Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Question
Chapter 42, Problem 32P
(a)
To determine
The bias voltage for which
(b)
To determine
The bias voltage for which
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes.
1
A bar of n-type germanium 10mmx1mmx1mm. The electron density in the bar is
7×10²1 m³ and B=0.2 Wb/m². If one millivolt is applied across the long ends of the bar,
determine the current through the bar, the Hall coefficient and the Hall voltage.
Assume n = 0.39 m²/v.s.
Ans: 43.6 μA; 8.9×10-4 m³/C; 7.76 μV.
In a Si semiconductor sample of 200 am length at 600 K the hole concentration as a' function
of the sample length follows a quadratic relation of the form p (x) = 1 x1015x, at equilibrium
the value of the electric field at 160 jum will be:
O 1.935 V/cm
O 3.250 V/cm
O 5805 V/cm
O 55.56 V/cm
O 6.450 V/cm
Chapter 42 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 42.1 - For each of the following atoms or molecules,...Ch. 42.2 - Prob. 42.2QQCh. 42.2 - Prob. 42.3QQCh. 42 - Prob. 1PCh. 42 - Prob. 2PCh. 42 - Prob. 3PCh. 42 - Prob. 4PCh. 42 - Prob. 5PCh. 42 - The photon frequency that would be absorbed by the...Ch. 42 - Prob. 8P
Ch. 42 - Prob. 9PCh. 42 - Prob. 10PCh. 42 - (a) In an HCl molecule, take the Cl atom to be the...Ch. 42 - Prob. 12PCh. 42 - Prob. 13PCh. 42 - Prob. 14PCh. 42 - Prob. 15PCh. 42 - Prob. 16PCh. 42 - Prob. 17PCh. 42 - Prob. 19PCh. 42 - Prob. 21PCh. 42 - Prob. 22PCh. 42 - Prob. 23PCh. 42 - Prob. 24PCh. 42 - Prob. 25PCh. 42 - Prob. 26PCh. 42 - Prob. 27PCh. 42 - Prob. 28PCh. 42 - Prob. 29PCh. 42 - Prob. 30PCh. 42 - Prob. 32PCh. 42 - Prob. 33PCh. 42 - Prob. 35PCh. 42 - Prob. 36APCh. 42 - Prob. 37APCh. 42 - Prob. 39APCh. 42 - Prob. 40APCh. 42 - As an alternative to Equation 42.1, another useful...
Knowledge Booster
Similar questions
- At a certain temperature, the electron and hole mobilities in intrinsic germanium are given as 0.43 and 0.21 m2/V s, respectively. If the electron and hole concentrations are both 2.3 x 10'® m, find the conductivity at this temperature.arrow_forwardWhat fraction of the electrons in a good conductor have energies between 0.90 EF and EF at T = 0?arrow_forwardGiven the resistivity of sodium 4.2 x10 Q2.m and the relaxation time 3.1x10-14 S, calculate the number of free electrons per cm³ and per atom.arrow_forward
- You put a diode in a microelectronic circuit to protect the system in case an untrained person installs the battery backward. In the correct forward-bias situation, the current is 200 mA with a potential difference of 100 mV across the diode at room temperature (300 K). If the battery were reversed, so that the potential difference across the diode is still 100 mV but with the opposite sign, what would be the magnitude of the current in the diode?arrow_forwardA diode is an electronic device that let the current flow in only one direction, that is the current through the diode is I=V/R (R is the diode resistance, a constant) if V > 0 and I=0 if V≤0. The diode is excited by an alternative voltage V(t) = cos(wot). (a) Sketch the diode response I(t) as a function of t. Indicate clearly the minimum values +to for which I(t) drops to zero. (b) Calculate the Fourier transform F(w) of the diode response. Hint: because the function is periodic, you only have to calculate the integral between ±to. (c) Schematically plot F(w) versus w. (d) A Lock-in amplifier is an electronic setup that can measure the response at integer frequency of the excitation frequency. What would a Lock-in measure at f = 2fo? (wo = 2nfo).arrow_forwardi provided a solution for the same problem but while using 5.26 T instead of 5.37 T. In the solution provided it gives the answer in eV but for my question, i need the answer to be in micro electron Volt (μeV).arrow_forward
- An abrupt silicon pn junction at zero bias has dopant concentrations of Nd = 5 X 1017 cm 3 and N₂ = 1 X 1017 cm-3 at T = a 300K. Determine the peak electric field for this junction for a reverse voltage of 5 V. Emax = O Emax O Emax 3.88 X 105 V/cm Emax 3.21 X 105 V/cm Emax = 1.70 X 105 V/cm 1.35 X 105 V/cm =arrow_forwardTwo diodes are connected in parallel as shown below: p(0.c.) = 0.15 p (s.c.) = 0.25 A diode may fail in one of the two ways: by short-circuiting or by open-circuiting. What is the probability that the two-diode arrangement will work as a diode?arrow_forwardX. Determine the resistivity of a germanium crystal at 300 K that has the simultaneous presence of (i) donor impurity of 1 in 10' and (ii) acceptor impurity of 1 in 10%. Given data: Atomic concentration in Germanium is 4.4 x 1022 atoms/cm³, n₁ = 2.5 × 10¹3 /cm³, He = 3800 cm² /Vs and h= 180 cm² /V s.arrow_forward
- In a certain biased limiter, the bias voltage is 5 V and the input is a 10 V peak sine wave. If the positive terminal of the bias voltage is connected to the cathode of the diode, the maximum voltage at the anode is O 10 V O 0.7 V O 5.7 V O 5V O 9.3 O Other:arrow_forwardA germanium diode is used to detect radiation with energy of 1.6 MeV. To do so, the anode (positive terminal) is exposed to the incoming radiation. Assume that the energy is absorbed entirely at the point of entry or the anode. With mobilities of 1,150 cm?/V.s for holes and 3,750 cm?/V.s for electrons, calculate the current through the diode with reverse bias: V = 24 V and d = 12 mm. Neglect the effects of electrodes and 4. of the n and p layers and assume a single radiation event, that is a single particle or a short burst of radiation. (Charge of an electron = 1.602 x 10-19 C; d is the distance between anode and cathode).arrow_forwardProblems Q1: Calculate the drift current density in silicon sample. If T=300 K, N=102/m*, Na=102/m’V, un=0.85m’/V.s, Up=0.04m/V.s, E=35 V/cm.. (Ans: 6.8×104A/m²).arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax