Skip to main content
Ch 42: Molecules and Condensed Matter
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 42: Molecules and Condensed MatterProblem 29a
Chapter 42, Problem 29a

At a temperature of 290 290 K, a certain p−np-n junction has a saturation current IS=0.500I_S = 0.500 mA. Find the current at this temperature when the voltage is (i) 1.001.00 mV, (ii) −1.00-1.00 mV, (iii) 100100 mV, and (iv) −100-100 mV.

Verified step by step guidance
1
Step 1: Understand the problem. The current through a p-n junction diode can be calculated using the diode equation: \( I = I_S (e^{\frac{qV}{kT}} - 1) \), where \( I \) is the current, \( I_S \) is the saturation current, \( q \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C), \( V \) is the applied voltage, \( k \) is Boltzmann's constant (\( 1.38 \times 10^{-23} \) J/K), and \( T \) is the temperature in Kelvin.
Step 2: Substitute the given values into the equation. The temperature \( T \) is 290 K, and the saturation current \( I_S \) is 0.500 mA (or \( 0.500 \times 10^{-3} \) A). For each voltage \( V \) (1.00 mV, -1.00 mV, 100 mV, -100 mV), substitute \( V \) into the equation.
Step 3: Convert the voltage \( V \) into volts for consistency. For example, 1.00 mV = \( 1.00 \times 10^{-3} \) V, -1.00 mV = \( -1.00 \times 10^{-3} \) V, 100 mV = \( 100 \times 10^{-3} \) V, and -100 mV = \( -100 \times 10^{-3} \) V.
Step 4: Calculate the exponential term \( e^{\frac{qV}{kT}} \) for each voltage. Use \( q = 1.6 \times 10^{-19} \) C, \( k = 1.38 \times 10^{-23} \) J/K, and \( T = 290 \) K. For example, for \( V = 1.00 \times 10^{-3} \) V, calculate \( \frac{qV}{kT} \) and then find \( e^{\frac{qV}{kT}} \). Repeat this for each voltage.
Step 5: Use the diode equation to calculate the current \( I \) for each voltage. For example, for \( V = 1.00 \times 10^{-3} \) V, substitute \( I_S \), \( e^{\frac{qV}{kT}} \), and \( -1 \) into the equation \( I = I_S (e^{\frac{qV}{kT}} - 1) \). Repeat this process for \( V = -1.00 \times 10^{-3} \) V, \( V = 100 \times 10^{-3} \) V, and \( V = -100 \times 10^{-3} \) V.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

p-n Junction

A p-n junction is formed by joining p-type and n-type semiconductors, creating a region where charge carriers (holes and electrons) can recombine. This junction is crucial in semiconductor devices, as it allows current to flow in one direction while blocking it in the opposite direction, which is fundamental for diodes and transistors.
Recommended video:
Guided course
04:08
Kirchhoff's Junction Rule

Saturation Current (IS)

Saturation current (IS) is the small current that flows through a diode when it is reverse-biased. It is primarily due to the minority charge carriers in the semiconductor and is temperature-dependent. In this context, IS provides a baseline for calculating the diode's behavior under different forward and reverse bias conditions.
Recommended video:
Guided course
05:38
Intro to Current

Diode Equation

The diode equation describes the current flowing through a diode as a function of the applied voltage. It is given by I = IS (e^(qV/kT) - 1), where I is the current, V is the voltage across the diode, q is the charge of an electron, k is Boltzmann's constant, and T is the absolute temperature. This equation is essential for determining the current at various voltage levels in the given problem.
Recommended video:
Guided course
08:25
Kinematics Equations
Related Practice
Textbook Question

A forward-bias voltage of 15.015.0 mV produces a positive current of 9.259.25 mA through a p−np-n junction at 300300 K. What does the positive current become if the forward-bias voltage is reduced to 10.010.0 mV?

1
views
Textbook Question

Suppose a piece of very pure germanium is to be used as a light detector by observing, through the absorption of photons, the increase in conductivity resulting from generation of electron–hole pairs. If each pair requires 0.670.67 eV of energy, what is the maximum wavelength that can be detected? In what portion of the spectrum does it lie?

2
views
Textbook Question

The maximum wavelength of light that a certain silicon photocell can detect is 1.11 mm. (b) Explain why pure silicon is opaque.

2
views
Textbook Question

At the Fermi temperature TFT_F, EF=kTFE_F = kT_F (see Exercise 42.2242.22). When T=TFT = T_F, what is the probability that a state with energy E=2EFE = 2E_F is occupied?

2
views
Textbook Question

Pure germanium has a band gap of 0.670.67 eV. The Fermi energy is in the middle of the gap. For temperatures of 250250 K, 300300 K, and 350350 K, calculate the probability f(E)f(E) that a state at the bottom of the conduction band is occupied.

2
views