Chapter 12, Problem 25P

(a)

To determine

The current in the loop using graphical method.

Answer to Problem 25P

The current in the loop, using graphical method, is $2.98\text{ mA}$ .

Explanation of Solution

Write the ideal diode equation for the current through the diode.

  $I_D=I_s\left(e^{e\Delta V/k_{B}T}-1\right)$

Here, $I_D$ is the current through the diode, $I_s$ is the reverse saturation current, $e$ is the magnitude of the electronic charge, $\Delta V$ is the voltage across the diode, $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

Substitute $1.00\text{ μA}$ for $I_s$ and $25.0\text{ meV}$ for $k_{B}T$ in the above equation.

  $\begin{array}{c}{I}_D=\left(1.00\text{ μA}\frac{{10}^{-6}\text{ A}}{1\text{ μA}}\right)\left(e^{e\Delta V/\left(25.0\text{ meV}\frac{1\text{ eV}}{{10}^3\text{ meV}}\right)}-1\right)\\ =\left({10}^{-6}\text{ A}\right)\left(e^{e\Delta V/0.025\text{ eV}}-1\right)\\ =\left({10}^{-6}\text{ A}\right)\left(e^{\Delta V/0.025\text{ V}}-1\right)\end{array}$        (I)

Write the equation for the current in the wire.

  $I_w=\frac{\epsilon -\Delta V}{R}$

Here, $I_w$ is the current through the wire, $\epsilon$ is the potential difference provided by the battery and $R$ is the value of the resistor.

Substitute $2.42\text{ V}$ for $\epsilon$ and $745\Omega$ for $R$ in the above equation.

  $I_w=\frac{2.42\text{ V}-\Delta V}{745\Omega }$        (II)

Equations (I) and (II) are plotted in figure 1 with the currents along the vertical axis and $\Delta V$ along the horizontal axis.

The two graphs meet at $\Delta V=0.200\text{ V}$ .

Conclusion:

Substitute $0.200\text{ V}$ for $\Delta V$ in equation (I) to find $I_D$ .

  $\begin{array}{c}{I}_D=\left({10}^{-6}\text{ A}\right)\left(e^{0.200\text{ V}/0.025\text{ V}}-1\right)\\ =2.98\times {10}^{-3}\text{ A}\frac{1\text{ mA}}{{10}^{-3}\text{ A}}\\ =2.98\text{ mA}\end{array}$

Substitute $0.200\text{ V}$ for $\Delta V$ in equation (II) to find $I_w$ .

  $\begin{array}{c}{I}_w=\frac{2.42\text{ V}-0.200\text{ V}}{745\Omega }\\ =2.98\times {10}^{-3}\text{ A}\frac{1\text{ mA}}{{10}^{-3}\text{ A}}\\ =2.98\text{ mA}\end{array}$

The currents agree to three digits.

Therefore, the current in the loop, using graphical method, is $2.98\text{ mA}$ .

(b)

To determine

The ohmic resistance of the diode.

Answer to Problem 25P

The ohmic resistance of the diode is $67.1\Omega$ .

Explanation of Solution

Write the equation for the ohmic resistance of the diode.

  $R_D=\frac{\Delta V}{I_D}$        (III)

Here, $R_D$ is the ohmic resistance of the diode.

Conclusion:

Substitute $0.200\text{ V}$ for $\Delta V$ and $2.98\text{ mA}$ for $I_D$ in equation (III) to find $R_D$ .

  $\begin{array}{c}{R}_D=\frac{0.200\text{ V}}{2.98\text{ mA}\frac{1\text{ A}}{{10}^3\text{ mA}}}\\ =67.1\Omega \end{array}$

Therefore, the ohmic resistance of the diode is $67.1\Omega$ .

(c)

To determine

The dynamic resistance of the diode.

Answer to Problem 25P

The dynamic resistance of the diode is $8.39\Omega$ .

Explanation of Solution

Write the equation for the dynamic resistance of the diode.

  $R_{dy}=\frac{d\left(\Delta V\right)}{d{I}_D}$

Here, $R_{dy}$ is the dynamic resistance of the diode.

Rearrange the above equation.

  $R_{dy}={\left[\frac{d{I}_D}{d\left(\Delta V\right)}\right]}^{-1}$

Put equation (I) in the above equation.

  $\begin{array}{c}{R}_{dy}=\frac{d\left(\left({10}^{-6}\text{ A}\right)\left(e^{\Delta V/0.025\text{ V}}-1\right)\right)}{d\left(\Delta V\right)}\\ =\left({10}^{-6}\text{ A}\right)\frac{d\left(e^{\Delta V/0.025\text{ V}}-1\right)}{d\left(\Delta V\right)}\\ =\left({10}^{-6}\text{ A}\right)\frac{e^{\Delta V/0.025\text{ V}}}{0.025\text{ V}}\end{array}$        (IV)

Conclusion:

Substitute $0.200\text{ V}$ for $\Delta V$ in equation (IV) to find $R_{dy}$ .

  $\begin{array}{c}{R}_{dy}=\left({10}^{-6}\text{ A}\right)\frac{e^{0.200\text{ V}/0.025\text{ V}}}{0.025\text{ V}}\\ =8.39\Omega \end{array}$

Therefore, the dynamic resistance of the diode is $8.39\Omega$ .