Emitter follower following a high impedance source

[wanna_eat_1kg_of_elecons]

Hello i am new here and have got a question about emitter followers.

I am now learning about electronics and i don't care about a circuit that works like copy it from the Internet and build it, but i am more interested in HOW it works. I know it is much easier to make a buffer with a opamp but then again i want to learn how transistors work. So i found a example and want to discus what is going on.

ok here is the question i have got a emitter follower with:
A BC337 transistor.
The collector is connected to the +15V.
The emitter is connected to a resistor (Re) this resistor is also connected to the -15V.
There is a 10k resistor between the base and a functiongenerator i name it Rb, the 10k resistor is here to simulate a high impedance of a previous stage. I am not interested in that stage it could be a voltage amplifier or a sensor or a lot of things but i don't care its just there.
There are 2 situations:
situation A: Re is 3k3 and situation B: where Re is 750Ω

When i change from situation A to situation B the output voltage drops.
When i Analyze this circuit i find:

the impedance looking into the base is Hfe*Re so i think can i say:
Vout = Vin* (Hfe*Re) / ((Hfe*Re) + Rb)

Or if i ad the diode drop and the intrinsic emitter resistance re i would get:

Vout = (Vin* ((Hfe*Re) / ((Hfe*Re) + Rb +re))) -0.6v


I know a emitter follower normally follows the input signal minus 0.6V but here there is probably a voltage drop over the Rb resistor causing a voltage drop on the output and i want to get a better idea how to calculate this situation.

can anyone with more experience please tell me if my findings are correct?

 

[alec_t]

Where in the circuit precisely are you measuring Vin and Vout?

 

[wanna_eat_1kg_of_elecons]

The input is the function generator and the output is the Emitter

 

[colin55]

Your findings are correct.
The current entering the base will be about 1/100th of the current through the emitter resistor. Obvioulsy more current will flow through the 750R compared with the 3k3, as the voltage on the base is increased. This current will be about 4 times greater. The current entering the base will also be 4 times greater than when the emitter resistor is 3k3 and thus the voltage drop across the 10k resistor will be greater. That's why the voltage on the emitter is slightly less when the emitter resistor is 750R.

 

[audioguru]

Now try your test with a darlington transistor or make your own darlington with two ordinary transistors.
Then try a Jfet.

 

[wanna_eat_1kg_of_elecons]

Thank you all for the effort it helped me a lot