Current Through a Diode. The current flowing through the semiconductor diode shown in Figure \(4.4\) is given by the equation $$ i_{\theta}=l_{e}\left(e^{\frac{w_{3}}{1 T}}-1\right) $$ where $$ \begin{aligned} &i_{D}=\text { the voltage across the diode, in volts } \\ &v_{n}=\text { the current flow through the diode, in amps } \\ &I_{e}=\text { the leakage current of the diode, in amps } \\ &q=\text { the charge on an electron, } 1.602 \times 10^{-19} \text { coulombs } \\ &k=\text { Boltzmann's constant, } 1.38 \times 10^{-23} \text { joule/K } \\ &T=\text { temperature, in kelvins }(\mathrm{K}) \end{aligned} $$ The leakage current \(I_{o}\) of the diode is \(2.0 \mu \mathrm{A}\). Write a program to calculate the current flowing through this diode for all voltages from \(-1.0 \mathrm{~V}\) to \(+0.6 \mathrm{~V}\), in \(0.1 \mathrm{~V}\) steps. Repeat this process for the following temperatures: \(75^{\circ} \mathrm{F}, 100^{\circ} \mathrm{F}\), and \(125^{\circ} \mathrm{F}\), Create a plot of the current as a function of applied voltage, with the curves for the three different temperatures appearing as different colors.

Step by step solution

01

Define constants

First, the constants given in the problem must be defined: - \(I_e = 2.0 \times 10^{-6} A\) (Leakage current) - \(q = 1.602 \times 10^{-19} C\) (Charge on an electron) - \(k = 1.38 \times 10^{-23} J/K\) (Boltzmann's constant)

02

Convert temperature to Kelvin

Next, the temperatures given in Fahrenheit should be converted to Kelvin: - \(T_1 = (75°F - 32) \times \frac{5}{9} + 273.15 K \approx 297.04 K\) - \(T_2 = (100°F - 32) \times \frac{5}{9} + 273.15 K \approx 310.93 K\) - \(T_3 = (125°F - 32) \times \frac{5}{9} + 273.15 K \approx 324.82 K\)

03

Calculate current through the diode

For each temperature, we will calculate the current through the diode for voltage values from -1.0 V to 0.6 V, in 0.1 V steps. To do this, we need to use the given equation: $$i_{D}=I_{e}\left(e^{\frac{qv_{D}}{kT}}-1\right)$$ For example, let's calculate the current for temperature \(T_1\) and voltage \(v_D = -1.0 V\): $$i_{D} = 2.0 \times 10^{-6}(e^{\frac{1.602 \times 10^{-19} \times -1.0}{1.38 \times 10^{-23} \times 297.04}} - 1) \approx -1.999 \times 10^{-6} A$$

04

Repeat for different temperatures and plot the results

Continue calculating the currents for each voltage step and temperature values, and plot these currents versus voltage. Use different colors to represent different temperatures on the graph.

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