Chapter 6, Problem 6.19P

Consider the circuit shown in Figure P6.19 where the signal−source is ${\upsilon }_s=4\sin \text{ω}t\text{mV}$ . (a) For transistor parameters of $\beta =80$ and $V_A=\infty$ , (i) find the small−signal voltage gain $A_{\upsilon }={\upsilon }_o/{\upsilon }_s$ and the transconductance function $G_f=i_o/{\upsilon }_s$ , and (ii) calculate ${\upsilon }_o\left(t\right)$ and $i_o\left(t\right)$ . (b) Repeat part (a) for $\beta =120$ .

Figure P6.19

To determine

The small signal voltage gain $A_v$ and function of transconductance $G_f$ , output voltage $v_o\left(t\right)$ and output current $i_o\left(t\right)$ for the given current gain $\beta$ .

Answer to Problem 6.19P

The small signal voltage gain $A_v$ is $-26.97$ , the transconductance function $G_f$ is $-5.39\text{mA}/\text{V}$ , the output current $i_o$ is $-21.6\sin \text{ω}t\text{ μA}$ and the output voltage $v_O$ is $-0.108\sin \text{ω}t\text{ V}$ .

Explanation of Solution

Given:

The sinusoidal source voltage $v_s$ is $4\sin \text{ω}t\text{ mV}$ .

The circuit for $npn$ transistor is shown in Figure 1.

  

The current gain $\beta$ is $80$ and the early voltage $V_A$ is $\infty$ .

The circuit parameters are shown below.

  $\begin{array}{c}{V}^{+}=5\text{ V}\\ R_E=10\text{ k}\Omega \\ R_S=2.5\text{ k}\Omega \\ R_C=5\text{ k}\Omega \\ R_L=5\text{ k}\Omega \\ V^{-}=-5\text{ V}\end{array}$

Concept used:

The expression for small signal voltage gain $A_v$ is written below.

  $A_v=-g_m\left(R_C\parallel R_L\right)\left(\frac{r_{\pi }}{r_{\pi }+R_S}\right)$ …… (1)

The expression for transconductance is written below.

  $G_f=\frac{i_o}{v_s}$ …… (2)

Calculation:

Apply dc analysis and KVL in base emitter $\text{B}-\text{E}$ loop.

  $\begin{array}{c}{V}^{+}-V_{BE}\left(\text{on}\right)-I_{BQ}{R}_S-I_{EQ}{R}_E=0\\ V^{+}-V_{BE}\left(\text{on}\right)-I_{BQ}{R}_S-\left(1+\beta \right)I_{BQ}{R}_E=0\\ I_{BQ}\left(R_S+\left(1+\beta \right)R_E\right)=V^{+}-V_{BE}\left(\text{on}\right)\\ I_{BQ}=\frac{V^{+}-V_{BE}\left(\text{on}\right)}{R_S+\left(1+\beta \right)R_E}\end{array}$

Substitute $5$ for $V^{+}$ , $0.7$ for $V_{BE}\left(\text{on}\right)$ , $2.5$ for $R_S$ , $80$ for $\beta$ and $10$ for $R_E$ in above equation.

  $\begin{array}{c}{I}_{BQ}=\frac{5-0.7}{2.5+\left(1+80\right)10}\\ \approx 5.292\text{ μA}\end{array}$

The quiescent collector current is calculated as,

  $\begin{array}{c}{I}_{CQ}=\beta I_{BQ}\\ =80\left(5.292\right)\\ =0.4234\text{ mA}\end{array}$

The transconductance is calculated as,

  $\begin{array}{c}{g}_m=\frac{I_{CQ}}{V_T}\\ =\frac{0.4234}{26}\\ =16.283\text{mA}/\text{V}\end{array}$

The diffusion resistance is calculated as,

  $\begin{array}{c}{r}_{\pi }=\frac{\beta V_T}{I_{CQ}}\\ =\frac{80\cdot 26}{0.4234}\\ \approx 4.913\text{ k}\Omega \end{array}$

Substitute $16.283$ for $g_m$ , $5$ for $R_C$ , $5$ for $R_L$ , $4.913$ for $r_{\pi }$ and $2.5$ for $R_S$ in equation (1).

  $\begin{array}{c}{A}_v=-16.283\left(5\parallel 5\right)\left(\frac{4.913}{4.913+2.5}\right)\\ \approx -26.97\end{array}$

Therefore, the small signal voltage gain $A_v$ is $-26.97$ .

The output current $i_o$ is obtained as,

  $\begin{array}{c}{i}_o=\frac{v_o}{R_L}\\ =\frac{A_v{v}_s}{R_L}\\ =\frac{-26.97v_s}{5}\\ =-5.39v_s\end{array}$

Substitute $-5.39v_s$ for $i_o$ in equation (2).

  $\begin{array}{c}{G}_f=\frac{-5.39v_s}{v_s}\\ =-5.39\text{mA}/\text{V}\end{array}$

Therefore, the transconductance function $G_f$ is $-5.39\text{mA}/\text{V}$ .

The output current is calculated as,

  $\begin{array}{c}{i}_o=-5.39\left(4\sin \text{ω}t\right)\\ =-21.6\sin \text{ω}t\text{ μA}\end{array}$

Therefore, the output current $i_o$ is $-21.6\sin \text{ω}t\text{ μA}$ .

The output voltage is calculated as,

  $\begin{array}{c}{A}_v=\frac{v_O}{v_S}\\ v_O=A_v{v}_S\end{array}$

Substitute $-26.97$ for $A_v$ and $4\sin \text{ω}t$ for $v_S$ in above equation.

  $\begin{array}{c}{v}_O=-26.97\left(4\sin \text{ω}t\right)\\ \approx -0.108\sin \text{ω}t\text{ V}\end{array}$

Therefore, the output voltage $v_O$ is $-0.108\sin \text{ω}t\text{ V}$ .

Conclusion:

Thus, the small signal voltage gain $A_v$ is $-26.97$ , the transconductance function $G_f$ is $-5.39\text{mA}/\text{V}$ , the output current $i_o$ is $-21.6\sin \text{ω}t\text{ μA}$ and the output voltage $v_O$ is $-0.108\sin \text{ω}t\text{ V}$ .

To determine

The small signal voltage gain $A_v$ and function of transconductance $G_f$ , output voltage $v_o\left(t\right)$ and output current $i_o\left(t\right)$ for the given current gain $\beta$ .

Answer to Problem 6.19P

The small signal voltage gain $A_v$ is $-30.5$ , the transconductance function $G_f$ is $-6.1\text{mA}/\text{V}$ , the output current $i_o$ is $-24.4\sin \text{ω}t\text{ μA}$ and the output voltage $v_O$ is $-0.122\sin \text{ω}t\text{ V}$ .

Explanation of Solution

Given:

The current gain $\beta$ is $80$ .

Calculation:

Apply dc analysis and KVL in base emitter $\text{B}-\text{E}$ loop.

  $\begin{array}{c}{V}^{+}-V_{BE}\left(\text{on}\right)-I_{BQ}{R}_S-I_{EQ}{R}_E=0\\ V^{+}-V_{BE}\left(\text{on}\right)-I_{BQ}{R}_S-\left(1+\beta \right)I_{BQ}{R}_E=0\\ I_{BQ}\left(R_S+\left(1+\beta \right)R_E\right)=V^{+}-V_{BE}\left(\text{on}\right)\\ I_{BQ}=\frac{V^{+}-V_{BE}\left(\text{on}\right)}{R_S+\left(1+\beta \right)R_E}\end{array}$

Substitute $5$ for $V^{+}$ , $0.7$ for $V_{BE}\left(\text{on}\right)$ , $2.5$ for $R_S$ , $120$ for $\beta$ and $10$ for $R_E$ in above equation.

  $\begin{array}{c}{I}_{BQ}=\frac{5-0.7}{2.5+\left(1+120\right)10}\\ \approx 3.546\text{ μA}\end{array}$

The quiescent collector current is calculated as,

  $\begin{array}{c}{I}_{CQ}=\beta I_{BQ}\\ =120\left(3.546\right)\\ =0.4256\text{ mA}\end{array}$

The transconductance is calculated as,

  $\begin{array}{c}{g}_m=\frac{I_{CQ}}{V_T}\\ =\frac{0.4256}{26}\\ =16.366\text{mA}/\text{V}\end{array}$

The diffusion resistance is calculated as,

  $\begin{array}{c}{r}_{\pi }=\frac{\beta V_T}{I_{CQ}}\\ =\frac{120\cdot 26}{0.4256}\\ \approx 7.33\text{ k}\Omega \end{array}$

Substitute $16.366$ for $g_m$ , $5$ for $R_C$ , $5$ for $R_L$ , $7.33$ for $r_{\pi }$ and $2.5$ for $R_S$ in equation (1).

  $\begin{array}{c}{A}_v=-16.366\left(5\parallel 5\right)\left(\frac{7.33}{7.33+2.5}\right)\\ \approx -30.5\end{array}$

Therefore, the small signal voltage gain $A_v$ is $-30.5$ .

The output current $i_o$ is obtained as,

  $\begin{array}{c}{i}_o=\frac{v_o}{R_L}\\ =\frac{A_v{v}_s}{R_L}\\ =\frac{-30.5v_s}{5}\\ =-6.1v_s\end{array}$

Substitute $-6.1v_s$ for $i_o$ in equation (2).

  $\begin{array}{c}{G}_f=\frac{-6.1v_s}{v_s}\\ =-6.1\text{mA}/\text{V}\end{array}$

Therefore, the transconductance function $G_f$ is $-6.1\text{mA}/\text{V}$ .

The output current is calculated as,

  $\begin{array}{c}{i}_o=-6.1\left(4\sin \text{ω}t\right)\\ =-24.4\sin \text{ω}t\text{ μA}\end{array}$

Therefore, the output current $i_o$ is $-24.4\sin \text{ω}t\text{ μA}$ .

The output voltage is calculated as,

  $\begin{array}{c}{A}_v=\frac{v_O}{v_S}\\ v_O=A_v{v}_S\end{array}$

Substitute $-30.5$ for $A_v$ and $4\sin \text{ω}t$ for $v_S$ in above equation.

  $\begin{array}{c}{v}_O=-30.5\left(4\sin \text{ω}t\right)\\ \approx -0.122\sin \text{ω}t\text{ V}\end{array}$

Therefore, the output voltage $v_O$ is $-0.122\sin \text{ω}t\text{ V}$ .

Conclusion:

Thus, the small signal voltage gain $A_v$ is $-30.5$ , the transconductance function $G_f$ is $-6.1\text{mA}/\text{V}$ , the output current $i_o$ is $-24.4\sin \text{ω}t\text{ μA}$ and the output voltage $v_O$ is $-0.122\sin \text{ω}t\text{ V}$ .