Chapter 6, Problem 6.54P

For the circuit in Figure P6.54, the parameters are $V_{CC}=5\text{V}$ and $R_E=500\text{Ω}$ . The transistor parameters are $\beta =120$ and $V_A=\infty$ . (a) Design the circuit to obtain a small−signal current gain of $A_i=i_o/i_s$ for $R_L=500\text{Ω}$ . Find $R_1,R_2$ , and also the small−signal output resistance $R_o$ . (b) Using the results of part (a), determine the current gain for $R_L=2\text{kΩ}$ .

Figure P6.54

To determine

The value of $R_1,R_2$ and also the small-signal output resistance $R_o$ .

Answer to Problem 6.54P

The values of resistances are:

  $R_1=30.1\text{kΩ}$

  $R_2=8.78\text{kΩ}$

  $R_o=29.93\text{Ω}$

Explanation of Solution

Given:

Given circuit:

  

Given Data:

  $\begin{array}{l}\beta =120\hfill \\ V_A=\infty \hfill \\ V_{CC}=5\text{V}\hfill \\ R_E=500\text{Ω}\hfill \end{array}$

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

  $I_{co}=\beta I_{ne}\dots \dots \dots \dots (2)$

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

DC analysis of given circuit

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

  

The Thevenin resistance $R_{TH}$ is

(Shorting the voltage source)

  $\begin{array}{l}{R}_m=R_1\parallel R_2\hfill \\ \therefore R_m=\frac{R_1{R}_2}{R_1+R_2}\mathrm{...........}(3)\hfill \end{array}$

  

Therefore, the Thevenin voltage from the above circuit,

  $V_{TH}=\frac{V_{\alpha C}}{R_1+R_2}{R}_2$

  $\Rightarrow V_{TH}=\frac{Vcc}{R_1}\left(\frac{R_1{R}_2}{R_1+R_2}\right)$

Using the equation (3),

  $\therefore V_{TH}=\frac{V_{CC}}{R_1}\times R_{TH}\dots \dots \dots \dots (4)$

Modified circuit as,

  

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

  $V_{CEQ}=V_{CC}-I_{EQ}{R}_E$

To calculate the value of $I_{CQ}$ , consider the quiescent collector-emitter voltage $V_{CEQ}=4.62\text{V}$

  $\begin{array}{l}\Rightarrow V_{CBQ}=V_{CC}-I_{CQ}{R}_E\left[\because I_{RQ}\cong I_{CQ}\right]\hfill \\ \Rightarrow 4.62=5-I_{CQ}\times 0.5\text{kΩ}\hfill \\ \therefore I_{CQ}=0.76\text{mA}\dots \dots \dots \mathrm{..}(5)\hfill \end{array}$

Small-signal analysis of given circuit

(Reducing the dc source to zero and the capacitors to short)

  

Diffusion resistance $r_{\pi }$ : [using equation 5] .

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

  $\Rightarrow r_{\pi }=\frac{120\times 0.026\text{V}}{0.76\times {10}^{-3}\text{A}}$

  $\therefore r_{\pi }=4.1\text{kΩ}\mathrm{..........}\text{(6)}$

The input resistance $R_{ib}$ is [using equation 6]

  $R_{bb}=r_x+(1+\beta )\left(R_E\parallel R_L\right)$

  $\left[\because r_o=\infty \right]$

  $\Rightarrow R_d=\{4.1+(1+120)(0.5\parallel 0.5)\}\text{kΩ}$

  $\left[\because R_L=0.5\text{kΩ}\right]$

  $\therefore R_s=34.35\text{kΩ}\dots \dots \dots \dots \text{(7)}$

Small signal current gain $A_i=[\text{using equation 7]}$ .

Given the small signal current gain $A_i=10$

  $A_i=\frac{(1+\beta )\left(R_1\parallel R_2\right)}{\left\{\left(R_1\parallel R_2\right)+R_B\right\}}\times \left(\frac{R_E}{R_E+R_L}\right)$

  $\begin{array}{l}\Rightarrow 10=\frac{(1+120)\times \left(R_1\parallel R_2\right)}{\left\{\left(R_1\parallel R_2\right)+34.35\text{kΩ}\right\}}\times \left(\frac{0.5\text{kΩ}}{0.5\text{kΩ}+0.5\text{kΩ}}\right)\hfill \\ \Rightarrow \left\{\left(R_1\parallel R_2\right)+34.35\text{kΩ}\right\}=\frac{121\times \left(R_1\parallel R_2\right)}{10}\times \left(\frac{0.5}{1}\right)\hfill \end{array}$

  $\Rightarrow \left\{\left(R_1\parallel R_2\right)+34.35\text{kΩ}\right\}=6.05\left(R_1\parallel R_2\right)$

  $\therefore R_1\parallel R_2=6.8\text{kΩ}\mathrm{............}\text{(8)}$

Modified circuit as,

  

The Thevenin network is shown in above figure and $I_{BQ}$ can be determined by applying

Kirchhoff's voltage law in the base-emitter loop as,

  $\begin{array}{l}-I_{BQ}{R}_m-V_{BE}(\text{on})-I_{EQ}{R}_E+V_m=0\hfill \\ \Rightarrow -I_{BQ}{R}_m-V_{BE}(\text{on})-(1+\beta )I_{BQ}{R}_E+V_{TH}=0\hfill \\ \therefore I_{BQ}=\frac{V_{TH}-V_{BE}(\text{on})}{R_m+(1+\beta )R_E}\dots \dots \dots \dots (9)\hfill \end{array}$

From equations (5), (8) and (9)

  $\begin{array}{l}\frac{I_{CQ}}{\beta }=\left[\frac{V_m-0.7}{\{6.8+(1+120)\times 0.5\}\times {10}^3}\right]\hfill \\ \Rightarrow \frac{0.76}{120}\times {10}^{-3}=\left[\frac{V_m-0.7}{\{6.8+(1+120)\times 0.5\}\times {10}^3}\right][\because \beta =120]\hfill \\ \Rightarrow \{6.8+(121\times 0.5)\}\times \frac{0.76}{120}=V_m-0.7\hfill \end{array}$

  $\therefore V_{TH}=1.13\text{V}\mathrm{.........}\text{(10)}$

Combining equations (3), (4) and (10)

  $V_{TH}=\frac{V_{CC}}{R_1}\times R_{TH}$

  $\begin{array}{l}\Rightarrow 1.13=\frac{5}{R_1}\times 6.8\text{kΩ}\hfill \\ \therefore R_1=30.1\text{kΩ}\mathrm{...........}\text{(11)}\hfill \end{array}$

From equations (3) and (11)

  $\begin{array}{l}{R}_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)\hfill \\ \Rightarrow 6.8\text{kΩ}=\left(\frac{30.1\text{kΩ}\times R_2}{30.1\text{kΩ}+R_2}\right)\hfill \\ \Rightarrow 30.1\text{kΩ}+R_2=\left(\frac{30.1\text{kΩ}\times R_2}{6.8\text{kΩ}}\right)\hfill \\ \Rightarrow 30.1\text{kΩ}+R_2=\left(4.43\times R_2\right)\hfill \\ \therefore R_2=8.78\text{kΩ}\hfill \end{array}$

Finding the output resistance $R_o:[\text{using equation}6]$

  $R_o=\left(\frac{r_{\pi }}{1+\beta }\right)\parallel R_E\parallel R_L$

  $\Rightarrow R_o=\left\{\left(\frac{4.1}{1+120}\right)\parallel 0.5\parallel 0.5\right\}\times {10}^3\text{Ω}$

  $\Rightarrow R_o=\{(0.034)\parallel 0.5\parallel 0.5\}\times {10}^3\text{Ω}$

  $\therefore R_o=29.93\text{Ω}$

To determine

The current gain.

Answer to Problem 6.54P

The current gain for $R_L=2\text{kΩ}$ is $A_i\cong 4$ .

Explanation of Solution

Given:

Given circuit:

  

Given Data:

  $\begin{array}{l}\beta =120\hfill \\ V_A=\infty \hfill \\ V_{CC}=5\text{V}\hfill \\ R_E=500\text{Ω}\hfill \end{array}$

  $R_L=2\text{kΩ}$

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

  $I_{co}=\beta I_{ne}\dots \dots \dots \dots (2)$

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

DC analysis of given circuit

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

  

The Thevenin resistance $R_{TH}$ is

(Shorting the voltage source)

  $\begin{array}{l}{R}_m=R_1\parallel R_2\hfill \\ \therefore R_m=\frac{R_1{R}_2}{R_1+R_2}\mathrm{...........}(3)\hfill \end{array}$

  

Therefore, the Thevenin voltage from the above circuit,

  $V_{TH}=\frac{V_{\alpha C}}{R_1+R_2}{R}_2$

  $\Rightarrow V_{TH}=\frac{Vcc}{R_1}\left(\frac{R_1{R}_2}{R_1+R_2}\right)$

Using equation (3),

  $\therefore V_{TH}=\frac{V_{CC}}{R_1}\times R_{TH}\dots \dots \dots \dots (4)$

Modified circuit as,

  

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

  $V_{CEQ}=V_{CC}-I_{EQ}{R}_E$

To calculate the value of $I_{CQ}$ , consider the quiescent collector-emitter voltage $V_{CEQ}=4.62\text{V}$.

  $\begin{array}{l}\Rightarrow V_{CBQ}=V_{CC}-I_{CQ}{R}_E\left[\because I_{RQ}\cong I_{CQ}\right]\hfill \\ \Rightarrow 4.62=5-I_{CQ}\times 0.5\text{kΩ}\hfill \\ \therefore I_{CQ}=0.76\text{mA}\dots \dots \dots \mathrm{..}(5)\hfill \end{array}$

Small-signal analysis of given circuit

(Reducing the dc source to zero and the capacitors to short)

  

Diffusion resistance $r_{\pi }$ : [using equation 5]

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

  $\Rightarrow r_{\pi }=\frac{120\times 0.026\text{V}}{0.76\times {10}^{-3}\text{A}}$

  $\therefore r_{\pi }=4.1\text{kΩ}\mathrm{..........}\text{(6)}$

The input resistance $R_{ib}$ is [using equation 6]

  $R_{bb}=r_x+(1+\beta )\left(R_E\parallel R_L\right)$

  $\left[\because r_o=\infty \right]$

  $\Rightarrow R_d=\{4.1+(1+120)(0.5\parallel 0.5)\}\text{kΩ}$

  $\left[\because R_L=0.5\text{kΩ}\right]$

  $\therefore R_s=34.35\text{kΩ}\dots \dots \dots \dots \text{(7)}$

Small signal current gain $A_i=[\text{using equation 7]}$ .

Given the small signal current gain $A_i=10$

  $A_i=\frac{(1+\beta )\left(R_1\parallel R_2\right)}{\left\{\left(R_1\parallel R_2\right)+R_B\right\}}\times \left(\frac{R_E}{R_E+R_L}\right)$

  $\begin{array}{l}\Rightarrow 10=\frac{(1+120)\times \left(R_1\parallel R_2\right)}{\left\{\left(R_1\parallel R_2\right)+34.35\text{kΩ}\right\}}\times \left(\frac{0.5\text{kΩ}}{0.5\text{kΩ}+0.5\text{kΩ}}\right)\hfill \\ \Rightarrow \left\{\left(R_1\parallel R_2\right)+34.35\text{kΩ}\right\}=\frac{121\times \left(R_1\parallel R_2\right)}{10}\times \left(\frac{0.5}{1}\right)\hfill \end{array}$

  $\Rightarrow \left\{\left(R_1\parallel R_2\right)+34.35\text{kΩ}\right\}=6.05\left(R_1\parallel R_2\right)$

  $\therefore R_1\parallel R_2=6.8\text{kΩ}\mathrm{............}\text{(8)}$

Modified circuit as,

  

The Thevenin network is shown in above figure and $I_{BQ}$ can be determined by applying

Kirchhoff's voltage law in the base-emitter loop as,

  $\begin{array}{l}-I_{BQ}{R}_m-V_{BE}(\text{on})-I_{EQ}{R}_E+V_m=0\hfill \\ \Rightarrow -I_{BQ}{R}_m-V_{BE}(\text{on})-(1+\beta )I_{BQ}{R}_E+V_{TH}=0\hfill \\ \therefore I_{BQ}=\frac{V_{TH}-V_{BE}(\text{on})}{R_m+(1+\beta )R_E}\dots \dots \dots \dots (9)\hfill \end{array}$

From equations (5), (8) and (9)

  $\begin{array}{l}\frac{I_{CQ}}{\beta }=\left[\frac{V_m-0.7}{\{6.8+(1+120)\times 0.5\}\times {10}^3}\right]\hfill \\ \Rightarrow \frac{0.76}{120}\times {10}^{-3}=\left[\frac{V_m-0.7}{\{6.8+(1+120)\times 0.5\}\times {10}^3}\right][\because \beta =120]\hfill \\ \Rightarrow \{6.8+(121\times 0.5)\}\times \frac{0.76}{120}=V_m-0.7\hfill \end{array}$

  $\therefore V_{TH}=1.13\text{V}\mathrm{.........}\text{(10)}$

Combining equations (3), (4) and (10)

  $V_{TH}=\frac{V_{CC}}{R_1}\times R_{TH}$

  $\begin{array}{l}\Rightarrow 1.13=\frac{5}{R_1}\times 6.8\text{kΩ}\hfill \\ \therefore R_1=30.1\text{kΩ}\mathrm{...........}\text{(11)}\hfill \end{array}$

From equations (3) and (11)

  $\begin{array}{l}{R}_{TH}=\left(\frac{R_1{R}_2}{R_1+R_2}\right)\hfill \\ \Rightarrow 6.8\text{kΩ}=\left(\frac{30.1\text{kΩ}\times R_2}{30.1\text{kΩ}+R_2}\right)\hfill \\ \Rightarrow 30.1\text{kΩ}+R_2=\left(\frac{30.1\text{kΩ}\times R_2}{6.8\text{kΩ}}\right)\hfill \\ \Rightarrow 30.1\text{kΩ}+R_2=\left(4.43\times R_2\right)\hfill \\ \therefore R_2=8.78\text{kΩ}\hfill \end{array}$

Finding the output resistance $R_o:[\text{using equation}6]$

  $R_o=\left(\frac{r_{\pi }}{1+\beta }\right)\parallel R_E\parallel R_L$

  $\Rightarrow R_o=\left\{\left(\frac{4.1}{1+120}\right)\parallel 0.5\parallel 0.5\right\}\times {10}^3\text{Ω}$

  $\Rightarrow R_o=\{(0.034)\parallel 0.5\parallel 0.5\}\times {10}^3\text{Ω}$

  $\therefore R_o=29.93\text{Ω}$

Determining the Small signal current gain $A_i:$

With load resistance $R_L=2\text{kΩ}$ and using above results,

  $\begin{array}{l}{A}_i=\frac{(1+\beta )\left(R_1\parallel R_2\right)}{\left\{\left(R_1\parallel R_2\right)+R_B\right\}}\times \left(\frac{R_E}{R_E+R_L}\right)\hfill \\ \Rightarrow A_i=\frac{(1+120)\times 6.8\text{kΩ}}{\{6.78+34.35\}\text{kΩ}}\times \left(\frac{0.5\text{kΩ}}{0.5\text{kΩ}+2\text{kΩ}}\right)\hfill \\ \therefore A_i\cong 4\hfill \end{array}$