**"If I sat down with a pen and paper and nothing else, how would I go about working out what was needed and how it all works to build any of those (or anything else)"**
Here is a way to learn the very basics of getting started in transistor circuit design.

First you need to be able to bias a transistor on to use it in a circuit.

Here is a good practice learning on how to do it.

The parameters found on a data sheet, they are given as typical values.

They are only approximate values, and designing transistor circuits, is all about making the circuit perform reliable, nondependant of those component parameters.

This will teach you how to simply calculate approximate values for the resistors needed to bias a transistor.

Biasing a transistor is very necessary when designing transistor circuits, it allows you to set its working properties, wether it is an amplifier of current or voltage or a electronic switch.

Voltages are refrenced from a ground point, in this case the negative side of the supply.

The component designations are the choice of the designer.

Every schematic has its own designations.

In this schematic:

VCC = the supply voltage.

VC is the voltage at the collector of the transistor, refrenced from ground, (neg. side) of the supply.

VE is voltage at the emitter term. ref. to ground.

VB is the voltage at the base term. ref. to ground.

VBE(sat)* Base-Emitter

Saturation Voltage:

Minimum = 0.65V

To bias this transistor so it has 1/2 the supply voltage at its collector would be written as

" VC = VCC/2"

VC=4V. Meaning 4V is dropped across Q1 and RE from ground.

That leaves the remaining 4V to be dropped across RC.

Using ohms law is very extensive in designing transistor circuits.

So (4V / RC) = IC where IC is the current flowing through the transistor on the collector emitter side.

(4V / 10Kohms.) = 400uA. = IC.

This same current is in the emitter resistor RE as well so now we can solve for RE.

There are several ways to choose a resistor value for RE, in this example make the voltage VE across RE be around 1/3 of VCC.

"VE = VCC/3" ~= 2.7V and "RE = VE/IC"

Therefore "RE = 2.7V/400uA" = 6.6K ohms, choose a value available close to it, 6.65K ohms would be a good choice.

There can be extensive calculations involved in solving for the base biasing resistors, for extreme accuracy in choosing component values, however most of the time to bias a transistor it is not needed, except in extreme cases, so the best way to approach this biasing scheme is to calculate ballpark (first approximation) values, then tweak them in the protobuild.

Therefore RB1 can be calculated as 10 to 20 times RE. So choosing 10 times RE would be "RB1 = 10RE" = 66.5K ohms.

Now we look at the data sheet above and see:

VBE(sat)* Base-Emitter

Saturation Voltage:

Minimum = 0.65V

So use that value to calculate the base voltage required.

the equation to solve for base voltage is "VB = (VE + VBEmin.)" = (2.7V + 0.65V) = 3.35V = VB

Now there is a base divider current "Idiv. that needs to be solved for so RB2 can be calculated.

"Idiv. = (VB / RB1) = (3.35V / 67K ohms) ~=49uA = Idiv.

Now this equation can be used to solve for RB2 value needed to bias the transistor to the set value chosen in step one.

"RB2 = (VCC - VB) / Idiv.) = (8V - 3.35V) / 49uA ~= 94K ohms a 95.3K value could be chosen.

Note:The resistor values are the ones available in LTspice:

As always,protobuild it and tweak any values as needed.

Now apply a small signal of 1KHz at 100mV to it and see what voltage waveform is produced at its collector terminal, referenced to ground.

The input signal is 200mV pk-pk and the output signal is 294mV pk-pk. That's a Voltage gain (Av) of (294 / 200) = 1.47 = (Av).

NOTICE: the ratio of the collector branch where we have (R1 and R2) they are designated as RC and RE, if we divide RE into Rc we get close to the same value. (RC / RE) = (10K / 6.65K) ~=1.5, this is very close to the Av calculation.

That's because the value of the RC and RE can be chosen to design for a specific gain.

Lets say RC has to remain at 10K for proper impedance matching, but we can change the value of RE to give a specific amplification gain.

Try to achieve a voltage amplification of around 10 times the input signal.

Redo the calculations keeping VC at 1/2 of VCC as before for symmertical output.

Only now to solve for RE use the Av factor instead of the emitter voltage as used before.

Av = (RC / RE) so rearranging "RE = RC / Av"

RE = 10K / 10 = 1K ohms.

Because RE was changed, that will also change the emitter voltage as well so new values for RB1 and RB2 need to be recalculated.

IC is still the same since RC was not changed, therefor VE is now (400uA x 1Kohms) =0.4V

VB is now (0.65v + 0.4v) =1.05V

make RB1= (10xRE)= 10K ohms

now (Idiv = (VB / 10K)= (1.05 / 10K) = 105uA.

solving for RB2 gives {(8v - 1.05v) / 105uA} ~=66K ohms. 66.5K is available.

Now protobuild it and see how it works.

So the DC bias voltage at the collector is good at around 3.97V.

And the Av is showing (1.856V p-p / 0.2V p-p) ~=9.28 very close to the ratio of RC/RE of 10.

This is an introductory of what's involved in transistor circuit design.

I gave you the very basic of basics, in this, once you can understand how to bias a transistor you can then advance to designing small signal amplifiers, power amplifiers digital circuits and the like, but first you need to learn how to turn on a transistor how to apply it in different classes of operation, also the 3 configurations CE, CC, CB, and switching characteristics as well as amplifying ect..

There's a lot of good utube video teachings on transistor biasing and designing small signal amps and switches.

Hope this helps in getting started learning this from the very basics.