Lecture 11 Midterm Review Lecture

Lecture 11 Midterm Review Lecture 1/128 Announcements Homework 3/6: Is online now. Due Monday 14 th May at 10:00am. 2/128 1

Lecture 11 Exam Format. Units. Example Questions. Note: this is not an exhaustive list of questions. These are here to give you practice and to inform as to what type of question you can expect to see. Do not use this lecture alone to prepare for the exam! 3/128 Exam Format 4/128 2

Front Page 5/128 Details Friday May 4 th at 10:00am in STAG113 Exam will last 45 minutes. The exam will start exactly at 10:00am! Closed book and closed notes. You can, and are expected to, use a calculator. Choose 2 out of 3 questions. If you answer 3 I will take best 2 scores. It will contribute 25% of overall grade for class. The exam will material covered in lectures 2-10 (inclusive). There will be 10 marks per question. 20 total. Each mark = 5% of exam grade. 6/128 3

Details All constants will be provided. All relevant formulae will also be provided. Parameters will be labeled as clearly as possible. 7/128 Units 8/128 4

Unit Conversions It is worth spending a few slides on units. In semiconductor characterization we tend to use often non-si units as standard parameters. In particular, the use of cm is common over m. For example: Concentration, n, p ~ cm -3 rather than m -3. Mobility, μ e, μ h ~ cm 2 /Vs rather than m 2 /Vs. You are expected to know how to convert between units without notes / references. This is a useful skill generally. 9/128 Unit Conversions In one dimension it is straightforward: 1 m = 100 cm 1 cm = 0.01 m It can be easier to go from a unit to meters before going to cm: 1 μm = 10 6 m 10 6 m = 10 4 cm 1 km = 10 3 m 10 3 m = 10 5 cm 10/128 5

Inverse Dimensions For inverse dimensions we go the other way: 1 m = 100 cm 1 m 1 = 0.01 cm 1 1 cm = 0.01 m 1 cm 1 = 100 m 1 E.g. If you can have a line that is one meter long, it can accommodate 100 items in this meter. I.e. 100 m - 1. If you shorten the length to 1 cm, it can now only hold 1 item. I.e. 1 cm -1. 11/128 Inverse Dimensions For inverse dimensions we go the other way: 1 m = 100 cm 1 m 1 = 0.01 cm 1 1 cm = 0.01 m 1 cm 1 = 100 m 1 Again, it is often easier to convert to meters first. 1 nm 1 = 10 9 m 1 10 9 m 1 = 10 7 cm 1 1 km 1 = 10 3 m 1 10 3 m 1 = 10 5 cm 1 12/128 6

Area When talking about a length dimension squared, you apply the original conversion twice: 1 m = 100 cm 1 m 2 = 100 100 cm 2 = 10 4 cm 2 1 cm = 0.01 m 1 cm 2 = 0.01 0.01 m 1 = 10 4 m 2 This should be self-evident from geometry: 1 m Not to scale 1 cm 1 cm 1 m 13/128 Area When talking about a length dimension squared, you apply the original conversion twice: 1 m = 100 cm 1 m 2 = 100 100 cm 2 = 10 4 cm 2 1 cm = 0.01 m 1 cm 2 = 0.01 0.01 m 1 = 10 4 m 2 Again it is often easier to convert to meters first. 1 nm 2 = 10 18 m 2 10 18 m 2 = 10 14 m 2 1 km 2 = 10 6 m 2 10 6 m 2 = 10 10 cm 2 14/128 7

Inverse Area For inverse area we apply the inverse conversion twice: 1 m = 100 cm 1 m 1 = 0.01 cm 1 1 m 2 = 10 4 cm 2 1 cm = 0.01 m 1 cm 1 = 100 cm 1 1 cm 2 = 10 4 m 2 Again it is often easier to convert to meters first. 1 nm 2 = 10 18 m 2 10 18 m 2 = 10 14 cm 2 1 km 2 = 10 6 m 2 10 6 m 2 = 10 10 cm 2 15/128 Volume For volume you apply the conversion 3 times 1 m = 100 cm 1 m 2 = 10 4 cm 2 1 m 3 = 10 6 cm 3 1 cm = 0.01 m 1 cm 2 = 10 4 m 2 1 cm 3 = 10 6 m 3 Again it is often easier to convert to meters first. 1 nm 3 = 10 27 m 3 10 27 m 3 = 10 21 cm 3 1 km 3 = 10 9 m 3 10 9 m 3 = 10 15 cm 3 16/128 8

Inverse Volume Finally, for inverse volume you apply the conversion in reverse 3 times 1 m = 100 cm 1 m 2 = 10 4 cm 2 1 m 3 = 10 6 cm 3 1 cm = 0.01 m 1 cm 2 = 10 4 m 2 1 cm 3 = 10 6 m 3 Again it is often easier to convert to meters first. 1 nm 3 = 10 27 m 3 10 27 m 3 = 10 21 cm 3 1 km 3 = 10 9 m 3 10 9 m 3 = 10 15 cm 3 17/128 Non-SI Units in Equations If all parameters in an equation are in SI units you are guaranteed to get the correct answer. You may have to convert input parameters to SI, depending on how they are provided. You also may have to convert the final answer from SI, if the question asks for a non-si final unit. However sometimes it is more convenient to just use the non SI unit. For example if input parameters are in cm, and final answer is in Ω cm. 18/128 9

Non-SI Units in Equations For example, the resistivity in terms of resistance and dimensions: ρ = RA L If we are told: R = 1Ω. A = 1cm 2. L = 1cm. We want answer in Ωcm Ωcm ρ = RA L So we can just use cm directly. Ω cm 2 cm 19/128 Non-SI Units in Equations However consider this example of photon energy: E = hc λ If for example we are told: λ = 1μm. We want the energy in J. Js 3.0 10 8 m/s E = hc λ μm Unit mismatch 20/128 10

Non-SI Units in Equations However consider this example of photon energy: E = hc λ If for example we are told: λ = 1μm. We want the energy in J. Js 3.0 10 14 μm/s J E = hc λ μm 21/128 Lecture 2 22/128 11

Example 1 Consider the following thin film of indium tin oxide (ITO): 10 μm 100 nm 100 μm If we measure a resistance between these two electrodes of 7Ω, what is the conductivity of the ITO film? 23/128 Example 1 Need to determine which direction current flows: 10 μm 100 nm 100 μm So we can consider the resistor to have the following dimensions: 10 μm 100 μm 100 nm 24/128 12

Example 1 Hence in this case the area is a very thin slab: 100 μm 10 μm A 100 nm So the dimensions are: A = 100 10 6 100 10 9 = 1 10 11 m 2 L = 1 10 5 m 25/128 Example 1 You would be given the equation for resistivity: ρ = RA L ρ = 7 10 11 10 5 = 7 10 6 Ωm = 7 10 4 Ω cm 26/128 13

Example 2 Consider intrinsic (undoped) silicon. If we approximate the hole and electron densities are room temperature as: n i = p i =10 10 cm -3. And the room temperature carrier mobilities as: μ e =1400 cm 2 /Vs. μ h =500 cm 2 /Vs. What is the room temperature conductivity of intrinsic silicon (in the dark)? 27/128 Example 2 Use our equation for conductivity (would be given): σ = q μ e n + μ h p Put in the numbers (work in cm): σ = 1.6 10 19 1400 10 10 + 500 10 10 σ = 1.6 10 19 1900 10 10 σ = 3.04 10 6 Scm 1 28/128 14

Lecture 3 29/128 Example 1 Consider a 4-probe measurement on a sample we can approximate as infinitely thick. I V I 1 2 3 4 S 1 S 2 S 3 I Let: S 1 = S 2 = S 3 = S 30/128 15

Example 1 If we drive a current of I = 1mA across the outer two probes, we measure a voltage drop of 40μV between the inner two. I 1 2 3 4 S S S If the probe separation is 500 μm, determine the resistivity of the sample. Give your answer in Ωcm. You would be given: I V I ρ = 2πS V I 31/128 Example 1 ρ = 2πS V I So we cam just put in numbers. We can work in m to start: 500 μm 40 μv 40 10 6 6 ρ = 2π 500 10 10 3 ρ = 0.0031 Ωm 1 ma ρ = 0. 31 Ωcm 32/128 16

Example 2 For a sample of finite thickness the resistivity is given by: ρ = 2πSF V I Where: ρ is the resistivity of the sample. S is the sample thickness. V is the voltage. I is the current. F is a correction factor. List 3 things which affect F. 33/128 Example 2 List 3 things which affect F. ρ = 2πSF V I F is comprised of several factors. F = F 1 F 2 F 3 F 1 is associated with the sample thickness. F 2 is associated with the lateral sample dimensions. F 3 is associated the probe locations relative to the sample edges. 34/128 17

Example 3 Below is an equation for sheet resistance. R sh = π V ln 2 I = 4.532 V I What units are conventional for sheet resistance? ΩΤ To indicate this is sheet resistance, it is normally expressed in ohms per square (Ω/ ), but is dimensionally a resistance. 35/128 Example 4 Consider a measurement on sample of arbitrary geometry: 1 4 2 3 If we apply a voltage of 100 mv between probes 3 and 4, we measure a current of 40 ma between probes 1 and 2. If we apply a voltage of 100 mv between probes 1 and 4, we measure a current of 20 ma between probes 2 and 3. 36/128 18

Example 4 If this sample is assumed to be uniformly 500 μm thick, homogeneous and isotropic, what is its resistivity? We would be given the following equation for an arbitrary geometry: Sample thickness ρ = π ln 2 t R 12,34 + R 23,41 2 R 12,34 = V 34 I 12 R 23,41 = V 41 I 23 1 2 F v 4 3 Correction factor associated with arbitrary geometry 37/128 Example 4 Evaluate resistances: R 12,34 = V 34 I 12 R 12,34 = 0.1 0.04 I 1 2 V 4 3 V R 23,41 = V 41 I 23 R 23,41 = 0.1 0.02 R 12,34 = 2.5Ω I R 23,41 = 5Ω Evaluate ratio of resistances: R r = R 12,34 R 23,41 R r = 2.5 5 R r = 0.5 38/128 19

Example 4 Evaluate correction factor. The correction factor is given by: R r 1 R r + 1 = F v ln 2 arccosh exp ln 2 /F v 2 R r = R 12,34 R 23,41 You would not be expected to calculate something 1.0 numerically. So, you would be told F v for a given R r Or a graph would be provided. F v 0.8 0.6 0.4 0.2 0.0 1 10 100 1000 10000 R r 39/128 Example 4 E.g. would be told for our value of R r that F v = 0.96. Then we can just put numbers in: ρ = π ln 2 t R 12,34 + R 23,41 2 F v ρ = π ln 2 500 10 6 2.5 + 5.0 2 0.96 ρ = 0.00816 Ωm ρ = 0. 816 Ωcm 40/128 20

Example 5 The correction factor for a van der Pauw measurement is given by: R r 1 R r + 1 = F v ln 2 arccosh exp ln 2 /F v 2 R r = R 12,34 R 23,41 Using this equation, explain why we often use a symmetric geometry (i.e. a geometry with equal distances) for a van der Pauw measurement. If distances are equal we can say: R 12,34 = R 23,41 R r = 1 41/128 Example 5 If R r = 1: 1 1 2 + 1 = 0 = F v ln 2 arccosh exp ln 2 /F v 2 arccosh exp ln 2 /F v 2 = 0 exp ln 2 /F v 2 = cosh 0 cosh x = ex + e x 2 cosh 0 = e0 + e 0 2 cosh 0 = 1 42/128 21

Example 5 cosh 0 = exp ln 2 /F v 2 = 1 exp ln 2 /F v = 2 Take logs: ln 2 F v = ln 2 F v = ln 2 ln 2 = 1 I.e. if R 12,34 = R 23,41, F v =1. Hence if the path lengths are equal, the analysis becomes much more straight forward. 43/128 Example 6 Consider the following hot-probe measurement. V 1 2 Hot Cold If we are measuring a p-type sample, what sign would the voltage V 12 = V 1 V 2 in such a measurement? 44/128 22

Example 6 In a hot-probe experiment, a concentration gradient will exist between the hot and cold regions: Low hole density h + V 1 2 h + h + h + h+ h + h + h + h + h + h + h + h + High hole density If holes are the majority carrier, the potential at probe 1 will be lower than at probe 2. Hence the potential between probe 1 and 2 (V 12 = V 1 V 2 ) will be negative. 45/128 Lecture 4 46/128 23

Example 1 Consider a silicon sample, doped with 10 20 cm -3 donors. At a temperature of T = 77K would we expect thermionic emission, field emission, or thermionicfield emission to dominate? Assume: ε r = 11.7 (silicon). m tun = 0.3m e. 47/128 Example 1 To solve this we will have to evaluate the characteristic energy. E 00 = eh 4π N D ε 0 ε r m tun If k B T E 00 Thermionic Emission (TE) dominates. If k B T E 00 Thermionic-Field Emission (TFE) dominates. If k B T E 00 Field Emission (FE) dominates. So basically we just calculate E 00 and compare it k B T. 48/128 24

Example 1 Put numbers in: E 00 = eh 4π N D ε 0 ε r m tun Vacuum permittivity has units of Fm -1, so we must use meters! N D = 10 20 cm 3 N D = 10 26 m 3 E 00 = 1.60 10 19 6.63 10 34 4π 10 26 8.85 10 12 11.7 0.3 9.11 10 31 49/128 Example 1 Put numbers in: Now evaluate k B T: E 00 = 1.59 10 20 J k B T = 1.38 10 23 100 k B T = 1.38 10 21 J So: E 00 k B T Field Emission (FE) dominates. 50/128 25

Example 2 Consider a sample with a circular contact of radius r = 200μm, a resistivity of ρ = 3 10 5 Ωcm, and a thickness t=2mm. SiO 2 Determine the spreading resistance (R SP ) at this contact. You would be given: R SP = t I r Sample thickness ρ Ohmic contact ρ 2t arctan 2πr r So this would just be an exercise in entering numbers. Contact radius V Metal Sample resistivity 51/128 Example 2 Just enter numbers, all in m: R SP = ρ 2t arctan 2πr r R SP = 2.4 10 6 arctan 20 R SP = 3.6 10 6 Ω 52/128 26

Example 3 We carry out a current-voltage measurement on structure to the right: I 1 V 2 Z If we are told that: I = 10 ma. L = 10 μm. Z = 50 μm. 0 L Z = 50 μm. R sh = 10 Ω/. ρ c =10-6 Ωcm 2. x What is the voltage under the middle of the electrode? 53/128 Example 3 In this case we would be given the expression for voltage under the contact: I 1 V 2 Z V x = I R shρ c Z L x cosh L T sinh L L T = ρ c R sh L T 0 L x Under the middle of the contact x = LΤ2. So we need to evaluate: V LΤ2 54/128 27

Example 3 First evaluate transfer length: L T = ρ c R sh ρ c =10-6 Ωcm 2 ρ c =10-10 Ωm 2 L T = 10 10 10 = 3.16 10 6 m 55/128 Example 3 Now evaluate V: V x = I R shρ c Z cosh sinh L x L T L L T V x = L 2 = I R shρ c Z cosh L L L T 2L T sinh L L T V L 2 = I R shρ c Z 2L L cosh 2L T sinh L V L 2 = I L T R shρ c Z cosh sinh L 2L T L L T 56/128 28

Example 3 V L 2 = I R shρ c Z cosh sinh L 2L T L L T Now just enter numbers (use SI): V L 2 = 10 10 10 10 3 10 10 10 cosh 6 2 3.16 10 6 50 10 6 10 10 6 sinh 3.16 10 6 V V L 2 = 10 2 10 9 cosh 1.58 5 10 5 sinh 3.16 L 2 = 6.32 10 3 cosh 1.58 sinh 3.16 57/128 Example 3 Note: coshθ = eθ + e θ 2 sinhθ = eθ e θ 2 V L 2 = 6.32 10 3 0.214 V L 2 = 1.36 10 3 V 58/128 29

Example 4 Consider the following transmission line measurement: d 1 d 2 d 3 d 4 d 5 Z W L We are told that W = Z. If we plot total resistance measured (R T ) as a function of electrode separation (d), we observe an extrapolated y-axis intercept of 50 Ω. What is the contact resistance? 59/128 Example 4 We would be given the equation for the total resistance between any two contacts: R T = R shd i Z + 2R c We can identify this as our equation for a straight line. Where 2R c is the y-axis intercept. I.e. 2R c = 50Ω R c = 25Ω 60/128 30

Lecture 5 61/128 Example 1 A capacitance-voltage measured is carried out on a sample. 1ΤC 2 is plotted against magnitude of voltage. 1 C 2 If the gradient of this plot is 10 21 F -2 V -1, determine the doping density in this V sample. Give you answer in cm -3. We are told the sample is a square pad of A = 100μm 100 μm, the relative permittivity is ε r =12. 62/128 31

Example 1 We would be given: N A W = 2 qa 2 ε 0 ε r d dv 1 C 2 We would be able to identify our gradient: d dv 1 C 2 = 10 21 F 2 V 1 The remainder of the question is just entering numbers. 63/128 Example 1 N A W = 2 qa 2 ε 0 ε r d dv 1 C 2 First get the area: A = 100 10 6 100 10 6 = 10 8 m 2 A 2 = 10 8 10 8 = 10 16 m 2 Now enter numbers: 2 N A W = 1.60 10 19 10 16 8.85 10 12 12 10 21 64/128 32

Example 1 N A W = 2 1.60 10 19 10 16 8.85 10 12 12 10 21 N A W = 1.18 10 24 m 3 N A W = 1.18 10 18 cm 3 65/128 Example 2 The equation for an ideal diode is given below: I V = I S exp qv nk B T 1 Where: q = Fundamental unit of charge. n = Ideality factor (~1). k B = Boltzmann Constant. T = Temperature. I S = Reverse saturation current. 66/128 33

Example 2 What is meant by reverse saturation current (I S )? This is the current that flows when a large reverse bias is applied. I V = I S exp qv nk B T 1 I V = I S I R F V I S 67/128 Example 3 The current flowing in a diode (I), with an ideality factor n = 1, was measured at a constant voltage of V = 0.2V, and as a function of temperature (T). The data was plotted as IΤT 2 vs 1000ΤT on a logarithmic y-axis scale. We are told the gradient (on log 10 scale) of this plot is -0.25 AK -2. From this figure, determine the injection barrier (φ B ). Give you answer in V. 68/128 34

Example 3 We would be given the following approximation: I = AA T 2 exp q V n φ B k B T And we would be given the Richardson constant: A = 32 Acm -2 K -2. Or we would be given the effective mass, and be expected to calculate A. A = 4πqk B 2 m h 3 69/128 Example 3 First we need to re-arrange the equation for I so that it matches our provided plot. I = AA T 2 exp q V n φ B k B T I AA T 2 = exp q V n φ B k B T Take (natural) logs: I ln AA T 2 = q V n φ B k B T 70/128 35

Example 3 ln Re-arrange: ln I AA T 2 = q V n φ B k B T I T 2 = ln AA q φ V B n k B T The y-axis is plotted on a log10 scale. So we need to use (would be given): log a x = ln x ln a log 10 x = ln x = 2.3log 10 x ln x ln 10 71/128 Example 3 ln x = 2.3log 10 x ln I T 2 = 2.3log 10 log 10 I T 2 ln I T 2 = ln AA q φ V B n k B T I T 2 = ln AA q φ V B n k B T ln AA = 2.3 q φ B V n 2.3k B T Hence the y-axis is now correct. y = mx + c 72/128 36

Example 3 log 10 I T 2 ln AA = 2.3 q φ B V n 2.3k B T The x-axis is: 1000ΤT y = mx + c x = 1000 T x 1000 = 1 T log 10 I T 2 ln AA = 2.3 q φ B V n 2300k B x 73/128 Example 3 log 10 I T 2 ln AA = 2.3 q φ B V n 2300k B x So we can identify the gradient as: Now re-arrange: m = q φ B V n 2300k B φ B = V n m2300k B q 74/128 37

Example 3 We are given V = 0.2, n = 1, m = -0.25. φ B = 0.2 1 φ B = 0.2 + φ B = V n m2300k B q 0.25 2300 1.38 10 23 1.60 10 19 7.935 10 21 1.60 10 19 φ B = 0.2 + 0.05 φ B = 0.25 V 75/128 Lecture 6 76/128 38

Example 1 Give three ways in which we can define a deep level impurity. Definition 1: A level is deep if its binding energy is greater than 0.1 ev. Definition 2: A level is deep if is non-effective-masslike. Definition 3: A level is deep if is important in nonradiative recombination, i.e., Shockley-Read-Hall recombination. 77/128 Example 2 The rate of electron capture in a semiconductor trap is given by: r a = σ n v th nn T 1 f T Where: r a = Rate of electron capture. σ n = Electron capture cross-section (e.g. cm 2 ). v th = Electron thermal velocity. n = Delocalized electron density in conduction band. N T 1 f T = Number density of empty traps. 78/128 39

Example 2 The rate of electron emission is given by: Where: r b = e n N T f T r b = Rate of electron emission. e n = Proportionality constant equal to the rate at which a single electron in an occupied trap will jump into the conduction band. N T f T = Number density of filled traps. 79/128 Example 2 If we can describe trap occupancy by Fermi-Dirac statistics, derive an expression for electron emission rate (e n ) under steady state conditions, in terms of the capture cross section (σ n ), thermal velocity (v th ), number density of conduction band states (N c ), trap energy (E T ), conduction band energy (E C ), and temperature (T). We would be given: f T = 1 1 + exp E T E F k b T n = N c exp E C E F k b T 80/128 40

Example 2 First, since we are in steady state, we can equate the rates: r a = r b σ n v th nn T 1 f T = e n N T f T e n = σ n v th n 1 f T f T We can also re-arrange the Fermi-Dirac equation: f T = 1 1 + exp E T E F k b T 1 f T f T = exp E T E F k b T 81/128 Example 2 Combine this with our equation for emission rate: 1 f T f T = exp E T E F k b T e n = σ n v th n 1 f T f T e n = σ n v th n exp E T E F k b T Substitute in our given equation for n: e n = σ n v th N c exp E C E F k b T exp E T E F k b T e n = σ n v th N c exp E C E T k b T 82/128 41

Lecture 7 83/128 Example 1 Briefly describe the steps involved in carrying out a deep level transient spectroscopy (DLTS) experiment. 1. Create a space charge region by applying a DC reverse bias across your measurement device. 2. Apply a repetitive voltage pulse in which electrons or holes are captured by deep levels. 3. When the pulse is terminated, the deep levels will attempt to reach their equilibrium occupancy by thermal emission. 84/128 42

Example 1 4. The return to equilibrium can be monitored by measuring capacitance (or conductance or current or charge). 85/128 Example 2 When carrying out a deep level transient spectroscopy (DLTS) experiment, describe with diagrams what is meant by the parameter T max. In a DLTS measurement we would measure capacitance as a function of time, at a range of temperatures. As shown to the right. The decay rate will depend on temperature C t, T t 86/128 43

Example 2 We define two samples times: t 1 and t 2, at which we consider the capacitance. We can then define the difference in capacitance sampled at these two times as: C t, T t 1 t 2 t δc = C t 1 C t 2 87/128 Example 2 When we plot δc as a function of temperature, we will observe a peak δc at some value of T. See figure to the right. We describe this maximum change in capacitance as δc max. δc = C t 1 C t 2 We define the temperature at which δc max is measured as T max. T T max 88/128 44

Example 3 When carrying out a deep level transient spectroscopy (DLTS) experiment it turns out that a certain T max occurs when we define the sample times t 1 = 5 ms and t 2 = 20 ms. Using this information, determine the emission lifetime for electrons at T max. We would be given: 1 e n T max = τ n,max = t 2 t 1 ln t 2 t1 89/128 Example 3 Just enter numbers: τ n,max = 0.020 0.005 ln 0.020 0.005 τ n,max = 0.015 ln 4 τ n,max = 2.89 ms 90/128 45

Example 4 We measure the electron emission probability (e n ) as a function of temperature. If we plot log e n versus 1ΤT we would get something like the following: log e n 1000ΤT 91/128 Example 4 If the gradient of this plot was found to be 2.5, determine the energy offset between the trap and the conduction band. Give your answer in ev You would be given: e n = σ n v th N c exp E C E T k b T e n = exp E C E T σ n v th N c k b T Take logs (base-10): log 10 e n = log σ n v th N 10 exp E C E T c k b T 92/128 46

Example 4 Use the following relationship for logarithms of different base (would be given): log a x = ln x ln a log 10 log 10 x = ln x ln 10 e n = 1 σ n v th N c 2.3 ln exp E C E T k b T ln x 2.3 log 10 e n σ n v th N c = E T E C 2.3k b T log 10 e n log 10 σ n v th N c = E T E C 2.3k b T 93/128 Example 4 log 10 e n = E T E C 2.3k b T + log 10 σ n v th N c We identify this as the equation for a straight line: Hence we can say: y = mx + c y = log 10 e n If the x-axis is 1000ΤT: mx = E T E C 2.3k b T mx = E T E C 2.3k b T 1000 T m = E T E C 2.3k b T m = E T E C 2300k b 94/128 47

Example 4 We are after the offset in energy: m = E T E C 2300k b E T E C = 2300k b m E T E C = 2300 1.38 10 23 2.5 E T E C = 7.94 10 20 J E T E C = 0.496 ev 95/128 Lecture 8 96/128 48

Example 1 Using diagrams, briefly describe the following processes and how energy is conserved in each case: 1). Shockley-Read-Hall recombination. 2). Radiative recombination. 2). Auger recombination. 97/128 Example 1 1). Shockley-Read-Hall recombination. In Shockley-Read-Hall (SRH) recombination a charge carrier (electron or hole) recombines with a trap. Energy is conserved through multiple-phonon emission (MPE). E T Carrier MPE Trap E C E V SRH 98/128 49

Example 1 2). Radiative recombination In radiative recombination an electron and a hole recombine (annihilate). Energy is conserved through the emission of a photon. hν Hole Electron E C hν E V 99/128 Example 1 2). Auger recombination In Auger recombination an electron and a hole recombine (annihilate). However rather than emit a photon, energy is carried away with via a third carrier. This can be an electron or a hole. Hole Momentum Transfer thermalizes Electron 3 rd carrier E C E V 100/128 50

Example 2 If the Shockley-Read-Hall (SRH) lifetime is τ SRH, as given below, show that for a p-type material under low light conditions ( n p 0 ), the lifetime reduces ll to: τ SRH τ n. τ SRH = τ p n 0 + n 1 + n + τ n p 0 + p 1 + p p 0 + n 0 + n τ p = 1 σ p v th N T τ n = 1 σ n v th N T 101/128 Example 2 If the sample is p-type: p = 0 n 0 = 0 ll τ SRH = τ p n 1 + n + τ n p 0 + p 1 p 0 + n We are also told that this is low level illumination: n p 0 n 0 ll τ SRH = τ pn 1 + τ n p 0 + p 1 p 0 102/128 51

Example 2 ll τ SRH = τ pn 1 + τ n p 0 + p 1 p 0 ll τ SRH = n 1 p 0 τ p + 1 + p 1 p 0 τ n p 1 : concentration of holes in the valence band when the Fermi level is positioned at the trap energy level. n 1 : concentration of electrons in the conduction band when the Fermi level is located at the trap energy. n 1, p 1 n 0, p 0 ll τ SRH 0 τ p + 1 + 0 τ n ll τ SRH τ n 103/128 Example 3 The band gap of GaAs is 1.43eV. Determine the wavelength (λ) of a photon emitted by radiative recombination in a GaAs device. Give your answer in nm. You would be given: E = hc λ λ = hc E Remember to convert energy to joules: E = 1.43 ev E = 2.29 10 19 J 104/128 52

Example 3 Enter values: λ = 6.63 10 34 3 10 8 2.29 10 19 λ = 8.69 10 7 λ = 869nm 105/128 Example 4 Briefly describe the concept of photon recycling. Photon recycling is the phenomenon of photons that are emitted by radiative recombination in a semiconductor being re-absorbed and generating more electron hole pairs. e - e - h + h + 106/128 53

Lecture 9 107/128 Example 1 A photoconductance decay (PCD) measurement is carried out on a range of wafers with various thicknesses (d). The data is shown below. 150 s -1 We are told that the surface recombination velocity is very small (s r 0 ). From this data, determine the bulk carrier lifetime. Give your answer in ms. 108/128 54

Example 1 You would be given the following equations: 1 = 1 + 1 τ eff τ B τ s 1 = 1 τ s Dβ 2 tan βd 2 = s r βd If s r is very small, we can make the small-angle approximation for a tan function: tan θ θ Dβ 2 = 2s r d tan βd 2 βd 2 = s r βd β 2 d 2 = s r D 1 = 1 τ s Dβ 2 τ s s r 0 = d 2s r 109/128 Example 1 τ s s r 0 = d 2s r Consider the effective lifetime (due to both surface and bulk recombination): 1 = 1 + 1 1 s τ eff τ B τ s τ r 0 = 2s r eff d + 1 τ B Identify this as the equation for a straight line: y = mx + c Plotting 1Ττ eff vs 1Τd will have a slope of 2s r and an intercept of 1Ττ B. 110/128 55

Example 1 We were told the intercept was 150 s -1. 1 τ B = 150 s 1 τ B = 6.66 ms 111/128 Example 2 A surface photovoltage (SPV) measurement is carried out a semiconducting sample, by adjusting the incident photon flux (Φ) and keeping the measured surface voltage (V spv ) constant. From the data in the figure to the right, determine the diffusion length of carriers in this semiconductor. -0.003 cm 112/128 56

Example 2 You would be given: Φ = C 1 L n + 1 α λ The intercept of the x-axis is when Φ=0. C 1 L n + 1 α λ = 0 L n + 1 α λ = 0 1 α λ = L n I.e. the intercept is just L n. L n = 0.003 cm 113/128 Example 3 In the context of electronic devices (e.g. transistors), what is meant by generation lifetime? The generation lifetime is the average time that it takes to thermally generate one electron-hole pair in the space charge region under extraction conditions. 114/128 57

Lecture 10 115/128 Example 1 We are told that in a certain material, in a certain direction, at a certain temperature, the mean scattering time is τ =300 fs, and the effective mass is m =0.067m e m e is the rest mass of an electron. Determine the charge carrier mobility under these conditions. Give your answer in cm 2 /Vs. You would be given: μ = q τ m 116/128 58

Example 1 Just enter numbers: μ = 1.60 10 19 300 10 15 0.067 9.11 10 31 μ = 0.787 m 2 /Vs μ = 7870 cm 2 /Vs 117/128 Example 2 Below is a schematic of the band structure in graphene. Energy E k Conduction band Valence band Momentum Explain why standard (non-relativistic) semiconductor models cannot be applied to determine the effective mass of carriers in such a system. 118/128 59

Example 2 You would be given the equation for effective mass: m = ħ2 d 2 E dk 2 Here we identify the 2 nd derivative of the band energy with respect to momentum: d 2 E dk 2 Notice, on the previous slide, the energy is linearly proportional to momentum: E = Ak A = Some constant 119/128 Example 2 In this case the first derivative would be: de dk = A And the second derivative would therefore be: d 2 E dk 2 = 0 And hence the effective mass would be: m = ħ2 d 2 E dk 2 m ħ2 0 m 120/128 60

Example 2 Hence the effective mass is infinite, or undefined for this band structure. This problem is solved by considering the effects of special relativity 121/128 Example 3 Consider the following Hall measurement. V ρ B L z W d I x y V H 122/128 61

Example 3 We are told the sample has the following dimensions: W = 1 mm. L = 5 mm. d = 1 mm. We are told the sample is p-type, and: The electron density is n = 0. The Hall Factor is r = 0. If we apply a voltage in the x direction of 10V, we measure a current of 500 μa. 123/128 Example 3 If we apply a magnetic field of 0.5 T, we observe a Hall voltage of V H = +30 mv. From this information, determine the hole mobility in this sample. Give your answer in cm 2 /Vs. First calculate the resistivity of our sample. We would be given: ρ = Wd V ρ L I Enter values (stay in meters for now): ρ = 1 10 3 1 10 3 10 5 10 3 = 4 Ωm 500 10 6 124/128 62

Example 3 Now calculate Hall Factor (R H ). You would be given: Enter values: R H = V Hd IB z R H = 0.03 10 3 500 10 6 0.5 = 0.12 m3 C 1 Finally, we can get the mobility from ρ and R H. This would also be given: μ H = R H ρ 125/128 Example 3 Enter values: μ H = 0.12 4 μ H = 0.03 m 2 /Vs μ H = 300 cm 2 /Vs 126/128 63

Example 4 Describe some of the problems with time-of-flight drift mobility measurements: In addition to transport in an electric field, carriers also undergo diffusion, broadening the signal in space and time. Injection barriers may significantly effect results. Carrier velocity may not be linear with electric field strength over the distances studied. 127/128 Good Luck! 128/128 64