Transistor distortion

[ibanezts808]

okay,  I am still new to this.  but I'm wondering, if I bias my transistor to say like 6v then could that produce an asymmetrical clipping?  or do I need diodes etc to clip?

[Steben]

okay,  I am still new to this.  but I'm wondering, if I bias my transistor to say like 6v then could that produce an asymmetrical clipping?  or do I need diodes etc to clip?

Oh, yes you can make transistors clip assymetrical. What about a fuzz face? That's assymetrical allright!
With 2 germanium trannies you can even get a very very sweet overdrive-like fuzz sound, let's say Led Zep II. Check this out (with sound samples):
http://www.angelfire.com/droid/burvenich/fat_germs.htm
Basic is a series of only two treble boosters.

[davebungo]

okay,  I am still new to this.  but I'm wondering, if I bias my transistor to say like 6v then could that produce an asymmetrical clipping?  or do I need diodes etc to clip?

There's a bit more to it than that.  To get clipping you also need quite a bit of gain in the circuit and to be honest, it all depends on the circuit design.

[niftydog]

a transistor is a glorified diode. Most anytime you're talking about semiconductors, there's diodes involved somewhere, often everywhere!

You seem to be getting the idea of transistor circuits. You are on the trail. I encourage you to check out this series of excellent articles by starting somewhere around here, for more detail (asymmetrical clipping is shown on the very next page!)

[ibanezts808]

okay,  I am still new to this.  but I'm wondering, if I bias my transistor to say like 6v then could that produce an asymmetrical clipping?  or do I need diodes etc to clip?

There's a bit more to it than that.  To get clipping you also need quite a bit of gain in the circuit and to be honest, it all depends on the circuit design.

Okay, I'm taking this one step at a time so I do apologize for my questions.  So when you say gain do you mean to tell me that the signal needs to be amplified before it is able to experience clipping from a transistor?

Niftydog  thank you also for your response.  I'll delve right into the articles now!

[niftydog]

usually, that is the case. The transistor has a finite amount of gain (amplification) that it can provide. When you try to exceed this limit, you can cause clipping.

[ibanezts808]

I'm reading those articles you linked and I'm having trouble with one section.  it is showing transistor coupling as such:



it says q1 is going straight into q2, which I entriely understand.  what is throwing me is the voltage divider i'm very sketchy when it comes to voltage dividers, and I know this is an important fundamental.  if it is too much to cover in this forum could you maybe direct me to a very thorough news tutorial that covers that.  I've read all of the major ones like by r.g. and just about anything, but so far I've had no luck getting an actual tutorial that says, this is a voltage divider, this is why you need it, this is what it does, here's an example of what it does, this is why it looks this way in a schematic.  sorry for all of the questions, and nifty, thank you so much for the help.  it is greatly appreciated.  I think this voltage divider is the prob I've been having with the whole understanding of pedal design.

[aron]

I know it doesn't explain exactly like your schematic but if you check out the DIY FAQ, there's a section on V.R. or search for divider in the FAQ.

[ibanezts808]

thanks aron i'll see if that can shed some light on this topic for me.

[niftydog]

that artical says; The network of R4, R5, and R6 is a voltage divider used to provide the bias and operating voltages for Q1 and Q2. The entire circuit provides two stages of amplification.

Have you read the section on biasing transistors? That's all it's doing.

A voltage divider is quite simple. The quick explanation is simply to say it divides a voltage by the ratio of the resistors.

The longer explanation is this; take two resistors, lets say a 1k and a 4k, connected in series with 10VDC accross them, like this;

Code Select Expand

10VDC
  |
 .-.
 | |
 | |1k
 '-'
  |
  o----o ­A
  |
 .-.
 | |
 | |4k
 '-'
  |
  |
 ===
 GND

the current through both resistors is the same, and it's equal to (ohms law) 10/5k = 0.002A. Now, the voltage accross each resistor is (ohms law again) 0.002x1k = 2VDC and 0.002x4k = 8VDC. Thus you can see that the 10VDC is divided by the ratio of the resistors 1k:4k = 2VDC:8VDC.

So, at point A it is 8VDC with respect to ground.

The circuit shown is not a common way of doing things. Usually stages are capacitively coupled, making the biasing circuitry much simpler to understand. Keep reading, you make sense of it all eventually.

[ibanezts808]

thank you very much for the explanation.  I understand how it works, I think I'm more or less lost onto the reasons as to why.  I don't quite get why 4.5v needs to be fed to the transistor.  and what happens to the other 4.5v?  I am so sorry.  Believe me, I'm scouring google right now as we speak.  I feel quite bad tieing you guys up like this.  but I'll be sure to dedicate my first working pedal to you guys.

[niftydog]

it's biasing the transistor. The tpub articles should explain it fairly well, let me see...

An external voltage applied to a PN junction is call BIAS. If, for example, a battery is used to supply bias to a PN junction and is connected so that its voltage opposes the junction field, it will reduce the junction barrier and, therefore, aid current flow through the junction. This type of bias is known as forward bias, and it causes the junction to offer only minimum resistance to the flow of current.

Since transistors are either PNP or NPN junctions, similar theory applies to transistor amplifiers as it does for basic diodes. The reason behind bias is to set up the DC conditions on the transistor so that an AC signal passes through the transistor in the desired way. For zero distiortion, you want the biasing set such that the signal passes through totally unaffected by the transistor. For distortion you want the biasing set "incorrectly" so that the transistor messes up the signal as it passes through.

I could rattle on, but it seems as though you have enough of an interest to keep reading the articles, and I would higly recommend that to you. It may take some time, but you will understand it much better and you won't be subject to my mistakes or assumptions.

[ibanezts808]

nifty thank you sooo much, you've made it entirely clear.  if I bias the transistor at 3v then the engative side will be more amplified thus causing asymmetrical dist.  if I bias the transistor at 6 it will skew the plus side of the signal also causing asymmetrical dist.  am I correct?  I think I understand now.  If I do, you're getting a shrine in my house. lol.  and if that's too creepy I'll just say thanks.

[niftydog]

well, close... nothing to do with the amount of amplification (gain), that stays more or less consistent once it's set.

the transistor has a max and min voltage that it can produce, dependent on supply volts, collector current etc etc.

Bias it low and the transistor will go into "cutoff" which will clip the valleys of a sine wave.

Bias it high and the transistor will go into "saturation" which will clip the peaks of a sine wave.

Essentially, the transistor cannot produce what you're asking of it but it still trys, resulting in clipping.