The Transistor — Question 8

Question 8

The Early effect adds a finite transistor output resistance

A supplied active-region model is $I_C=I_Se^{V_{BE}/V_T}(1+V_{CE}/V_A)$, with $V_A=100\,\mathrm V$. At $V_{CE}=5.00\,\mathrm V$, measured $I_C=1.00\,\mathrm{mA}$. Hold $V_{BE}$ fixed.

1. Derive and evaluate the incremental output resistance $r_o=(\partial I_C/\partial V_{CE})^{-1}$.

2. With emitter ac grounded, $g_m=.0400\,\mathrm S$ and $R_C=10.0\,\mathrm{k\Omega}$, find voltage gain $-g_m(R_C\parallel r_o)$ and the percentage reduction from ignoring $r_o$.

Question 8 - Solution

The Early effect adds a finite transistor output resistance

(a) Method and calculation

$$\frac{\partial I_C}{\partial V_{CE}}=\frac{I_Se^{V_{BE}/V_T}}{V_A}=\frac{I_C}{V_A+V_{CE}}.$$$$\boxed{r_o=(V_A+V_{CE})/I_C=105\,\mathrm{k\Omega}}.$$

(b) Interpretation and further calculation

$$\boxed{R_C\parallel r_o=9.1304\,\mathrm{k\Omega},\quad A_v=-365.22}.$$$$A_{v,\infty}=-400\Rightarrow\boxed{1-|A_v|/400=.08696=8.696\%}.$$

Physical check

The exact resistance in this specified model uses VA+VCE; VA/IC is its common large-VA approximation. Output conductance reduces gain by shunting the collector load.