Voltage and transistors
So if I understand this correctly you have current flowing from the base -> emitter. If it is enough current it will allow current flowing from the collector -> emitter in a certain proportion (the factor being beta on the fact sheet). At some point the transistor is saturated, meaning that the current from c->e will no longer proportinally increase but has reached it's maximum.
That reads correct!
Does a saturated transistor have resistance on the C->E route?
yes. That's its "on resistance". It's typically small, but cannot be zero for anything that is not a superconductor! Typical values range from milliohms to Ohms.
In this schematic, why is 5V -> R2 -> Output not a working circuit in itself? Are digital inputs not grounded?
Whether these outputs have "pull down" or "pull up" resistors or are "high-z" (ie. "floating") depends on the output.
However, you're right, when the Collector-Emitter resisitivity is very high (i.e. no base current is flowing), then this circuit will emit a "high" signal. For many microcontroller buses, a "high by default" is right. For others, it isn't.
Why connect the Output to the collector and not the emitter? (emitter -> emitting something (current/signal)?)
The emitter in your circuit is constantly at ground level, and there's nothing to change that. So there'd nothing you'd do when feeding current into the base.
If Q1 is saturated and R2 is the only resistor in the cuircit (assuming transisors really dont have resistance) wouln't R2 cause a 5V Voltagedrop, meaning there is no more Voltage (or less if there is stuff happening with the current at the output, I think that would be a voltage divider) at the output?
Exactly that is the idea: By inducing a base current in Q1, you pull down the output from 5V level. (where it was before; since we assume no significant current \$I_\text{output}\$ flows into the output, the voltage \$U_2\$ over \$R_2\$ is \$U_2 = I_\text{output} \cdot R_2 = 0 \cdot R_2 = 0\$, and thus, output is at 5V when no current flows into Q1's base, and at (nearly) 0V when the transistor saturates.
The same thing with R1: as I understand it current flows from B->E, meaning the circut is 3.3V -> R1 -> GND. Isn't there 0V after R1?
Typically, the voltage you're looking at is the "base-emitter voltage in on state". It's typically a diode's forward voltage – something like 0.7 V or 0.2 V or so, depending on the transistor and the base current – see \$U_\text{BE}\$ in the datasheet.
What would the Voltage be at E in this schematic?
Strange question. Your transistor's emitter is tied to ground – so it's a constant 0V.
Sort of. It's more like a voltage source of tens or hundreds of mV with some series resistance, both of which vary with base current. A MOSFET behaves more like a pure resistance. You can think of it approximately as a short or very low resistance between C and E.
If you remove the transistor the output will be 5V. When the transistor is off it behaves like that.
The current flows CE in the direction of the emitter arrow. You could connect the collector to 5v and a resistor from emitter to ground, but the voltage would then be 0V for 0V in and about 2.6V for 3.3V in. The reason for that is that the output voltage subtracts from Vbe so the transistor will never saturate at all.
Yes, that's what we want to happen. Note that this circuit inverts, so 3.3V in is 0V out, and 0V in is 5V out.
As above
0V by definition