A pn diode is a semiconductor device in which, unlike in a metal, the current flow is due not only tonegative 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 withEr = 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 equilibriumfollows in which a surplus of negative charges will accumulate in the (thin) p region, and a surplus of positive chargeswill 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 negative charges in the p region. Ifboth sides have the same surface area S, then the equilibrium condition is just13) is normally dielectric but can become conductive when small amounts of%3DP+ S. Xn = p- · S. xp,(1)where xn and xp are the thickness of the n and p regions, respectively.In a one sided step junction, xn > xp. We can model the accumulation of negative charges on the p side as a sheetof negligible thickness with surface charge density o = p_ · Xp. On the n side, the charges are effectively distributeduniformly in a constant volume, i.e. p+ is constant.N.Q-Xp-Xp--N_Q(a)(b)O =p dx+p = N+Qdxo = - N. Qxp.EEEX=0X = 0X = 0(c)a) For the slab of positive charge only (ignoring the sheet of negative charge of the p-region), show the electricfield at depth x inside a slab of thickness rn but infinite in the other two directions and containing a constantcharge density P+, is given by(- - ).PtXnEslab(2)%3DOne can do this in many ways, but I suggest you consider a thin slice of thickness dæ containing surface chargedensity p+ dx. From this, one can carefully integrate to obtain the net field. (Beware: there are layers on theleft and on the right of the layer located at depth x.)b) Show the net field between the plates at position a in the n-region is just Eout = P+(x – xn)/e.%3Dc) Find the potential difference between the thin p side and the end of the n side. (Hint: you may assume thatis negligible, as per the left figure of part (c).)X'pd) Find the capacitance of the device. (Hint: C = 2eS/xn where S is the cross sectional area of the diode, in thedirection perpendicular to x in the figure.)%3DIII 14 IIIIIIIII

According to bartleby only three subparts are to be answered.

(c) a) For the slab of positive charge only (ignoring the sheet of negative charge of the p-region), show the electric field at depth a inside a slab of thickness a, but infinite in the other two directions and containing a constant charge density p+, is given by Pt Elab = (2) One can do this in many ways, but I suggest you consider a thin slice of thickness dx containing surface charge density p4 dx. From this, one can carefully integrate to obtain the net field. (Beware: there are layers on the left and on the right of the layer located at depth r.) b) Show the net field between the plates at position a in the n-region is just Eout = P+(x – an)/e. c) Find the potential difference between the thin p side and the end of the n side. (Hint: you may assume that a, is negligible, as per the left figure of part (c).)

(c) a) For the slab of positive charge only (ignoring the sheet of negative charge of the p-region), show the electric field at depth a inside a slab of thickness a, but infinite in the other two directions and containing a constant charge density p+, is given by Pt Elab = (2) One can do this in many ways, but I suggest you consider a thin slice of thickness dx containing surface charge density p4 dx. From this, one can carefully integrate to obtain the net field. (Beware: there are layers on the left and on the right of the layer located at depth r.) b) Show the net field between the plates at position a in the n-region is just Eout = P+(x – an)/e. c) Find the potential difference between the thin p side and the end of the n side. (Hint: you may assume that a, is negligible, as per the left figure of part (c).)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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