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ECE 875: Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

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Presentation on theme: "ECE 875: Electronic Devices Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University"— Presentation transcript:

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

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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:

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