Chapter 17, Problem 17.11EP

(a)

To determine

The value of the current $i_D,{i^{\prime }}_B$ and ${i^{\prime }}_C$ .

Answer to Problem 17.11EP

The value of the current $i_D$ is $1.747\text{mA}$ , ${i^{\prime }}_B$ is $0.253\text{mA}$ and ${i^{\prime }}_C$ is $3.791\text{mA}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Apply KVL in the above circuit.

  $5\text{V}=i_C\left(2.25\text{k}\Omega \right)-V_{\gamma }\left(\text{SD}\right)+V_{BE}\left(\text{on}\right)$

Substitute $0.3\text{V}$ for $V_{\gamma }\left(\text{SD}\right)$ and $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{l}5\text{V}=i_C\left(2.25\text{k}\Omega \right)-0.3\text{V}+0.7\text{V}\\ i_C=2.0444\text{mA}\end{array}$

The expression for the value of the current ${i^{\prime }}_C$ is given by,

  ${i^{\prime }}_C=\frac{i_B+i_C}{1+\left(\frac{1}{\beta }\right)}$

Substitute $2\text{mA}$ for $i_B$ , $2.0444\text{mA}$ for $i_C$ and $15$ for $\beta$ in the above equation.

  $\begin{array}{c}{i^{\prime }}_C=\frac{2\text{mA}+2.0444\text{mA}}{1+\left(\frac{1}{15}\right)}\\ =3.791\text{mA}\end{array}$

The expression for the value of the base current ${i^{\prime }}_B$ is given by,

  ${i^{\prime }}_B=\frac{{i^{\prime }}_C}{\beta }$

Substitute $3.791\text{mA}$ for ${i^{\prime }}_C$ and $\text{15}$ for $\beta$ in the above equation.

  $\begin{array}{c}{i^{\prime }}_B=\frac{3.791\text{mA}}{15}\\ =0.253\text{mA}\end{array}$

The expression for the value of the drain current is given by,

  $i_D=i_B-{i^{\prime }}_B$

Substitute $2\text{mA}$ for $i_B$ and $0.253\text{mA}$ for ${i^{\prime }}_B$ in the above equation.

  $\begin{array}{c}{i}_D=2\text{mA}-0.253\text{mA}\\ =1.747\text{mA}\end{array}$

(b)

To determine

The value of the current $i_D$ , ${i^{\prime }}_B$ and ${i^{\prime }}_C$ .

Answer to Problem 17.11EP

The value of the current $i_D$ is $1.122\text{mA}$ , ${i^{\prime }}_B$ is $0.878\text{mA}$ and ${i^{\prime }}_C$ is $13.166\text{mA}$ .

Explanation of Solution

Calculation:

Apply KVL in the above circuit.

  $5\text{V}=i_C\left(2.25\text{k}\Omega \right)-V_{\gamma }\left(\text{SD}\right)+V_{BE}\left(\text{on}\right)$

Substitute $0.3\text{V}$ for $V_{\gamma }\left(\text{SD}\right)$ and $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ in the above equation.

  $\begin{array}{l}5\text{V}=i_C\left(2.25\text{k}\Omega \right)-0.3\text{V}+0.7\text{V}\\ i_C=2.0444\text{mA}\end{array}$

The expression for the value of the current ${i^{\prime }}_C$ is given by,

  ${i^{\prime }}_C=i_D+i_C+i_L$ ……. (1)

The expression for the value of the current $i_B$ is given by,

  $i_B={i^{\prime }}_B+i_D$ ……. (2)

The expression for the value of the current ${i^{\prime }}_C$ is given by,

  $\begin{array}{l}{i^{\prime }}_C=\beta {i^{\prime }}_B\\ {i^{\prime }}_B=\frac{{i^{\prime }}_C}{\beta }\end{array}$ ……. (3)

From equation (2) and equation (3), the value of the current $i_D$ is given by,

  $i_D=i_B-{i^{\prime }}_B$

Substitute $\frac{{i^{\prime }}_C}{\beta }$ for ${i^{\prime }}_B$ in the above equation.

  $i_D=i_B-\frac{{i^{\prime }}_C}{\beta }$

Substitute $i_B-\frac{{i^{\prime }}_C}{\beta }$ for $i_D$ in the above equation

  $\begin{array}{l}{i^{\prime }}_C=i_B-\frac{{i^{\prime }}_C}{\beta }+i_C+i_L\\ {i^{\prime }}_C=\frac{i_B+i_C+i_L}{1\left(+\frac{1}{\beta }\right)}\end{array}$

Substitute $10\text{mA}$ for $i_L$ , $2\text{mA}$ for $i_D$ , $2.0444\text{mA}$ for $i_C$ and $15$ for $\beta$ in the above equation.

  $\begin{array}{c}{i^{\prime }}_C=\frac{2\text{mA}+2.0444\text{mA}+10\text{mA}}{1+\left(\frac{1}{15}\right)}\\ =13.166\text{mA}\end{array}$

Substitute $15$ for $\beta$ and $13.166\text{mA}$ for ${i^{\prime }}_C$ in equation (3).

  $\begin{array}{c}{i^{\prime }}_B=\frac{13.166\text{mA}}{15}\\ =0.878\text{mA}\end{array}$

The expression for the value of the drain current is given by,

  $i_D=i_B-{i^{\prime }}_B$

Substitute $2\text{mA}$ for $i_B$ and $0.878\text{mA}$ for ${i^{\prime }}_B$ in the above equation.

  $\begin{array}{c}{i}_D=2\text{mA}-0.878\text{mA}\\ =1.122\text{mA}\end{array}$

Conclusion:

Therefore, the value of the current $i_D$ is $1.122\text{mA}$ , ${i^{\prime }}_B$ is $0.878\text{mA}$ and ${i^{\prime }}_C$ is $13.166\text{mA}$ .

(c)

To determine

The value of the maximum load current.

Answer to Problem 17.11EP

The maximum value of the load current is $28\text{mA}$ .

Explanation of Solution

Calculation:

The expression for the value of the $V_{CE}$ is given by,

  $V_{CE}\left(\text{sat}\right)=V_{BE}\left(\text{on}\right)-V_{\gamma \left(\text{SD}\right)}$

Substitute $0.7\text{V}$ for $V_{BE}\left(\text{on}\right)$ and $0.3\text{V}$ for $V_{\gamma }\left(\text{SD}\right)$ in the above equation.

  $\begin{array}{c}{V}_{CE}\left(\text{sat}\right)=0.7\text{V}-0.3\text{V}\\ =0.4\text{V}\end{array}$

The expression for condition of the edge saturation is given by,

  $V_{CC}-\left({i^{\prime }}_C-i_D-i_L\right)R_C\le V_{CE}\left(\text{sat}\right)$

Substitute $0.4\text{V}$ for $V_{CE}\left(\text{sat}\right)$ in the above equation.

  $V_{CC}-\left({i^{\prime }}_C-i_D-i_L\right)R_C\le 0.4\text{V}$

Substitute $0$ for $i_D$ , $5\text{V}$ for $V_{CC}$ , $2.25\text{k}\Omega$ for $R_C$ in the above equation.

  $\begin{array}{l}5\text{V}-\left({i^{\prime }}_C-i_L\right)\left(2.25\text{k}\Omega \right)\le 0.4\text{V}\\ \frac{4.6\text{V}}{2.25\text{k}\Omega }\le {i^{\prime }}_C-i_L\\ \left({i^{\prime }}_C-i_L\right)\ge 2.0444\text{mA}\end{array}$

The expression for the value of the load current is given by,

  $\begin{array}{l}\frac{i_B+i_C+i_L}{1+\left(\frac{1}{\beta }\right)}-i_L\ge 2.044\text{mA}\\ i_B+i_C+\frac{i_L}{\beta }\ge 2.044\text{mA}\left(1+\frac{1}{\beta }\right)\end{array}$

Substitute $15$ for $\beta$ , $2.044\text{mA}$ for $i_C$ , $2\text{mA}$ for $i_B$ in the above equation.

  $\begin{array}{l}2\text{mA}+2.044\text{mA}+\frac{i_L}{15}\ge 2.044\text{mA}\left(1+\frac{1}{15}\right)\\ i_L\approx 28\text{mA}\end{array}$

Conclusion:

Therefore, the maximum value of the load current is $28\text{mA}$ .