Chapter 6, Problem 6.14TYU

(a).

To determine

The small-signal transistor parameters $g_m,r_{\pi },r_o$ .

Answer to Problem 6.14TYU

The values of small-signal parameters are,

  $\begin{array}{l}{g}_m=7.73\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\\ r_{\pi }=15.5k\Omega ,r_o=\infty \end{array}$

Explanation of Solution

Given Information:

The circuit diagram is shown below.

  

  $\begin{array}{l}\beta =120,V_{BE\left(on\right)}=0.7V,V_A=\infty \\ V_{CC}=V_{EE}=3.3\text{V,}{R}_S=500\Omega ,\\ R_L=6\text{kΩ},R_B=100\text{kΩ},R_E=12\text{kΩ},R_C=12\text{kΩ}\end{array}$

Calculation:

The coupling and bypass capacitor act like open circuit for DC analysis. The AC voltage source ( v_s ) is short circuited for DC analysis.

The modified figure is,

  

Applying Kirchhoff’s voltage law in base-emitter loop,

  $\begin{array}{l}{I}_{BQ}{R}_B+V_{BE\left(on\right)}+I_{EQ}{R}_E+V_{EE}=0\\ I_{BQ}{R}_B+V_{BE\left(on\right)}+\left(1+\beta \right)I_{BQ}{R}_E+V_{EE}=0\\ I_{BQ}=\frac{-V_{EE}-V_{BE\left(on\right)}}{R_B+\left(1+\beta \right)R_E}\\ I_{BQ}=\frac{-\left(-3.3\right)-0.7}{100\text{k}+\left(1+120\right)12\text{k}}\\ I_{BQ}=\frac{3.3-0.7}{1552\text{k}}\\ I_{BQ}=1.675\text{μA}\end{array}$

Determining the collector current,

  $\begin{array}{l}{I}_{BQ}=1.675\text{μA}\\ I_{CQ}=\beta I_{BQ}\\ I_{CQ}=\left(120\right)\left(1.675\text{μA}\right)\\ I_{CQ}=0.201\text{mA}\end{array}$

Determining the Trans-conductance,

  $\begin{array}{l}{g}_m=\frac{I_{CQ}}{V_T}\\ g_m=\frac{0.201}{0.026}\\ g_m=7.73\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.\end{array}$

Determining the diffusion resistance $r_{\pi }$ ,

  $\begin{array}{l}{g}_m{r}_{\pi }=\beta \\ r_{\pi }=\frac{\beta }{g_m}\\ r_{\pi }=\frac{120}{7.73\raisebox{1ex}{$\text{mA}$}\!\left/ \!\raisebox{-1ex}{$\text{V}$}\right.}\\ r_{\pi }=15.5\text{kΩ}\end{array}$

Determining the small-signal transistor output resistance $r_o$ ,

  $\begin{array}{l}{r}_o=\frac{V_A}{I_{CQ}}\\ r_o=\frac{\infty }{0.201\times {10}^{-3}}\\ r_o=\infty \end{array}$

(b).

To determine

The small signal current gain $\left(A_i=\frac{i_o}{i_i}\right)$ and small signal voltage gain $\left(A_v=\frac{v_o}{v_s}\right)$ .

Answer to Problem 6.14TYU

The values of current and voltage gain,

  $\begin{array}{l}{A}_i=0.654\\ A_v=6.25\end{array}$

Explanation of Solution

Given Information:

The circuit diagram is shown below.

  

  $\begin{array}{l}\beta =120,V_{BE\left(on\right)}=0.7V,V_A=\infty \\ V_{CC}=V_{EE}=3.3\text{V,}{R}_S=500\Omega ,\\ R_L=6\text{kΩ},R_B=100\text{kΩ},R_E=12\text{kΩ},R_C=12\text{kΩ}\end{array}$

Calculation:

The coupling and bypass capacitors are short circuited for small-signal analysis. The DC voltage source is short circuited for small−signal analysis.

The modified figure is,

  

Applying current division rule in output,

  $I_o=-\left(\frac{R_C}{R_L+R_C}\right)\left(g_m{V}_{\pi }\right)\mathrm{...}\left(1\right)$

Considering input circuit,

  

  $V_{\pi }=I_b\left(\frac{r_{\pi }}{1+\beta }\right)$

Applying current division rule in input,

  $\begin{array}{l}{V}_{\pi }=-\left(\frac{r_{\pi }}{1+\beta }\right)\left(\frac{R_E}{R_E+\frac{r_{\pi }}{1+\beta }}\right)I_i\\ V_{\pi }=-\left(\frac{r_{\pi }}{1+\beta }\left|\right|R_E\right)I_i\end{array}$

From equation (1),

  $\begin{array}{c}{I}_o=\left(\frac{R_C}{R_L+R_C}\right)\left(g_m\right)\left(\left(\frac{r_{\pi }}{1+\beta }\left|\right|R_E\right)I_i\right)\\ A_i=\frac{I_o}{I_i}\\ =g_m\left(\frac{R_C}{R_L+R_C}\right)\left(\frac{r_{\pi }}{1+\beta }\left|\right|R_E\right)\\ =\left(7.73\frac{\text{mA}}{\text{V}}\right)\left(\frac{\text{12}\text{k}}{12\text{k}+6\text{k}}\right)\left(\frac{15.5\text{k}}{1+120}\left|\right|12\text{k}\right)\\ =\left(5.153\right)\left(0.127\right)\\ =0.654\end{array}$

Determining the small-signal voltage gain,

  $V_o=-\left(R_c\left|\right|R_L\right)\left(g_m{V}_{\pi }\right)$

Applying voltage division rule in input,

  $V_{\pi }=-\left(\frac{\left(R_E\left|\right|\frac{r_{\pi }}{1+\beta }\right)}{R_s+\left(R_E\left|\right|\frac{r_{\pi }}{1+\beta }\right)}\right)V_s$

Plugging the value of $V_{\pi }$ ,

  $\begin{array}{c}{V}_o=\left(R_c\left|\right|R_L\right)\left(g_m\right)\left(\left(\frac{\left(R_E\left|\right|\frac{r_{\pi }}{1+\beta }\right)}{R_s+\left(R_E\left|\right|\frac{r_{\pi }}{1+\beta }\right)}\right)V_s\right)\\ A_v=\frac{V_o}{V_s}\\ =g_m\left(R_c\left|\right|R_L\right)\left(\frac{\left(R_E\left|\right|\frac{r_{\pi }}{1+\beta }\right)}{R_s+\left(R_E\left|\right|\frac{r_{\pi }}{1+\beta }\right)}\right)\\ =\left(7.73\frac{\text{mA}}{\text{V}}\right)\left(12\text{k}\left|\right|6\text{k}\right)\left(\frac{\left(12\text{k}\left|\right|\frac{15.5\text{k}}{1+120}\right)}{0.5\text{k}+\left(12\text{k}\left|\right|\frac{15.5\text{k}}{1+120}\right)}\right)\\ =7.73\times 4\times 0.202\\ =6.25\end{array}$

(c).

To determine

The values of input and output resistance.

Answer to Problem 6.14TYU

The values of input and output resistance are,

  $\begin{array}{l}{R}_i=127\Omega \\ R_o=12\text{kΩ}\end{array}$

Explanation of Solution

Given Information:

The circuit diagram is,

  

  $\begin{array}{l}\beta =120,V_{BE\left(on\right)}=0.7V,V_A=\infty \\ V_{CC}=V_{EE}=3.3\text{V,}{R}_S=500\Omega ,\\ R_L=6\text{kΩ},R_B=100\text{kΩ},R_E=12\text{kΩ},R_C=12\text{kΩ}\end{array}$

Calculation:

Considering the small signal equivalent circuit,

  

The value of input resistance is,

  

  $\begin{array}{c}{R}_i=R_E\left|\right|\frac{r_{\pi }}{1+\beta }\\ =12\text{k}\text{||}\frac{15.5\text{k}}{1+120}\\ =12\text{k}\left|\right|0.128\text{k}\\ \text{=}\frac{12\times 0.128}{12+0.128}\text{k}\\ \text{=127}\text{Ω}\end{array}$

Determining the output resistance,

  $\begin{array}{c}{R}_o=R_c\\ =12\text{kΩ}\end{array}$