Basic opamp/transistor theory questions:

[pott]

If I understand well, an opamp or transistor just makes the signal louder, right? (yes, simplistic I know, but you gotta start somewhere eh...)

The opamp is basically a very very powerful transistor? (staying with the uber basics there again).

Basically I understood that if you were to take the diodes off most boost or light OD circuits, you'd get rid of the clipping. But transistors and opamps contain a diode don't they? So technically there always is some clipping?

If you know any intuitive, rather simple websites, it'd help a lot thanks...

[petemoore]

If I understand well, an opamp or transistor just makes the signal louder, right? (yes, simplistic I know, but you gotta start somewhere eh...)
  They can be used to amplify voltage..
  A buffer stage can be used to provide altered impedance.
  The opamp is basically a very very powerful transistor? (staying with the uber basics there again).
  Looking into an opamps insides, the schematics show many transistors are used.
  Basically I understood that if you were to take the diodes off most boost or light OD circuits, you'd get rid of the clipping. But transistors and opamps contain a diode don't they? So technically there always is some clipping?
  It depends on if they're clipping...EZ short expl. when conditions are present which cause clipping, they clip, they can also be made to operate within their 'linear range', in which case lifting diodes would get rid of the clipping. Actually most 'boost' circuits don't use clipping diodes. many 9v Od's will still clip or distort a little even with diodes lifted.
  Transistors and opamps can be made to clip, simply make the output try to exceed the power supply voltage, they'll clip. Misbiasing can make clipping [or harder clipping] on one side of the waveform.             
Opamps are known for a harsher clipping when pushed to do so [see Black Cat OD], transistors can clip, and can have a 'smoother' edge to the signal wave, or not.
If
  you know any intuitive, rather simple websites, it'd help a lot thanks...
  Look in links, FAQ, GEO, AMZ, DIY FAQ, Runoff Groove, Fuzzcentral..too many to mention, I haven't searched lately but there are sites which 'dissect/inspect/explain' transistors and other components.

[GibsonGM]

Well, you're pretty right, Pott...for most of our uses, opamps and transistors are used as amplifiers.  They can also do other things, such as switching, filtering, comparing voltages, and so on, too.  Opamps are collections of multiple transistors and the parts that help them operate, all etched onto a very small chip.  Picture having 10+ transistors amplifying; that's what's in an IC opamp!  That's why it's "powerful".  And easy to use, too! )  Opamps are more voltage-based, and transistors are current devices...a difference that makes more useful sense as you get more into designing with them.

There will always be a little clipping with transistors or opamps, yes.  If you lower their required voltage and/or operate out of their linear range, each will distort - for example, feed a 1V input signal to a BJT circuit designed to operate at 100mV, and there you go!  A look at a characteristic curve for any device such as an opamp or transistor will show this effect.

  Even in their ideal operating range, there is SOME clipping that goes on, primarily in an uneven way that is frequency-based.  That's why there is a freq. specified on the data sheet.  Generally, with good design, that clipping/distortion is minimized and we can't even hear it...be we can see it with an oscilloscope! 

Any basic electronics site should help you to understand more...try Googling "Basic electronics tutorial", that helped me when I was new!

[R.G.]

It's never been finished, but read some of this:

http://geofex.com/Article_Folders/How_It_Works/hiw.htm

[MetalGod]

It's never been finished, but read some of this:

http://geofex.com/Article_Folders/How_It_Works/hiw.htm

hey RG, I got the end of that article and the Noisless Biasing sounds interesting.  I've been playing round with single BJT booster circuits with different biasing methods to observe the tonal differences and to see how stable they all are.  Using the simple self-bias method on a common emitter setup gives a fair amount of noise if the transistor is biased with the collector around 1/2 Vs (this is where it 'sounds' best to my ears).  Reducing the collector voltage to around 3v gets rid of the noise, but also looses some edge off the tone (softer treble response).

Anyway, I'd like to try your Noiseless Biasing setup but a few noob questions... (1) what is the purpose of the cap bypassing Rb2?, (2) why the series resistor between the base and voltage divider? - I've seen this combination bias setup before, but not with the cap and series resistor so I'd be interested to know the reasoning behind it.

As always, many thanks.

[R.G.]

(1) what is the purpose of the cap bypassing Rb2?,

You have to step back to a bit of info about components and noise first.

Resistors and active devices make noise. Reactances - inductance and capacitance - do not make noise.
Resistors have thermal noise proportional to their absolute temperature (degrees Kelvin), current noise proportional to the current through them, and excess noise depending on the material they're made from.

R1 and R2 have their inescapable thermal noise, but the thermal noise at their junction is shunted to ground by that cap bypassing R2 to ground. That's what the cap is for. It effectively makes R1 and R2 noiseless, for thermal, current, and excess noise. By the way, the noise made by R1 and R2 is RMS-additive, so you get the noise of two resistors in parallel.

(2) why the series resistor between the base and voltage divider?

That's what lets you use the cap to attenuate the noise. It carries ONLY the base bias current, which is much smaller than the current through R1 and R2.

It has its own thermal noise, but the current noise is small because the current is small. The cap "grounds" the R1/R2 end of this resistor for AC purposes, so they can't contribute noise while they set the bias voltage. The series resistor carries only the base current, so it can be high valued, and in fact can be much higher than the parallel value of R1 and R2 for equal bias stability, so bigger input impedance is possible if you do the rest of the circuit right. And you get the noise of only one resistor, not two.

If you pick a resistor with low excess noise - like metal film - for this one, you approach only the thermal noise of this resistor for the biasing resistor noise in the stage, little current noise and no additional resistors to contribute. It's about as low as you can go.

[MetalGod]

Thanks RG, that all makes alot of sense (which is a miracle in itself for my hard-head, heh).  Here's the circuit I've been playing with this afternoon (I just started with the Rangemaster but with a 2N3906). 



I couldn't get it to bias with the collector trimpot setup like the RM where the output is taken from the wiper, so I just set the trim up like above to get the thing biased. 

I had a play round with the voltage divider and emitter resistor values and could get different tones by doing so and resetting the bias with the collector trimpot.  What I'm curious about about is how to 'design' this stage rather than plug in resistors till it works.  What approach 'should' I be using to get the transistor biased?

[R.G.]

OK. Good question.

To design a bipolar transistor stage like this one.

1. Decide how you want the total power supply voltage apportioned between the
a. collector resistor
b. collector-to-emitter across the transistor
e. emitter resistor
in the absence of signal.

In this case we'll choose 7V on the collector, 0.56V on the emitter for a 9.0V power supply. You are free to choose these anywhere within the power supply voltage range, subject only to the power supply being able to supply enough current if you make silly choices and the transistor not conducting any signal if you make Vce too small or too large. If Vce is a significant fraction of the power supply, it will work fine. Most apps use Vce of 1/3 to 2/3 of the power supply.

2. Decide how much current you want to flow in the transistor. This is usually 10uA to 1ma. This decision is somewhat arbitrary. You can pick this to be whatever you like, bearing in mind that if you pick currents too large or too small, you won't be able to get components to make your picks come true. If you for instance pick a current of 1A (1000ma), you can still do this, but the transistor choice will be limited to power transistors. If you pick 1uA, you may not be able to get resistors big enough to do the biasing right.

Let's pick 100uA and see where it leads us.

3. with the current picked, we know the collector and emitter resistors, since we know the voltage across them and we picked the current through them. The collector current is what we picked. The emitter current is equal to the collector current plus the base current. In most modern transistors, the gain is 100 or more, so ignoring the base current introduces only 1% error in the emitter resistor.

So Rcollector = (Vsupply-Vcollector)/Icollector = (9V-7V)/100uA = 2V/0.1ma = 20k
and Remitter = 0.56V/0.1ma = 5.6k

Cool. We have reasonable values. Next we have to make the base voltage and current be right to make these picks come true.

4. Determine the base voltage. For silicon, the base-emitter voltage will be in the range of 0.45 to 0.7V for small signal devices. Generally, it's 0.5V to 0.6V. Let's guess at 0.5V. The emitter voltage is 0.56V, the base-emitter adds 0.5V, so the base voltage has to be 1.06V plus or minus about 0.05V because we guessed a base-emitter voltage instead of measuring our transistor. This turns out not to be too big a problem.

5. We have to make the base voltage be 1.06V, and there is some current going into the base. How much? We can only guess because hfe is ill-defined. If we guess again that it's about 100 or over, we can further surmise that the base current will be in the range of 1uA.  Using the rule of thumb that a resistor divider needs to conduct much more than ten times the current you pull out of the middle of the divider, we decide that the R1/R2 base divider must conduct much more than 1uA. Let's pick 25uA.

We then have 9V/(R1+R2) = 25uA. We also want 9V * (R2/(R1+R2)) to be 1.06V.

We solve the left equation for (R1+R2) = 9V/25uA = 360k and sub this into the right equation to get R2=(1.06/9)*360k = 42.4k.

So R1 =360k, R2 = 42.4k, Re=5.6k and Rc = 20k.

I made a few major assumptions in doing that.
First: hfe>100 . This is important to choosing the emitter resistor and to the base current calculations. It is almost always true, and if hfe was only 50 the error introduced would be pretty neglegible too.
Second: The base current is neglegible (i.e. less than 1/10th) compared to the resistor bias string current so it doesn't mess up the string too much.

Does that help, and where is more info needed?

[MetalGod]

Thanks for that RG, thats really MUCH more than I was hoping for - you're a real gent.

I've read through that and understand it for the most part.  I'll have to play around with that in a practical sense and I'll come back with specific questions (and there will be questions, lol)

[PerroGrande]

R.G. :

What a great post!  Well written, explained nicely, absolutely accurate. 

The only thing I could possibly think of to add would be a brief discussion of why you selected the voltages you did for the design and the various implications therein.