Chapter 6, Problem 6.26P

For the transistor in the circuit in Figure P6.26, the parameters are $\beta =100$ and $V_A=\infty$ . (a) Determine the Q−point. (b) Find the small−signal parameters $g_m,r_{\pi }$ , and $r_o$ . (c) Find the small−signal voltage gain $A_{\upsilon }={\upsilon }_o/{\upsilon }_s$ and the small−signal current gain $A_i=i_o/i_s$ . (d) Find the input resistances $R_{ib}$ and $R_{is}$ . (e) Repeat part (c) if $R_S=0$ .

Figure P6.26

To determine

The quiescent $Q$ point for the $npn$ transistor.

Answer to Problem 6.26P

The quiescent $Q$ point $\left(V_{CEQ},I_{CQ}\right)$ for the $npn$ transistor is $\left(5.36\text{ V},1.693\text{ mA}\right)$ .

Explanation of Solution

Given:

The current gain $\beta$ is $100$ and the early voltage $V_A$ is $\infty$ .

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

  

The circuit parameters are written below.

  $\begin{array}{c}{V}^{+}=16\text{ V}\\ V^{-}=-6\text{}\text{ V}\\ R_S=0.5\text{ k}\Omega \\ R_B=10\text{ k}\Omega \\ R_E=3\text{ k}\Omega \\ R_C=6.8\text{ k}\Omega \\ R_L=6.8\text{ k}\Omega \end{array}$

Concept used:

The expression for quiescent collector current is written below.

  $I_{CQ}=\beta I_{BQ}$  ..........(1)

Calculation:

Apply dc analysis and KVL in base emitter loop.

  $\begin{array}{c}-V^{-}-I_{EQ}{R}_E+V_{BE}-I_{BQ}{R}_B=0\\ -V^{-}-\left(1+\beta \right)I_{BQ}{R}_E+V_{BE}-I_{BQ}{R}_B=0\\ I_{BQ}\left[\left(1+\beta \right)R_E+R_B\right]=V^{-}-V_{BE}\\ I_{BQ}=\frac{V^{-}-V_{BE}}{\left(1+\beta \right)R_E+R_B}\end{array}$

Substitute $6$ for $V^{-}$ , $0.7$ for $V_{BE}$ , $100$ for $\beta$ , $3$ for $R_E$ and $10$ for $R_B$ in above equation.

  $\begin{array}{c}{I}_{BQ}=\frac{6-0.7}{\left(1+100\right)\cdot 3+10}\\ =16.93\text{ μA}\end{array}$

Substitute $16.3$ for $I_{BQ}$ and $100$ for $\beta$ in equation (1).

  $\begin{array}{c}{I}_{CQ}=100\left(16.93\right)\\ =1.693\text{ mA}\end{array}$

Therefore, the quiescent collector current $I_{CQ}$ is $1.693\text{ mA}$ .

Apply KVL in base collector loop.

  $\begin{array}{c}{V}^{+}-V_{CEQ}-I_C{R}_C-I_{EQ}{R}_E-V^{-}=0\\ V^{+}-V_{CEQ}-I_C{R}_C-\left(1+\beta \right)I_{BQ}{R}_E-V^{-}=0\\ V_{CEQ}=V^{+}-I_C{R}_C-\left(1+\beta \right)I_{BQ}{R}_E-V^{-}\end{array}$

Substitute $16$ for $V^{+}$ , $1.693$ for $I_C$ , $6.8$ for $R_C$ , $100$ for $\beta$ , $16.93$ for $I_{BQ}$ , $3$ for $R_E$ and $-6$ for $V^{-}$ in above equation.

  $\begin{array}{c}{V}_{CEQ}=16-\left(1.693\right)\left(6.8\right)-\left(1+100\right)\left(16.93\right)\left(3\right)-\left(-6\right)\\ =5.36\text{ V}\end{array}$

Therefore, the $Q$ point $\left(V_{CEQ},I_{CQ}\right)$ is $\left(5.36\text{ V},1.693\text{ mA}\right)$ .

Conclusion:

Thus, the quiescent $Q$ point $\left(V_{CEQ},I_{CQ}\right)$ for the $npn$ transistor is $\left(5.36\text{ V},1.693\text{ mA}\right)$ .

To determine

The hybrid $\pi$ parameters: transconductance $g_m$ , diffusion resistance $r_{\pi }$ and output resistance $r_o$ .

Answer to Problem 6.26P

The transconductance $g_m$ is $65.115\text{mA}/\text{V}$ , the diffusion resistance $r_{\pi }$ is $1.535\text{ k}\Omega$ and the output resistance $r_o$ is $\infty$ .

Explanation of Solution

Concept used:

The expression for transconductance $g_m$ is written below.

  $g_m=\frac{I_{CQ}}{V_T}$  ..........(2)

The expression for diffusion resistance $r_{\pi }$ is written below.

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$  ..........(3)

The expression for output resistance $r_o$ is written below.

  $r_o=\frac{V_A}{I_{CQ}}$  ..........(4)

Calculation:

Substitute $1.693$ for $I_{CQ}$ and $26$ for $V_T$ in equation (2).

  $\begin{array}{c}{g}_m=\frac{1.693}{26}\\ =65.115\text{mA}/\text{V}\end{array}$

Therefore, the transconductance $g_m$ is $65.115\text{mA}/\text{V}$ .

Substitute $100$ for $\beta$ , $26$ for $V_T$ and $1.693$ in equation (3).

  $\begin{array}{c}{r}_{\pi }=\frac{100\cdot 26}{1.693}\\ =1.535\text{ k}\Omega \end{array}$

Therefore, the diffusion resistance $r_{\pi }$ is $1.535\text{ k}\Omega$ .

Substitute $\infty$ for $V_A$ and $1.693$ for $I_{CQ}$ in equation (4).

  $\begin{array}{c}{r}_o=\frac{\infty }{1.693}\\ =\infty \end{array}$

Therefore, the output resistance $r_o$ is $\infty$ .

Conclusion:

Thus, the transconductance $g_m$ is $65.115\text{mA}/\text{V}$ , the diffusion resistance $r_{\pi }$ is $1.535\text{ k}\Omega$ and the output resistance $r_o$ is $\infty$ .

To determine

The small signal voltage gain $A_v$ and current gain $A_i$ .

Answer to Problem 6.26P

The small signal voltage gain $A_v$ is $-1.06$ and the small signal current gain $A_i$ is $-1.59$ .

Explanation of Solution

Concept used:

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

  $A_v=\frac{-\beta \left(R_C\parallel R_L\right)R_B}{R_B{R}_{ib}+R_S\left(R_B+R_{ib}\right)}$  ..........(5)

The expression for small signal current gain $A_i$ is written below.

  $A_i=-\beta \left(\frac{R_C}{R_C+R_L}\right)\left(\frac{R_B}{R_B+r_{\pi }+R_{ib}}\right)$  ..........(6)

Calculation:

The input resistance $R_{ib}$ is calculated below.

  $\begin{array}{c}{R}_{ib}=r_{\pi }+\left(1+\beta \right)R_E\\ =1.535+\left(1+100\right)\cdot 3\\ =304.535\text{ k}\Omega \end{array}$

Substitute $100$ for $\beta$ , $6.8$ for $R_C$ and $R_L$ , $10$ for $R_B$ , $304.535$ for $R_{ib}$ and $0.5$ for $R_S$ in equation (5).

  $\begin{array}{c}{A}_v=\frac{-100\left(6.28\parallel 6.28\right)\left(10\right)}{10\left(304.535\right)+0.5\left(10+30.535\right)}\\ \approx -1.06\end{array}$

Therefore, the small signal voltage gain $A_v$ is $-1.06$ .

Substitute $100$ for $\beta$ , $6.8$ for $R_C$ and $R_L$ , $10$ for $R_B$ , $1.535$ for $r_{\pi }$ and $304.535$ for $R_{ib}$ in equation (6).

  $\begin{array}{c}{A}_i=-100\left(\frac{6.8}{6.8+6.8}\right)\left(\frac{10}{10+1.535+304.535}\right)\\ \approx -1.59\end{array}$

Therefore, the small signal current gain $A_i$ is $-1.59$ .

Conclusion:

Thus, the small signal voltage gain $A_v$ is $-1.06$ and the small signal current gain $A_i$ is $-1.59$ .

To determine

The input resistance $R_{ib}$ to the base and input resistance $R_{is}$ to the source.

Answer to Problem 6.26P

The input resistance $R_{ib}$ to the base is $304.535\text{ k}\Omega$ and the input resistance $R_{is}$ to the source is $10.18\text{ k}\Omega$ .

Explanation of Solution

Concept used:

The expression for input resistance $R_{ib}$ is written below.

  $R_{ib}=r_{\pi }+\left(1+\beta \right)R_E$  ..........(7)

The expression for input resistance $R_{is}$ to the input is written below.

  $R_{is}=R_S+R_B\parallel R_{ib}$  ..........(8)

Calculation:

Substitute $1.535$ for $r_{\pi }$ , $100$ for $\beta$ and $3$ for $R_E$ in equation (7).

  $\begin{array}{c}{R}_{ib}=1.535+\left(1+100\right)\cdot 3\\ =304.535\text{ k}\Omega \end{array}$

Therefore, the input resistance $R_{ib}$ is $304.535\text{ k}\Omega$ .

Substitute $0.5$ for $R_S$ , $10$ for $R_B$ and $304.535$ for $R_{ib}$ in equation (8).

  $\begin{array}{c}{R}_{is}=0.5+10\parallel 304.535\\ =10.18\text{ k}\Omega \end{array}$

Therefore, the input resistance $R_{is}$ to the source is $10.18\text{ k}\Omega$ .

Conclusion:

Thus, the input resistance $R_{ib}$ to the base is $304.535\text{ k}\Omega$ and the input resistance $R_{is}$ to the source is $10.18\text{ k}\Omega$ .

To determine

The small signal voltage gain $A_v$ and current gain $A_i$ for no source resistance.

Answer to Problem 6.26P

The small signal voltage gain $A_v$ with no source resistance is $-1.03$ and the small signal current gain $A_i$ is $-1.59$ .

Explanation of Solution

Concept used:

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

  $A_v=\frac{-\beta \left(R_C\parallel R_L\right)}{R_{ib}}$  ..........(9)

Calculation:

Substitute $100$ for $\beta$ , $6.28$ for $R_C$ and $R_L$ , and $304.535$ for $R_{ib}$ in equation (9).

  $\begin{array}{c}{A}_v=\frac{-100\left(6.28\parallel 6.28\right)}{304.535}\\ =-1.03\end{array}$

Therefore, the small signal voltage gain $A_v$ with no source resistance is $-1.03$ .

Since, small signal current gain is independent of source resistance, so it is same as obtained in part (c).

Therefore, the small signal current gain $A_i$ with no source resistance is $-1.59$ .

Conclusion:

Thus, the small signal voltage gain $A_v$ with no source resistance is $-1.03$ and the small signal current gain $A_i$ is $-1.59$ .