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ECE 874: Physical Electronics

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ECE 874:Physical Electronics

Prof. Virginia Ayres

Electrical & Computer Engineering

Michigan State University

VM Ayres, ECE874, F12

Answers I can find:

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

1. You can find y(x) by inspection whenever the Schroedinger equation takes a form with a known solution like and exponential. The standard form equation will also give you one relationship for kx.

2. Matching y(x) at a boundary puts a different condition on kx and setting kx = kx enables you to also solve for E in eV.

VM Ayres, ECE874, F12

Or equivalent Aexpikx + Bexp-ikx form

Infinite potential well

VM Ayres, ECE874, F12

With B = 0: tunnelling out of a finite well

VM Ayres, ECE874, F12

(eV)

Electron energy: E > U0

Electron energy: E < U0

(nm)

Regions:

-∞ to 0

0 to a

a to +∞

VM Ayres, ECE874, F12

(eV)

Electron energy: E < U0

(nm)

Region:

0 to a

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

(eV)

Electron energy: E < U0

(nm)

Regions:

-∞ to 0

a to +∞

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

New: e- goes away at ±∞

New boundary matching condition

New: e- exists outside of well region

VM Ayres, ECE874, F12

Gives a decreasing exponential e-a|x| in this region

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

y-(x) and y0(x) are done to within A0. If you need A0, use Working Tool 3: the existence theorem, in the easy region: 0 < x < a.

VM Ayres, ECE874, F12

To find B+ in terms of A0 to complete y+(x) add 2.41b and 2.41d, and re-arrange to get B+:

(2.41a)

(2.41b)

(2.41c)

(2.41d)

VM Ayres, ECE874, F12

y-(x), y0(x) and y+(x) are done to within A0. If you need A0, use Working Tool 3: the existence theorem, in the easy region: 0 < x < a.

Wave functions that represent e- are found.

Now find its total energy E in eV.

VM Ayres, ECE874, F12

.42)

VM Ayres, ECE874, F12

.42)

This is basically the solution for E in eV.

VM Ayres, ECE874, F12

VM Ayres, ECE874, F12

Red: LHS curve

Blue: RHS curve

Solve graphically:

LHS = tan(…E)

RHS = polynomial (…E)

Where they intersect is the value for E in eV

E in eV

VM Ayres, ECE874, F12

Red: LHS curve

Blue: RHS curve

Solve graphically:

LHS = tan(…E)

RHS = polynomial (…E)

Where they intersect is the value for E in eV

Quantized E1, E2, E3,.. for the finite well too, since tan(…E) repeats itself in multiples of p/2

E in eV

VM Ayres, ECE874, F12

En in eV

These are the energies En for the e- in the well, but the values are consistent with the physical situation that the well has a finite height U0 and that the e- can tunnel into the out of well regions on either side.

(eV)

Electron energy: E < U0

(nm)

Regions:

-∞ to 0

0 to a

a to +∞

VM Ayres, ECE874, F12

Advantage is: you scale to well height U0 and width a.

Note that width a only affects the LHS: the number/spacing of tan curves.

VM Ayres, ECE874, F12

Red: LHS curve

Blue: RHS curve

Solve graphically:

LHS = tan(…E/U0)

RHS = polynomial (…E/U0)

Where they intersect are the values for En/U0 in eV

VM Ayres, ECE874, F12

(a) Shallow well U0, single intersection for E1

(b) Deeper well U0, more intersections for E1, E2, E3,….

(c) Comparison of finite (solid) and infinite (dotted) well energy levels En shows that the infinite well solution progressively over-estimates the higher En

VM Ayres, ECE874, F12