Looking at the equations of the common base amplifier circuit shown below:
The input current πΌπ should be given by the equation: $$I_e = (V_{in} – 0.7)/R_e$$
Then, the output current πΌπ should be: $$I_c = \frac{\beta}{\beta + 1} I_e$$
Taking KVL on the loop at the right would give: $$V_{cc} – V_{cb} – I_cR_{load}= 0$$
So, this gives us that πππ’π‘ is directly propotional to (πππβ0.7), but it is also limited by
πππ such that it can't increase more than it to keep the third equation valid without getting a negative πππ. Now, my question is, I'm wondering what will happen if a big πππ(or a small π
π) were present such that πΌπ is high and in turn it wants to make a voltage drop across π
ππππ which is bigger than πππ(considering πππ to be approximately 0), what will happen after that if πΌπ was further increased?
Best Answer
You are correct : those equations are the small signal analysis.
When you run out of VCC, the transistor will saturate, and in that case Ib will increase (or an alternative view, beta will decrease, amounts to the same thing) until the equations work again.
Ultimately Ic will be controlled by Vcc and Rc and all the rest is base current.