The Transistor — Question 1

Question 1

Current continuity connects common-base and common-emitter gain

An NPN bipolar transistor in forward active operation has $I_C=\alpha I_E$ with $\alpha=.990$. Currents denote positive magnitudes and satisfy $I_E=I_C+I_B$. Let $I_E=2.00\,\mathrm{mA}$.

1. Derive common-emitter gain $\beta=I_C/I_B$ in terms of alpha.

2. Find collector and base currents, then find the emitter current required for $I_C=5.00\,\mathrm{mA}$ at the same alpha.

Question 1 - Solution

Current continuity connects common-base and common-emitter gain

(a) Method and calculation

$$I_B=I_E-I_C=(1-\alpha)I_E.$$$$\boxed{\beta=\alpha/(1-\alpha)=99.0}.$$

(b) Interpretation and further calculation

$$\boxed{I_C=1.980\,\mathrm{mA},\quad I_B=.0200\,\mathrm{mA}=20.0\,\mu\mathrm A}.$$$$\boxed{I_E=I_C/\alpha=5.0505\,\mathrm{mA}\quad\text{when }I_C=5.00\,\mathrm{mA}}.$$

Physical check

Large beta arises because nearly all emitter current reaches the collector. It does not imply creation of charge: emitter current still equals the sum of the other two.