Chapter 6, Problem 6.12TYU

(a)

To determine

The quiescent values $I_{EQ}$ and $V_{CEQ}$ .

Answer to Problem 6.12TYU

The values are

$I_{EQ}=0.164\text{mA}$

  $V_{CEQ}=4.14\text{V}$

Explanation of Solution

Given:

Given circuit:

  

Given Data:

  $\begin{array}{l}\beta =120\hfill \\ V_A=\infty \hfill \\ V_{BE}(\text{on})=0.7\text{V}\hfill \end{array}$

Calculation:

Considering the BJT (Bipolar Junction Transistor) as single node, then, by Kirchhoff's current law, the quiescent emitter current $I_{EQ},$ the quiescent base current $I_{BQ}$ and the quiescent collector current $I_{cQ}$ are related as

  $I_{EQ}=I_{BQ}+I_{CQ}\dots \dots \dots \dots (1)$

In CE mode:

The quiescent collector current $I_{cQ}$ and the quiescent base current $I_{BQ}$ are related as

  $I_{CQ}=\beta I_{BQ}\dots \dots \dots \dots (2)$

Here, $\beta$ is the DC common emitter current in CE configuration.

From equations (1) and (2),

  $\begin{array}{l}{I}_{EQ}=I_{sQ}+I_{CQ}\hfill \\ \Rightarrow I_{EQ}=I_{BQ}+\beta I_{BQ}\hfill \\ \therefore I_{EQ}=(1+\beta )I_{BQ}\dots \dots \dots \dots \dots (3)\hfill \end{array}$

  $\text{DC}$ analysis of given circuit:

(Reducing the ac source $v_s$ to zero and the capacitor to open)

  

  $I_{BQ}$ can be determined by applying Kirchhoff s voltage law in the base-emitter loop as,

  $-I_{BQ}{R}_B-V_{BE}(\text{on})-I_{EQ}{R}_E+V^{-}=0$

Using equation (3),

  $\begin{array}{l}\Rightarrow -I_{BQ}{R}_B-V_{BE}(\text{on})-(1+\beta )I_{BQ}{R}_E+V^{-}=0\hfill \\ \Rightarrow I_{BQ}=\frac{V^{-}-V_{BE}(\text{on})}{R_B+(1+\beta )R_E}\hfill \end{array}$

  $\begin{array}{cc}\Rightarrow I_{BQ}=\frac{3.3-0.7}{100\text{kΩ}+(1+120)15\text{kΩ}}& \\ \Rightarrow I_{BQ}=\left(1.3577\times {10}^{-3}\right)\times {10}^{-3}\text{A}& \\ \therefore I_{BQ}=1.3577\mu \text{A}\mathrm{..................}\text{(4)}& \end{array}$

From equation (3),

$I_{EQ}=(1+\beta )I_{BQ}$

Using equation (4),

$\Rightarrow I_{EQ}=(1+120)\times 1.3577\mu \text{A}$

  $\Rightarrow I_{EQ}=164.28\mu \text{A}$

  $\therefore I_{EQ}=0.164\text{mA}\mathrm{..........}\text{(5)}$

Applying Kirchhoff’s voltage law around the collector-emitter loop as,

  $\begin{array}{l}{V}_{CEQ}=V^{+}+V^{-}-I_{EQ}{R}_E\hfill \\ \Rightarrow V_{CEQ}=3.3+3.3-(0.164\text{mA})(15\text{kΩ})\hfill \\ \therefore V_{CEQ}=4.14\text{V}\hfill \end{array}$

From equation (2),

$I_{CQ}=\beta I_{BQ}$

Using the equation (4),

  $\begin{array}{l}\Rightarrow I_{CQ}=120\times 1.3577\mu \text{A}\hfill \\ \Rightarrow I_{CQ}=163\mu \text{A}\hfill \end{array}$

  $\therefore I_{CQ}=0.163\text{mA}\mathrm{..........}\text{(6)}$

(b)

To determine

The small-signal voltage gain and small-signal current gain

Answer to Problem 6.12TYU

The values of small signal voltage and small signal current gain are:

  $A_v=0.892$

  $A_i=36.37$

Explanation of Solution

Given:

Given circuit:

  

Given Data:

  $\begin{array}{l}\beta =120\hfill \\ V_A=\infty \hfill \\ V_{BE}(\text{on})=0.7\text{V}\hfill \end{array}$

Calculation:

Small-signal analysis of given circuit:

[Reducing the dc source to zero and shorting all capacitors]

  

Determining the Diffusion resistance $r_{\pi }$ :

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

Using equation (6),

$\Rightarrow r_{\pi }=\frac{120\times 0.026\text{V}}{0.163\text{mA}}$

  $\therefore r_{\pi }=19.14\text{kΩ}\mathrm{...........}\text{(7)}$

Let $R_{E1}$ is the parallel combination of $R_E$ and $R_L$

  $\begin{array}{l}{R}_{E1}=R_E\parallel R_L\hfill \\ \Rightarrow R_{E1}=(15\parallel 2)\text{kΩ}\hfill \\ \therefore R_{E1}=1.765\text{kΩ}\dots \dots \dots \dots (8)\hfill \end{array}$

The input resistance $R_{ib}$ is

  $\begin{array}{l}{R}_{ib}=r_{\pi }+(1+\beta )\left(R_E\parallel r_o\parallel R_L\right)\hfill \\ \Rightarrow R_{ib}=r_{\pi }+(1+\beta )\left(R_E\parallel R_L\right)\left[\because r_o=\infty \right]\hfill \\ \Rightarrow R_{ib}=r_{\pi }+(1+\beta )R_{E1}\hfill \end{array}$

Using equations (7) and $(8),$

  $\Rightarrow R_{ib}=\{19.14+(1+120)(1.765)\}\text{kΩ}$

  $\therefore R_{ib}=232.7\text{kΩ}\mathrm{........}\text{(9)}$

Let $R_i$ is the small signal resistance

  $R_i=R_B\parallel R_{ib}$

Using the equation (9),

  $\begin{array}{l}\Rightarrow R_i=100\text{kΩ}\parallel 232.7\text{kΩ}\hfill \\ \therefore R_i=69.94\text{kΩ}\dots \dots \dots \dots (10)\hfill \end{array}$

Finding the Small signal voltage gain $A_v:$

  $A_v=\frac{(1+\beta )\left(R_E\parallel R_L\right)}{r_n+(1+\beta )\left(R_E\parallel R_L\right)}\times \left(\frac{R_i}{R_i+R_s}\right)$

Using equations (7) and $(10),$

  $\begin{array}{l}\Rightarrow A_v=\frac{(1+120)\times 1.765\text{kΩ}}{19.14\text{kΩ}+(1+120)\times 1.765\text{kΩ}}\times \left(\frac{69.94\text{kΩ}}{69.94\text{kΩ}+2\text{kΩ}}\right)\hfill \\ \Rightarrow A_v=\frac{(1+120)\times 1.765}{19.14+(1+120)\times 1.765}\times \left(\frac{69.94}{69.94+2}\right)\hfill \end{array}$

  $\therefore A_v=0.892$

Finding the Small signal current gain $A_i:$

  $\begin{array}{l}{A}_1=(1+\beta )\times \frac{R_B}{R_B+R_b}\times \frac{r_o}{r_o+R_s}\hfill \\ \Rightarrow A_i=(1+\beta )\times \frac{R_B}{R_B+R_m}\left[\because r_o=\infty \right]\hfill \end{array}$

Using equation (9),

$\Rightarrow A_i=(1+120)\times \frac{100\text{kΩ}}{100\text{kΩ}+232.7\text{kΩ}}$

  $\Rightarrow A_1=(1+120)\times \frac{100}{100+232.7}$

  $\therefore A_i=36.37$

(c)

To determine

The small-signal input resistance $R_{ib}$ and the small-signal output resistance $R_o$ .

Answer to Problem 6.12TYU

The values of small-signal input resistance $R_{ib}$ and the small-signal output resistance $R_o$ are:

  $R_{ib}=232.7\text{kΩ}$

  $R_o=172.3\Omega$

Explanation of Solution

Given:

Given circuit:

  

Given Data:

  $\begin{array}{l}\beta =120\hfill \\ V_A=\infty \hfill \\ V_{BE}(\text{on})=0.7\text{V}\hfill \end{array}$

Calculation:

Small-signal analysis of given circuit:

[Reducing the dc source to zero and shorting all capacitors]

  

Determining the Diffusion resistance $r_{\pi }$ :

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

Using the equation (6),

$\Rightarrow r_{\pi }=\frac{120\times 0.026\text{V}}{0.163\text{mA}}$

  $\therefore r_{\pi }=19.14\text{kΩ}\mathrm{...........}\text{(7)}$

Let $R_{E1}$ is the parallel combination of $R_E$ and $R_L$

  $\begin{array}{l}{R}_{E1}=R_E\parallel R_L\hfill \\ \Rightarrow R_{E1}=(15\parallel 2)\text{kΩ}\hfill \\ \therefore R_{E1}=1.765\text{kΩ}\dots \dots \dots \dots (8)\hfill \end{array}$

The input resistance $R_{ib}$ is

  $\begin{array}{l}{R}_{ib}=r_{\pi }+(1+\beta )\left(R_E\parallel r_o\parallel R_L\right)\hfill \\ \Rightarrow R_{ib}=r_{\pi }+(1+\beta )\left(R_E\parallel R_L\right)\left[\because r_o=\infty \right]\hfill \\ \Rightarrow R_{ib}=r_{\pi }+(1+\beta )R_{E1}\hfill \end{array}$

Using the equations (7) and $(8),$

  $\Rightarrow R_{ib}=\{19.14+(1+120)(1.765)\}\text{kΩ}$

  $\therefore R_{ib}=232.7\text{kΩ}\mathrm{........}\text{(9)}$

Determining the input resistance $R_{\text{a}}$

From the equation (9), we get

$\therefore R_{ib}=232.7\text{kΩ}$

The output resistance $R_o$ of the amplifier

  $R_o=\left(\frac{r_n+\left\{R_B\parallel R_s\right\}}{1+\beta }\right)\parallel R_E$

Using the equation (7),

  $\begin{array}{l}\Rightarrow R_o=\left\{\left(\frac{19.14\text{kΩ}+\{100\parallel 2\}\text{kΩ}}{1+120}\right)\parallel 15\text{kΩ}\right\}\hfill \\ \Rightarrow R_o=\left\{\left(\frac{19.14+\{100\parallel 2\}}{1+120}\right)\parallel 15\right\}\text{kΩ}\hfill \\ \Rightarrow R_o=\left\{\left(\frac{21.1}{121}\right)\parallel 15\right\}\text{kΩ}\hfill \\ \Rightarrow R_o=0.1723\text{kΩ}\hfill \end{array}$

  $\therefore R_o=172.3\Omega$