6.012 - Electronic Devices and Circuits

Lecture 4 - Non-uniform Injection (Flow) Problems - Outline• Announcements

Handouts - 1. Lecture Outline and Summary; 2. Thermoelectrics• Review

Thermoelectricity: temperature gradients as a driving forceThe 5 basic equations

• Non-uniform injection (flow) problemsFive assumptions:

1. Uniform doping: r = e ∂E/∂x ≈ q(p' – n'); ∂n = ∂n', ∂p = ∂p'2. Low level injection: [np - ni2]r(T) ≈ n'/tmin3. Quasi-neutrality: n' ≈ p', ∂n'/∂x ≈ ∂p'/∂x4. Negligible minority carrier drift: Je ≈ qDe∂n'/∂x (assumes p-type)5. Quasi-static: ∂/∂t ≈ 0

The diffusion equation for minority carriers• Getting Je(x), Jh(x), E(x), and p'(x), knowing n'(x)• Solving the diffusion equation for n'(x) (using p-type example)

Homogeneous solutions; minority carrier diffusion length, LminParticular solutionBoundary conditions: ohmic contact, reflecting boundary,

internal boundaryTotal solution

Clif Fonstad, 9/03 Lecture 4 - Slide 1

Consider our electron current density expression with the drift termrewritten in terms of the potential and including the term that mustbe added if we have a temperature gradient*:

This way of writing the current emphasizes the fact that the carriersrespond to gradients in various quantities. Drift is the response to agradient in the electrostatic potential, diffusion is the response to agradient in the carrier concentration, and the Seebeck effect is theresponse to a gradient in temperature.

The Seebeck effect is interesting because it can be used to generateelectric power from a temperature difference. Conversely, a temp-erature difference can be generated by passing a current through asuitably designed semiconductor structure. This reverse effect iscalled the Peltier effect.

Thermoelectric effects - the Seebeck and Peltier effects (current fluxes caused by temperature gradients, and visa versa)

Clif Fonstad, 9/03 Lecture 4 - Slide 2

†

Je = -q -men( ) -dfdx

Ê

Ë Á

ˆ

¯ ˜ + -q( )De -

dndx

Ê

Ë Á

ˆ

¯ ˜ + -q( )Sen -

dTdx

Ê

Ë Á

ˆ

¯ ˜

* Don’t panic. We will only consider isothermalsituations in 6.012 exams and problem sets.

Two examples:Right - The hot point probe, an

apparatus for determining thecarrier type of semiconductorsamples.

Below - A thermoelectric coolerarray like that found in solid-state refrigerators.

Thermoelectric effects - the Seebeck and Peltier effects (current fluxes caused by temperature gradients, and visa versa)

Clif Fonstad, 9/03 Lecture 4 - Slide 3Ref.: Appendix B in the course text.

Non-uniform material with non-uniform excitations (laying the groundwork to model diodes and transistors)

A. General descriptionTo model devices we must understand semiconductors in which we

have (1) spatial variation of the doping and (2) spatial andtemporal variation of the generation function:

Because the doping and excitation are not uniform, we anticipate thatthe carrier population, currents, and electric field will also varywith space and time:

To determine these five quantities, assuming we know the dopingprofile and generation function, we will need five equationsrelating these quantities. We have:

2 continuity equations (divergence of fluxes, generation, and recombination)

2 current equations (drift plus diffusion)1 charge balance equation (Poisson's Eq.)

Clif Fonstad, 9/03 Lecture 4 - Slide 4

†

Nd (x), Na (x), gL (x, t)

†

n(x, t), p(x,t), Je (x,t), Jh (x,t), E(x,t)

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

B. The Five EquationsElectron population

Hole population

Electron current density

Hole current density

Poisson's equation

Clif Fonstad, 9/03 Lecture 4 - Slide 5

†

∂n(x,t)∂t

-1q

∂Je (x, t)∂x

= gL (x, t) - n(x, t) ⋅ p(x, t) - no(x) ⋅ po(x)[ ]r(T)

†

∂p(x,t)∂t

+1q

∂Jh (x, t)∂x

= gL (x, t) - n(x, t) ⋅ p(x, t) - no(x) ⋅ po(x)[ ]r(T)

†

Je(x, t) = qmen(x, t)E(x,t) + qDe∂n(x, t)

∂x

†

Jh (x, t) = qmh p(x, t)E(x,t) - qDh∂p(x, t)

∂x

†

∂E(x,t)∂x

= -qe

p(x, t) - n(x, t) + Nd+(x) - Na

-(x)[ ]

C. Solving the five equations: easy to do in several important specialcases

1. Uniform doping, thermal equilibrium:

Lecture 12. Drift in uniformly doped semiconductor

Lecture 2

3. Uniform optical injection in uniformly doped semiconductor,photoconductivity

Lecture 2

4. Flow problems: uniform doping, non-uniform injection

Today

5. Junction problems: non-uniform doping in thermal equilibrium

Lecture 5Clif Fonstad, 9/03 Lecture 4 - Slide 6

†

∂∂x

= 0, ∂∂t

= 0, gL (x,t) = 0, Je = Jh = 0

†

∂∂x

= 0, ∂∂t

= 0, gL (x,t) = 0, Ex constant

†

∂∂x

= 0, Ex constant, n'<< po

†

∂Nd

∂x=

∂Na

∂x= 0, n'ª p', n'<< po, Je ª qDe

∂n'∂x

, ∂∂t

ª 0

†

∂∂t

= 0, gL (x, t) = 0, Je = Jh = 0

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

C. Solving the five equations, cont.: where the special cases we cansolve appear in important semiconductor devices

Junction diodes, LEDs:

Bipolar transistors:

MOS transistors:

Clif Fonstad, 9/03 Lecture 4 - Slide 7

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

p-type n-type

Flow problemJunctiion problem

Flow problem

n-type p n-typeE

B

C

Flow problems

Junction problem

n+n+

S G D

p-type

Diodes Depletion approximationDrift

D. Flow Problems: uniformly doped material, non-uniform carrierinjection, and boundary conditions; no applied voltageFive unknowns, five equations, five flow problem assumptions:

1. Uniform doping

2. Quasineutrality

3. Low level injection (in p-type, for example)

4. Negligible minority carrier drift

Note: It is also always true that

Clif Fonstad, 9/03 Lecture 4 - Slide 8

†

dno

dx=

dpo

dx= 0 fi

∂n∂x

=∂n'∂x

, ∂p∂x

=∂p'∂x

†

n'ª p', ∂n'∂x

ª∂p'∂x

†

n'<< po fi np - ni2( )r ª n' por =

n't e

†

Je(x, t) ª qDe∂n'(x,t)

∂x

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

∂n∂t

=∂n'∂t

, ∂p∂t

=∂p'∂t

†

no - po + Na - Nd = 0 fi n - p + Na - Nd = n'-p'

D. Flow Problems, cont. (where we are after imposing four conditions)Combining everything we have to date, with these four

conditions our five equations become:

Before continuing to our fifth condition, we note that alreadyEquations 1 and 3 can be combined to give one equation inn'(x,t):

Clif Fonstad, 9/03 Lecture 4 - Slide 9

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

1,2 : ∂n'(x,t)∂t

-1q

∂Je (x, t)∂x

=∂p'(x,t)

∂t+

1q

∂Jh (x,t)∂x

= gL (x,t) -n'(x,t)

t e

3 : Je(x, t) ª +qDe∂n'(x, t)

∂x 4 : Jh (x, t) = qmh p(x, t)E(x,t) + qDh

∂p'(x, t)∂x

5 : ∂E(x,t)∂x

=qe

p'(x, t) - n'(x, t)[ ]

†

∂n'(x,t)∂t

- De∂ 2n'(x,t)

∂x 2 = gL (x,t) -n'(x,t)

t e

The time dependent diffusion equation

D. Flow Problems, cont. (the fifth constraint)The time dependent diffusion equation, which is repeated

below, is in general still very difficult to solve

Consequently we impose a final, fifth constraint.

5. Quasi-static excitation

With this constraint the above equation becomes a secondorder linear differential equation:

which in turn becomes, after rearranging the terms :

Clif Fonstad, 9/03 Lecture 4 - Slide 10

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

∂n'(x,t)∂t

- De∂ 2n'(x,t)

∂x 2 = gL (x,t) -n'(x,t)

t e

†

d2n'(x)dx 2 -

n'(x)Det e

= -1

De

gL (x)

†

gL (x, t) such that all ∂∂t

ª 0

†

-Ded2n'(x)

dx 2 = gL (x) -n'(x)

t e

The steady state diffusion equation

D. Flow Problems, cont. (the other unknowns)Solving the steady state diffusion equation gives n’. Finding p’,

Je, Jh, and Ex knowing n’:

First find Je:

Then find Jh:

Next find Ex:

Then find p’:

Finally, go back and check that all of the five conditions are metby the solution.

Clif Fonstad, 9/03 Lecture 4 - Slide 11

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

Je(x) ª qDedn'(t)

dx

†

Jh (x) = JTot - Je (x)

†

Ex(x) ª1

qmh po

Jh (x) -Dh

De

Je (x)È

Î Í

˘

˚ ˙

†

p'(x) ª n'(x) +eq

dEx (x)dx

D. Flow Problems, cont. (General solutions)The homogeneous solution of the steady state diffusion

equation,

is

Le is the called the “minority carrier diffusion length” and it isthe characteristic length over which returns to its equilibriumvalue. It is thus the spatial equivalent of the minority carrierlifetime, te. “A” and “B” are unknowns at this point.

The particular solution of this 2nd order differential equation“looks” functionally like gL(x). Beyond that it is difficult tosay much in general without specifying gL(x).

The unknowns “A” and “B” are found by matching the sum ofthe homogeneous and particular solutions to the boundaryconditions.

Clif Fonstad, 9/03 Lecture 4 - Slide 12

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

d2n'(x)dx 2 -

n'(x)Det e

= -1

De

gL (x)

†

nHS' (x) = Aex / Le + Be-x / Le where Le ≡ Det e

D. Flow Problems, cont. (Boundary conditions - sample ends)At sample ends we have three possible situations we deal with

in 6.012, each of which imposes a different condition on eitherthe excess minority carrier population or its gradient:

An ohmic contact: here there is infinite recombination and theexcess population is identically zero

A reflecting surface: here there is no additional recombinationand thus the flux into the surface is zero

An injecting contact: here we can either have the excessminority carrier population specified

or the minority carrier flux specified

Clif Fonstad, 9/03 Lecture 4 - Slide 13

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

Option I : n'(x) specified at injecting contact†

dn'dx

= 0 at a reflecting surface†

n'(x) = 0 at an Ohmic contact

†

Option II : dn'dx

specified at injecting contact

D. Flow Problems, cont. (Boundary conditions - internal)At any internal point we must always have continuity of the

carrier population:

and, if the generation function, gL, is finite at that point, thegradient of the population is also continuous:

If the generation function is infinite at that point, as is thecase if it is modeled as an impulse, then we have:

Clif Fonstad, 9/03 Lecture 4 - Slide 14

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

At x = X : n'(X +) = n'(X -)

†

gL(X) finite fi dn'dx x= X +

= dn'dx x= X -

†

gL (X) = Md(x - X) fidn'dx x= X +

= dn'dx x= X -

-MDe

D. Flow Problems, cont. (Special solutions relevant to devices)1. “Infinite lifetime” or “long base” situations: These terms

refer to situations in which the minority carrier diffusionlength is much greater than the sample. In this case we willalways have Lmin >> x, and the n’/Lmin term can be neglected inthe steady state diffusion equation, leaving:

Integrating twice we arrive immediately at:

2. “Impulse excitation”: When gL can be modeled as a singleimpulse*, then n’(x) is only the homogeneous solution:

Clif Fonstad, 9/03 Lecture 4 - Slide 15

Non-uniform material with non-uniform excitations, cont. (laying the groundwork to model diodes and transistors)

†

d2n'(x)dx 2 ª -

1De

gL (x)

†

n'(x) ª -1De

gL (x)dxdxÚÚ + Ax + B

†

gL (x) = Gd(X) fi n ' (x) = Ae(x-X ) / Le + Be-(x-X ) / Le

* Note: Multiple impulses are be handled using superposition.

6.012 - Electronic Devices and Circuits

Lecture 4 - Non-uniform Injection (Flow) Problems - Summary• Review

Thermoelectric soda cooler• Non-uniform injection (flow) problems

Five assumptionsThe diffusion equation for minority carriers

∂2n'/∂x2 – n'/Le2 = – gL(x)/De (assumes p-type)

• Getting Je(x), Jh(x), E(x), and p'(x), knowing n'(x)Je(x) ≈ q De dn'/dxJh(x) = Jtot - Je(x)E(x) ≈ [Jh(x) + (Dh/De)Je(x)]/q µh pop'(x) ≈ r(x)/q + n'(x), and r(x) = e dE/dx

• Solving the diffusion equation for n'(x)Homogeneous solutions: exp ±x/Lmin where Lmin ≡(Dmintmin)1/2

Particular solution: "looks" like driveBoundary conditions: ohmic contact (n' = p' = 0); reflecting

boundary (dn'/dx = dp'/dx = 0); internal boundary (n', p' alwayscontinuous; dn'/dx, dp'/dx continuous unless impulse generation)

Total solution: HS + PS fit to BCsClif Fonstad, 9/03 Lecture 4 - Slide 16