Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
7th Edition
ISBN: 9780199339136
Author: Adel S. Sedra, Kenneth C. Smith
Publisher: Oxford University Press
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Chapter 3, Problem 3.8P
To determine
The current that flows in the silicon bar.
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Chapter 3 Solutions
Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
Ch. 3.1 - Prob. 3.1ECh. 3.2 - Prob. 3.2ECh. 3.2 - Prob. 3.3ECh. 3.3 - Prob. 3.4ECh. 3.3 - Prob. 3.5ECh. 3.3 - Prob. 3.6ECh. 3.4 - Prob. 3.7ECh. 3.4 - Prob. 3.8ECh. 3.4 - Prob. 3.9ECh. 3.5 - Prob. 3.10E
Ch. 3.5 - Prob. 3.11ECh. 3.5 - Prob. 3.12ECh. 3.5 - Prob. 3.13ECh. 3.6 - Prob. 3.14ECh. 3.6 - Prob. 3.15ECh. 3.6 - Prob. 3.16ECh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Prob. 3.3PCh. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Prob. 3.9PCh. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Prob. 3.12PCh. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Prob. 3.24PCh. 3 - Prob. 3.25PCh. 3 - Prob. 3.26PCh. 3 - Prob. 3.27PCh. 3 - Prob. 3.28PCh. 3 - Prob. 3.29P
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- i need the answer quicklyarrow_forwardA pn diode is a semiconductor device in which, unlike in a metal, the current flow is due not only to negative electron charge flow but also due to positive charges. The positive charges are usually referred to as holes, as they are really "deficit" in the negative charge distribution. The semiconduction material (usually silicon with Er = 11 or gallium arsenide with e, impurities are added. Initially, the physically important region on the p side contains a density p+ of positive charges per unit volume, whereas the n region contains a negative charge density p-. When they are brought in contact, a static equilibrium follows in which a surplus of negative charges will accumulate in the (thin) p region, and a surplus of positive charges will accumulate in the (much thicker) n region. These two layers produce an electric field similar to that of a capacitor with infinite parallel plates. In equilibrium, the total number of positive charges in the n region is equal to the number of…arrow_forwardTwo silicon diodes, with a forward voltage drop of 0.7 V, are used in the circuit shown in the figure. The range of input voltage Vi for which the output voltage V0 = Vi is - Vo R D₁ Z -1 V DC( -0.3 V< Vi < 1.3 V O -1.0 V< Vi < 2.0 V -0.3 V< Vi < 2 V O -1.7 V < Vi < 2.7 V .D₂ 2 V Voarrow_forward
- Please do not copy from other websites Correct and detailed answer will be Upvoted else downvoted. Thank you!arrow_forwardmodels Diodes-Piece-wise Problem #1 In the circuit shown below, the voltage source Vin is given by the sketch. Assuming an ideal diode, sketch the waveform resulting at Vout. Use the axis at the bottom to help you sketch the voltage output. 10Ω Vin +7V -7V + Vin Vin AA 1 3 4 5 1 2 D 3 5 V 4 + Vout 5 6 ➡t (ms) +++t (ms) 6arrow_forwardCan you help solve this and explainarrow_forward
- 21. An ideal Schottky junction is formed by p-Si with Na = 10 cm 1/C ? 3 and an unknown metal. The results of the C-V measurements of this structure together with a straight-line fit are shown in the figure. Determine the work function of the metal. Answer: 4.73 eV - 0.2 V Show the workings to get to the provided answer pleasearrow_forwardA silicon diode at room temperature has a cross sectional area of 10µm². It is doped such that: Na = 4 x10¹4 cm³³ and Nd = 10¹5 cm³³ and biased such that VD = .2V. Determine the current through the diode. Show your work! Parameters to use K = 1.38 × 10−23 J/K q= 1.6 × 10-19 coulombs For Silicon at room temperature(T = 300°K): EG = 1.12eV ni = 1 × 10¹0cm-3 €Si = K sit0 = (11.8) (8.85 × 10-14)F/cm Dp = 10cm²/sec fn = 1230.77cm²/V - sec, Dn = 32cm²/sec Tn = 2.0 × 10−4sec, Tp=9.0 × 10-5 sec En = 8.0 × 10−²cm, Łp = 3.0 × 10-²cm Hp = 384.62cm²/V – sec, -arrow_forwardFind the value of the current across both diodesarrow_forward
- Using drawing and Explanation. Effect of light: What will happen if you shine a light on three parts of the p-n junction. Show the effect of light separately on p-si, n-si, and depletion regions. ww.arrow_forwardConsider a silicon P- N step junction diode with Nd = 1018 cm-3 and Na = 5 × 1015 cm-3 . Assume T=300K. Calculate the capacitance when it is under reverse biased at 1.5V. Assume a cross sectional area of 1um2. If you want to make the capacitance decrease by factor of 3 what should be the width of the depletion layer?arrow_forwardcan you solve this question handwritten solution step by step how did we obtain the v0 graph? my question is not a part of a graded assignmentarrow_forward
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Diodes Explained - The basics how diodes work working principle pn junction; Author: The Engineering Mindset;https://www.youtube.com/watch?v=Fwj_d3uO5g8;License: Standard Youtube License