ECE 874: Physical Electronics - PowerPoint PPT Presentation

About This Presentation
Title:

ECE 874: Physical Electronics

Description:

ECE 874: Physical Electronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv_at_msu.edu Lecture 25, 26 Nov 12 Chp. 05 ... – PowerPoint PPT presentation

Number of Views:143
Avg rating:3.0/5.0
Slides: 27
Provided by: virgini171
Learn more at: https://www.egr.msu.edu
Category:

less

Transcript and Presenter's Notes

Title: ECE 874: Physical Electronics


1
ECE 874Physical Electronics
  • Prof. Virginia Ayres
  • Electrical Computer Engineering
  • Michigan State University
  • ayresv_at_msu.edu

2
Lecture 25, 26 Nov 12Chp. 05
Recombination-Generation Processes
3
(No Transcript)
4
pn junction in Si at equilibrium ( no bias)
Recombination-generation (R-G) via a trap (a
local defect) why this is important
- - - -

p p 1019 cm-3
n 1015 cm-3
Si
W
5
Same pn junction in Si in reverse bias - 5V
- Vrev




- - - -
- - - -
- - - -
- - - -
p 1019 cm-3
n 1015 cm-3
Si
W
Reverse bias goal turn the device OFF no
current flowing.
6
You didnt turn your device OFF as well as you
thought you did by four orders of
magnitude. What happened trap-mediated
recombination-generation (R-G) processes act to
restore what ever the previous steady state was.
The trap released more carriers to try to restore
the previous neutral region concentrations.
7
Two key relationships/equations
Note that n1 and p1 look just like ordinary
dopant concentrations n and p except that they
originate at a trap energy level ET
8
Two key relationships/equations in HW05
Prs. 5.3, 5.4, 5.5
Prs. 5.2, 5.3, 5.4, 5.5 All
9
Prs. 5.2, 5.3, 5.4, 5.5 All
Pr. 5.2 ET Ei is given Pr. 5.3 ET near Ei
is given Pr. 5.4 ET Ei is worst case
scenario the one to check Pr. 5.5 ET Ei is
given
Obviously there is something important about this
condition
10
(No Transcript)
11
(No Transcript)
12
(No Transcript)
13
3b)
14
Points p. 149
para- live
NT
n or p
15
Pr. 5.2
Pr. 5.3, 5.4, 5.5
3b)
16
Pr. 5.2
(a) Given a condition on the capture cross
sections cp and cn (c) Given a condition on the
emission cross sections ep and en But question
is about a concentration nT not an R-G rate
R Saw nT in Lecture 24 in derivation of rates rN
and rP under equilibrium conditions. nT is
concentration of electrons in traps, pT is
concentration of holes. Total concentration of
filled traps NT nT pT. Note in Lecture 24,
we worked through an equilibrium derivation but a
steady state example. Time to connect these two
ideas. Prs. 5.2, 5.3, 5.4, 5.5 are all steady
state problems.
17
Lecture 24 General info
Processes the change the e- headcount
Processes the change the hole headcount
18
Lecture 24 Each one of these processes happens
with better or worse efficiencies
Hole capture
Hole emission
19
Lecture 24 General info
Processes the change the e- headcount
Processes the change the hole headcount
20
Lecture 24 Equilibrium
0 0
Pierret says same things the definition for
equilibrium is that the rates for changing
electrons and holes concentrations via a trap
individually sum to 0.
0a) 0b)
21
Lecture 24 Equilibrium
Under equilibrium conditions you can solve for
the emission coefficients in terms of the capture
coefficients (p. 145). Then, assuming that even
away from equilibrium, the capture/emission
coefficient values dont change too much
rN and rP are not 0, so this is not equilibrium
22
Steady State
Under equilibrium conditions you can solve for
the emission coefficients in terms of the capture
coefficients (p. 145). Then, assuming that even
away from equilibrium, the capture/emission
coefficient values dont change too much
The definition of the steady state condition
is when rN rP
23
Equilibrium/Steady State
Under equilibrium conditions you can solve for
the emission coefficients in terms of the capture
coefficients (p. 145). Then, assume that even
away from equilibrium, the capture/emission
coefficient values dont change too much.
.13a)
.13b)
These connectors between emission and capture
coefficients should be used in Pr. 5.2 (c).
24
Steady State
Set rN rP Use NT nT pT to eliminate
pT Solve for nT
25
nT for steady state conditions
.21)
Use this definition for nT under steady state
conditions to answer Pr. 5.2.
26
R for steady state conditions
Can use the steady state nT definition and pT
NT - nT to eliminate nT and pT in the above. The
result is
(5.24)
Use this R in Prs. 5.3, 5.4 and 5.5
Write a Comment
User Comments (0)
About PowerShow.com