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CatsEyes
1 month ago

I would start like this: U1+U2=18V-Ur4-Ube

Where Ube can never be 2V, it would result in immediate smoke signs.

So, m. E., this would be like this:

U1+U2=18V-3V-0,7V=14,3V (also corrected – instead of + )

So you can calculate the current through R1+R2.

Update: Can use 2V instead of 0.7V, but this is definitely the end of the transis. 😉

If you then know the R1+R2 current, you can take the Bn for the collector current.

CatsEyes
1 month ago
Reply to  CatsEyes

Here is the entry code field Ube:

https://www.leifiphysik.de/elektronik/transistor/tests/input characteristic of the transistor

HansWurst45
1 month ago

First of all: not at all, because at 2V base-emitter voltage, each bipolar transistor is simply broken. So, as the creator of the task apparently has a damp throat around reality, one can easily count on it.

  • With the mesh rule, you calculate the voltage at the series connection of the two resistors R1 +R2 (Ucc, Ube and Ur4 are given.
  • with the voltage you calculate the current through the resistors (Rges = R1 + R2). This is then also the basic current Ib
  • With the ominous value Bn, which probably the current gain ⁠β to be at work, you can now calculate Ic
Lutz28213
1 month ago
Reply to  HansWurst45

Yes – the taskman is not very close to reality. The series and parallel circuits are already arbitrary. Not to mention the 2V between B and E. The term “reinforcement factor” is also quite unusual.