SELF-BIASED STAGE CHARACTERISTICS
The self-biased stage allows the transistor to be turned on by a very small current from the bias resistor and this allows you to increase and decrease the current via an input signal and the resistance of the transistor changes in a very linear way to produce a distortion-free stage.
You will learn the transistor is simply a variable resistor and it is the transistor in combination with the collector resistor that produces a varying voltage on the collector that is passed to the next stage. 

How the stage works and why you only get a gain of 70.
When a rising input signal enters the base, the transistor amplifies this about 200 times and the collector voltage falls. The current through the base-bias resistor falls and it reduces the effect of the incoming signal. That's why the transistor does not get the full effect and the gain is not as high as expected.
When the input signal falls, the transistor is turned off and the collector voltage rises. But now the base-bias resistor supplies more current to the base to turn it ON and this works against what the signal is trying to do. The end result is a gain of 70.  
The following diagram shows the input signal is falling and at the same time the output signal is larger but rising. They are "out of phase" with each other.

The output is "OUT OF PHASE" with the input

The following circuit shows an electret microphone connected to the input of the self-biased stage and the component values to get a gain of 70.

Component values and
Connecting an electret microphone

Matching the impedance of the electret microphone to the input of the amplifier.
This is something you cannot work out.  You just have to take my advice.
The LOAD for the microphone stage is 22k. The input of the self-biased stage is about 3k to 5k as this is the effect of the base-emitter junction.
You can see these two values are an enormous mismatch and that's why we get the figure of 70 for the gain of the circuit.
The 22n also produces a considerable loss in the transfer of the signal and you can get a CRO to see the losses.
 
When you see the signal on the electret is 20mV, this voltage is not passed to the base of the transistor as a voltage.  It is converted into a current.
What happens is this: The electret turns off a small amount and the voltage rises 20mV. This rise is passed through the coupling capacitor to the base of the transistor.
The current associated with this rise is passed to the base of the transistor. The current is determined by the value of the load resistor on the microphone. There is no way we know how much current is associated with this rise as the capacitor is charging at the same time and removing some of the energy. Whatever current is detected by the transistor is amplified 200 times and some of the effect of this is delivered to the output capacitor and to the next stage. But the output capacitor and the next stage absorbs some of the energy that the transistor is able to deliver and that's why, in the end, the result we see is a gain of about 70.

INPUT -  OUTPUT CAPACITORS
The value of an output capacitor should be 10 times larger than the input capacitor because the output will be handling a higher current and the resistors around this part of the circuit will be much lower than the value of resistors on the input.

2 FACTS
In the end, the value of the input capacitor does not really matter, as long as it delivers the amplitude. It can be higher than calculated and will not affect the performance.
Secondly, the biasing of the stage does not really matter as the energy entering the stage via the input capacitor will swamp the energy provided by the biasing components. Suppose you have biased the stage so the collector is at mid rail. It will require very little current-reduction to turn the transistor OFF but a lot more current-increase to turn the transistor ON fully.
How are you going to provide that requirement from a previous stage?   You can't !!
All this requirement about biasing a stage is completely redundant and "falls over" when you see how a stage actually behaves when a signal is processed. When a decreasing signal enters a stage, only the first part of the signal turns the stage off and then the base becomes "open circuit" and it does not remove and of the incoming energy.  The only energy that is removed is done by the biasing component and this removes very little. The charge across the input capacitor hardly changes.
That's why you should pay very little attention to the mathematics of designing these stages, as the stage will create its own results and they will be completely different to anything you have predicted.

THE LOAD-LINE
Let's go over the hype of designing a circuit, again.
So much time is wasted on talking about the LOAD LINE for a stage.
This is the concept of placing the transistor in a mid-bias state so the positive and negative excursions of the input signal will be equally amplified.
But let's go into the truth.
The transistor is sitting at mid-rail and the transistor is receiving this "turn-ON" current from the biasing components.
But here's the point that no-one mentions.
It only takes a small current to get the transistor to "half-turn ON"  It takes a lot more current to turn ON fully.
An input signal has a lot of current-capability when it starts at zero, and as the input signal rises, the current capability decreases.
This is because the input signal is generated from the LOAD RESISTOR of the previous stage and as the voltage across this resistor decreases, the current it can supply DECREASES.
The voltage across the resistor decreases, mainly because the stage-coupling-capacitor charges.
So we have two conflicting situations.
The signal finds it very easy to start to turn the transistor ON and as the signal rises, it has less energy to deliver this energy into the base, while the base wants more and more energy to turn the transistor on HARDER and HARDER.
In other words we have two situations that do not match each other AT ALL.
The effect is this: When the signal starts to rise it has a very big effect on turning the transistor ON and when the signal reaches its peak, it has very little effect on turning the transistor ON.
So, the curve of the input signal is not reflected in the output of the amplifier AT ALL.
That's why you can't do any mathematical calculation. You have no parameters, no values, nothing to go-on. 
Even if the inter-stage coupling capacitor does not charge and reduce the transfer effect, the "turning-ON" of the transistor in the stage we are designing takes more current for each percentage increase in the "turning-ON process"  and this is far from a linear relationship.
It's amazing the amplifier works at all, but because the whole thing is so sloppy and so tolerant and so non-demanding, you get an output signal that is recognisable. 
Once you realise this, you will see how far off the mark the instructors are. Things DO NOT MATTER. Just about any values will work and all you have to do is listen to the output and adjust the values until you get a clear reception. 
Let's mention another point that has never been covered.
As the input signal rises, the input capacitor may or may not charge. As long as the input signal is higher than the voltage on the base and higher than the input lead of the capacitor, it has "headroom" and will deliver current to the base. When the signal reduces, it also delivers energy to the base. The base will only rise and fall a few millivolts, and as soon as it falls below 550mV, the transistor does not accept any more current.
This means the input signal only has to be 50mV in amplitude and if it can supply current, the circuit will work perfectly.
That's why the amplitude of the input signal is quite unimportant. It is the current-carrying ability of the signal that the transistor detects.

WHAT REALLY HAPPENS TO THE INPUT SIGNAL
The input signal to a transistor stage might be 1mV in amplitude or several volts.
Once a transistor is "turned on" via the components in a stage, the base does not want or need an increased voltage. It just need an increased CURRENT.
So, what happens to the 1v input signal?
The input signal has a current-capability associated with it. This is a difficult fact to understand, but for the moment we will just say that all input signals have a current capability. If the current-capability is very small, the 1v signal will not be amplified very much. If it has a lot of current capability, the 1v signal will be amplified the greatest amount.
I am not going into the amount of current needed, but the signal will be able to deliver a higher current into the base of the transistor when the input signal is a maximum.
The transistor does not know the voltage of the signal but it detects the increased CURRENT.  In other words, the amplitude (voltage) of the input signal is converted to an increased CURRENT and the transistor will amplify this current 100 to 200 times.
The voltage on the base may increase a small amount but we do not know how much, and this increase NEVER comes into any calculation. 
So, the VOLTAGE is CONVERTED to CURRENT.
The signal is like a car battery. The battery might be capable of supplying 400 amps, but a car globe will only consume 10 amps.
The same with an input signal, it might be capable of delivering 100mA, but the transistor will only "ask for" (require) 2mA. 
But if the signal can only supply 2mA, its voltage will drop by about 50%. The theory behind this is a whole new discussion.
That's why it is important to know the current-capability of the input signal, so that it is not reduced by the transistor stage.
And that's why it is important to have a high impedance input for the transistor stage, so little or no attenuation is produced.
You cannot find and you cannot work out the impedance of a transistor amplifier because it will vary according to the surrounding components and the current entering the base.
The highest input impedance applies to the self-biased stage while the H-Bridge design has the input reduced by up to 90% due to the bridging components. This means the effectiveness of the input signal reduces to 10% and the stage has to amplify the current 10 times just to regain is effectiveness.
This means the amplitude of the input signal REDUCES TO ZERO AT THE BASE and only the varying current enters the transistor.
Now you know why all the text books, University lectures, and YouTube videos are so inaccurate, misleading and uninformed.
This is the most important factor to understand with transistor circuits.
All the diagrams show voltage amplification, but the transistors only see CURRENT VARIATIONS.
If you could inject the base with a voltage having NO amplitude; but a varying current, it would work exactly the same.
But it is the voltage that allows the current to increase and this is what the transistor wants.
 
HOW DOES A TRANSISTOR WORK?
I have seen so many complex discussions on how a transistor works, that no beginner will understand how it actually works, when the explanation is so complex.
So, here are the simple facts:

The diagram shows a transistor with a resistor connected between the collector and the 12v rail. The resistor is called the LOAD.
The base is the input and the collector is the output. The emitter just keeps the transistor "in place."
The transistor is just a variable resistor. That's all it is.  Just a variable resistor.
When nothing is connected to the base, the transistor has a very high resistance. We mean it has a very high resistance between the collector-emitter terminals.
When you connect a voltage to the base, the resistance of the transistor decreases.
But it is not just the voltage that the transistor sees. It sees the current you are supplying.
You can supply the transistor with a small current by adding a "safety resistor" to the input voltage and as you reduce the value of this resistor, the current into the base will increase.

Supplying a small current to the base

This will make the transistor turn on "more and more" and its effective resistance will be less and less.  This means more and more current will flow through the LOAD resistor.
All transistors have an AMPLIFICATION FACTOR called GAIN and it is normally greater than 100. This means the current supplied to the base will be increased 100 times through the LOAD. 
We are not going into any mathematics but the voltage on the collector will rise and fall when the base current decreases and increases and this is how a signal is amplified by the transistor. 
A signal has a certain amplitude and a certain "strength."   Its "strength" is the amount of current it can deliver. But a transistor does not "deliver" a current.   It simply reduces resistance between the collector-emitter terminals and this allows more current to flow through the LOAD. As a signal moves through each stage of amplification, it increases in amplitude and also increases in "strength."
The transistor is just a variable resistor. It is the surrounding components that create the increase in amplitude and strength of the signal.
We say this to keep the discussion simple.
Later you will learn how the surrounding components create the amplitude and strength of the signal.
At the moment you just need to understand the CURRENT into the base decreases the resistance of the transistor. The transistor is a CURRENT controlled device. There are other types of "transistors" that are voltage controlled (FETs).
But the transistor (shown above) does not increase the current. It allows a higher current to flow through the LOAD. This is because its resistance decreases and a higher voltage will appear across the load and thus more current will flow through the LOAD. This current also passes through the transistor. It is the LOAD that "asks-for" or "demands" the extra current and the transistor allows the current to flow. 
It is the LOAD that does all the "demanding." The transistor simply does its job and allows the current to flow.
You have got to understand, the transistor "sits back" and does very little of the "decision-work"  It is merely a variable resistor and the surrounding components create all the voltages and gain and even the quality of the signal.
That's why everything you have been told in any lecture and text book has "side-stepped" the real way things work and that's why you have little understanding of how to design things.
It's like them telling you SANTA is real.  You are living in a total miss-conception.
  

3 simple questions:

1. If the transistor in a self-biasing stage is replaced with one having higher gain, what will happen to the collector voltage???

2. If the base-bias resistor in a self-biased stage is increased, what will happen to the collector voltage?

3. What is the ratio of input to output capacitor values? 

As you can see, the real way the circuit works is completely different to anything that has been explained in any text book. Unless you know how a circuit works, you cannot find and fix a problem.

Answers:
1. The collector voltage will fall.   2. The collector voltage will rise   3. Output capacitor is 10 times larger.

0.7v BASE VOLTAGE
There are so many misconceptions and false discussions about the base voltage being 0.7v.
Firstly you cannot deliver 0.7v to the base and expect the transistor to work.
Secondly you cannot provide ANY voltage to the base. You have to provide a higher voltage via a current limiting resistor and allow the transistor to create its own voltage on the base.
And thirdly, you have to place the transistor in a circuit so the transistor and surrounding components all the transistor to produce the EXACT voltage it needs.
In other words, you should do very little or PRACTICALLY NOTHING about setting the base-emitter voltage.
In fact, the operation of the transistor has NOTHING to do with the base-emitter voltage.
The base-emitter voltage is VERY critical and one transistor may have 635mV on the base when it is quiescent mode (idle mode) and and it rises to 645mV when operating. You might think this is a small rise, but the rise have nothing to do with the current-capability of the transistor.
The current may be 0.1mA when the base-emitter is 635mV and rises to 645mV when 10mA flows.  This is less than 1% rise in voltage but the current increases 100 times !!
The operation of a transistor is completely NON-LINEAR and I have never mentioned the base-emitter voltage in any of my discussions because it has absolutely no bearing on designing a circuit. 
As long as the transistor gets its "turn-on" voltage, the transistor and surrounding components will do the rest.
All the discussions that add 0.7v or take off 0.7v, is like me saying I gamble $3,500.32 each day.

THE TRUTH
Now we come to the truth.
You cannot design a circuit without knowing the characteristics of the stage before and after the self-biased stage you are designing.
All the instructors on YouTube don't have a clue about designing a stage. They have never designed a circuit in their life.
The effectiveness of an input signal depends on its CURRENT-CAPABILITY. In other words, the IMPEDANCE of the signal - the strength of the signal being supplied by the electret microphone stage.
You will notice the electret microphone does not "drive" the signal into the self-biased stage, but the microphone turns OFF and the current IT DOES NOT TAKE is passed to the stage via the LOAD resistor, through the capacitor to the base we are designing.
It's a bit like you earning $1,000 a week and you spend as much as you want and the odd change at the end of the week is given to your family to buy food.   They won't get much. And the amplifier does get much either. It only gets a 20mV rise via a 22k resistor. This energy passes through a capacitor that absorbs 30% (as it gets charged during this time). There is very little left for the transistor amplifier.
The incoming signal sees the base of the transistor as a 5k resistor (resistance - impedance).
In simple terms the 22k becomes 40k (due to the losses incurred by the capacitor) and now it is passed into 5k. This becomes an enormous voltage mismatch and that's why we can only deal with it as an "energy package."
But if the 22k load resistor was 1k, a lot more energy would be transferred. That's why you really cannot make any calculation because you don't know the real impedance of the base-emitter junction and you don't know how much energy it gets.
The same applies with the output.
If the stage (following the one we are designing),  is removed, the amplitude of the signal on the collector of the self-biased transistor may be as high as 2,000mV. But when the following stage is connected, it will drop to 1,400mV or even lower. 
This is because the base-emitter junction of the following stage and the effect of the output capacitor create a voltage-divider with the collector load resistor.
So, the gain of the stage we are designing depends on the input impedance of the stage that follows.
We can increase the gain by reducing the value of the load resistor on the collector and reducing the value of the base-bias resistor so the stage sits at half-rail voltage but it draws a higher current.  Or we can reduce the impedance of the input of the stage that follows.

But this is only half the story.  You cannot start to design a stage until you know: "what you are driving into."   You need to know the input impedance of the stage that will follow.
That's why none of the videos on the web have any value.
They are just a JUNK presentation.

A SURPRISE!
How do you think I can see a circuit working "in my head" and solve problems and design things without any paperwork?
Because I see how a transistor REALLY WORKS.  It works completely differently to anything you have read or viewed and anything we have covered above.
If you really want to know how a transistor works, FORGET EVERYTHING you learnt.
A transistor is just a variable resistor. It does NOT produce the output waveform. It does not produce the gain or the size of the signal.   All it does is allow a current to flow in the collector-emitter circuit. It does not have anything to do with voltage . . .. it does not control the voltage.
The item that produces the output waveform is the LOAD RESISTOR and you are going to learn one of the most important concepts of electronics, that has NEVER been discussed.
It is called CONVERSION.
And it all revolves around 2 resistors IN SERIES that create a VOLTAGE DIVIDER.
A Voltage Divider consists of 2 resistors in series and when they are equal, the voltage at the join is half the supply voltage.
If you decrease the value of the lower resistor, the voltage at the join decreases.
There is one more concept you have to understand. If you connect something to the join, you will mess up the voltage and that's why we are going to assume you do not connect anything. The transistor is the lower resistor. When you don't supply any current to the base of the transistor is has a very high resistance.
This means the join of the two resistors will be very close to rail voltage.
When you supply a lot of current to the base, the transistor will have a very low resistance and the join will be very close to the 0v rail.
That's how the output signal rises and falls.
We supply CURRENT to the circuit and the LOAD RESISTOR converts this to a rise and falling voltage output.
I never mentioned the input VOLTAGE. I don't need it.
So, a varying input CURRENT produces a varying output VOLTAGE. This is called CONVERSION.  
The transistor cannot do this without a LOAD RESISTOR.
But in a practical circuit the output signal is passed to the next stage, and the next stage wants CURRENT to do the same thing as the stage we have described.
Each stage in a circuit normally wants 10 to 100 times more CURRENT than the previous.
This is called "Bleed-off" and upsets our discussion about voltage at the mid-point.
So far we have a lower resistor can can go from a low to high value and change the voltage at the midpoint from a low value to a high value. But the next stage is like putting a resistor from the mid-point to the 0v rail.
You have to learn about a voltage divider with a fixed resistor.
Suppose the load is 10k and the resistor on the output is 10. These two create a voltage divider in which the voltage at the join is half-rail voltage.
This means the transistor can only reduce the voltage to 0v and the signal is immediately less than half the expected.
Use the same reasoning to see the voltage will never rise as high as before.
This is what I see when I look at a circuit. I see voltage dividers and the results have nothing to do with the voltage-amplitude of the input signal, the gain of the transistor or the output current of a stage.
With this simplified approach you will be able to mentally diagnose a stage to 90% success and only an examination of the stage "in action" will get a closer result.
All the mathematics you have been lumbered with "flies out the door" when you understand how the stage really works and you can see why a University could not charge $19,000 for our way to do things. Not only would they produce a capable, competent student, but show up the text books for their worthlessness.
 

Page1A. more on the stages
Page-2 emitter-follower - common collector stage
Page-2A common base stage
Page 3  Coupling stages
Page 4  More on Coupling Stages
Page-5 The  FLIP FLOP
Page-6 All the "YouTube" mistakes and the Common Base Circuit
The transistor Potentiometer
The REAL Transistor Amplifier

Contact Colin Mitchell if you want any help

Go to Talking Electronics website