Normally, one stage is connected to next with a capacitor and most of the calculations are done by determining the resistance (capacitive resistance - called capacitive reactance) of the capacitor at a specified frequency.
But no-one has explained what is actually going on with this connection and why this interfacing is very inefficient. The only reason why most circuits work is due to the high gain of the transistors and even though the biasing and coupling of stages removes a large amount of energy, the final result of two stages is quite impressive.

The first thing you have to understand is this: The transistor in the first stage does not "push" or "drive" energy to the next stage.  
When the transistor turns OFF, the voltage on the collector LOAD rises and this rise, associated with the current that can be supplied by the load resistor is passed to the next stage.
For instance, when a 10k resistor has 5v across it, the current capability of the resistor will be 0.5mA.   But if the stage is sitting at mid rail, this current is 0.25mA.
As the collector voltage rises, the voltage across the load resistor reduces to say 1v, and the current capability reduces to 0.1mA
You can see the real current from the stage is a lot less than expected.
What we are talking about is a package of energy from one stage to the next.  This has never been discussed before and that's why no-one understands how "combining stages" works.
Of course it is much better to DIRECTLY COMBINE STAGES, but this is sometimes not possible and other times it requires a lot of circuit design.
This packet of energy is transferred to the next stage via a capacitor and the capacitor will always have a voltage across it as one stage is at mid-voltage and the other will have a low voltage on the base.
As the voltage on the load resistor decreases, less and less current is available from the load resistor and this means the capacitor gets charged slower and slower.
But at the same time it gets charged and so the rise on the top plate of the capacitor does not pull the bottom plate as high. So a lot of the energy goes into charging the capacitor and the following stage does not see much of the energy.
When the transistor in the stage we are studying is turned ON, the voltage on the load resistor increases and the voltage across the coupling capacitor reduces and the capacitor is  put into a "discharging condition." If you slowly reduce the voltage on the top of the capacitor, it will discharge by delivering its energy into the base of the transistor. But as soon as the voltage on the lower plate of the capacitor reaches 0.55v, the current into the base ceases as the base does not accept any current when the base falls below 0.55v.
This means you might still have voltage (and energy) across the capacitor that has not been removed and when the next cycle takes place, the capacitor is already partially charged.
This means the capacitor will not be able to transfer the same amount of energy as in the first cycle and that's why the capacitor becomes such a poor transfer component. A resistor between base and zero volt rail will improve this is the answer to why they have been included.
And that's why you will find some of your circuits DON'T WORK AT ALL.
The only reason why the circuits work AT ALL is due to the high gain of the transistor and all these losses bring the final gain of many stages to about 70.
You have to be able to visualise this action to see why your circuit is not performing as well as expected. Especially when high currents are involved.
Most data sheets show transistor gain at 1mA or 10mA.  When you increase this to 100mA, the gain drops considerably. That's why NOTHING can be calculated.

It is pointless doing any calculations as you have no idea of the gain you will get from each transistor or the value of the current-capability of the input signal or the impedance of the stage that follows.
If I could design a circuit with pen and paper, I would be a multimillionaire.
Things can work out much better than expected or not work at all. And it's when they don't work, that skill comes in - something that no University course has ever touched-on, because they have never had to design ANYTHING.
There is no skill in designing a circuit. The skill is getting it to WORK.

MORE ON THE H-BRIDGE
I did not want to go further into discussing the H-Bridge but the instructors on YouTube are so incompetent at designing this stage that I want to cover the steps you need to follow so you see how inept the instructors are.
You cannot design a stage UNTIL you get some external values. They have NEVER designed a circuit in their life and their approach is completely impractical and shows no understanding.

The first thing you need to find out is the impedance of the stage you are driving as it will affect all the values of the H-Bridge stage.
This is also called the input impedance and/or the base-emitter impedance or some value in the range of 500 ohms to 5,000 ohms.
The self-bias stage we are designing is called a small-signal stage but what this really means it is a low current stage.
It can be from 0.1mA to 5mA.
You cannot design the stage and then connect it to a following stage as the output of the self-biased common emitter stage will be reduced and you will be very disappointed.

The collector LOAD resistor has to be equal or less than the impedance (resistance) of the following stage.
That's because the load resistor delivers the amplitude and if it has the same resistance, only about 50% of amplitude will be delivered. This is because the collector resistor and the coupling capacitor and the input impedance of the following stage forms a voltage divider.
But we cannot make any predictions because the real way to look at the transfer from one stage to the other is via ENERGY TRANSFER. And this involves analysing the circuit.
Next you need to work out the value of the emitter resistor and base-bias resistors to get the transistor to sit with the collector at mid-rail.
The bleed-current through the base resistors can ne 1/100th the collector current.
Next you need to work out the value of the emitter-bypass capacitor (electrolytic) to give a gain of 50 to 70 for the stage (to make it worthwhile). To do this you need to  know the frequency you will be amplifying.
Normally it is at least 1/10th the resistance of the emitter resistor.  If the emitter resistor is 1k, the capacitive reactance of the by-pass electrolytic should be 100 ohm at the desired frequency.
Next you need to know the current-capability of the input signal and it has to be at least 1/50th the collector current of the stage we are designing..
This is just the beginning of the parameters you need to start designing.
If you think the values are acceptable, I would not even waste my time doing any calculations.
Time is far too valuable.  
Build the circuit and adjust the values to get the required results.
You can try working out the values mathematically and when you build the circuit you will see how incorrect your values were.

Here is a typical problem.

We have two stages. Each stage has been correctly designed with the first stage allowing the collector voltage to rise to 12v and fall to almost zero when the stage is not connected to the emitter-follower stage.
In other words we have full rail voltage amplitude and we know the emitter-follower stage has a high impedance.
But the term "high impedance is relative. It has 100 times higher impedance than the 8R load.
This might not be a high impedance as far as the first stage is concerned.
We have already mentioned the first stage does not "push" energy to the second stage.  The transistor turns OFF and the 1k load resistor passes energy to the emitter-follower.
When the transistor turns OFF, a voltage-divider is formed with the 1k, the 47u electrolytic and the input impedance of the emitter follower.
The 47u is about 2 ohms and the input of the emitter-follower is 800 ohms.
Let's say it is 1k and 1k. This means half rail voltage will appear on the base and thus the emitter follower will only rise to 6v. We know the speaker will receive up to twice this swing but if you are expecting something like 24v, you will be mistaken.
The two stages have not been correctly matched and even though the first stage is taking more than 10mA, it is not only the current-flow that has to be correct, but the value of the collector LOAD.
Matching these two stages involves a PNP stage:

The first stage uses a PNP transistor and it can raise the voltage to 12v and when it is turned OFF, the voltage will fall to 0v. The next thing you have to work on is discharging the 47u on the collector of the PNP transistor.  

That's a topic for another discussion. 
If you want to see the solution . . .  click HERE

All circuits work on the principle of VOLTAGE DIVISION. Components are "dividing up" the available voltage and that is how and why I can "see" a circuit working. The real way to lean and understand "circuit design" is completely different from any lecture, text book or discussion.
You have to be "guided" into seeing a circuit working and see that a particular circuit WILL NOT WORK. This does not involve mathematics. It does not involve calculations, other than a simple understanding.

Here's what you need to remember:
1mA flowing through a 1k resistor will produce a voltage drop of 1 volt.
From there, you just multiply and divide:
10mA will produce a voltage drop of 10 volt.
1mA through 10k will produce a voltage drop of 10v.
Similarly: A 1k resistor on a 10v supply will pass 10mA.
10k on 10v will pass 1mA
And once you remember the above, you continue with voltage division.
A 10k and 10k in series on 10v will produce 5v at the join.
1k and 1k on 10v will produce 5v at the join.
3k3 and 3k3 and 3k3 in series on 10v will produce 3v3 and 6v6 at the joins.
  

And there is another area of electronics that Universities do not cover.  It's how to fix things.
All my projects include a section: IF IT DOES NOT WORK.   I teach how to fix a project.
Universities don't cover this topic because they have absolutely no idea how to fix anything. That's why the piece of paper they hand you at the end of 4 years is ABSOLUTELY WORTHLESS.  One University graduate came for a job application and said:  Oh, I don't do soldering !!!
 

The circuit above is a 3-transistor amplifier and the first thing you will want to know is the output wattage.
The limiting factor of the amplifier is the 680R resistor. The PNP transistor will be pulled down to the negative rail as hard as possible via the driver transistor, so this part of the cycle will work. It will be like pulling down with a 10R resistor. But the NPN transistor will be pulled high via the 680R and this effect will be a lot weaker than the other section.
We have explained that a transistor effectively converts the base resistor to a smaller resistance by a factor equal to the gain of the transistor. The gain in this circuit will be about 50, for the power transistor, and the 680R will be converted to 13 ohms.
We will assume the 1,000u has an impedance of one ohm.
The circuit now become a voltage divider of 14:8  so that the speaker gets 36% of the supply voltage.  The way to find this percentage is to add the total resistance in the circuit (14 + 8 = 22) and the speaker will get 8 parts of this total.  The equation is: 8/22 = 36%
You can see the speaker gets nothing like 100% of the energy of the amplifier and although the output will be quite loud, the amplifier is a very inefficient design.
This is how you look at a circuit and instantly work out its performance. A simple voltage-divider equation will get a very accurate value.

You can see this amplifier will have NO output AT ALL. The BC337 is being pulled up by a 4k7 and this will be converted to 470R or maybe 200R via the gain of the transistor and this value will be part of the voltage divider network with the 8R speaker.
Even if the BC338 acts like an 8 ohm resistor, the speaker will get only 50% of rail voltage and so both parts of the push-pull arrangement have to be very low impedance to get the full energy from the circuit.

Here is a circuit that the designer thought would give a good output.
Look at the data sheet and you will find the transistor are RUBBISH.  The collector-emitter voltage is 4v and the gain is less than 50.
The TIP31A is going to be effectively 1,000/50 = 20 ohms and you are going to lose 4v. So you have 8v divided between a voltage-divider of 20:16 and you might get 4v across the speaker.
That's why power amps start with a high voltage  - to get the output wattage.

Here's a 3-Transistor amplifier and the first thing you ask:  "Where is the fault?"
The fault lies in the 330R across the 220u electrolytic.
The transistor can charge the 220u and drive the speaker, but when the transistor turns OFF, the speaker, 220u and 330R are in a "loop" and the 330R is going to take a very long time to discharge the 220u.
This means the transistor will turn ON in the next cycle and the 220u will already be 90% charged from the previous cycle and the transistor will only be able to deliver 10% or less of its energy.
Without doing any complex maths, the capacitor will take 8 milliseconds to discharge to half voltage and if the frequency of the amplifier is 1kHz, the cycle-time will be 1 millisecond.  You can see the capacitor will discharge only a very small amount in this time.

Another faulty circuit.
You can see Q3 is being pulled HIGH by Q2.  But what is pulling Q4 LOW??

To make it easy we have removed all the components that are effectively "out of circuit" as the top transistor is not turned ON.

But the 100u electrolytic is charged to about 6v, and it effectively becomes a battery, so we have changed it to a battery:

Now you can see the base and emitter are both connected to the same voltage and nothing is turning ON the transistor!!

You can now see the fault in this design. The 25u is going to take a long time to discharge through 660 ohms of resistance.

Don't worry about the old-type of transistors.
The fault lies in the direct-connection of the two transistors across the supply. When the first transistor turns ON, the current through it will be very high and this current will flow through the emitter-base junction of Q2 and will be much higher than the transistor wants or requires.  The transistor may be damaged.
You must always put a current-limiting resistor in any path with these features.   A 1k resistor will limit the current and not affect the operation of the circuit.

It does not matter if you use a power transistor, the output of a circuit will be controlled by the surrounding components.
The 4k7 is going to be converted to a 50 ohm resistor by the gain of the 2N3055 (or even a higher value). This mean the speaker will see less than 1v across it when in the idle state. The rest of the circuit can only have a very small effect on turning ON the transistor via the two 470R resistors. The base is at the join of these resistors so the effect will be very small. You would see less than 2v swing on the output. The circuit is so inefficient that it is not worth working out any actual values.

And now we have a circuit that has a different fault.
The 22u capacitor connected to the base of the driver transistor will turn the transistor ON when it gets charged by the 3k3, but when Q2 turns ON, the whole 22u will drop a small amount and when it drops by 0.7v, the voltage on the base of the output transistor becomes zero and the transistor is turned OFF. The 22u can drop further and further but the output transistor is already turned OFF and this will have no effect on the circuit.
As it drops further and further, it will be discharged a very small amount by the 15k base resistor but this will leave it partly charged and the following cycle will have very little capability to charge it.
This means the electrolytic will pass very little energy to the output stage and the circuit will be almost worthless. 

3-TRANSISTOR AUDIO AMPLIFIER

This circuit has a surprising feature. The base of the upper transistor is connected to the output of the circuit and when the bottom transistor turns ON, the 470u electro will charge and this charging current will flow through the speaker, The top transistor is turned off  during this time by the action of the two diodes. The 470u is now fully charged and the current through the speaker is zero.
The driver transistor now turns OFF and releases its grip on the whole of the output and the bottom output transistor is allowed to rise. The "pulling-down" effect of the two diodes disappears and we just have the top output transistor, the fully charged 470u and the 1k base resistor. The 470u is like a 9v battery

PUSH PULL AMPLIFIER

THE BLOCKING OSCILLATOR
This circuit uses a capacitor that is not connected to either rail and continues our discussion.

BLOCKING OSCILLATOR

The circuit is called a BLOCKING OSCILLATOR to give it a name everyone can remember.
It is really a FEEDBACK OSCILLATOR as the winding from the transformer provides POSITIVE FEEDBACK that turns the transistor ON fully - actually more than fully - and it becomes SATURATED and even SUPERSATURATED.
This is all part of the first half of the cycle and now the important thing to understand is this:
How does the cycle produce the second half of the cycle?
It is the 100u electrolytic.
But first we have to understand how the transformer works,
The transformer has two windings and we say the winding connected to the LED and collector is the primary and the winding connected to the capacitor is the secondary.  It does not matter how many turns on either winding as long as the transformer works in the circuit.
Here is the point to understand:
When the transistor turns ON, the centre tap will see 9v and the winding connected to the LED will have a lower voltage.  It does not matter how much lower but you can see the voltage decreases as you move down the turns on the diagram.
This means the voltage on the secondary winding will have a higher voltage as you move up the diagram.
When the switch is closed, the 100u will be uncharged and the positive lead of the electro will see 9v. This will pull the negative lead of the electro UP and because the base cannot rise higher than 0.7v, the positive lead will not rise higher than slightly more than 0.7v. This means the secondary winding will not be higher than 0.7v but it will deliver a current to the electro that will charge it.
This charging current will pass through the transistor to turn it ON and illuminate the LED. This current will be very high and as there is no current limiting resistor in the circuit, the LED will be damaged. That's why this circuit is so disastrous.  However let's continue.
The electro will keep charging and as it charges, the current into the base of the transistor reduces.
A point is reached where the current is not able to keep the transistor fully turned ON and it turns OFF a small amount.
Now, here's the point to understand.
The magnetic flux in the core of the transformer is a maximum and now the current is reducing. When the current reduces, the magnetic flux starts to collapse and this produces a voltage in the secondary winding that is of opposite polarity. This means the positive lead of the 100u sees a slightly lower voltage and since the electro is charged, the negative lead falls by the same amount.
This puts a lower voltage on the base and it only has to drop by 0.15v and the transistor is fully turned OFF.
That's what happens. The transistor turns OFF. The transistor  disappears from the circuit and there is nothing to discharge the 100u. That's where the 1M resistor come in. It discharges the electro very slightly so that the negative lead has 0.7v on it and this turns the transistor ON again.
This puts a high voltage spike on the positive lead that makes the electro charge and this turns the transistor ON more to create the next cycle.

ooooo00000ooooo
 

These circuit faults give you some idea of what to avoid when designing an amplifier. You must build the circuit and test it before putting in on the web for everyone else to criticise. 

Page-1 Common-Emitter stage
Page1A. more on the stages
Page-2 emitter-follower - common collector stage
Page-2A common base stage
Page 4  More on Coupling Stages
Page-5 The  FLIP FLOP
Page-6 All the "YouTube" mistakes

The REAL Transistor Amplifier

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