Transcript Lecture 15, 12 Feb 14 - Michigan State University

ECE 875:
Electronic Devices
Prof. Virginia Ayres
Electrical & Computer Engineering
Michigan State University
[email protected]
Lecture 15, 12 Feb 14
Hw 04: FRI: Pr. 2.07
Chp. 02: pn junction:
Experimental measurements for concentration:
Hall effect – Chp. 01: material:
measure VAB, and I, choose dimensions and Bext
C-V – Chp. 02: pn junction
Two realistic configurations beyond abrupt linear pn junction:
Linearly graded junction
Double layer junction: important, develops at interfaces
VM Ayres, ECE875, S14
Lecture 15, 12 Feb 14
Hw 04: FRI: Pr. 2.07
Chp. 02: pn junction:
Experimental measurements for concentration:
Hall effect – Chp. 01: material:
measure VAB, and I, choose dimensions and Bext
C-V – Chp. 02: pn junction
Two realistic configurations beyond abrupt linear pn junction:
Linearly graded junction
Double layer junction: important, develops at interfaces
VM Ayres, ECE875, S14
Example:
Sweep
the
voltage
Instrument reads out C
typically in Farads
Example: Sze Fig:
V = Vbattery
Where is C: depletion region of a pn junction:
Can show
equivalence to
parallel plate
capacitor:
- Qtotal
es
+ Qtotal
Where is C: depletion region of a pn junction:
CD =
Where you measure: C-V = same as I-V:
= SMU
+Vext-
Sweep
the
voltage
p+
n
WD
= WDp + WDn
Lecture 15, 12 Feb 14
Hw 04: FRI: Pr. 2.07
Chp. 02: pn junction:
Experimental measurements for concentration:
Hall effect – Chp. 01: material:
measure VAB, and I, choose dimensions and Bext
C-V – Chp. 02: pn junction
VM Ayres, ECE875, S14
pn junction at equilibrium: ECE 474: Streetman & Bannerjee
p
n
p
n
VM Ayres, ECE875, S14
pn junction at equilibrium: ECE 474: Streetman & Bannerjee
p
n
Q = charge density r x Vol
r = q with sign (ND+ or NA-)
Poisson equation relates charge to
electric field E :
dE /dx = r/ese0
(material is not polarized or magnetic)
VM Ayres, ECE875, S14
Abrupt pn junction at equilibrium: ECE 875: Sze:
p
n
Q = charge density r x Vol
r = q with sign (ND+ or NA-)
Poisson equation
dE /dx = r/ese0
Solve for E
Solve for built in potential ybi  V0
Any potential:
= Area
VM Ayres, ECE875, S14
Abrupt pn junction at equilibrium: Sze: names:
p
Potential V0  ybi
Potential barrier qV0  qybi
p-side: Ei – EF  qyBp
n-side: EF – Ei  qyBn
p-side: EF – EV  qfp
n-side: EC – EF  qfn
n
Potential drop across depletion region WD=
Pot’l drop across p-side of WD + pot’l drop across nside of WD : ybi = yp + |yn|
Potential drop from 0 to x in WD: yi(x)
VM Ayres, ECE875, S14
Important questions are:
What is the magnitude and direction of the internal electric field?
What are the values of the various potential drops that matter?
Can I get an experimental measure of anything?
VM Ayres, ECE875, S14
What is the magnitude and direction of the internal electric field?
Can I get an experimental measure of anything?
An antenna probe won’t work inside a solid
-no direct experimental measure of E (x)
-x
E
-WDp =
direction
= WDn
VM Ayres, ECE875, S14
What is the magnitude and direction of the internal electric field?
Can I get an experimental measure of anything?
-x
E
About the directions:
E (x) direction = -x
dx from/to direction = +x
x is + from 0 to WDn
x is – from –WDp to 0
E
-WDp =
direction
magnitude: f(x)
= WDn
What are the values of the various potential drops that matter?
Can I get an experimental measure of anything?
Yes: can get an experimental measure of potential. The loop will
include the built-in potential ybi and Vext and any IR drops.
Total potential drop Vp-to-n is mainly across W
ybi
Vext
= SMU
VM Ayres, ECE875, S14
Potential energy barrier and built-in potential in terms of dopants:
But what if doping concentrations
are not what you think?
VM Ayres, ECE875, S14
Internal electric field E (x): in WD:
VM Ayres, ECE875, S14
Internal electric field E (x):
Note: Linear:
VM Ayres, ECE875, S14
Internal electric field E (x):
Can solve for maximum value of E -field:
VM Ayres, ECE875, S14
Internal electric field E (x):
VM Ayres, ECE875, S14
Internal electric field E (x):
Note: Linear:
VM Ayres, ECE875, S14
Go from electric field E (x) to potential yi(x). Why: you may be able to
measure a potential drop.
+
Can integrate this!
=
E0x +C
VM Ayres, ECE875, S14
Must separate this into p-side and n-side of depletion region answers:
p-side of depletion region:
VM Ayres, ECE875, S14