Output Swing of An LPB1

[chromesphere]

Hey guys, recently got an oscilloscope for my bday (yay!) and I'm learning a lot with it!

I'm currently tracing through an LPB1 boost with an input square wave of 1khz (all I have at the moment) and I noticed something on the output.

Schematic: http://www.beavisaudio.com/schematics/Electro-Harmonix-LPB-1-Schematic.htm

Everything made sense up until I probed the output side of the output capacitor.

Basically, on the transistor side of the capacitor, the output amplified square wave oscillates between 0v and 8.32v.  But on the other side of the output capacitor, the output swings between +/- 4.24v.   The square wave has been lowered so the zero volts is at its center instead of at its base.  I'm sure this has something to do with DC blocking of the capacitor, but im a little confused because the input signal squarewave doesn't act this way on the input capacitor (it stays at 0 volts at the base of the square wave on both sides of the input cap).

Edit: A couple of screenshots that hopefully illustrate what I'm trying to say.  Note the position of the yellow cursor "2" on the left side of the screenshot.  Red wave is on the input.

Before Cap


After Cap


Your thoughts are appreciated
Paul

[Mike Burgundy]

the output amplified square wave oscillates between 0v and 8.32v.  But on the other side of the output capacitor, the output swings between +/- 4.24v.   The square wave has been lowered so the zero volts is at its center instead of at its base.  I'm sure this has something to do with DC blocking of the capacitor, but im a little confused because the input signal squarewave doesn't act this way on the input capacitor (it stays at 0 volts at the base of the square wave on both sides of the input cap).

I think you've answered your own question.
The DC offset voltages are part of transistor bias. Emitter sits a little bit above ground (because of bias CE current through the emitter resistor), base some 0.6-0.7V above that (because that's how a transistor works) and collector at a certain voltage determined by Rc and CE current. For maximum voltage swing collector voltage is usually chosen close to 1/2 V+.
So, C sits at 4.15V, signal swings *around* that. After the cap, the DC offset is removed, DC level is set at 0V by VR1, and signal swings around 0V.

Same goes for the input, but with a much smaller offset (0.6V). I can't spot where 0V reference is for the red wave in your pictures? If there is an offset voltage at both ends of the cap (check with no signal applied) something's wrong.

[gjcamann]

With a transistor amplifier like the LPB1, the first thing you do is bias it with DC voltage so the transistor will work. That's basically what is going on between C1 and C2. Those 2 capacitors "keep the DC in the pedal".

Now you want you guitar signal to be amplified and send it to your amp, but you still can't let the DC out. So you put your guitar signal in through C1, whereyou guitar signal (AC) gets added to the bias voltage (DC) and then amplified by the transistor and the amplified AC signal goes out the collector. Then it goes through C2 that let's the amplified AC signal out but keeps the DC in.
So on the left side of C2, you're looking at DC+AC and on the right side of C2 you're just looking at AC. DC is about 4.5V, so that's why you "loose" 4.5V on the right side of C2.

Good luck, happy learning!

[chromesphere]

Aha!  Excellent explanations gents! Ill mess around with this some more today!

Also, Mike, the red cursor on the side of the screen is zero volts for th e input signal.  You can see that the squarewave is floating about 400mv above zero volts.  This happens when I connect the signal source (which is the probe compositie on the oscilloscope, is that a bad idea?  only signal source I have at the moment...) to the lpb1, the signal raises 400mv's.  Might be normal, im not sure at this point.

Paul