Download presentation
Presentation is loading. Please wait.
Published byKory Stone Modified over 9 years ago
1
ECE 875: Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu
2
VM Ayres, ECE875, S14 Chp. 01 Concentrations Nondegenerate Degenerate Nondegenerate Intrinsic Contributed by impurities Wanted impurities: “dopants” Unwanted impurities: “traps” Lecture 10, 03 Feb 14 Effect of temperature Concentrations n
3
VM Ayres, ECE875, S14 Example Pr. 1.12:
4
VM Ayres, ECE875, S14 Example Pr. 1.12: Use: ✔ ✔ Need this
5
VM Ayres, ECE875, S14 Centripetal force = Coulomb force Average is defined as the harmonic mean not the geometric mean. Used when E-field is present P and B are shallow impurities in Si General Info: will use figure next slide to do Pr. 1.12:
6
VM Ayres, ECE875, S14 Tabulated: can get E C – E D @ 300 K from Sze Fig 10 (useful): For P in Si: E C – E D = 0.046 eV This is the “ionization energy” Pr. 1.12: Given: ionization energy doesn’t change as a function of T
7
VM Ayres, ECE875, S14 Have: E C – E D Need: E F – E D Note: (E C – E D ) – (E C – E F ) = E F – E D Given: N D = 10 16 cm -3 = nondegenerate in Si Therefore: use Sze (21): Need: N C @ 77K In App G: N C @ 300K = 2.8 x 10 19 cm -3
8
VM Ayres, ECE875, S14 Use the ratio to something you know method: 2
9
VM Ayres, ECE875, S14
10
Also: kT at 77K: Plug and chug method: Ratio to something you know method: OR
11
VM Ayres, ECE875, S14 Thus far: Sze (21): @ 77K = 3.64 X 10 18 cm -3 ✔ Problem: n ≠ N D + = 10 16 cm -3 at 77K because not all N D are ionized
12
VM Ayres, ECE875, S14
14
Important part: NOT EQUAL
15
VM Ayres, ECE875, S14 Important step: COMPARE:
16
VM Ayres, ECE875, S14 Chp. 01 Concentrations Nondegenerate Degenerate Nondegenerate Intrinsic Contributed by impurities Wanted impurities: “dopants” Unwanted impurities: “traps” Lecture 10, 03 Feb 14 Effect of temperature Concentrations n
17
VM Ayres, ECE875, S14 Summary: concentration as a function of temperature and doping: For: nondegenerate doping (n-type Si shown in figure):
18
VM Ayres, ECE875, S14 DONORS: Neutral N D Electron occupies a local energy level E D, h = 1, g D = 2 Ionized N D + A local energy level E D is empty and available Summary: Freeze-out range: concentration partially ionized dopants: ACCEPTORS: Neutral N A A local energy level E A is empty and available Ionized N A - Electron occupies a local energy level E A, h = 1, g A = 4 F(E) is probability of an electron occupying an energy level E. If energy level E = local level E D (neutral N D ) or local level E A (ionized N A - ) use:
19
VM Ayres, ECE875, S14 Summary: Intrinsic range: concentration partially ionized dopants: n ≈ p ≈
20
VM Ayres, ECE875, S14 Summary: Saturation range: concentration fully ionized dopants AND some probability of valence-to-conduction band transitions: Fully ionized dopants: single doping:n-type: p = n i 2 /n, solve quadratic for n p-type: n = n i 2 /p, solve quadratic for p
21
VM Ayres, ECE875, S14 Summary: Intrinsic range: concentration fully ionized dopants AND high probability of valence-to-conduction band transitions: n ≈ p i = n i
22
VM Ayres, ECE875, S14 Summary: For all ranges: E gap = f(T); also f(P):
23
VM Ayres, ECE875, S14 Fig. 14: (a) Silicon; (b – not shown: GaAs)
24
Example: VM Ayres, ECE875, S14 Sun-side for a LEO (low-earth orbit) satellite is 200 o C = 473 K. What doping concentrations can’t be used in Si electronics because all pn junctions will act like intrinsic Si, causing device inoperability?
25
VM Ayres, ECE875, S14 473 K Answer: n and p doping concentrations 10 13 cm -3 and below can’t be used.
26
VM Ayres, ECE875, S14 Chp. 01 Concentrations Nondegenerate Degenerate Nondegenerate Intrinsic Contributed by impurities Wanted impurities: “dopants” Shallow dopants Deep level dopants Unwanted impurities: “traps” Lecture 10, 03 Feb 14 Effect of temperature Concentrations n
27
VM Ayres, ECE875, S14 Donors and acceptors don’t have to be “shallow”. Many atoms can get into Si: reason for cleanrooms
28
DONORS: Neutral N D Electron occupies a local energy level E D, h = 1, g D = 2 Ionized N D + A local energy level E D is empty and available ACCEPTORS: Neutral N A A local energy level E A is empty and available Ionized N A - Electron occupies a local energy level E A, h = 1, g A = 4 How to read graph: Above E i : read E C – E D Below E I : read E A - E V Generally for single substitutional impurities: Donor Levels/two charge states: E D (neutral 0, positive +1) Acceptor levels/two charge states: E A (neutral 0, negative -1)
29
VM Ayres, ECE875, S14 Can have donor levels below E i and acceptor levels above E i :
30
VM Ayres, ECE875, S14 Pr. 1.16 (a): gold dopant/impurity: Au: The acceptor level at E A-0.54 has two charge states: neutral 0 to start and -1 if it gets an e-. The donor level at E D-0.29 has two charge states: neutral 0 to start and +1 if it loses an e- At start: “state of charge of the Au levels in Si”:E A-.54 = neutral 0 E D-.29 = neutral 0
31
VM Ayres, ECE875, S14 Pr. 1.16 (a): Add neutral B to Au dopant/impurity: The acceptor level at E A-0.044 has two charge states: neutral 0 to start and -1 if it gets an e-. It does get e- from the E D-0.29 donor level. The acceptor level at E A-0.54 has two charge states: neutral 0 to start and -1 if it gets an e-. The donor level at E D-0.29 has two charge states: neutral 0 to start and +1 if it loses an e-
32
VM Ayres, ECE875, S14 Pr. 1.16 (a): After neutral B addition to Si with Au dopant/impurities: The acceptor level at E A-0.54 is neutral 0 at start and WHAT at finish The donor level at E D-0.29 is neutral 0 at start and WHAT at finish The acceptor level at E A-0.44 is neutral 0 at start and WHAT at finish Neutrality is maintained overall.
33
VM Ayres, ECE875, S14 Pr. 1.16 (b): effect of Au impurities on electron concentration n and hole concentration p, both from Si: Consider what the acceptor level at E A-0.54 can do with n from Si Consider what he donor level at E D-0.29 can do with p in Si Result: Si e- and holes that would have participated in current I are both being tied up (“trapped”) instead by the Au impurity levels. They are NOT contributing to current I while they are in traps.
34
VM Ayres, ECE875, S14 Effect on currents: 1.5.4: Will show: recombination and generation due to deep level dopants has greatest effect on ordinary current Ordinary current: Current I is in an assumed pn junction device
35
VM Ayres, ECE875, S14 1.5.4: Rate U is related to current I:
36
VM Ayres, ECE875, S14 What is U: Net transition rate driven by powerful pn = n i 2 at thermal equilibrium Fig 25 (a) matches Fig. 4 (b) Recombination with emission of photon Has a rate R e Generation with creation of e- hole pair Has a rate G thermal Note: Recombining e- must have a momentum value that matches the crystal momentum of the hole it is dropping into. Direct bandgap = OK all the way to the valence band
37
VM Ayres, ECE875, S14 What is U: Net transition rate driven by powerful pn = n i 2 at thermal equilibrium Fig 25 (a) matches Fig. 4 (b) Recombination decreases both n and p by 1 Now: pn < n i 2 Think: what if just recombination kept going?
38
VM Ayres, ECE875, S14 What is U: Net transition rate driven by powerful pn = n i 2 at thermal equilibrium
39
VM Ayres, ECE875, S14 What is U: Net transition rate driven by powerful pn = n i 2 at thermal equilibrium Fig 25 (a) matches Fig. 4 (b) Recombination rate R e depends on having: - Concentration of electrons in E C, - Concentration of holes in E V to take the e- - Probability of spontaneous recombination R ec Therefore: R e = R ec np = R ec ni 2
40
VM Ayres, ECE875, S14 What is U: Net transition rate driven by powerful pn = n i 2 at thermal equilibrium
41
VM Ayres, ECE875, S14 pn junction What is U: Net transition rate driven by powerful pn = n i 2 at thermal equilibrium General: Specific:
42
VM Ayres, ECE875, S14 Evaluate U: within 1 diffusion length of the junction on the n-side of a pn junction:
43
VM Ayres, ECE875, S14 n n0 ≈ N D + p n0 ≈ n i 2 /n=N D + excess holes: p LpLp Review: For a pn junction with low level injection p on n-side and n on p-side: p p0 ≈ N A - n p0 ≈ n i 2 /p=N A - electrons: n LnLn Within 1 diffusion length L p of the junction on the n-side of a pn junction:
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.