Chapter 17, Problem 17.9EP

The parameters of the TIL NAND circuit in Figure 17.24 are: $R_1=12\text{k}\Omega$ , $R_2=4\text{k}\Omega$ , $R_B=2\text{k}\Omega$ , and $R_C=6\text{k}\Omega$ . Assume ${\beta }_F\equiv \beta =25$ and ${\beta }_R=0.1$ (for each input emitter). For a no−load condition, determine the base and collector currents in each transistor for: (a) ${\upsilon }_X={\upsilon }_Y=0.1\text{V}$ and (b) ${\upsilon }_X={\upsilon }_Y=5\text{V}$ . (Ans. (a) $i_1=i_{B1}=0.342\text{mA}$ , $i_{C1}\cong 0$ , $i_{B2}=i_{C2}=0$ , $i_{Bo}=i_{Co}=0$ ; (b) $i_1=i_{B1}=0.225\text{mA}$ , $i_{B2}=\left|i_{C1}\right|=0.27\text{mA}$ , $i_2=i_{C2}=1.025\text{mA}$ $i_{Bo}=0.895\text{mA}$ , $i_{Co}=0.8167\text{mA}$ )

To determine

The value of the base and the collector current in each of the transistor.

Answer to Problem 17.9EP

The value of the currents are $i_{C1}$ is $0$ , $i_{C2}$ is $0$ , $i_{B2}$ is $0$ , $i_{BO}$ is $0$ , $i_{CO}$ is $0$ , $i_{B1}$ is $0.342\text{mA}$ and $i_1$ is $0.342\text{mA}$ ,

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

  

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

  $v_{B2}=v_X+v_{CE}\left(\text{sat}\right)$

Substitute $0.1\text{V}$ for $v_X$ and $0.1\text{V}$ for $v_{CE}\left(\text{sat}\right)$ in the above equation.

  $\begin{array}{c}{v}_{B2}=0.1\text{V+}0.1\text{V}\\ =0.2\text{V}\end{array}$

The second transistor and the transistor at the output are both cut off, then the base voltage $v_{B1}$ is given by,

  $v_{B1}=v_X+v_{BE}\left(\text{sat}\right)$

Substitute $0.1\text{V}$ for $v_X$ and $0.8\text{V}$ for $v_{BE}\left(\text{sat}\right)$ in the above equation.

  $\begin{array}{c}{v}_{B1}=0.1\text{V}+0.8\text{V}\\ =\text{0}\text{.9V}\end{array}$

The expression for the value of the current $i_1$ is given by,

  $i_1=\frac{5\text{V}-v_{B1}}{12\text{k}\Omega }$

Substitute $0.9\text{V}$ for $v_{B1}$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{5\text{V}-0.9\text{V}}{12\text{k}\Omega }\\ =0.342\text{mA}\end{array}$

The expression to determine the value of the base current $i_{B1}$ is given by,

  $i_{B1}=i_1$

Substitute $0.342\text{mA}$ for $i_1$ in the above equation.

  $i_{B1}=0.342\text{mA}$

The transistor two and the output transistor are cutoff, therefore the value of the current are given by,

  $\begin{array}{l}{i}_{C1}=0\\ i_{C2}=0\\ i_{B2}=0\\ i_{BO}=0\end{array}$

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

  $i_{CO}=0$ .

Conclusion:

Therefore, the value of the currents are $i_{C1}$ is $0$ , $i_{C2}$ is $0$ , $i_{B2}$ is $0$ , $i_{BO}$ is $0$ , $i_{CO}$ is $0$ , $i_{B1}$ is $0.342\text{mA}$ and $i_1$ is $0.342\text{mA}$

To determine

The value of the base and the collector current in each of the transistor.

Answer to Problem 17.9EP

The value of the currents are $i_{C1}$ is $0.27\text{mA}$ , $i_2$ is $1.025\text{mA}$ , $i_{C2}$ is $1.025\text{mA}$ , $i_{B2}$ is $0.27\text{mA}$ , $i_{BO}$ is $0.895\text{mA}$ , $i_{CO}$ is $0.8167\text{mA}$ , $i_{B1}$ is $0.255\text{mA}$ and $i_1$ is $0.255\text{mA}$ .

Explanation of Solution

Calculation:

The input transistor is biased with inverse active mode when the input voltage is $5\text{V}$ .

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

  $v_{B1}=v_{BE}{\left(\text{sat}\right)}_{Q_O}+v_{BE}{\left(\text{sat}\right)}_{Q_1}+v_{BE}{\left(\text{sat}\right)}_{Q_1}$

Substitute $0.7\text{V}$ for $v_{BE}{\left(\text{sat}\right)}_{Q_1}$ , $0.8\text{V}$ for $v_{BE}{\left(\text{sat}\right)}_{Q_O}$ and $0.8\text{V}$ for $v_{BE}{\left(\text{sat}\right)}_{Q2}$ in the above equation.

  $\begin{array}{c}{v}_{B2}=0.8\text{V}+0.8\text{V}+0.7\text{V}\\ =2.3\text{V}\end{array}$

The expression to determine the value of the collector voltage is given by,

  $v_{C2}=V_{BE}{\left(\text{sat}\right)}_{Q_O}+V_{CE}{\left(\text{sat}\right)}_{Q_2}$

Substitute $0.1\text{V}$ for $V_{CE}{\left(\text{sat}\right)}_{Q_2}$ and $0.8\text{V}$ for $v_{BE}{\left(\text{sat}\right)}_{Q_O}$ in the above equation.

  $\begin{array}{c}{v}_{C2}=0.1\text{V}+0.8\text{V}\\ =0.9\text{V}\end{array}$

The expression to determine the value of the current $i_1$ is given by,

  $i_1=\frac{5\text{V}-v_{B1}}{12\text{k}\Omega }$

Substitute $2.3\text{V}$ for $v_{B1}$ in the above equation.

  $\begin{array}{c}{i}_1=\frac{5\text{V}-2.3\text{V}}{12\text{k}\Omega }\\ =0.255\text{mA}\end{array}$

The expression to determine the value of the base current $i_{B1}$ is given by,

  $i_{B1}=i_1$

Substitute $0.255\text{mA}$ for $i_1$ in the above equation.

  $i_{B1}=0.255\text{mA}$

The expression to determine the value of the current $i_{B2}$ is given by,

  $i_{B2}=\left(1+2{\beta }_R\right)i_1$

Substitute $0.1$ for ${\beta }_R$ and $0.225\text{mA}$ for $i_1$ in the above equation.

  $\begin{array}{c}{i}_{B2}=\left(1+2\left(0.1\right)\right)\left(0.225\text{mA}\right)\\ =0.27\text{mA}\end{array}$

The expression to determine the value of the collector current $i_{C1}$ is given by,

  $i_{C1}=i_{B2}$

Substitute $0.27\text{mA}$ for $i_{B2}$ in the above equation.

  $i_{C1}=0.27\text{mA}$

The expression to determine the value of the current $i_2$ is given by,

  $i_2=\frac{5\text{V}-v_{C2}}{4\text{k}\Omega }$

Substitute $0.9\text{V}$ for $v_{C2}$ in the above equation.

  $\begin{array}{c}{i}_2=\frac{5\text{V}-0.9\text{V}}{4\text{k}\Omega }\\ =1.025\text{mA}\end{array}$

The expression to determine the value of the base current $i_{C2}$ is given by,

  $i_{C2}=i_2$

Substitute $1.025\text{mA}$ for $i_2$ in the above equation.

  $i_{C2}=1.025\text{mA}$

The expression to determine the value of the base current $i_{E2}$ is given by,

  $i_{E2}=i_2+i_{B2}$

Substitute $1.025\text{mA}$ for $i_2$ and $0.27\text{mA}$ for $i_{B2}$ in the above equation.

  $\begin{array}{c}{i}_{E2}=0.27\text{mA}+1.025\text{mA}\\ =1.295\text{mA}\end{array}$

The expression to determine the value of the current $i_4$ is given by,

  $i_4=\frac{V_{BE}\left(\text{sat}\right)}{2\text{k}\Omega }$

Substitute $0.8\text{V}$ for $V_{BE}\left(\text{sat}\right)$ in the above equation.

  $\begin{array}{c}{i}_4=\frac{0.8\text{V}}{2\text{k}\Omega }\\ =0.4\text{mA}\end{array}$

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

  $i_{BO}=i_{E2}-i_4$

Substitute $1.295\text{mA}$ for $i_{E2}$ and $0.4\text{mA}$ for $i_4$ in the above equation.

  $\begin{array}{c}{i}_{BO}=1.295\text{mA}-0.4\text{mA}\\ =0.895\text{mA}\end{array}$

The expression to determine the value of the current $i_{CO}$ is given by,

  $i_{CO}=\frac{5\text{V}-V_{CE}\left(\text{sat}\right)}{6\text{k}\Omega }$

Substitute $0.1\text{V}$ for $V_{CE}\left(\text{sat}\right)$ in the above equation.

  $\begin{array}{c}{i}_{CO}=\frac{5\text{V}-0.1\text{V}}{6\text{k}\Omega }\\ =0.8167\text{mA}\end{array}$

Conclusion:

Therefore, the value of the currents are $i_{C1}$ is $0.27\text{mA}$ , $i_2$ is $1.025\text{mA}$ , $i_{C2}$ is $1.025\text{mA}$ , $i_{B2}$ is $0.27\text{mA}$ , $i_{BO}$ is $0.895\text{mA}$ , $i_{CO}$ is $0.8167\text{mA}$ , $i_{B1}$ is $0.255\text{mA}$ and $i_1$ is $0.255\text{mA}$ .