Exercise 5.7 An amplfier using a single NPN

a)

An answer:
You cannot ignore the base current as an $\mathit{\beta fe}$ is explicitly specified.

$$\begin{array}{rcll}& I_C& =\mathit{\beta fe}{I}_B& \\ & & I_B=\frac{V_C-V_B}{R_B}& \\ & & V_B=0.6& \\ & & V_C=V_{\mathit{CC}}-R_C\cdot (I_C+I_B)& \\ & & I_B=\frac{V_{\mathit{CC}}-R_C\cdot (I_C+I_B)-0.6}{R_B}& \\ & & I_B=\frac{V_{\mathit{CC}}-0.6}{R_B+(\mathit{\beta fe}+1)R_C}& \\ & I_C& =\mathit{\beta fe}\cdot \frac{V_{\mathit{CC}}-0.6}{R_B+(\mathit{\beta fe}+1)R_C}& \end{array}$$

b)

An answer:

c)

An answer:

$$\begin{array}{rcll}& a_V& =\frac{v_{\mathit{out}}}{v_{\mathit{in}}}& \\ & & v_{\mathit{out}}=R_C\cdot (-g_m{v}_{\mathit{be}}-\frac{v_{\mathit{out}}-v_b}{R_B})& \\ & & v_e=0& \\ & & v_b=v_{\mathit{in}}& \\ & & v_{\mathit{out}}=R_C\cdot (-g_m{v}_{\mathit{in}}-\frac{v_{\mathit{out}}-v_{\mathit{in}}}{R_B})& \\ & & v_{\mathit{out}}=v_{\mathit{in}}\cdot \frac{(-g_m{R}_C+\frac{R_C}{R_B})}{1+\frac{R_C}{R_B}}& \\ & a_V& =-R_C\cdot \frac{-g_m+\frac{1}{R_B}}{1+\frac{R_C}{R_B}}& \end{array}$$

If $R_C<<R_B$ this could be simplified to $a_v\approx -g_m{R}_C$

d)

An answer:

$$\begin{array}{rcll}& R_B& =\frac{V_C-V_B}{I_B}& \\ & & V_C=V_{\mathit{CC}}-R_C{I}_C& \\ & & V_B\approx 0.6& \\ & & I_B=\frac{I_C}{\mathit{\beta fe}}& \\ & R_B& =\frac{V_{\mathit{CC}}-R_C{I}_C-0.6}{\frac{I_C}{\mathit{\beta fe}}}& \\ & & =40k\mathrm{\Omega }& \end{array}$$

e)

An answer:
If the transistor is replaced and $\mathit{\beta fe}=40$ the collector current and voltage gain will both decrease,but by far not by the amount by which $\mathit{\beta fe}$ has decreased. Rewriting the previously derived equation directly yields $I_C\approx 0.8\mathit{mA}$.

So $I_C$ decreases with $20\%$ as does the small signal parameter $g_m$.