Yesterday somone helped me out a lot desinging a level shifter from 3.3V -> 5V. (here)
At first I thought I basically understand what he did, but the more I read up the more I think my understanding of transitors is wrong.
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.
Now I have a few questions I just didn't find too much about.
- Does a saturated transistor have resistance on the C->E route?
- In this schematic, why is 5V -> R2 -> Output not a working circuit in itself? Are digital inputs not grounded?
- Why connect the Output to the collector and not the emitter? (emitter -> emitting something (current/signal)?)
And in terms of Voltage:
- 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?
- 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?
- What would the Voltage be at E in this schematic?