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
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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
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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
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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
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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
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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
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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
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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 ]
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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
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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
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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
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Electrostatic Potential Distribution
Potential is the negative integral of electric field Choose constant of integration to give = 0 at left hand side
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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
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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
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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
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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
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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.
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Example 11.1 - Solution
The first step is to calculate the built in potential from (recall that at 30...