1M and 100k, which, according to the math, also form an ouput voltage of 818.182 milivolts.

You are right.  I brought this to Dano's attention several months back (not that I'm a genius guys!!!) and he sighed a robust "D'oh."  Wasn't sure if he should leave it alone because 80 gazillion people had already downloaded/viewed it, or should correct it.  It looks like he tried to correct some of it at least, but may have missed a sentence somewhere.  Rest assured though that his stuff is still very, very solid and offers lots of excellent advice. 

[DWBH]

Well, I wasn't critisizing Dano at all. I think he is super cool and makes great stuff, besides providing us with THAT website.
I did know, however, that someone or something wasn't right. Just didn't know what or who exactly.

The pedal's up and running, so case closed