Undestanding the flow of AC & DC in Common Emitter Amp

[fuzzy645]

My questions here are not about any particular pedal build.  My goal is to attemt to  try to understand the "flow of traffic" in a basic NPN Common emitter amp circuit with regards to both AC and DC.  I have a generic image below that can be used for reference. 

To oversimplify for a moment, lets assume the pedal is "fired up" but nobody is playing guitar at the moment, therefore no AC signal is in the mix (just yet).

Background Points
----------------------
1.  I undestand that in essence a transistor is a pair of diodes.  I alse get that in a NPN transistor, base current flows from positive to negative, through the base-emitter diode.

2.  I also undestand (I think), that when a small current flow into the base, this "opens up the proverbial flood gates" for a larger amount of collector current to flow and then pass down through the emitter.  I believe we can calculate the collector current as being the product of base current and the HFE of the transistor.

Questions
-----------
Once we bring AC into the mix I'm totally lost.  So now we plug in our guitar and start Jamming on Voodoo Chile, bringing some AC into the equation.

     a). Does AC flow through the entire circuit. and in the same direction as DC, or is it JUST into the base and out the of the collector, and thats it? I know the caps block DC, so I understand that DC won't make it past our output cap. 

    b). When does the AC guitar signal actually get "amplified"

Any enlightenment is appreciated!!

[defaced]

If I understood it more firmly, I'd type up an explanation.  However, your figure reminded me of the style of figures used on this site (not sure if it's pulled from there, don't matter really).  AC amplification is illustrated in the figure toward the bottom of the page titled "Output Characteristic Curves"  http://www.electronics-tutorials.ws/amplifier/amp_2.html  An image just like this one for tubes are what made load lines and such make sense to me, maybe this will help you too.  

Edit: typo

[R.G.]

Background Points
----------------------
1.  I undestand that in essence a transistor is a pair of diodes.  I alse get that in a NPN transistor, base current flows from positive to negative, through the base-emitter diode.

Yes. The collector-base diode is reverse biased, the base-emitter is forward biased.

2.  I also undestand (I think), that when a small current flow into the base, this "opens up the proverbial flood gates" for a larger amount of collector current to flow and then pass down through the emitter.  I believe we can calculate the collector current as being the product of base current and the HFE of the transistor.

Yes. Current into the base "poisons" the ability of the collector-base to hold off current flow. The amount of "weakness" introduced is proportional to the amount of current flow into the base. As an aside, the "poisoning" is not perfectly linear, and varies with temperature, current flow and phase of the moon, so the Hfe varies a bit, about 2-3 to 1 over current changes for any given transistor.

     a). Does AC flow through the entire circuit. and in the same direction as DC, or is it JUST into the base and out the of the collector, and thats it? I know the caps block DC, so I understand that DC won't make it past our output cap. 

    b). When does the AC guitar signal actually get "amplified"

It's all DC to the transistor.

The DC current flows are modulated on an instant-by-instant basis by the signal. The base current is fixed at some point by the biasing arrangement when there is no signal. This trickles in a fixed amount of DC current into the base. This, in turn, lets a fixed amount of collector current flow. When you put in a signal through a capacitor, the AC signal coming in adds to/subtracts from the DC bias current. So on an instant-by-instant basis, the base current changes upwards/downwards, and so in turn does the collector current. This state of affairs lasts until the incoming signal gets so big that it either (a) runs the base current down to zero on downward-going alternations, or (b) runs the collector current up so high that there is no more available. When (a) happens, the base and collector are both cut off, and nothing flows. When (b) happens, the base current may increase, but the collector is already doing all it can. Ignore this for the moment until you understand where the AC and DC current flows.

On the base, AC current is fed in and mixed with the DC bias. So it adds to (both positive and negative) the DC condition and wobbles the base up and down. This makes the collector current change, amplified by the gain of the transistor.

On the collector, the collector current is changing. This makes the collector *voltage* change if there happens to be a resistor between it and the power supply voltage. This is just a restatement of Ohm's law, that V = I*R, where the "V" is the voltage that results in collector resistor R because collector current I flows.

The collector current changes up and down with the base current changes. This makes the collector's DC voltage change. This voltage change is "tapped" by a capacitor which blocks the collector's DC voltage, but lets the wobbling AC part out through the cap.

Capacitors block DC, but let AC through. The input cap lets the AC input signal in and mixes it with the DC bias current. The output capacitor blocks the collector DC voltage/current, but lets the AC current out.

That may have made things muddier. Ask where it raises more questions.

[fuzzy645]

That may have made things muddier. Ask where it raises more questions.

Thank you RG.  You have confirmed my (simple) understanding of the DC part as being reasonably accurate.

You have also confirmed there is a lot more to understanding the AC side of things as I suspected, which explains why i have felt lost.

With regards to your explanation, you have enlightened me on quite a few points, however your answer has inspired some more questions. First, let me restate what I now understand (thanks to your explanation):

• AC Current feeds into the base and is mixed in with the DC causing what would have been "fixed" (if dealing with DC only) to now modulate.  To the transistor, this is all considered to be DC.
• This modulating current, causes corresponding modulations to collector current.
• All these current changes lead to corresponding voltage changes due to Ohm's law V=I/R (as I goes up, so does V)

Follow up questions:

1. Is the AC signal "amplified" on the path directly above/before the collector, which is why we send the output at that point?  For example, I have noticed consistently on other "buffer" type circuits, the output appears after the emitter. My assumption is the signal is not "amplified" after the emitter, but rather only before the collector.  In contrast, on these amplifier type circuits I have noticed consistently the output is always above (before) the collector?  This is what is causing me to make the assumption the gain occurs only above/before the collectors and not after/below the emitter?  I'm sure I'm missing something here....  

2. How does one "measure" the amplification as a proportion to the current gain?  So for example, if current gain of the transistor is 100, how does one calculate/measure/quantify the amplification gain (as related to question 1 above)?  What I mean here is from an "audio" standpoint (such as DB etc...)

Also, I have an additional follow up quesiton  regarding one of your statements in quotes.

This state of affairs lasts until the incoming signal gets so big that it either (a) runs the base current down to zero on downward-going alternations, or (b) runs the collector current up so high that there is no more available. When (a) happens, the base and collector are both cut off, and nothing flows. When (b) happens, the base current may increase, but the collector is already doing all it can.

Is the condition you are describing above a form of "overload" of the transistor?  Would that cause some kind of undesirable clipping?

Thanks again![/list]

[R.G.]

1. Is the AC signal "amplified" on the path directly above/before the collector, which is why we send the output at that point?  For example, I have noticed consistently on other "buffer" type circuits, the output appears after the emitter. My assumption is the signal is not "amplified" after the emitter, but rather only before the collector.  In contrast, on these amplifier type circuits I have noticed consistently the output is always above (before) the collector?  This is what is causing me to make the assumption the gain occurs only above/before the collectors and not after/below the emitter?

You're overthinking it. Gain occurs in and around the base. It's intimately tied to the base-emitter voltage and current.

The base current is controlled by the base-to-emitter voltage. It's a diode, so the base-emitter diode conducts almost nothing up to some threshold voltage, then above the threshold conducts exponentially more current for each micro-increase in voltage. There are two things which limit the base current. One is that each signal source has some built-in resistance, and so when the base begins conducting at its threshold, it can't just suddenly go to infinity; the source current softens and limits that.

Also, it's the base-to-EMITTER voltage that causes current to flow in the base. If there is any resistance in the emitter-to-ground, the base current flows through it, but also the many-times-more collector flows with it. That additional current from the collector flowing in any emitter resistance causes the emitter voltage to rise, which reduces the base-to-emitter voltage and therefore reduces the current which flows in the base. It's a simple negative feedback mechanism.

Even if there is no obvious resistor between the emitter and ground, there is a small internal (and, predictably, nonlinear) resistance that acts.

So the way this works is that one sticks a little bit more current into the base by raising the base voltage temporarily with a signal through a capacitor. That makes more base current flow, and that makes maybe 100X or more collector current flow. The sum of base current plus collector current goes out the emitter.

If there is an emitter resistor, the huge gain in current from base to emitter (that is Hfe+1) raises the emitter current, which raises the emitter voltage on the emitter resistor, which takes reduces the base-emitter voltage which produces it. This simple feedback mechanism makes the VOLTAGE gain from base to emitter be just slightly less than unity.  The emitter follows the base up and down in VOLTAGE. Just for understanding's sake, don't go off on "but what if the emitter's grounded" on me yet, OK? 

A biased base plus an emitter resistor gives an emitter FOLLOWER. This is because the emitter FOLLOWS the base on very nearly a 1:1 basis in terms of voltage. The high current gain from injecting current into the base and that controlling an Hfe-times bigger collector current is what makes this simple unity gain follower work.

It does not matter what the collector resistor is for this to work. As long as there is enough current and voltage at the collector to keep it from limiting some other way, the collector just supplies the current which the base current controls, and the emitter uses the current gain to follow the base up and down with a gain of 0.98, or 0.99 or ... so the emitter follows the base, gain of one in terms of voltage. The current is much higher, but voltage is what we normally look at.

You can take a signal output at the emitter, and that's what an "emitter follower" is. You can do this for any transistor circuit which has an emitter resistor.

Meanwhile back at the ranch... er, back at the collector 
The collector-base is a reverse biased diode that lets this current through. It's current comes from some power supply. If the resistance to that power supply is zero, then a voltage appears across it according to ohm's law, V = I*R, but R=0, so the voltage is zero. The collector always stays at the power supply. Some emitter followers do this: tie the collector to the power supply.

But you can insert a resistor there, between collector and power supply. Now the current does exactly what it did before, be the right amount to let the emitter VOLTAGE follow the base VOLTAGE. But the voltage difference between the collector now has to follow that V=I*R with R being more than zero.

With a big current gain - say, about 100 - the base current in the emitter is 1% of the current that flows from the collector into the emitter. Let's ignore it. So we'll say the SAME current flows in collector and emitter. If the emitter has a resistance between emitter and ground of Re, and the collector has a resistance between it and power supply of Rc, and the same current flows in both of them (within what little error we've chosen to ignore) then the voltage which appears across the resistors has to be in the ratio of the resistors; that is, Vc/Ve = Rc/Re. (Notice, I've ignored phasing and DC voltages; this is the CHANGE in voltage on an AC basis).

That's voltage gain. The gain in a biased NPN transistor from base to emitter resistor is so close to unity that you can just say it's one. The gain from base to collector is Rc/Re. You can take out the output at either emitter or collector OR BOTH AT THE SAME TIME. Transistor phase splitters with Rc = Re do this exact thing.

2. How does one "measure" the amplification as a proportion to the current gain?  So for example, if current gain of the transistor is 100, how does one calculate/measure/quantify the amplification gain (as related to question 1 above)? 

One does not do that. The current gain has meaning for voltage gain only in that it makes that approximation that the base current is negligible come true. There *is* a picky, detailed set of equations for how close to unity gain the emitter gets per amount of current gain, but it's bad design practice to count on it. Good design practice says "get a transistor with ENOUGH current gain to make the approximations come true."

Is the condition you are describing above a form of "overload" of the transistor?

In gross terms, yes. You're driving it into some form of distortion.

Would that cause some kind of undesirable clipping?

It causes some kind of clipping, yes. Whether that's *undesirable* or not is a tough question. A lot of circuits do just this to GET the distortion.

[PRR]

> Once we bring AC into the mix I'm totally lost.

So forget the transistor!!

Forget electricity.

We can find "DC" and "AC" concepts in other places.

Take R.G.'s words: "made things muddier".

I have a stream across my land. Water flows steady.

I have a vibration. I want it bigger.

I can use the small vibration to control a valve in the stream.



The output has _both_ a steady flow AND a vibration. DC and AC.

An observer may ignore the steady flow and enjoy the vibration.

The water vibration can be much stronger than the string vibration because it leverages the power of the steady stream.

[fuzzy645]

Thanks RG and PRR, this is starting to make alot more sense to me now.  Particularly your exlanation of voltage gain, as well as your clarifcation about the emitter follower....that really helps.

You mentioned that the gain from base to collector is Rc/Re.  So if the collector resistor is (lets say) 50k, and the the emitter resistor is (lets say) 2K, that would be a voltage gain of 50/2 = 25, correct?  Is that true regardless of the HFE of the transtor?   Also, there must be some kind of "saturation point" of gain would get out of hand, no?  I mean if you had Rc= 500K and Re = 1K we couldn't possibly have a gain of 500, I would assume at some point with a 9V batter we "max out" (so to speak).

Thanks!

[R.G.]

You mentioned that the gain from base to collector is Rc/Re.  So if the collector resistor is (lets say) 50k, and the the emitter resistor is (lets say) 2K, that would be a voltage gain of 50/2 = 25, correct?  Is that true regardless of the HFE of the transtor?   Also, there must be some kind of "saturation point" of gain would get out of hand, no?  I mean if you had Rc= 500K and Re = 1K we couldn't possibly have a gain of 500, I would assume at some point with a 9V batter we "max out" (so to speak).

Now you're onto it. Good thinking, all counts. You will get away with gains of 10 to 25-30 maybe from this "Rc/Re" thinking. As you try to get larger, you do get more voltage gain, but less than you'd expect from how much the emitter resistor goes down. In particular, when Re=0, the gain is not infinite, as you guessed. It's limited by internal factors. You get less than expected over gains of 20-30, maybe 20-40, and a lot less at either very high or very low currents where hfe falls off because of internal device physics.

The hfe is the raw engine of gain which is hidden underneath the voltage gain. That business about "if hfe is high enough, the emitter follows the base" is the biggie. In general, the hfe has to be high enough to make the base current negligible compared to the collector and emitter currents, and also high compared to the gain you're trying to make it give you. In general, gains of under 50 mean that you can't blithely make the assumptions, and the actual gains at emitter and collector will be less than expected.

To illustrate one very simplistic way to look at it, if the hfe is only 10, then you get ten units of collector current for each one unit of base current. And since the base current adds to collector current to make the emitter current, then the emitter moves around by a voltage of eleven units of current times the emitter resistor, and the collector only moves around by ten units of current times the collector resistor. So the voltage gain can at best be (10/11)*Rc/Re, not Rc/Re. The actual equations get more complicated, and there is that internal emitter resistance to worry about too.

In addition, the AC load on the collector is more than just the collector resistor. The collector actually drives both the collector resistor and any load outside the collector capacitor in series. So that reduces the effective gain at the collector.

The "maxing out" is not hard and fast, but a soggy diminishing-returns kind of thing.

Note that it is *possible* to get gains as high as a couple of thousand out of a single NPN, but you have to do really fancy other stuff to do it, like current source loading, buffering the output, and emitter biasing, which can be much more complex than the NPN you're trying to get gain out of.

[fuzzy645]

To illustrate one very simplistic way to look at it, if the hfe is only 10, then you get ten units of collector current for each one unit of base current. And since the base current adds to collector current to make the emitter current, then the emitter moves around by a voltage of eleven units of current times the emitter resistor, and the collector only moves around by ten units of current times the collector resistor. So the voltage gain can at best be (10/11)*Rc/Re, not Rc/Re. The actual equations get more complicated, and there is that internal emitter resistance to worry about too.

Thanks.  All good stuff.

Now, a question about this paragraph quoted above. It seems to me as if that 10/11 fraction in your example (which is roughly .9) will start to approach 1 as HFe increases, but never can be greater than 1.   Restating this 10/11 fraction is collector current (10) over the sum of collector(10) + base corrent (1).  I might be misinterpreting this, but if we convert this example to an example wth high HFe of 500 (lets say), then the fraction becomes 500/501  (or .998), that is collector current of 500 divided by the sum of collector current (500) and base current (1).  It seems as if this type of fraction will approach the number 1, but never quite get there.  So in the first example with HFe of just 10, this view of voltage gain can be summarized by .9 * Rc/Re, but with a higher HFe of 500 (as  in the 2nd example) the equation is 1 * Rc/Re.

If what I'm summarizing is true, then quadrupling HFe in the 1st example from 10 to 40 would be reasonably significant, but quadrupling HFe in the 2nd example from 500 to 2000 would do absolutely nothing (point of diminishing returns).

I'm guessing the moral of the story is then if HFe is high, then the Rc/Re gain formula is as true as it can be (within reason).  If however HFe is low (as you have qouted 10 in the example), then Rc/Re is not reliable in giving us a picture of the gain. 

Am I barking up the wrong tree?

BTW, I totally appreciate your disclaimer that this is a simplistic view, and the actual equation(s) is more complicated.

[R.G.]

You are correct.The gain of an emitter follower will approach but never quite equal 1.00000 as gain goes up.

You are correct in that the step from 10 to 50 is much, much more important than the step from 100 to 500 in hfe. You also have a grasp of the right reasons.

Good thinking.

[PRR]

> true regardless of the HFE of the transtor?

Yes.

A low-hFE part has a lower input resistance. That does not directly affect Voltage Gain IF your source has even lower resistance. But in many cases the low input resistance of a transistor (and associated bias parts) dominates the gain figurings.

> gain would get out of hand, no?

Of course. If you have a 1V signal and 10V battery, gain over 10 will just distort.

In many audio systems, distortion must be avoided. ("Here" it may be welcomed.)

> if you had Rc= 500K and Re = 1K we couldn't possibly have a gain of 500

You would... except the _load_ is in parallel with the collector resistor and typical audio loads are under 500K.

OTOH, with a fully bypassed emitter resistor, "Re" is the internal intrinsic impedance of the emitter junction. Shockley's Law: Re ~= 30 ohms / mA.

Say you run the transistor at 1mA. Say you have 10V supply. Rc could be about 5V/1mA or 5K. Typical audio loads tend to be higher, neglect that for now. Voltage Gain will be something less than 5K/30 or 5000/30 or 166.

Peak undistorted output voltage may be 4V. Peak undistorted input voltage is therefor 4V/166 or 0.025V or 25mV.

Guitar levels are "usually" over 20mV. Therefore a max-gain transistor amp will "usually" distort guitar.

Further: input resistance (neglecting bias issues) is that 30 ohms times hFE. Typical hFE may be 200. 30*200= 6,000 ohms input. Guitar output impedance is typically higher. The naked max-gain transistor loads-down the guitar.

Further: with just naked emitter resistance, the distortion will rise above 20% THD before clipping starts.

Mostly we do NOT want maximum gain from a transistor. The input overload is way lower than most reasonable sources, the THD gets high fast. Further the high gain also amplifies bias, so a slight mis-bias slams the amp to one extreme or another. Most transistor design is about trading-off raw gain for stability. A significant external emitter resistor is a key tool.

---
> with HFe of just 10, this view of voltage gain can be summarized by .9 * Rc/Re, but with a higher HFe of 500 (as  in the 2nd example) the equation is 1 * Rc/Re.

As R.G. hints, this is over-simplified. It makes a point: high hFE is not "good", you want "enough" hFE to simplify design and variation problems.

The voltage gain of a nicely-loaded emiter follower is the ratio of the external resitances to the sum (the external reistances + intrinsic emitter resistance (30 ohms at 1mA). And unless the emitter resistor drop is much less than 1V, this comes SO close to 0.999 that we can write "unity".

What hFE buys in the emitter follower is input resistance. Take 1K emitter resistor. If hFE is 20, input Z is 20K. If hFE=500, input Z is 500K. If source is a 50-ohm signal generator, this makes no difference. If source is a 100K guitar it makes a real difference.

[fuzzy645]

On a related topic, I was interested in a statement made on "self bias" on this website ---> http://www.tpub.com/neets/book7/25d.htm

Just as I thought I was getting a handle on things, this "self bias" arrangement seems most confusing. Here is a statement from the website:

"....method of biasing is obtained by inserting the bias resistor directly between the base and collector, as shown in figure 2-13. By tying the collector to the base in this manner, feedback voltage can be fed from the collector to the base to develop forward bias. This arrangement is called SELF-BIAS. Now, if an increase of temperature causes an increase in collector current, the collector voltage (VC) will fall because of the increase of voltage produced across the load resistor (RL). This drop in VC will be fed back to the base and will result in a decrease in the base current. The decrease in base current will oppose the original increase in collector current and tend to stabilize it. The exact opposite effect is produced when the collector current decreases."

They also mention:

"....the collector and base signals are 180 degrees out of phase (opposite in polarity) and the part of the collector signal that is fed back to the base cancels some of the input signal. This process of returning a part of the output back to its input is known as NEGATIVE FEEDBACK. "


Some questions:
1.  Regarding the 180 degree out of phase issue, I assume they are talking about AC only, correct?  

2. When they say "negative voltage can be fed from collector to base" - this confuses me because is it my understanding in NPN,  the base collector diode is reverse biased, so NOTHING can travel from collector to base.  I assume then they are talking about BEFORE the collector we feed this voltage into the base. I guess it is just the semantics that bother me and I'm being a bit anal.

3.  They are referring to sending PART of the output back into the input.  How then does one control how much/what percentage of the input gets fed back into the output?  Are we talking 100%, 50%, 25%?  I would think that would have a big impact on the tone produced by the amplfier.

4.  When/how does the original signal become 180 degrees out of phase.  I assume the original signal first goes through the base/emitter diode? Is it when it emerges out of the base emitter diode that it becomes out of phase?

5.  Finally, I would assume this out of phase thing would be bad.  For example, I know on a guitar when mixing pickups from different manufacturers sometimes a "nasty" thing happens and you can find 2 pickups that are "out of phase" with one another.  This is a bad thing, in that the 2 pickups sound very thin an nasal when combined.  I am therefore assuming this business about mixing the original signal with an out of phase signal might cause problems, unless controlled in some way.

Thanks!

[R.G.]

1.  Regarding the 180 degree out of phase issue, I assume they are talking about AC only, correct?  

Yes. Any movement of voltage on the base causes a movement of voltage on the collector in the opposite direction (positive - negative).

2

. When they say "negative voltage can be fed from collector to base" - this confuses me because is it my understanding in NPN,  the base collector diode is reverse biased, so NOTHING can travel from collector to base.  I assume then they are talking about BEFORE the collector we feed this voltage into the base. I guess it is just the semantics that bother me and I'm being a bit anal.

They said that at least poorly. A much more accurate way to say it would be "a current can be fed to the base to bias it; the direction of this current is such that it biases the base with forward current, but the AC variations of the collector will be inverted from the AC variations of the base, and so will introduce negative AC signal feedback."

3.  They are referring to sending PART of the output back into the input.  How then does one control how much/what percentage of the input gets fed back into the output?  Are we talking 100%, 50%, 25%?  I would think that would have a big impact on the tone produced by the amplfier.

The value of that collector-base feedback resistor sets the DC feedback portion; it does this by converting the voltage between the collector and base to a current via Ohm's law. The emitter in this circuit is tied to ground, so it only has the small internal resistance we talked about between the real emitter and ground. We'll ignore that for the moment and call it grounded. The base therefore has to have a voltage of one diode drop above ground. This does vary a bit, but is almost fixed. It's 0.5 to 0.7 for a silicon transistor in most conditions. So the base effectively does not move compared to the collector. The current let into the base by the collector-base resistor is then (Vcollector-Vbase) divided by the resistor.

As the collector voltage rises, this gets bigger and puts more current into the base, which acts through the hfe to pull the collector back down. As the collector falls, it puts less current into the base, and lets the collector rise. With proper values of resistances for the collector resistor and the collector base resistors, this eventually balances at some point. The DC conditions depend on the hfe, the "gain" of the collector resistor in converting collector current into voltage change, and the "gain" of the collector-base resistor in converting collector voltage into base current change.

I have a textbook somewhere that describes the process of designing this bias arrangement, but I don't use the arrangement all that often, so I'm less familiar with the ins and outs of it than I am some of the better-stabilized later forms of biasing.

The AC gain gets tricky. There's a part that helps set AC gain that's not shown in your diagram. That's the source impedance. All real signal sources act like a voltage source hidden behind a resistance (impedance actually, but think resistance right now). The feedback from the collector to base is better thought of as a current, as the base voltage change is very small. The source voltage acts through a source resistance to feed current to the base, too. The math and explanation gets long, but the voltage gain which results from the source current in and collector feedback current makes the voltage gain approach the ratio of the collector-base resistor divided by the source impedance. Again, it's the internal current gain that makes this come true, so the higher the current gain, the better this approaches Rcb/Rsource. And loading of the collector by an AC load through a capacitor "steals" some of the current that would go into the collector resistor otherwise, so it also tends to lower the internal gain that makes the gain approach Rcb/Rsource. It's trickier to design this circuit well than it is the four-resistor stabilized-bias circuit, but it does have its advantages.

To finally answer your question, you feed back just enough DC to make the collector sit at the voltage you want it to, also computing the effect of the collector resistor and feedback resistor in this, and accepting the changes that variations of hfe force on you. Then you hope that you get the AC gain you need from the signal source you have. Often it requires two or more other resistors to get things set up. One of these is a series resistor into the base to add more source resistance to lower gain. Another is a resistor from base to ground to "eat" some of the feedback current to let the bias be different than a single feedback resistor would do.

As an aside, this is the circuit used in the Fuzz Face. It has the embellishment that it uses a second transistor as a follower to buffer the first collector and add more gain, but the input of the Fuzz Face is as shown in the schemo. The gain depends on the source impedance. This is also why Fuzz Faces sound harsh when buffered. The low impedance of the buffer drives the base HARD.

Fuzz Face aside as a special case, the tone of an amplifier set up like this may not vary all that much until you hit clipping, because it inherently uses feedback to linearlize its voltage gain. Negative feedback is the Great Linearizer.

4.  When/how does the original signal become 180 degrees out of phase.  I assume the original signal first goes through the base/emitter diode? Is it when it emerges out of the base emitter diode that it becomes out of phase?

Trying to push more voltage onto the base means more current flows into the base. More current into the base causes more current to flow in the collector. More current in the collector causes a bigger voltage drop in the collector resistor RL, which is tied to the power supply. So more voltage on the base causes the collector voltage to drop. The voltage on the collector then moves in the opposite direction to the voltage on the base. The signal is inverted, 180 degrees out of phase.

5.  Finally, I would assume this out of phase thing would be bad.  For example, I know on a guitar when mixing pickups from different manufacturers sometimes a "nasty" thing happens and you can find 2 pickups that are "out of phase" with one another.  This is a bad thing, in that the 2 pickups sound very thin an nasal when combined.  I am therefore assuming this business about mixing the original signal with an out of phase signal might cause problems, unless controlled in some way.

It's not bad in this instance. It's what lets you have some predictability of gain and other performance measures. The reputation for "out of phase" with pickups and speakers is a consequence of near-cancellation of the base sound. For now, set that aside as a separate case. That case has a whole other series of lectures about what's happening and why it's good or bad, depending on what you want to do. The out of phase and negative feedback is very good for biasing transistors.

[PRR]

> I would assume this out of phase thing would be bad. ...when mixing pickups from different manufacturers sometimes ... the 2 pickups sound very thin an nasal when combined.  I am therefore assuming this business about mixing the original signal with an out of phase signal might cause problems

You say it yourself. The out-o-faze problem happens with TWO signals which are very similar in amplitude.

In amplifiers, the output is "much" larger than input. The output wins.

There are cases where the phase inversion makes us think. The collector-base resistor negates some of the input signal and specifically leads to a lower input impedance. As shown in another essay, this may not be a big deal.

FWIW, "all" 3-leg linear voltage-amplifying devices in "normal good-working" connection seem to invert. I can't think of a big exception. The "yeahbuts" relate to some extreme cases where a small non-inverting output is possible.

Why does it invert? One leg is common to input and output. Output is take from a resistor to a battery. Raise the input. Current increases. Increased current in resistor drops the output voltage toward common. This is true for tubes, BJTs, FETs.

[PRR]

> the process of designing this bias arrangement, but I don't use the arrangement all that often, so I'm less familiar with the ins and outs of it than I am some of the better-stabilized later forms of biasing.

It is plenty stable enuff for any small-signal work, and dead easy to design.

Pick your collector resistor, guided generally by load demands.

In many cases you want an emitter resistor to limit gain and add NFB. For DC biasing, lump that in with collector resistance.

Pick a transistor, look-up its hFE range, and pick the middle of that range.

Multiply collector resistance by nominal hFe, that's your collector-base resistor.

With nominal hFE it will bias-up halfway between rails (minor adjustments for Vbe and Re drops). With other hFE it will bias somewhat off halfway, but is robustly self-correcting.

Say 10K collector load, 1K in emitter. Say typical hFE of 200. 11K*200 is 2.2Meg collector-base resistor. Bias is done.



Designed for hFE=200, but 2:1 either way leaves 3V peak swing (1/3rd of supply) and current varies only 2:1. If you work it out with transistors, at Re=1K the gain changes only 10% (but Re=0 gives 2:1 change of gain).

Audio input resistance is (Rb divided by audio gain) in shunt with base resistance. "Hih voltage gain" leads to very low input impedance. (However you can also design entirely for current-gain, where low Rin is best; or cascade voltage-gain and current-gain stages ala Cherry/Hooper.)

Gain is 10, so Rcb looks like 2.2Meg/10 or 220K. With 9V supply (note we have not needed this value until now) the current is near 0.4mA, Rie is near 70 ohms, plus the 1K is 1K (so we didn't need Vcc for this example), times hFe is 200K. Input looks like 100K. 0.1uFd cap gives full bass.

An assumption is that Vcc is much greater than Vbe. When this is violated you simply take Vbe into account. Here is a plan for modest gain for sub-milliVolt signals:



Transistor would be specced for low Rb, not for hFE. Device current is optimized for lo-Z noise, and appropriate gain-set resistors led to an awkwardly low 1V supply. So I added a dropper (with filter) and moved Rcb up there. Still plenty stable, clean, and no NFB via Rcb.