Find the collector current in Fig 2.
- The base current is negligible
- Si diode.
I have solved it, but I don't know if it's right or wrong.
Furthermore I have considered Ic=Ie+Ib which in wrongly written.
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Assume that the collector is connected to a potential such that the transistor is in active mode (eg. grounded). Assume junctions have 0.7V across them.
The current through the 10K resistors and diodes is (20V-1.4V)/20K = 0.93mA (so far so good)
So the base voltage is 10.7V above the -20V rail and therefore the emitter is at 10.0V above the -20V rail so the emitter current is ~10.0V/4.7K = 2.1mA.
Since base current is said to be negligible, collector current is also 2.1mA. Doing a quick simulation with real parts:
We can see that that's quite accurate.
If the -20V rail was much lower (say -3V) the exact magnitude of the Vbe and diode drops would be far more important, but here most of the voltage is dropped across the 10K resistors.
20V / (10k+10k) minus diode drops is approx 1mA . This means all Si diodes will be 0.6V not 0.7 @1mA
Therefore the two 10k voltage divider gives;
Vb= -10V + 2x0.6. = -8.6V which controls the emitter voltage.
Thus Ve = -8.6 - 0.6 = - 9.2V.
Now you can compute Ie which = Ic within 1% for hFE >100.