Chapter 12, Problem 1E

(a)

To determine

The voltage $V_{cb}$.

Answer to Problem 1E

The voltage $V_{cb}$ is $8.35\text{V}$.

Explanation of Solution

Given data:

The voltage $V_{ec}$ is $-9\text{V}$.

The voltage $V_{eb}$ is $-0.65\text{V}$.

Calculation:

The voltage $V_{cb}$ using KVL is given by,

$\begin{array}{c}{V}_{cb}=V_{ce}+V_{eb}\\ =V_{eb}-V_{ec}\end{array}$

Substitute $-9\text{V}$ for $V_{ec}$ and $-0.65\text{V}$ for ${\text{V}}_{eb}$ in the above equation.

$\begin{array}{c}{V}_{cb}=\left(-0.65\text{V}\right)-\left(-9\text{V}\right)\\ =8.35\text{V}\end{array}$

Conclusion:

Therefore, the voltage $V_{cb}$ is $8.35\text{V}$.

(b)

To determine

The power dissipated in the junction $b-e$.

Answer to Problem 1E

The power dissipated in the junction $b-e$ is $0.65\mu \text{W}$.

Explanation of Solution

Given data:

The current flowing into the terminal $b$ ${\text{I}}_b$ is $1\mu \text{A}$.

Calculation:

The power dissipated in the junction $b-e$ $P_{be}$ is given by,

$\begin{array}{c}{P}_{be}=V_{be}{I}_b\\ =-V_{eb}{I}_b\end{array}$

Substitute $-0.65\text{V}$ for $V_{eb}$ and $1\mu \text{A}$ for $I_b$ in the above equation.

$\begin{array}{c}{P}_{be}=-\left(-0.65\text{V}\right)\left(1\mu \text{A}\right)\\ =0.65\mu \text{W}\end{array}$

Conclusion:

Therefore, the power dissipated in the junction $b-e$ is $0.65\mu \text{W}$.