If I am using the 'beta' (dc gain) value of a transistor to calculate Ic and Ib in a bjt transistor , does that mean I have already assumed that the transistor is in active region?
\$\begingroup\$
\$\endgroup\$
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
Yes you have assumed this, but this is still not a bad first stab analysis technique. For example, by assuming the BJT is in its active region of operation, you can do easy back of the envelope calculations to determine parameters like the voltage drop between collector and emitter to easily tell by contradiction if the transistor could not possibly be in its active region.
It is worth pointing out however that the inverse of this technique is not strictly true, of course. In practical engineering, the datasheet of a transistor will have guidelines on how to ensure reliable active operation of the transistor part over its rated environmental tolerances.
\$\begingroup\$
\$\endgroup\$
Yes, and that includes the non-linear region where Vce becomes saturated and beta drops to 10 or worse.
Since BJT's are symmetrical, they work in reverse, but not very well, since the emitter is heavily doped much more than the collector, we say for the "forward operation" it is "Active", when collector current is flowing and thus the base must be forward biased and there is dc gain.
