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### ELEC3908_Lect_11

Course: ELEC 3908, Fall 2009
School: ECCD
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Word Count: 1609

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3908, ELEC Physical Electronics, Lecture 11 pn-Junction Electrostatics Lecture Outline Initial discussion of diode equilibrium in lecture 8 showed that an electric field is built up when carriers diffuse across the metallurgical junction forming the depletion region Need to examine in more detail the relationship between excess carrier density, electric field and potential through the device using Poisson's...

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3908, ELEC Physical Electronics, Lecture 11 pn-Junction Electrostatics Lecture Outline Initial discussion of diode equilibrium in lecture 8 showed that an electric field is built up when carriers diffuse across the metallurgical junction forming the depletion region Need to examine in more detail the relationship between excess carrier density, electric field and potential through the device using Poisson's equation Result will be the ability to predict the maximum electric field, important for reverse breakdown (lecture 12), and the depletion width as a function of applied bias, which will be important in small signal model (lecture 13) ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-2 Poisson's Equation Poisson's equation relates three variables the excess charge density , in C/cm3 the electric field E, in V/cm the electrostatic potential , in V For quantities changing over one dimension (x), Poisson's equation is given by d 2 ( x ) dE ( x ) ( x ) - = = 2 dx dx the permittivity in F/cm, is constant for a given material for silicon, Si = 11.7 = 11.78.854x10-14 F/cm ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-3 Excess Charge Density In Poisson's equation, the excess charge density is determined by the charge imbalance at a given point Charge in semiconductor devices arises from electrons (-ve charge) and protons (+ve charge) A hole is considered a +ve charge because protons locally outnumber electrons (recall lecture 3) If both electrons and protons are present in balanced numbers (true for intrinsic and extrinsic material in equilibrium), then is zero ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-4 Solution of Poisson's equation for (x) = 0 The plots to the right illustrate the solution for Poisson's equation when the excess charge density (x) is zero The electric field E(x) is the integral of (x), and hence is constant, with the exact value set by an integration constant (here assumed to be positive) The potential (x) is the negative of the integral of E, and is therefore linear and decreasing, again to within an arbitrary constant d 2 ( x ) dE ( x ) - = =0 2 dx dx ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-5 Solution of Poisson's equation for (x) = constant The plots to the right illustrate the solution for Poisson's equation when the excess charge density (x) is constant and positive The electric field E(x) is the integral of (x), and hence is linear, with the exact value set by an integration constant (assumed to give a positive value of E(0)) The potential (x) is the negative of the integral of E, and is therefore quadratic and decreasing, again to within an arbitrary constant d 2 ( x ) d E ( x ) - = = const . 2 dx dx ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-6 Equilibrium Junction n and p - log Scale If numerical techniques are used to solve for the actual distributions of n and p throughout the pn-junction at equilibrium and plotted on a log scale, the plot to the right is obtained The centre transition or depletion region reflects the charge transfer necessary to achieve equilibrium The log scale compresses the very wide difference between the values on either side ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-7 Equilibrium Junction n and p - Linear Scale If the same data is plotted on a linear scale, it is clear that n and p are small over most of the depletion region ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-8 Excess Charge Density in Semiconductor Material Although only electrons and protons contribute charge, it is convenient to define four sources of charge in semiconducting material mobile electrons (n) with charge -q holes (p) with charge +q ionized donors (ND) with charge +q (arising from proton charge) ionized acceptors (NA) with charge -q (arising from electron charge) The excess charge is then given by = q[ p ( x ) - n ( x ) + N D - N A ] ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-9 The Depletion Approximation Since n and p are small over most of the depletion region, a first order approximation is to ignore these quantities in solving Poisson's equation - this is the depletion approximation Over the depletion region, the excess charge density is then ( x ) = q[ p ( x ) - n ( x ) + N D - N A ] q ( N D - N A ) In the n-type neutral region NA = 0, n ND, p very small 0, as expected for the neutral region In the p-type neutral region ND = 0, p NA, n very small 0, again as expected for the neutral region ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-10 Accurate vs. Depletion Approximation Plot to the right shows (x) computed from actual n(x), p(x) and approximate from depletion approximation Sharp edges of depletion approx. distribution arise from discontinuity of n(x) or p(x) On lhs, is determined by NA, on rhs by ND - sign change and discontinuity indicate metallurgical junction Charge requires balance the areas of the two rectangular regions to be equal ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-11 Electric Field Distribution Electric field is the integral of excess charge density Choose integration constant to give E = 0 at left hand edge Integral of constant in depletion region gives linear E(x) Slope of E(x) changes sign when sign of (x) changes Charge balance requires areas of to balance, therefore integral goes back down to zero after depletion region ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-12 Electrostatic Potential Distribution Potential is the negative integral of electric field Choose constant of integration to give = 0 at left hand side ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-13 Depletion Width and Built-in Potential Using the depletion approximation and Poisson's equation, the width of the depletion region W (in cm) can be shown to be W= 2 Si q 1 1 N + N Vbi D A Where Vbi is the built-in potential (in V), given by kT N A N D ln Vbi = 2 q ni ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-14 Depletion Width Contour Plot The contour plot to the right shows constant-W lines in the (NA,ND) plane The dashed lines connect points where the doping is 10x larger on one side - "outside" the lines the doping is more than an order of magnitude greater When the doping is more than 10x larger on one side, W is determined by the lighter doping W= 2 Si q 1 1 N + N Vbi D A ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-15 Depletion Extent into p and n Materials Using the charge balance (i.e. area) condition and the depletion region expression, the extents of the depletion region into the n-type (xn) and p-type (xp) materials are NA xn = W NA + ND ND xp = W NA + ND Note that the region of higher doping has the lower depletion region extent - a higher doping requires less depletion width for a given charge ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-16 Maximum Electric Field E(x) is the integral of (x)/Si, so the maximum electric field is minus the area of either side of the (x) distribution qN A qN D x =- x Edepl max = - Si p Si n This expression can be simplified to Edepl max 2q N D N A =- Vbi Si N D + N A Note that the negative sign indicates direction if the junction were reversed, the quantity would be positive ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-17 Example 11.1 Compute the depletion region width and extents of the depletion region into each type of material as well as the maximum depletion region electric field for the device used in Figures 11.2 (etc.), i.e. a junction with NA=4x1016 /cm3 and ND=1016 /cm3. The temperature is 300K. ELEC 3908, Physical Electronics: pn junction Electrostatics Page 11-18 Example 11.1 - Solution The first step is to calculate the built in potential from (recall that at 30...

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