The Transistor — Question 3

Question 3

Emitter feedback makes a transistor bias calculation self-consistent

An NPN transistor has $\beta=100$, $V_{BE}=.700\,\mathrm V$, $R_B=100\,\mathrm{k\Omega}$ from a 2.00 V base supply, and $R_E=1.00\,\mathrm{k\Omega}$ to ground. The collector connects through $R_C=2.00\,\mathrm{k\Omega}$ to $V_{CC}=10.0\,\mathrm V$. Assume forward active operation and then check it.

1. Use the base-emitter loop to find base, collector, and emitter currents.

2. Find $V_E$, $V_C$, and $V_{CE}$, and verify that the collector-base junction is reverse biased.

Question 3 - Solution

Emitter feedback makes a transistor bias calculation self-consistent

(a) Method and calculation

$$V_{BB}=I_BR_B+V_{BE}+(\beta+1)I_BR_E\Rightarrow\boxed{I_B=\frac{2.00-.700}{100000+101000}=6.468\,\mu\mathrm A}.$$$$\boxed{I_C=\beta I_B=.6468\,\mathrm{mA},\quad I_E=(\beta+1)I_B=.6532\,\mathrm{mA}}.$$

(b) Interpretation and further calculation

$$\boxed{V_E=.6532\,\mathrm V,\quad V_C=10.0-I_CR_C=8.7065\,\mathrm V,\quad V_{CE}=8.0532\,\mathrm V}.$$$$V_B=V_E+.700=1.3532\,\mathrm V\Rightarrow\boxed{V_C>V_B}\quad\text{consistent with forward active operation}.$$

Physical check

Emitter voltage rises with current, reducing the available base drive. Using IC in the emitter resistor would neglect the finite base-current contribution.