STARTING
In this article you are going to learn how to really design a transistor amplifier and forget everything you have been shown and told and learnt. 
All the YouTube Videos, University Video course and everything you have seen is totally irrelevant, distracting, inaccurate and worthless.
They all treat you live a cave dweller without the knowledge of electricity whereas the fact is different.  Everything you need is available as examples on the web and all you have to do is copy a circuit diagram and try it.  You don't nead any mathematics and very little understanding.
All you need is a set of components and a simple multimeter and be prepared to experiment.
That's how I leant. 3 years of an electronics course taught me NOTHING and when I entered an electronics repair shop, I suddenly realised I had been taught nothing. All the lectures were "smoke and mirrors" and "befuddlement" and "complication."  I didn't have a clue how to fix anything.  But I am very smart and there is nothing faster than "self teaching."
Simple transistor circuits are NOT designed like you see in any text book, video or University lecture.
They are designed completely differently but none of the lecturers have designed a circuit in their life and and have no idea what they are teaching.

You are going to learn that you cannot design a circuit (using mathematical equations) to produce mid-rail biasing as the actual set-point depends on the gain of the transistor and the biasing values are very very critical. Any instructor that says he can set the value does not have a clue about circuits. I have not found a single YouTube video that teaches the correct approach or anything that you can apply to real-life designing, so you need to start with a completely fresh approach and realise everything is actually done completely differently to what you have been taught.

We are going to start with the SELF BIASED TRANSISTOR AMPLIFIER.  Commonly called the COMMON EMITTER AMPLIFIER. (The other circuit is called the H-Bridge configuration)

Here is the circuit. It uses real values because that is the only way to learn. And it does produce a fairly accurate mid-rail biasing.

Fig1: The self-biased Transistor Amplifier

Whistle into the microphone and the circuit will produce a 4 volt waveform. You can change the LOAD resistor from 22k to 47k or down to 10k and the output will alter very little. You can change the base bias resistor from 1M to 2M with very little change in the output waveform.
These changes hardly affect the performance of the circuit because the transistor is creating its own OPERATING POINT. You are not forcing any conditions on the base (the H-Bridge has a tighter hold on the biasing of the transistor).
To create this circuit you simply use 22k for the load resistor for 3v or 6v supply (47k for 9v to 12v supply) and start with 1M for the base-bias resistor. Increase the value until you get half-rail-voltage on the collector. You must use only a digital voltmeter for measuring the voltage.

Now we come to the COMMON EMITTER CIRCUIT described on the web in so many YouTube videos, as the H-Bridge design:

Fig2: The H-Bridge Transistor Amplifier
330k and 47k work best

Build the circuit and you will get a shock.  It produces a small output signal. And the values of base-bias can be critical. Unless the lower base bias resistor is between 16k and 19k the circuit may not produce an output.
We used 330k and 47k to produce a 4v output and when the 47k was reduced to 39k, the output dropped to less than 2v.  The biasing components are critical and there is no way to design the circuit mathematically.  You need to experiment with the values to see the effect.
This is something that not one of the videos tell you. None of the instructors have built the circuit and none of them know what they are talking about.

By changing the biasing components you are allowing the transistor to set its own conditions of BIASING. The base biasing resistors set the condition called QUIESCENT or REST condition. But you do not know what biasing values to use.
150k and 47k did not work, 330k and 47k worked perfectly. What mathematical formula could you use to achieve the best result????  You cannot use any equations. The values of resistor will vary enormously because the gain of the transistor can be from 150 to 450.
The only way is to build the circuit and experiment.
 

CONCLUSION
Both the above circuits were built in December 2023.
Using 330k and 47k for the second circuit, the electret microphone delivered a 20mV output with 22k load resistor. This signal was not appreciably attenuated when connected to the two transistor amplifiers and the output at the collect was about 2v p-p. (There was a small attenuation but I could not measure it.)
We concluded that both circuits produced the same results.

The electret microphone detects a pin dropped on the floor and you can listen to voices in the next room with a clarity that is better than normal hearing.

The two circuit have not increased the current-capability of the signal as the impedance of the electret microphone is 22k and the impedance of the amplifying is 22k (as determined by the value of the collector load resistor).
However the amplitude of the signal has increased from 20mV to 2,000mV and it is an increase of 100 times.  So the stage has achieved something.
A stage can increase just the amplitude, or just the current, or both. And if the answer is about 100 we have been successful.
We now have to connect the output of this stage to a second stage.
The first amplifying stage is called a PRE-AMPLIFIER and the second amplifying stage is called a SMALL SIGNAL AMPLIFYING STAGE as it must be designed to have a high impedance input, so it does not load the first stage.
So far, we have not increased the "strength" of the signal. It is just as weak as the signal emerging from the electret microphone. All we have done is is increase its amplitude.
You just cannot connect it to "anything" (such as an earpiece or speaker) because if you reduce the amplitude, you will have lost some of the benefit of the first stage. And it is too weak to drive anything such as an earphone or speaker.  .
But, unfortunately, the second stage will require some of the energy (signal) from the first stage, for its operation.
Our skill is to lose as little of the amplitude as possible. More accurately, we want to lose as little ENERGY as possible.
In other words, the input of the second stage must have a high input impedance.
If we connect the stage to an input impedance of 22k, the signal will be attenuated by 50%. This is because the 22k LOAD resistor will be in series with 22k of the second stage and it will form a voltage divider of 50%.
If the impedance of the second stage is 10k, the signal will be reduced to 30% or less.
To have the smallest attenuation we need the input impedance to be 100k or higher.

Let's say we want to drive an 8R speaker. 
We will look at 4 different amplifying stages and see what results we obtain.

THE FIRST ARRANGEMENT
The first is called an EMITTER FOLLOWER STAGE. It increases the current by 100 times (or more.)
The following diagram shows this arrangement:

Fig3: Connecting an EMITTER FOLLOWER stage

The BC 338 emitter-follower transistor will have a gain of say 100.  This means it will amplify the resistance of speaker 100 times to get an impedance of 800 ohms on the base of the transistor.
You don't need any mathematics, just an understanding of how the circuit works.
The 22k collector LOAD resistor and the 800 ohms impedance of the speaker will form a voltage divider and the voltage across the 800 ohms will be less than 5% of rail voltage.
This is almost useless for a speaker and shows the stage cannot drive a speaker.
The first secret you have learnt is the emitter follower stage multiplies the impedance of the speaker by a value equal to the gain of the transistor. The gain may be 150, 250 or slightly more.
Secondly, the first transistor does not deliver any energy to the emitter-follower stage. The 22k LOAD resistor delivers the energy. The first transistor simply turns OFF and the 22k delivers the energy.
The transistor turns ON to produce the lower portion of the wave.   But it is the 22k that delivers the rising portion of the signal. That's why a single emitter-follower stage will not work.

THE SECOND ARRANGEMENT
If we add another emitter-follower transistor we will increase the impedance by 100 times and now the ratio of the 22k to the impedance of the output stage will be 80,000 ohms and it will not reduce the amplitude of the signal produced by the 22k.  That's because the voltage division has a ratio of 22,000:80,000 or about 25%. In reality the ratio will be a lot more beneficial and the the two transistors will deliver almost full amplitude from the first circuit.

Fig4: Connecting two EMITTER FOLLOWER stages

Just because the output impedance is 22k and the driver stage is 80,000 ohms, does not mean there is any miss-match. You have to realise the impedance will be so high that it does not reduce the amplitude of the signal of the first stage.
However the maximum amplitude will be reduced by about 1.5v due to the two base-emitter drops and the output will be about 4v. This is always the case with emitter-follower stages using two transistors.
The current delivered by the circuit will be:  4v/8R = 0.5A  and this will only occur when the transistor is not conducting and the 22k is pulling the two transistors HIGH.
You can use a single Darlington transistor in the output to get the same results.

THE THIRD ARRANGEMENT
The third way to drive an 8R speaker is to use a Darlington transistor in place of the two individual transistors as an EMITTER-FOLLOWER.
This is shown in the following diagram:

Fig5: Connecting a Darlington transistor

This has one major drawback because the collector of the upper transistor (within the Darlington transistor) is not connected to the supply rail but to the speaker.  This makes an enormous difference to the way the transistor works. I am not going into the technical aspects except to say the collector-emitter voltage is never less than about 1.5v to 2.5v due to the way the Darlington transistor operates. The end result is about 4v across the speaker and this is no better than the previous arrangement.
Here's another way to see the circuit: If the Darlington transistor has a gain of 80,000 it will divide the effectiveness of the 22k by 80,000.  If the gain was 22,000 the 22k divided by 22,000 would result in one ohm. But if the Darlington is 80,000 the result is 1/4 ohm.
In other words the Darlington gets converted to a 1/4 ohm resistor with a characteristic voltage across the collector-terminals of about 1.5v to 2.5v This is how you diagnose circuits quickly and easily.

THE FOURTH ARRANGEMENT
The fourth way to drive an 8R speaker is to place the BC547 and BC338 in a SUPER-ALPHA arrangement.  
This is shown in the following diagram:

Fig6: Connecting a SUPER-ALPHA pair.

The BC547 and BC338 transistors are connected "on top" of each other and they form a single transistor with a very high amplification factor.
This time we cannot say the impedance is 80,000 ohms but we can say the pair will amplify the current provided by the 22k by a factor of about 100 x 100 = 10,000 times. We don't know how much current will be supplied by the 22k but suppose it has 1v across it, the current will be 1/22,000 amp.  If we multiply this by 100 x 100 we get 100/22 amp and this is obviously more than the battery will deliver, so the circuit will certainly drive the speaker.
There will be a very small voltage across the collector-emitter junction of the BC338.
By re-positioning the two transistors we improve the signal from 4v to nearly 6v.

This is the best arrangement and we have shown the disadvantage of a Darlington transistor.

IMPEDANCE MATCHING
These 4 arrangements are called IMPEDANCE MATCHING.  Impedance Matching is the art of using the available current and voltage from a stage and not reduce it.  In other words the stage you are adding has to provide the energy to drive the LOAD. If the stage you are adding does not do this, the circuit will take energy from the first stage and reduce the amplitude of the signal.
That is the basics of Impedance matching. Sometimes, the amplitude of the signal will be reduced but the circuit will adjust and an accurate waveform will appear in the LOAD. So, there is no clearly defined way to connect stages and the only success is listening to the result. You can hear much more of the quality than any signal on an oscilloscope. 
Some output stages need a large voltage swing and some don't.  Some only need 50mV.  But they all require a certain amount of current and each transistor in the output stage amplifies this current 100 to 300 times.

How do you work out if arrangement #4 will work?
Maximum voltage across the 22k is:   6v  minus 0.7v  minus 0.7v  =  4.6v
The max current that will flow in the 22k resistor will be  4.6 / 22,000 amp
The first transistor in the driver stage will amplify this current by about 250 to 300 times.
And the second transistor in the driver stage will amplify about 100 times.
This means the circuit is capable of delivering: 4.6 / 22,000  x 300 x 100 = 4.6 x 3 x 10 / 22 =
46 x 3 / 22  =  2 x 3 = 6 amps.
This is obviously much more than required by the speaker  (5.8 / 8 = less than 1 amp) and so the driver stage will be suitable.

Now you can see the uselessness and stupidity of all the common-emitter discussions in text books and YouTube demonstrations.
You cannot design a circuit without knowing the current capability and amplitude of the input signal and the characteristics of the stage you will be connecting to the output. These two affect the operation of the stage enormously.

           ooooooooooooo0000000000000000000oooooooooooooooooooo

A REVOLUTIONARY UNDERSTANDING
Now we come to the next piece of revolutionary understanding.
I can see the operation of every circuit "in my mind."
Forget all the RUBBISH you have seen and learnt and read in text books and on the web.
None of it will help you understand and design and "see" how a circuit works.
I bet you have never been told a transistor does not deliver energy to the following stage. It is the LOAD resistor that delivers the energy. I call it: "failure by omission." None of the writers have ever designed or repaired a project. They simply repeat what they have read and been told.
There is an old saying.   If you can:  "Do."   If you can't.  "Teach."

It will be years before you can do the same, but we can start with the principles of how a transistor REALLY WORKS.
A transistor is just a VARIABLE RESISTOR - a POTENTIOMETER

Here is the circuit of a REAL TRANSISTOR:

Fig3: The Transistor is simply an adjustable resistor

Simply replace the transistor with a POTENTIOMETER. Look at how we have changed the transistor into a "pot."

Fig4: Two resistors

You now have two resistors in series.
Two RESISTORS IN SERIES is the basis of a VOLTAGE DIVIDER. And that is what each section of a circuit becomes. It becomes a VOLTAGE DIVIDER.
Now I can tell you the most outstanding comment you will ever hear. The potentiometer cannot produce any amplification. No voltage amplification and no current amplification.  And a transistor cannot produce any voltage or current amplification either.  You have  been fooled.
It is the LOAD RESISTOR that provide and produces and assists and is essential in creating these features.
And only when you realise the LOAD RESISTOR creates all the magic, will you understand how a circuit REALLY WORKS.
Just because the transistor increases the base current by 100 or 200, does not make the circuit "move" or "work."
The whole circuit works on the VOLTAGE DIVIDER principle.

Fig5: The VOLTAGE DIVIDER

The VOLTAGE DIVIDER is simply two resistor in series. When the lower resistor is a VARIABLE RESISTOR the voltage at the join will INCREASE or DECREASE.

When the resistance of the variable resistor increases, the voltage at the join increases.
That means:  When the resistance of the transistor increases, the output voltage increases.
But you must understand this as:  The top resistor PULLS the voltage high.   The transistor merely ALLOWS the LOAD to pull the output voltage HIGH.
Now, depending on the value of the LOAD RESISTOR, and the ability of the transistor to pull the output voltage LOW, the amplitude of the output is created. The transistor PULLS the output LOW and the LOAD RESISTOR pulls the output voltage HIGH.
But the LOAD RESISTOR can only pull the output voltage HIGH when the transistor stops pulling it LOW.
If the transistor can "pull-down" very strongly, the LOAD resistor can be a small value of resistance and this means the current in the circuit will he high. 

HOW DOES THE CIRCUIT WORK?
If the LOAD RESISTOR is a high value of resistance, the transistor will find it very easy to pull the output down to a low value. And this means only a small current will be present.
If the LOAD RESISTOR is a very LOW value of resistance, the transistor will not be able to pull the output down AT ALL (if the transistor is weak).
The input lead of the transistor is the BASE and it needs CURRENT. Current has MUSCLE and it will turn the shaft of the variable resistor to produce a LOW VALUE of resistance. 
In actual fact, the transistor increases the "strength" of the input current 100 times (or more) to turn the potentiometer shaft. 

THE NEXT POINT
If the value of resistance of the LOAD RESISTOR is HIGH, only a very small current will flow in the voltage divider and only a very small current will be available on the OUTPUT.
If the values of resistance of the LOAD RESISTOR is SMALL, a HIGH current will flow in the voltage divider and a HIGH current will be available on the OUTPUT.
This means the output current will depend on the value of resistance of the LOAD RESISTOR. As long as the transistor can pull the output voltage down to a very small value, the value of current depends only on the  LOAD RESISTOR.
You also have to understand the output voltage depends ONLY on the LOAD RESISTOR, (when the transistor is able to pull the output voltage LOW).
So, all we have to do make sure the transistor can pull the output voltage LOW and the amplitude and output current capability will depend on the value of the LOAD RESISTOR.

Now, THE NEXT POINT
Here's the next amazing thing you will learn:

Fig6: The Self Biased Transistor Amplifier

We will take the Self Biased Transistor Amplifier and see how it works.
When the stage is sitting with no audio input, nothing will be processed.
But a tap on the microphone will reduce is resistance (impedance) and the voltage to the 100n coupling capacitor will fluctuate up and down and when the cycle produces a rising voltage, the first 22k LOAD RESISTOR will do the "pulling up."
Now, what we say is:   22k is entering the transistor amplifier and the transistor reduces this value by 100 to 200 - as this the gain of the transistor - and it becomes 22,000/100 = 220 ohms.
That's right!!   The transistor deals with the input and turns it into a 220 ohm resistor, exactly like fig7:

Fig7: 22k:220R voltage divider

All you have to do is work out what voltage will appear at the join and you will find it only a very small voltage and that's how the circuit has the potential to produce a very good output waveform. And if I go into further investigation I will get a very accurate outcome without any mathematics or equations or frustration.
I will get a much closer result than any mathematician because I KNOW WHAT I AM DOING.
Technically we have not achieved much in this circuit because the first load resistor is the same value as the second. But we have created an increase in amplitude.  But the main aim for each stage is to create an increase in current. Otherwise the stage is called an INTERFACE STAGE to connect two items with different impedances.

Fig8: The REAL circuit

When designing and diagnosing the Self Biased Stage (with electret microphone), Fig 8 converts all the active components into variable resistors as shown by the symbol that will be new to you..
That's all I see.   The electret microphone can pull the 22k down a small amount and then the capability of the 22k will pass to the transistor via the 100n and get converted to 220R and this value, combined with 22k LOAD resistor will produce a HIGH and LOW output equal to about the 6v supply.

5 Transistor Radio

Now we come to a 5 Transistor Radio.
We have to convert a 5mV signal with 1microamp capability into a 3v signal with about 50milliamp capability.
The first stage consists of 2 transistors "on top of each other."
This is simply a single transistor with a very high input resistance (impedance), so it does not put a load on the tuned circuit.
Basically it will multiply the 1 microamp by a factor of about 100 and the other transistor will increase this 100 times.  So 1 microamp becomes  100 microamps  x 100 = 10,000 microamps which is 10 milliamps.  But lets reduce this to 1 milliamp. One milliamp flowing through 10k, resistor will produce a voltage drop of 10v. But we only have 3v, so the output will be quite sufficient for the radio.
So, now we have theoretically plenty of current and a high voltage swing.
The next stage simply increases the swing slightly and brings the current up to 1mA.
The final stage consists of two directly-coupled transistors.
This is the job or function result they must perform or carry-out.
The signal into this section has an impedance of about 10k and a current of less than 0.3mA and maybe a voltage swing of 2v.
The stage is required to deliver a current to a load that comprises the primary of speaker transformer and it may be 100 ohms. The maximum current that this load will allow to flow is 3/100 = 30mA.
So, we basically have sufficient amplitude by need a current gain of 30/0.3 = 100.
But the part that is very understood is the IMPEDANCE MATCHING.
The input impedance to the stage is 10k and the output impedance is 100 ohms. This is 10,000/100 = 100:1
In simple terms this means that if you try to drive a signal with 10k impedance into 100 ohms, you will only be able to get 1% success.
This circuit is called an IMPEDANCE MATCHING circuit and does the opposite job to a normal amplifier circuit.
A normal circuit matches a 10k input to a 1k output. One of the best ways to get impedance matching is DIRECT COUPLING and the two transistors are directly coupled to each other.
There are very few ways to this and a PNP NPN pair is the best.
The two transistors do these two things: They match the high impedance input to a low impedance output.
They increase 0.3mA current capability to about 30mA

The two transistors are replaced with variable resistors and the top variable resistor will be reduced in resistance by the signal through the 100n capacitor.
It will then turn ON the lower transistor with a large current because the two are connected with a 1k resistor. This large current will allow a high current to flow in the primary of the speaker transformer to produce a loud output. The 1k resistor tells you this will be possible.
 
CONSTANT VOLUME HEARING AID

We mentioned at the beginning that you cannot force conditions on the base of a transistor.
Here is a circuit that automatically adjusts the voltage on the base of the first transistor and this effectively changes the gain of the transistor to produce a constant volume, no matter how loud or quiet the surroundings. 

We have covered all the stages except the third transistor.
When  the circuit is quiet the transistor is not turned on at all and the 10u is charged via the 100k resistor. This put a voltage on the feedback line to the base of the first transistor.
The circuit is now very sensitive and will pick up the slightest sound.
This faint sound will pass through the first two stages and into the base of the third transistor.

ANOTHER EXAMPLE
If you don't follow my reasoning on SEEING A TRANSISTOR AS A VARIABLE RESISTOR, how are you going to design this type of circuit:
The transistor receives Infrared light from a LED and the resistance between the collector and emitter leads reduces to a point where pin 5 of the microcontroller sees a low of less than 1.3v
The characteristics of the IR transistor are:  1M in darkness and 10k or less when detecting IR light.
But this circuit not very sensitive with the 10k collector resistor and the IR LED has to be very close to the IR transistor for the microcontroller to register a LOW. .
How do you improve the sensitivity?
The IR transistor does not have a base lead, so all your theory about biasing the base of a transistor is worthless. It does not apply.

That's why you have to see the IR transistor as a variable resistor ,, ,  as shown in the following diagram. When the potentiometer shaft is rotated to full resistance, a voltage-divider is set-up between the pot and 10k load resistor. . . . . . . 10k is in series with 1Meg ohms. You need to understand voltage dividers and very little voltage will be "dropped" across the 10k.  If the supply is 5v, pin5 of the micro will see 4.95v
The IR transistor can on be turned on by the IR light and when it is turned ON by very bright light, the resistance across the collector-emitter leads is 10k.
The voltage divider becomes 10k in series with 10k and the voltage at the join is 2.5v.   This is not LOW enough for the microcontroller to register a LOW and the circuit DOES NOT WORK.

The solution is to increase the LOAD resistor to 100k.  The voltage at the join of the resistors becomes 100k:10k  and to keep things simple, the voltage across the 100k is 4.5v and the input of the microcontroller sees 0.5v. This is low enough for the micro to register a LOW.

This is the IR transistor and 100k LOAD:

Here's the way to look at the circuit. The IR transistor can only pass a certain amount of current.
If the 10k LOAD resistor is connected across the power rails, it will pass twice the current. But if the LOAD resistor is connected to a voltage that is HALF the 5v supply, it will take half the current. This current is the same as the capability of the IR transistor and so, when they are connected in series, they drop the same voltage across each component. That's why the lowest voltage in the first circuit is 50% of rail voltage or 2.5v.
If the IR transistor tries to "pull down" the 10k LOAD resistor, it will want more current from the circuit and the IR transistor cannot pass this extra current and so the two remain at 50%.

If the 10k LOAD is replaced by 100k LOAD, the current through the 100k when connected across the power rails is a lot less than the capability of the IR transistor and so the transistor will be able to "pull down" the 100k until the current wanted by the 100k is equal to the capability of the IR transistor. The transistor is capable of "pulling it down"  to almost the zero volt rail and we are not going into the exact value but it is low enough for the microcontroller to detect a LOW.

This discussion has shown how I "see" circuits working in my mind.

A lot more enlightening discussions and concepts can be found HERE.
https://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTrAmp/TheTrAmpP1.html

More on the Transistor Amplifier can be found HERE.
https://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTransistorAmplifier-P1.html

100 Transistor Circuits can be found HERE.
https://www.talkingelectronics.com/projects/200TrCcts/200TrCcts.html

Another 100 Transistor Circuits can be found HERE.
https://www.talkingelectronics.com/projects/200TrCcts/101-200TrCcts.html

After you read and study and understand what I have produced, you will be able to design many types of circuits.  These are called  BUILDING BLOCKS and by connecting blocks together you create circuits that do magical things.

Contact Colin Mitchell if you want any help

Go to Talking Electronics website

Let me know if I have missed a spelling mistake or forgotten something. I am proof-reading things all the time and this has reached more than 10 revisions. It takes an enormous amount of brain-power to write technical explanations and that's why no-one has gone into the depths of describing the technicalities of circuit-design.