Chapter 17, Problem 17.38P

(a)

To determine

The value of the collector current, base current and the node voltages for the given values.

Answer to Problem 17.38P

The value of the base voltage $v_{B1}$ is $1.1\text{V}$ , value of $i_{B1}$ is $1.39\text{mA}$ , $v_{B2}$ is $0.8\text{V}$ , $i_{C1}$ is $0\text{A}$ , $i_{C2}$ is $0\text{A}$ , $i_{C3}$ is $0\text{A}$ , $i_{C5}$ is $0\text{A}$ , $i_{CO}$ is $0\text{A}$ , $i_{B4}$ is $0.0394\text{mA}$ , $i_{C4}$ is $1.18\text{mA}$ and $v_{B4}$ is $4.97\text{V}$ ,

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression for the voltage $v_{B1}$ is given by,

  $v_{B1}=v_X+V_{BE{\left(on\right)}_{Q_2}}$

Substitute $0.4\text{V}$ for $v_X$ and $0.7\text{V}$ for $V_{BE{\left(on\right)}_{Q_2}}$ in the above equation.

  $\begin{array}{c}{v}_{B1}=0.4\text{V}+0.7\text{V}\\ =\text{1}\text{.1}\text{V}\end{array}$

The expression for the base current of the first transistor is given by,

$i_{B1}=\frac{5\text{V}-v_{B1}}{2.8\text{k}\Omega }$

Substitute $\text{1}\text{.1}\text{V}$ for $v_{B1}$ in the above equation.

  $\begin{array}{c}{i}_{B1}=\frac{5\text{V}-\text{1}\text{.1}\text{V}}{2.8\text{k}\Omega }\\ =1.39\text{mA}\end{array}$

The expression for the voltage $v_{B2}$ is given by,

$v_{B2}=V_{CE\left(\text{sat}\right)Q_1}+V_{CE\left(\text{sat}\right)Q_2}$

Substitute $0.4\text{V}$ for $V_{CE\left(\text{sat}\right)Q_2}$ and $0.4\text{V}$ for $V_{CE\left(\text{sat}\right)Q_1}$ inn the above equation.

$\begin{array}{c}{v}_{B2}=0.4\text{V}+0.4\text{V}\\ =0.8\text{V}\end{array}$

The first transistor are forward biased and the other transistor are operating in the cut off region therefore the value of the different collector current is given by,

  $\begin{array}{l}{i}_{C1}=0\text{A}\\ i_{C2}=0\text{A}\\ i_{C3}=0\text{A}\\ i_{C5}=0\text{A}\end{array}$

The expression for the current $i_{CO}$ is given by,

  $i_{CO}=0\text{A}$

The conversion from $\Omega$ to $\text{k}\Omega$ is given by,

$1\Omega ={10}^{-3}\text{k}\Omega$

The conversion from $760\Omega$ to $\text{k}\Omega$ is given by,

$760\Omega =760\times {10}^{-3}\text{k}\Omega$

The expression for the value of the current $i_{B4}$ is given by,

$i_{B4}=\frac{5\text{V}-V_{BE}}{0.76\text{k}\Omega +\left(1+{\beta }_F\right)3.5\text{k}\Omega }$

Substitute $0.7\text{V}$ for $V_{BE}$ and $30$ for ${\beta }_F$ in the above equation.

$\begin{array}{c}{i}_{B4}=\frac{5\text{V}-0.7\text{V}}{0.76\text{k}\Omega +\left(1+30\right)3.5\text{k}\Omega }\\ =0.0394\text{mA}\end{array}$

The expression for the value of the current $i_{C4}$ is given by,

$i_{C4}={\beta }_F{i}_{B4}$

Substitute $30$ for ${\beta }_F$ and $0.0394\text{mA}$ for $i_{B4}$ in the above equation.

  $\begin{array}{c}{i}_{C4}=\left(30\right)\left(0.0394\text{mA}\right)\\ =1.18\text{mA}\end{array}$

The expression for the base voltage of the fourth transistor is given by,

$v_{B4}=5\text{V}-i_{B4}\left(3.5\text{k}\Omega \right)$

Substitute $1.18\text{mA}$ for $i_{C4}$ in the above equation.

$\begin{array}{c}{v}_{B4}=5\text{V}-\left(1.18\text{mA}\right)\left(3.5\text{k}\Omega \right)\\ =4.97\text{V}\end{array}$

Conclusion:

Therefore, the value of the base voltage $V_{B1}$ is $1.1\text{V}$ , value of $i_{B1}$ is $1.39\text{mA}$ , $v_{B2}$ is $0.8\text{V}$ , $i_{C1}$ is $0\text{A}$ , $i_{C2}$ is $0\text{A}$ , $i_{C3}$ is $0\text{A}$ , $i_{C5}$ is $0\text{A}$ , $i_{CO}$ is $0\text{A}$ , $i_{B4}$ is $0.0394\text{mA}$ , $i_{C4}$ is $1.18\text{mA}$ and $v_{B4}$ is $4.97\text{V}$

(b)

To determine

The value of the collector current, base current and the node voltages for the given value.

Answer to Problem 17.38P

The value of the base voltage $v_{B1}$ is $\text{1}\text{.7}\text{V}$ , value of $i_{B1}$ is $1.39\text{mA}$ , $v_{B2}$ is $1.4\text{V}$ , $i_{B2}$ is $1.41\text{mA}$ , $i_{C1}$ is $1.41\text{mA}$ , $v_{B3}$ is $0\text{V}$ , $i_{C2}$ is $5.13\text{mA}$ , $i_{C3}$ is $0\text{A}$ , $i_{C5}$ is $0\text{A}$ , $i_{CO}$ is $0\text{A}$ , $i_{B4}$ is $0.0394\text{mA}$ , $i_{C4}$ is $1.18\text{mA}$ , $i_{B4}$ is $0.00369\text{mA}$ , $v_{B4}$ is $1.1\text{V}$ , $v_{BO}$ is $0.7\text{V}$ , $i_{BO}$ is $6.54\text{mA}$ .

Explanation of Solution

Calculation:

The expression for the value of voltage $v_{B1}$ is given by,

  $v_{B1}=V_{BE{\left(on\right)}_{Q_O}}+V_{BE{\left(on\right)}_{Q_2}}+V_{\gamma }\left(\text{SD}\right)$

Substitute $0.3\text{V}$ for $V_{\gamma }\left(\text{SD}\right)$ , $0.7\text{V}$ for $V_{BE{\left(on\right)}_{Q_O}}$ and $0.7\text{V}$ for $V_{BE{\left(on\right)}_{Q_2}}$ in the above equation.

  $\begin{array}{c}{v}_{B1}=0.7\text{V}+0.7\text{V+}0.3\text{V}\\ =\text{1}\text{.7}\text{V}\end{array}$

The expression for the base current of the first transistor is given by,

$i_{B1}=\frac{5\text{V}-v_{B1}}{2.8\text{k}\Omega }$

Substitute $\text{1}\text{.7}\text{V}$ for $v_{B1}$ in the above equation.

  $\begin{array}{c}{i}_{B1}=\frac{5\text{V}-\text{1}\text{.7}\text{V}}{2.8\text{k}\Omega }\\ =1.39\text{mA}\end{array}$

The expression for the voltage $v_{B2}$ is given by,

$v_{B2}=V_{BE\left(on\right)Q_O}+V_{BE\left(on\right)Q_2}$

Substitute $0.7\text{V}$ for $V_{BE\left(on\right)Q_O}$ and $0.7\text{V}$ for $V_{BE\left(on\right)Q_2}$ inn the above equation.

  $\begin{array}{c}{v}_{B2}=0.7\text{V}+0.7\text{V}\\ =1.4\text{V}\end{array}$

The expression for the value of the second transistor is given by,

$i_{B2}=i_{B1}\left(1+2{\beta }_R\right)$

Substitute $1.39\text{mA}$ for $i_{B1}$ in the above equation.

$\begin{array}{c}{i}_{B2}=\left(1.39\text{mA}\right)\left(1+2{\beta }_R\right)\\ =1.41\text{mA}\end{array}$

The expression for the value of the current $i_{C1}$ is given by,

$i_{C1}=i_{B2}$

Substitute $1.41\text{mA}$ for $i_{B2}$ in the above equation.

$i_{C1}=1.41\text{mA}$

The expression for the voltage $v_{B4}$ is given by,

$v_{B4}=V_{BE\left(on\right)Q_4}+V_{CE\left(\text{sat}\right)Q_2}$

Substitute $0.7\text{V}$ for $V_{BE\left(on\right)Q_4}$ and $0.4\text{V}$ for $V_{CE\left(\text{sat}\right)Q_2}$ inn the above equation.

  $\begin{array}{c}{v}_{B4}=0.7\text{V}+0.4\text{V}\\ =1.1\text{V}\end{array}$

The expression for the value of the current $i_{B4}$ is given by,

$i_{B4}=\frac{v_{B4}-V_{BE\left(on\right)Q_4}}{\left(1+{\beta }_F\right)3.5\text{k}\Omega }$

Substitute $1.1\text{V}$ for $v_{B4}$ , $0.7\text{V}$ for $V_{BE\left(on\right)Q_4}$ and $30$ for ${\beta }_F$ in the above equation.

  $\begin{array}{c}{i}_{B4}=\frac{1.1\text{V}-0.7\text{V}}{\left(1+30\right)3.5\text{k}\Omega }\\ =0.00369\text{mA}\end{array}$

The expression for the value of the voltage $v_{B3}$ is given by,

$v_{B3}=0\text{V}$

The expression for the value of the current $i_{B3}$ is given by,

$i_{B3}=0\text{A}$

The expression for the value of the current $i_{C3}$ is given by,

$i_{C3}=0\text{A}$

The expression for the value of the voltage $v_{BO}$ is given by,

$v_{BO}=V_{BE\left(on\right)Q_O}$

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

$v_{BO}=0.7\text{V}$

The expression for the value of the current $i_{C2}$ is given by,

$i_{C2}=\frac{5\text{V}-v_{B4}}{0.76\text{k}\Omega }$

Substitute $1.1\text{V}$ for $v_{B4}$ in the above equation.

$\begin{array}{c}{i}_{C2}=\frac{5\text{V}-1.1\text{V}}{0.76\text{k}\Omega }\\ =5.13\text{mA}\end{array}$

The expression for the value of the current $i_{BO}$ is given by,

$i_{BO}=i_{B2}+i_{C2}$

Substitute $1.41\text{mA}$ for $i_{B2}$ and $5.13\text{mA}$ for $i_{C2}$ in the above equation.

$\begin{array}{c}{i}_{BO}=1.41\text{mA}+5.13\text{mA}\\ =6.54\text{mA}\end{array}$

Conclusion:

Therefore, the value of the base voltage $v_{B1}$ is $\text{1}\text{.7}\text{V}$ , value of $i_{B1}$ is $1.39\text{mA}$ , $v_{B2}$ is $1.4\text{V}$ , $i_{B2}$ is $1.41\text{mA}$ , $i_{C1}$ is $1.41\text{mA}$ , $v_{B3}$ is $0\text{V}$ , $i_{C2}$ is $5.13\text{mA}$ , $i_{C3}$ is $0\text{A}$ , $i_{C5}$ is $0\text{A}$ , $i_{CO}$ is $0\text{A}$ , $i_{B4}$ is $0.0394\text{mA}$ , $i_{C4}$ is $1.18\text{mA}$ , $i_{B4}$ is $0.00369\text{mA}$ , $v_{B4}$ is $1.1\text{V}$ , $v_{BO}$ is $0.7\text{V}$ , $i_{BO}$ is $6.54\text{mA}$ .