What does that bit there do? Learning to understand circuits.

[philbinator1]

^^ agree ^^ thanks for bumping it mate!  i'm really exited to read this, i think Dano pretty much has the title of "Best noob info provider" although there are plenty of other contenders out there.  anyway just wanted to say thanks (no need to reply to this), can't wait to devour the info after band practice   

[rbrower]

Bump this!  Nice thread- Much appreciated!

I would like to propose that someone who understands tube amps might help describe the basic functions of components for a fender champ, for instance. 
I don't know much myself, but I have been fascinated to read the tube driven aesthetic and various adjectives used to describe soft clipping, asymmetrical distortion, compression, sustain, carbon comp dirt, coupling cap magic and etc, and would love to hear how the stompbox community thinks some of these can be emulated with analogue  electronics.

Perhaps I could edit some of these into a sticky...

Thanks in advance- Roger

[rosscocean]

Hi Dano

I just want to say that the stuff on your site is great. It's helped me a lot.

http://www.beavisaudio.com/

Cheers
Ross

[scnative]

OK.  Nice thread.  I think this is a good, relevant question.

I wanted to relate the information in this thread to the schematic for the beginner project.

Assuming the beginner project is based on a the common emitter amplifier design (if it is not, then please forgive my stupidity) given by:



Which resistors in the beginner project schematic:



correspond to R1, R2, RL and RE in the common emitter amplifier diagram.

I'm trying to understand what I am building before I build it.

Thanks in advance for any insight.

SJ

[bonaventura]

im no expert,but heres my take.

RL is the 10k at the collector

RE is the 5k at the emitter

these are two different designs in the way they are biased.

so no R1 R2 in the circuit at the bottom.

the circuit on the top is biased by R1 R2. while the NPN boost by Gus is biased by the network tied to the base which suppose to provide better/higher input impedance (?).

i think i read that here in the forum but till now i still cant really understand it. somebody can chime in?

[scnative]

Thanks Bonaventura.  That's what I figures as well. 

Really what I want to know more about how this thing is biased with those 3 resistors. It really is sort of a different animal than the first few schematics discussed here.

I'll re-read this this thread to see if I can figure it out but if anyone else knows or has any insight, please don't hesitate to explain.

Thanks,

SJ

[flicker]

Hi. I am hoping that someone could provide a schematic for Hemmo's "Standard Fuzz". I am trying to follow along with Dano's explanation on page one of this thread about how each part works and what is does. Any help would be much appreciated as I am trying hard to learn how a two transistor pedal works.

Thanks

[DrAlx]

these are two different designs in the way they are biased.

so no R1 R2 in the circuit at the bottom.

Not so.  This is a game of spot the difference.
You can see that the boost circuit has an extra 22 uF and an extra 10k resistor at the transistors base.
For bias levels (i.e. DC levels) you should mentally remove all capacitors from the circuit (i.e treat them as broken wires) since capacitors block DC.
So that leaves us with the extra 10k resistor at the base.  If you lowered the value of the 10k resistor all the way to zero (i.e. turned it into a wire) then you would have the same biasing scheme as the simpler circuit.

Therefore we have R1= 100k and R2 = 47k.
So the voltage at the base is  ( 47 /147 ) * 9V  = 2.87 V.

The difference is that the simple scheme connects that 2.87V directly to the base with a wire.
Gus' circuit connects it to the base through a "higher resistance wire" i.e. a 10k resistor.  The voltage at the base will be pretty much the same.

So why use the 10k instead of a regular wire.  My guess is that the 10k was chosen to give a particular low pass filtering at the input.  Wired up this way the audio signal at the input sees an RC filter where the C is 0.1 uF and the R has a value of 10k + (100k || 47k) =  10k + 1/(1/100k + 1/47k) = 10k + 32k = 42k.
So the RC product for the filter is 0.1uF * 42k,

There are other ways of wiring things up to get that RC product at the input (e.g. replace the 10k resistor with a wire and use higher values of R1 and R2 instead).  So maybe the 10k does something else that I can't think of, but I am pretty sure it does not effect biasing.

[GibsonGM]

I ran this thru LT Spice, just for curiosity...With the 10K in place, you get a shelving filter...actually, HP.

Without it, you get a LP response.

You're correct, the biasing stays essentially the same.