Chapter 6, Problem 6.1EP

The circuit parameters for the circuit in Figure 6.3 are $V_{CC}=3.3\text{V}$ , $V_{BB}=0.850\text{V}$ , $R_B=180\text{kΩ}$ , and $R_C=15\text{kΩ}$ . The transistor parameters are $\beta =120$ and $V_{BE}\text{(on)}=0.7\text{V}$ . (a) Determine the Q−point values $I_{CQ}$ and $V_{CEQ}$ . (b) Find the small−signal hybrid− $\pi$ parameters $g_m$ and $r_{\pi }$ . (c) Calculate the small−signal voltage gain. (Ans. (a) $I_{CQ}\text{=0}\text{.1mA}$ , $V_{CEQ}=1.8\text{V}$ ; (b) $g_m=3.846\text{mA/V}$ , $r_{\pi }=31.2\text{kΩ}$ ; (c) $A_{\upsilon }=-8.52$ ).

To determine

The quiescent collector current $I_{CQ}$ and the Q -point value $V_{CEQ}$ for the given transistor.

Answer to Problem 6.1EP

The quiescent collector current $I_{CQ}$ is $0.1\text{ mA}$ and Q -point $V_{CEQ}$ is $1.8\text{ V}$ .

Explanation of Solution

Given:

The circuit for common emitter is shown in Figure 1.

  

The circuit parameters for the transistor circuit shown in Figure 1 are as follows:

  $\begin{array}{l}{V}_{CC}=3.3\text{ V}\\ R_C=15\text{ k}\Omega \\ R_B=180\text{ k}\Omega \\ V_{BE}\left(\text{on}\right)=0.7\text{ V}\\ V_{BB}=0.85\text{ V}\end{array}$

The value of current gain $\text{β}$ is $120$ .

Concept used:

The expression for quiescent collector current is written below:

  $I_{\text{CQ}}=\text{β}{I}_{BQ}$ ...... (1)

The expression for quiescent value $V_{CEQ}$ is written below.

  $V_{CEQ}=V_{CC}-I_{CQ}{R}_C$ ...... (2)

Calculation:

From DC analysis the ac voltage source is reduced to zero and the equation can be written as,

  $I_{BQ}=\frac{V_{BB}-V_B\left(\text{on}\right)}{R_B}$ ...... (3)

Substitute $0.85$ for $V_{BB}$ , $0.7$ for $V_B\left(\text{on}\right)$ and $180\times {10}^3$ for $R_B$ in equation (3).

  $\begin{array}{c}{I}_{BQ}=\frac{0.85-0.7}{180\times {10}^3}\\ =\frac{1}{1200}\text{ mA}\end{array}$

Substitute $120$ for $\text{β}$ and $\frac{1}{1200}$ for $I_{BQ}$ in equation (1).

  $\begin{array}{c}{I}_{CQ}=120\cdot \frac{1}{1200}\text{ mA}\\ =\frac{1}{10}\text{ mA}\\ =0.1\text{ mA}\end{array}$

Therefore, the quiescent collector current $I_{CQ}$ is $0.1\text{ mA}$ .

Substitute $3.3$ for $V_{CC}$ , $0.1\times {10}^{-3}$ for $I_{CQ}$ and $15\times {10}^3$ for $R_C$ in equation (2).

  $\begin{array}{c}{V}_{CEQ}=3.3-\left(0.1\times {10}^{-3}\right)\left(15\times {10}^3\right)\\ =3.3-1.5\\ =1.8\text{ V}\end{array}$

Therefore, the Q -point $V_{CEQ}$ is $1.8\text{ V}$ .

Conclusion:

Thus, the quiescent collector current $I_{CQ}$ is $0.1\text{ mA}$ and Q -point $V_{CEQ}$ is $1.8\text{ V}$ .

To determine

The transconductance $g_m$ and the diffusion resistance $r_{\pi }$ for small signal analysis.

Answer to Problem 6.1EP

The transconductance $g_m$ is $3.846\text{mA}/\text{V}$ and the diffusion resistance $r_{\pi }$ is $31.2\text{ k}\Omega$ .

Explanation of Solution

Concept used:

The expression for transconductance $g_m$ is written below.

  $g_m=\frac{I_{CQ}}{V_T}$ ...... (4)

Here, $V_T$ is thermal voltage and its value is $26\text{ mV}$ .

The expression for diffusion resistance is written below.

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$ ...... (5)

Calculation:

Substitute $0.1\times {10}^{-3}$ for $I_{CQ}$ and $26\times {10}^{-3}$ for $V_T$ in equation (4).

  $\begin{array}{c}{g}_m=\frac{0.1\times {10}^{-3}}{26\times {10}^{-3}}\\ =3.846\text{mA}/\text{V}\end{array}$

Therefore, the transconductance $g_m$ is $3.846\text{mA}/\text{V}$ .

Substitute $120$ for $\beta$ , $26\times {10}^{-3}$ for $V_T$ and $0.1\times {10}^{-3}$ for $I_{CQ}$ in equation (5).

  $\begin{array}{c}{r}_{\pi }=\frac{120\left(26\times {10}^{-3}\right)}{0.1\times {10}^{-3}}\\ =31.2\times {10}^3\Omega \end{array}$

Therefore, the diffusion resistance $r_{\pi }$ is $31.2\text{ k}\Omega$ .

Conclusion:

Thus, the transconductance $g_m$ is $3.846\text{mA}/\text{V}$ and the diffusion resistance $r_{\pi }$ is $31.2\text{ k}\Omega$ .

To determine

The value of small signal voltage gain $A_v$ .

Answer to Problem 6.1EP

The value of small signal voltage gain $A_v$ is $-8.52$ .

Explanation of Solution

Concept used:

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

  $A_v=-\left(g_m{R}_C\right)\cdot \left(\frac{r_{\pi }}{r_{\pi }+R_B}\right)$ ...... (6)

Calculation:

Substitute $3.846$ for $g_m$ , $15$ for $R_C$ , $31.2$ for $r_{\pi }$ and $180$ for $R_B$ in equation (6).

  $\begin{array}{c}{A}_v=-\left[\left(3.846\right)\left(15\right)\right]\cdot \left(\frac{31.2}{31.2+180}\right)\\ \approx -8.52\end{array}$

Therefore, the voltage gain $A_v$ is $-8.52$ .

Conclusion:

Thus, the value of small signal voltage gain $A_v$ is $-8.52$ .