Chapter 6, Problem 6.52P

The transistor current gain $\beta$ in the circuit shown in Figure P6.52 is in the range $50\le \beta \le 200$ . (a) Determine the range in the dc values of $I_E$ and $V_E$ . (b) Determine the range in the values of input resistance $R_i$ and voltage gain

Figure P6.52

To determine

The range in the dc values of $I_E\text{and }{V}_E.$

Answer to Problem 6.52P

Range of emitter current $I_{EQ}$ is,

  $2.8\text{mA}\le I_{EQ}\le 5.427\text{mA for}50\le \beta \le 200$

and range of emitter current $V_E$ is,

  $2.8\text{V}\le V_E\le 5.427\text{V for}50\le \beta \le 200$

Explanation of Solution

Given:

Range of transistor current gain $\beta$ is $50\le \beta \le 200$ .

The circuit is given below:

  

From above circuit, considering BJT s single node, then by KCL, Quiescent emitter current $I_{EQ}$ , the quiescent base current $I_{BQ}$ and the quiescent collector current $I_{CQ}$ are related as

  $I_{EQ}=I_{BQ}+I_{CQ}\mathrm{..........}\left(1\right)$

In CE mode, $I_{CQ}$ and $I_{BQ}$ are related as

  $I_{CQ}=\beta I_{BQ}\mathrm{..........}\left(2\right)$

From equation (1) and (2),

  $\begin{array}{l}{I}_{EQ}=I_{BQ}+\beta I_{BQ}\\ \therefore I_{EQ}=\left(1+\beta \right)I_{BQ}\mathrm{...............}\left(3\right)\end{array}$

Now, DC analysis of the given circuit:

Reduce source V_s to zero and open the capacitor as shown below:

  

For $\beta =50$

Applying KCL in the base-emitter loop to determine $I_{BQ}$

  $\begin{array}{l}-I_{BQ}{R}_B-V_{BE}\left(\text{on}\right)-I_{EQ}{R}_E+9=0\\ \Rightarrow -I_{BQ}{R}_B-V_{BE}\left(\text{on}\right)-\left(1+\beta \right)I_{BQ}{R}_E+9=0\\ \therefore I_{BQ}=\frac{9-V_{BE}\left(\text{on}\right)}{R_B+\left(1+\beta \right)R_E}\mathrm{..........}(4)\\ \Rightarrow I_{BQ}=\left[\frac{9-0.7}{\left\{100+\left(1+50\right)\times 1\right\}\times {10}^3}\right]\text{A}\\ \therefore I_{BQ}=0.055\text{mA}\mathrm{.........}(5)\end{array}$

From equation (3)

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

Using equation (5) give,

  $\begin{array}{l}{I}_{EQ}=\left(1+50\right)\times 0.055\text{mA}\\ \Rightarrow I_{EQ}=2.8\times {10}^{-3}\text{A}\\ \therefore I_{EQ}=2.8\text{mA with }\beta =50\end{array}$

From dc analysis of the circuit, the emitter voltage is,

  $\begin{array}{l}{V}_{EQ}=I_{EQ}{R}_{EQ}\\ \Rightarrow V_{EQ}=2.8\text{mA}\times 1\text{k}\Omega \\ \therefore V_{EQ}=2.8\text{V}\text{with}\beta \text{=50}\end{array}$

For $\beta =200$

From equation (4)

  $\begin{array}{l}{I}_{BQ}=\frac{9-V_{BE}\left(\text{on}\right)}{R_B+\left(1+\beta \right)R_E}\mathrm{..........}(4)\\ \Rightarrow I_{BQ}=\left[\frac{9-0.7}{\left\{100+\left(1+200\right)\times 1\right\}\times {10}^3}\right]\text{A}\\ \therefore I_{BQ}=0.027\text{mA}\mathrm{.........}(6)\end{array}$

From equation (3)

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

Using equation (6) give,

  $\begin{array}{l}{I}_{EQ}=\left(1+200\right)\times 0.027\text{mA}\\ \Rightarrow I_{EQ}=5.427\times {10}^{-3}\text{A}\\ \therefore I_{EQ}=5.427\text{mA with }\beta =200\end{array}$

From dc analysis of the circuit, the emitter voltage is,

  $\begin{array}{l}{V}_E=I_{EQ}{R}_E\\ \Rightarrow V_E=5.4\text{mA}\times 1\text{k}\Omega \\ \therefore V_E=2.8\text{V}\text{with}\beta \text{=200}\end{array}$

So, final range of emitter current $I_{EQ}$ is,

  $2.8\text{mA}\le I_{EQ}\le 5.427\text{mA for}50\le \beta \le 200$

and final range of emitter current $V_E$ is,

  $2.8\text{V}\le V_E\le 5.427\text{V for}50\le \beta \le 200$

To determine

Range in the values of input resistance $R_i$ and voltage gain $A_V=\frac{V_0}{V_S}$

Answer to Problem 6.52P

Final range of input resistance $R_i$ is,

  $20.62\text{k}\Omega \le R_i\le 50.36\text{k}\Omega \text{ for}50\le \beta \le 200$

and final range of input resistance $A_V$ is,

  $0.66\le R_i\le 0.826\text{ for}50\le \beta \le 200$

Explanation of Solution

Given:

Range of transistor current gain $\beta$ is $50\le \beta \le 200$ .

The circuit is given below:

  

For $\beta =50$

From equation (2)

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ \Rightarrow I_{CQ}=50\times 0.055\times {10}^{-3}\text{A}\\ \therefore I_{CQ}=2.75\text{mA}\mathrm{..........}\left(7\right)\end{array}$

Now, small signal analysis of the given circuit:

Reduce dc voltage sources to zero, dc current source to open and capacitors to short.

  

Diffusion resistance $r_R$ [using equation (7)]

  $\begin{array}{l}{r}_R=\frac{\beta V_T}{I_{CQ}}\\ \Rightarrow r_R=\frac{50\times 0.026\text{V}}{2.75\times {10}^{-3}\text{A}}\\ \Rightarrow r_R=0.4727\times {10}^3\Omega \\ \therefore r_R=0.4727\text{k}\Omega \mathrm{..........}\left(8\right)\end{array}$

Input resistance  $R_{ib}$ with $\beta =50$

  $\begin{array}{l}{R}_{ib}=r_R+\left(1+\beta \right)\left(R_E\left|\right|R_L\right)\\ \Rightarrow R_{ib}=\left\{0.4727+\left(1+50\right)\left(1\left|\right|1\right)\right\}\text{k}\Omega \\ \therefore R_{ib}=25.97\text{k}\Omega \mathrm{.........}\left(9\right)\end{array}$

Input resistance $R_i$ is

  $\begin{array}{l}{R}_i=R_B\left|\right|R_{ib}\\ \Rightarrow R_i=\left(100\left|\right|25.97\right)\text{k}\Omega \\ \therefore R_i=20.62\text{k}\Omega \text{with}\beta =50\mathrm{........}\left(10\right)\end{array}$

For $\beta =200$

From equation (2)

  $\begin{array}{l}{I}_{CQ}=\beta I_{BQ}\\ \Rightarrow I_{CQ}=200\times 0.027\times {10}^{-3}\text{A}\\ \therefore I_{CQ}=5.4\text{mA}\mathrm{..........}\left(11\right)\end{array}$

Diffusion resistance $r_R$ [using equation (11)]

  $\begin{array}{l}{r}_R=\frac{\beta V_T}{I_{CQ}}\\ \Rightarrow r_R=\frac{200\times 0.026\text{V}}{5.4\times {10}^{-3}\text{A}}\\ \Rightarrow r_R=0.9629\times {10}^3\Omega \\ \therefore r_R=0.963\text{k}\Omega \mathrm{..........}\left(12\right)\end{array}$

Input resistance  $R_{ib}$ with $\beta =200$

  $\begin{array}{l}{R}_{ib}=r_R+\left(1+\beta \right)\left(R_E\left|\right|R_L\right)\\ \Rightarrow R_{ib}=\left\{0.963+\left(1+200\right)\left(1\left|\right|1\right)\right\}\text{k}\Omega \\ \therefore R_{ib}=101.46\text{k}\Omega \mathrm{.........}\left(13\right)\end{array}$

Input resistance $R_i$ is

  $\begin{array}{l}{R}_i=R_B\left|\right|R_{ib}\\ \Rightarrow R_i=\left(100\left|\right|101.46\right)\text{k}\Omega \\ \therefore R_i=50.36\text{k}\Omega \text{with}\beta =200\mathrm{........}\left(14\right)\end{array}$

So, final range of input resistance $R_i$ is,

  $20.62\text{k}\Omega \le R_i\le 50.36\text{k}\Omega \text{ for}50\le \beta \le 200$

For $\beta =50$

Small signal voltage gain $A_V$ is with equation (9) and (10)

  $\begin{array}{l}{A}_V=\frac{\left(1+\beta \right)\left(R_E\left|\right|R_L\right)}{R_{ib}}\times \left(\frac{R_i}{R_i+R_S}\right)\\ \Rightarrow A_v=\frac{\left(1+50\right)\left(1\left|\right|1\right)\text{k}\Omega }{25.97\text{k}\Omega }\times \left(\frac{20.62\text{k}\Omega }{20.62\text{k}\Omega +10\text{k}\Omega }\right)\\ \therefore A_V=0.66\text{with }\beta =50\end{array}$

For $\beta =200$

Small signal voltage gain $A_V$ is with equation (13) and (14)

  $\begin{array}{l}{A}_V=\frac{\left(1+\beta \right)\left(R_E\left|\right|R_L\right)}{R_{ib}}\times \left(\frac{R_i}{R_i+R_S}\right)\\ \Rightarrow A_v=\frac{\left(1+200\right)\left(1\left|\right|1\right)\text{k}\Omega }{101.46\text{k}\Omega }\times \left(\frac{50.36\text{k}\Omega }{50.36\text{k}\Omega +10\text{k}\Omega }\right)\\ \therefore A_V=0.826\text{with }\beta =200\end{array}$

So, final range of input resistance $A_V$ is,

  $0.66\le R_i\le 0.826\text{ for}50\le \beta \le 200$