Electrostatics: The Key to Understanding Electronic Devices. Boundary Conditions. Physics approach: vector calculus, highly symmetrical problems

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1 lectrostatics: The Key to Understanding lectronic Devices Boundary Conditions Physics approach: vector calculus, highly symmetrical problems 1. Potential: φ( = - ) = φ( = + ) φ() GaussÕs Law: ( ε) = ρ 1 2 Def. of Potential: = Ð φ PoissonÕs qn.: ( ε( Ð φ) ) = ε 2 φ = ρ Device physics 2. lectric Field: ε 1 = ( - ) + Q = ε 2 = ( + ) Real problems (not symmetrical, complicated boundary conditions) GaussÕs Law: DeÞnition of Potential: d( ε) = ρ d d φ = Ð d where Q is a surface charge (units, C/cm 2 ) located at the interface for the case where Q = : () ε 1 = 3ε 2 common materials: silicon, ε s = 11.7 ε o 1 2 silicon dioide (SiO 2 ), ε o =3.9 ε o PoissonÕs quation: d ε d φ Ð d d = Ð ε d2φ d 2 = ρ

2 Intuition for lectrostatics ample I: Applied lectrostatics ÒRules of ThumbÓ for sketching the solution BFOR doing the math: charge distribution ρ() ρ() lectric Þeld points from positive to negative charge lectric Þeld points ÒdownhillÓ on a plot of potential lectric Þeld is conþned to a narrow charged region, in which the positive charge is balanced by an equal and opposite negative charge Use boundary conditions on potential or electric Þeld to ÒpatchÓ together solutions from regions having different material properties GaussÕs law in integral form relates the electric Þeld at the edges of a region to the charge inside. Often, the Þeld on one side is known to be zero (e.g., because itõs on the outside of the charged region), which allows the electric Þeld at an interface to be solved for directly Charge density functions: only two cases needed for basic device physics ρ = > constant --> φ linear ρ = ρ o = constant --> linear --> φ quadratic φ( << ) =.5 V Sketch the electric Þeld and the charge. φ( >> ) = -.4 V φ.5 surface or sheet charge Q is sometimes present (e.g., on the surface of good conductors); the effect on electric Þeld can be incorporated through the boundary condition -.5

3 ample II: Applied lectrostatics * Given the electric Þeld, () Sketch the charge density and the potential ρ() Application of GaussÕs Law At a point, the electric Þeld can be found as the charge enclosed, divided by the permittivity of the material... caveats (warnings): (i) the Þeld must be zero at the other side of the charged region (ii) the sign of the Þeld can be found by keeping track of the + direction and the one-dimensional equivalent of the Òoutward normal;ó however, the best approach is to know the sign of the Þeld from the distribution of charge in the problem ample: metal-oide-silicon structure ρ() Find ( = - t o / 2) - t o φ() φ( >>) =.3 V Q G Find ( = + )... just inside the silicon

4 Boundary Condition on (cont.) Sketch () from = - t o to = () Potential and Carrier Concentration in Silicon What is a convenient reference for the electrostatic potential in silicon? Thermal equilibrium: no eternal stimulus --> must have: J p = and J n =. - t o dn o dφ = qn o µ n + qd n = qn d o µ n Ð d + qd dn o n d Sketch φ() through the structure, given that φ( ) = 4 mv φ().4 V - t o D n dφ dn o kt dn o dn o = = = V µ n n o q n th o n o where we have used insteinõs relation. D n kt = = V mv at ÒcoolÓ room temperature to µ n q th = mv at ÒwarmÓ room temperature V th is called the thermal voltage.

5 The Intrinsic Potential Reference By integrating the equation relating potential to the electron concentration from a position a to position, we Þnd that: φ() Ðφ ( a ) V th ln n o = ( () ) n o ( a ) The 6 mv Rule The hole concentration can also be related to the potential, by repeating the derivation starting with J p = or by substituting p o = n i 2 / n o into the 6 mv rule for electrons. The result is: We can choose any reference; one convenient choice is to set: φ V th ln n i p o p o = (----- ) = ( Ð26mV)ln( 1)log p o 1 1 = ( Ð6mV)log φ( a ) = when n o ( a ) = n i = 1 1 cm -3 at room temperature. p o, equilibrium hole concentration (cm 3 ) Using this reference, the potential in thermal equilibrium can be found, given the electron concentration: p-type intrinsic n-type φ V th ln n o (----- ) ( 26mV)ln( 1)log n o n i 1 1 ( 6mV)log n o = = = φ (mv) φ (mv) Donor concentrations from 1 13 to 1 19 cm -3 therefore correspond to potentials of (6 mv) 3 = 18 mv to (6 mv) 9 = 54 mv (at room temperature) p-type intrinsic n-type n o, equilibrium electron concentration (cm 3 )