Transcript Lecture 20, 24 Feb 14 - Michigan State University

ECE 875:
Electronic Devices
Prof. Virginia Ayres
Electrical & Computer Engineering
Michigan State University
[email protected]
Lecture 20, 24 Feb 14
Chp. 02: pn junction:
Info: Linearly graded junction
Multiple charge layers
Example
Chp. 03: metal-semiconductor junction: Schottky barrier
Review
Examples
New
VM Ayres, ECE875, S14
Linearly graded junction: Why:
Why: get a more uniform E (x) over a bigger x region
VM Ayres, ECE875, S14
Linearly graded junction: How:
Q and r
E (x)
yi(x)
VM Ayres, ECE875, S14
Linearly graded junction: How:
Q and r ~ a x
E (x) ~ ax2 + B
yi(x) ~ ax2 + Bx +C
Linearly graded junction: How:
VM Ayres, ECE875, S14
VM Ayres, ECE875, S14
Practical:
1/
The curvature of the initial C-V
curve is different from that for
an abrupt junction
The slope gives the grading
constant a
The intercept gives the
equilibrium built in potential ybi
VM Ayres, ECE875, S14
Lecture 20, 24 Feb 14
Chp. 02: pn junction:
Info: Linearly graded junction
Multiple charge layers
Example
Chp. 03: metal-semiconductor junction: Schottky barrier
Review
Examples
New
VM Ayres, ECE875, S14
Example: Set up the answer:
(a) Find ybi at equilibrium fro the following doping profile in si at 300 K
(b) Draw the energy band-bending diagram
p
n
p
1017 cm-3
1016 cm-3
1015 cm-3
VM Ayres, ECE875, S14
VM Ayres, ECE875, S14
1st
2nd
3rd
VM Ayres, ECE875, S14
q
q
VM Ayres, ECE875, S14
Lecture 20, 24 Feb 14
Chp. 02: pn junction:
Info: Linearly graded junction
Multiple charge layers
Example
Chp. 03: metal-semiconductor junction: Schottky barrier
Review
Examples
New
VM Ayres, ECE875, S14
@ Interconnects:
• Use energy band diagrams to describe what is happening
• One question to answer: is it an Ohmic contact or a Schottky
barrier contact?
• Interconnect contacts are key for nanotechnology:
– MOSFET: Ohmic contact = good
– NanoFET: SB contact = good
Individual energy band diagrams:
Different nature of a metal
Lots of e- and NO Egap
EC = at EF
Need 2 descriptions:
Electron affinity qcs = where is
EC relative to Evac
Need 1 description:
Work Function of the metal qFm:
where is EF = EC relative to Evac
Work Function qFs = where is EF
relative to Evac
When in physical contact EFm and EFs align:
Four cases = the same approach:
1. metal with small work function/n-type semiconductor: Ohmic (barrier)
2. metal with big work function/n-type semiconductor: Schottky barrier
3. metal with small work function/p-type semiconductor: Schottky barrier
4. metal with big work function/p-type semiconductor: Ohmic (barrier)
In every case, use logic: do I need to make the metal more n-type (add e- from
semiconductor) or less n-type (e- move into semiconductor)
Four cases = the same approach:
1. metal with small work function/n-type semiconductor: Ohmic (barrier)
2. metal with big work function/n-type semiconductor: Schottky barrier
3. metal with small work function/p-type semiconductor: Schottky barrier
4. metal with big work function/p-type semiconductor: Ohmic (barrier)
In every case, use logic: do I need to make the metal more n-type (add e- from
semiconductor) or less n-type (e- move into semiconductor)
2. metal with big work function/n-type semiconductor
• electrons move to metal side leaving Nd+
behind
• Size of n-side strip is set by doping
concentration and can be large
--N + N + n
--N + N +
To bring the Fermi energy level of the metal up:
make the metal more n-type
d
d
d
d
Schottky Barrier:
ND+ on n-side
--N + N + n
--N + N +
d
d
d
d
Schottky Barrier:
Very narrow region with high
concentration of e- similar to
ionized NA = large
--N + N + n
--N + N +
d
d
d
d
3. metal with small work function/p-type semiconductor
• electrons move to p-side and recombine with
its large hole population. This leaves Na- strip
• Size of p-side strip is set by doping
concentration and can be large
++ Na- Na-
++ Na- Na-
To bring the Fermi energy level of the metal down:
make the metal less n-type
p
Schottky Barrier:
NA- on p-side
++ Na- Na-
++ Na- Na-
p
Schottky Barrier:
Very narrow region = high
concentration exposed + nuclei
similar to ionized ND = large
NA- on p-side
++ Na- Na-
++ Na- Na-
p
Lecture 20, 24 Feb 14
Chp. 02: pn junction:
Info: Linearly graded junction
Multiple charge layers
Example
Chp. 03: metal-semiconductor junction: Schottky barrier
Review
Examples
New
VM Ayres, ECE875, S14
Example from Exam:
Answer:
Ei
EC – EF =
Egap/2
– (EF – Ei)
=
EF – Ei =
kT ln(ND/ni)
Streetman ni
EC – EF =
--N + N + n
--N + N +
d
d
d
d
Made the metal more ntype to bring EFm up to EFs
Electrons left the
semiconductor and went
into the metal.
The semiconductor is ntype: Nd+ left behind.
• Size WD of n-side depletion region is set
by doping concentration and can be large
Example:
(a) Evaluate the energy barrier qV0 = q ybi for previous problem
(b) Draw the band-bending diagram
Answer:
(a) qV0 =
--N + N + n
--N + N +
d
d
d
d
q ybi =qV0 = 0.057 eV
(a) Band-bending diagram:
Find W:
Junction
Equilibrium: metal contact to n-type Si when work functions qFm > qFs
metal
n0= 1017 cm-3
Although the charges are
balanced, the layer on the
metal side is very thin,
similar to ionized NA = large
EF
qV0
--
P+ P+ P P P P P P P
EF
Ei
E (x)
Neutral region
n-side
Depletion region W
= 1.14 x 10-5 cm
= 0.14 mm
Answer:
(a) qV0 =
--N + N + n
--N + N +
Also: qFB = 4.0 -3.8 eV = 0.2 eV
(a) Band-bending diagram:
W = 0.14 mm
d
d
d
d
q ybi =qV0 = 0.057 eV