Semiconductor Materials and Devices Flashcards
(240 cards)
1
Q
Give two expressions for the classical kinetic energy of a particle.
A
Check
2
Q
Give the equations for angular frequency and wavevector.
A
Check
3
Q
Give the expression for quantum mechanical KE of a free electron.
A
Check
4
Q
State the TISE.
A
Check
5
Q
Solve the TISE for a free election gas in volume L^3 stating an expression for the quantised energy states.
A
Check
6
Q
State the Bragg reflection condition for electrons incident on crystal planes
A
n lambda = 2dsin(theta)
7
Q
How can the Bragg condition be used to define the Brillion zone boundary?
A
If an electron travels normal to the plane, the Bragg reflection condition occurs when k=nπ/d. This means the electron wave vector will constantly be reflected back and forth (standing wave).
8
Q
What is the state of the electron energy in a system when the standing waves of electron density has high density overlapping positive ion cores?
A
Low energy state.
9
Q
What is the state of the electron energy in a system when the standing waves of electron density has high density between positive ion cores?
A
High energy state
10
Q
What is the band gap at the Brillouin zone boundary?
A
For the same k vector, the electron states have different energies. The difference in these energies is the band gap.
11
Q
Sketch the extended zone scheme and the reduced zone scheme band structure for electrons in the nearly free electron model.
A
Check slide 7
12
Q
Why do energy gaps form at the Brillouin zone boundary in the NFE band structure?
A
The periodic potential of the crystal lattice causes energy gaps to appear in the FE model.
13
Q
Define the 1st BZ in terms of reciprocal lattice vectors.
A
The 1st BZ is a closed volume about the origin in reciprocal space formed by bisecting near neighbour reciprocal lattice vectors.
14
Q
Sketch band structures for the free electron, NFE, tight binding and free atom with energy on the y axis.
A
Check slide 9
15
Q
Derive an expression for density of states.
A
Check slide 10
16
Q
Define density of states.
A
The number of states with an energy between E and E+dE per unit volume.
17
Q
What is true of the density of states at T=0K
A
The states are fully occupied up to the Fermi energy, Ef
18
Q
What distribution defines the occupation of the density of states?
A
The Fermi-Dirac distribution function.
19
Q
State the Fermi-Dirac distribution
A
Check
20
Q
State how filled the bands are in a metal.
A
Electron density is such that the highest occupied band is partially filled.
21
Q
State how filled the bands are in a semiconductor.
A
Highest band is completely filled at low T and the difference in energy to the next band is relatively small.
22
Q
State how filled the bands are in an insulator.
A
Highest occupied band is completely filled at low T and the difference in energy to the next band is large.
23
Q
What does conduction require?
A
A net electron momentum - changing the average k values. This causes a shift of the Fermi surface in k-space.
24
Q
Why can’t insulators conduct (momentum)?
A
There are no empty states for the electrons to move into hence the electrons can gain no net momentum.
25
Each Brillouin zone contains how many electrons?
2 electrons per primitive unit cell in real space.
26
Atoms with odd numbers of electrons form what types of materials?
Metals.
27
Atoms with even numbers of electrons for what type of materials?
They may form semiconductors or insulators. This depends of the band overlap since directionality must be included
28
Can there be an energy discontinuity at each Brillouin zone but be no overall bandgap?
Yes it is possible. Check slide 19
29
Give the bandgaps in Si, GaAs and GaN
1. 1 eV
1. 4 eV
3. 4 eV
30
What structure does GaAs have?
A zincblende cubic structure.
31
Define group velocity.
The group velocity of a wave is the velocity with which the variations in the shape of the wave's amplitude (envelope) propagate through space.
32
Give an expression for group velocity.
Check
33
Give an expression for effective mass of an electron.
Check
34
When do electrons respond as if they have an effective momentum and an effective mass?
Inside a crystal electrons respond to outer forces as if they have an effective momentum hk (where k is the position of the electron along the band) and near the band edges they respond as if they have an effective mass m*.
35
For N free carrier of charge e, find the effective number of free electrons when an electric field of E is applied to the crystal.
Check slide 27
36
What is the effective number of electrons for a completely full band and when is the effective number of electrons at its maximum?
For a completely full band, the effective number of electrons is 0.
The effective number of electrons is at its maximum when the band is half full (metals).
37
What is a hole?
An empty state in the valence band.
38
How do holes behave differently to electrons?
The holes move in the direction of the electric field, thus they behave as if they are positively charged.
39
What is removing an electron with a negative effective mass from a full band equivalent to?
Introducing a single +ve charged particle of +ve mass.
40
Why is the effective mass of an electron in the valence band -ve?
The curvature at the top of the valence band is negative therefore electrons have a negative effective mass.
41
Why is the electron mass in the conduction band usually smaller than the hole mass in the valence band?
The curvature of the conduction band is usually greater.
42
What can be said about the positioning of a conduction band minimum?
It will always be at k=0 or at some point along symmetry directions.
43
Why can bands separate with split off energy ∆?
Bands can separate due to spin orbital coupling with split off energy ∆.
44
What kind of band gap does GaAs have?
Direct.
45
What kind of band gap does Si have?
Indirect
46
True or false, a direct band gap semiconductor such as GaAs has strong optical transmissions and can be used for light emission.
True
47
What is an intrinsic material?
A pure material where all carriers are thermally excited so the number of electrons (ni) is equal to the number of holes (pi)
48
Derive an expression for the concentration of electrons (n) in the conduction band of an intrinsic material at T, stating any approximations made.
Check slide 39-45
1) Use Boltzmann distribution for f(E)
2) integrate from E=Ec to E=∞ (ignoring finite width of conduction band)
49
State an expression for the hole concentration in the valence band of an intrinsic material.
Check
50
What are Nc and Nv and give expressions for them in terms of effective mass and h.
Nc is the effective density of states in the conduction band.
Nv is the effective density of states in the valence band.
Expressions can be checked on slide 45/6
51
State the law of mass action giving two expressions.
Check slide 46
52
Derive an expression for the position of the Fermi level in an intrinsic material.
Check slide 47
53
What is an extrinsic material?
A material that has been doped to increase carrier concentration.
54
What do donor atoms do in extrinsic materials?
Donate negative carriers making the material n-type.
55
What do acceptor atoms do in an extrinsic material?
Produce positive carriers making the material p-type
56
Why are small and medium band gap semiconductors easy to dope?
They produce shallow levels (close to the energy levels of the intrinsic material).
57
What happens when a group VI atom is added to a group III-V semiconductor (GaAs)?
The group VI dopant will go into the As sites so dope n-type.
58
What happens when a group II atom is added to a group III-V semiconductor (GaAs)?
The group II dopant will go into the Ga sites so dope p-type.
59
What happens when a group IV atom is added to a group III-V semiconductor (GaAs)?
The group IV dopant will go into the Ga sites so dope n-type.
60
What limits are there on dopant concentrations in extrinsic semiconductors?
The minimum carrier conc cannot be lower than the intrinsic carrier conc of the material.
The maximum carrier conc cannot be greater than the solubility of the dopant in the material.
61
In what materials are donor and acceptor atoms in?
Donors in n-type
| Acceptors in p-type
62
When is Boltzmann statistics a good approximation to Fermi statistics?
When Ec-Ef ≥ 2kT
63
Find expressions for the position of the Fermi energy in an n or p doped material.
Check slide 61
64
What happens to the Fermi energy when a material is n-doped?
It tends towards the conduction energy.
65
What happens to the Fermi energy when a material is p-doped?
It tends towards the valence energy.
66
What is true in a neutral p-type material about the number of holes and acceptors?
They are equal.
67
What is true in a neutral n-type material about the number of electrons and donors?
They are equal.
68
What happens to the Fermi energy of an extrinsic material as temperature increases?
It tends towards the intrinsic Fermi energy.
69
Give an expression for the total KE of an electron considering it is free to wander in 3 dimensions.
Check slide 67
70
In a perfect periodic material, electrons suffer no scattering but in real materials, what can electrons be scattered by?
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Lattice vibrations (phonons)
Impurities, defects
Electron-electron, electron-hole interactions
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71
What is the motion of carriers in a semiconductor when no electric field is applied?
Random with frequent changes in direction.
72
What is the motion of carriers in a semiconductor when an electric field is applied?
Random but a net motion along the direction of the field.
73
Give equations for the drift velocity and mobility of electrons and holes.
Check slide 69
74
At high fields in Si, there is a limit to the speed of carriers, what is this?
Fast moving carriers generate phonons and so lose energy.
75
At high fields in GaAs, there is a maximum velocity that falls away, why is this?
The electrons get excited into a secondary minimum in the GaAs band structure that is close to the first. The effective mass is larger in this minimum thus the electrons slow down.
76
Why is the mobility of electrons increased at low temperatures from lattice vibration scattering (electron-phonon interactions)?
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Little thermal energy
Few phonons
Carrier velocity (thermal) is small
Mean scattering times are large
Hence µ is high
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77
Why is the mobility of electrons lowered at high temperatures from lattice vibration scattering (electron-phonon interactions)?
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More thermal energy
Many phonons
Large thermal velocities
Mean scattering times are small
Hence µ is low
```
78
What is the approximate relationship between µ and temperature as a result of electron-phonon scattering?
µ proportional 1/T^1.5
79
Why is the mobility of electrons lowered at low temperatures from impurity scattering (electron-phonon interactions)?
The carrier velocity (thermal) is small so electrons are easily deflected by ionised impurities.
The time between scattering events is small so µ is small
80
Why is the mobility of electrons increased at high temperatures from impurity scattering (electron-phonon interactions)?
The carrier velocity (thermal) is high so electrons are less easily deflected by ionised impurities.
The time between scattering events is greater so µ increases.
81
What is the approximate relationship between µ and temperature as a result of ionised impurity scattering?
µ proportional T^1.5
82
State a relationship between overall µ, µ(lattice) and µ(impurity).
1/µ = 1/µ(lattice) + 1/µ(impurity)
83
Which of µ(lattice) and µ(impurity) controls the overall µ?
Which ever is smaller is there one that controls the overall µ
84
In an n-ype material, how does concentration affect the type of scattering?
At low conc of dopant, the scattering is mostly lattice scattering whereas at high conc it is mostly ionised impurity scattering.
85
In a p-ype material, how does concentration affect the type of scattering?
At low conc of dopant, the scattering is mostly lattice scattering whereas at high conc it is mostly ionised impurity scattering.
86
Does mobility vary the same amount with concentration in both n and p-type materials?
No, the variation is much greater in n-type materials as holes have a lower effective mass so have lower mobilities and stay low.
87
Derive an expression for the conductivity of a material in terms of number of carriers and mobility.
Check slide 79
88
Sketch how conductivity varies with temperature and concentration of dopant for an n-type material giving reasons for the shape.
Check slide 81
89
What happens in recombination?
An electron and hole recombine (electron from conduction band and hole in valence band) releasing energy equivalent to the band gap.
90
What happens in generation?
Energy is input (minimum is band gap energy) promoting an electron to the conduction band and generating a hole in the valence band. It is the generation of an electron-hole pair.
91
Justify the law of mass action.
Check slide 83
92
What happens to the law of mass action under non-equilibrium conditions when there is carrier injection?
np > n(i)^2
93
What happens to the law of mass action under non-equilibrium conditions when there is carrier extraction?
np < n(i)^2
94
If there is a non-equilibrium variation in carrier concentration, and conc varies as a function of position, what will happen?
A diffusion crurent will be set up.
95
State expressions for the diffusion currents for electrons and holes.
Check slide 85
96
Under equilibrium conditions will the diffusion current will flow?
No. The diffusion current will be matched by an equal and opposite drift current.
97
What happens to the Fermi level when separate specimens are joined (eg p neutral n)?
Provided at equilibrium, the Fermi level is independent of position thus the Fermi levels of the materials will align.
98
At a p-n junction, why is there a built in electric field?
There is a conc gradient across the junction.
99
Derive the Einstein relationships for mobility of electrons and holes.
Check slide 89.
100
Derive an expression for the change in carrier conc due to difference in currents in and out of a point.
Check slide 91
101
Derive an expression for the change in carrier conc due to difference in currents in and out of a point when there is carrier injection at rate G and recombination rate U.
Check slide 93
102
What is direct recombination?
Normally dominates in direct band gap materials and is where the electron and hole have the same k. It is accompanied by the emission of a photon or a shower of phonons.
103
What is indirect recombination?
Normally in indirect band gap materials. Requires a state in the forbidden gap known as deep level. Recombination by emission of phonons and sometimes a photon. 2 transitions therefore not as effective as direct recombination.
104
How are deep levels introduced into a material?
With certain impurity atoms (transition metals in Si) or by crystal defects such as interstitials, vacancies, clusters, ion implantation damage (dislocations).
105
What is Auger recombination?
An Auger electron is emitted when the indirect electron and hole recombine with the emitted electron taking away excess energy and momentum.
Dominates at very high carrier concs.
106
In the depletion region of a p-n junction, what can be said of the carrier concs?
That the free carrier conc is much less than dopant conc.
107
Derive Poisson's Equation
Check slide 107
108
Derive an expression for the built in voltage across a p-n junction in terms of depletion region width.
Check slide 112
109
Derive an expression for the electric field across a p-n junction
Check slide 111
110
Derive an expression for built in voltage across a p-n junction in terms of donor and acceptor concs.
Check slide 116 and slide 120
111
What happens to the energy bands in a semiconductor when a bias is applied?
The energy bands tilt as the electron has to have work done to move down the field.
112
Can the concept of Fermi level be used in a semiconducting material with an electric field applied?
No, it is an equilibrium concept.
113
What happens to a p-n junction when a forward bias is applied?
The width of the depletion region decreases
Smaller potential barrier = V(BI) - V(F)
Smaller built in field.
114
What happens to a p-n junction when a reverse bias is applied?
Depletion region width increases (more space charge)
Larger potential barrier = V(BI) - V(R)
Greater built in field
115
How can equations for unbiased p-n junctions be turned into ones for biased?
Replace V(BI) with V(BI) + V
where for
no bias V = 0
reverse bias V = V(R)
forward bias V = -V(F)
116
State equations for the width of the depletion region and maximum electric field for a p-n junction under any bias.
Check slide 131
117
Derive an expression for the capacitance of a p-n junction.
Check slide 132
118
Under no bias, is there a current across a p-n junction?
No as the diffusion current of electrons from n to p is equal to the drift current of electrons from p to n. Similarly for holes, thus they cancel and there is no net current flow.
119
What is minority carrier injection?
Where electrons are added to the p-type and holes to the n-type.
120
What happens in a p-n junction to holes and electrons under forward bias.
Under forward bias the potential barrier is lower so majority carriers diffuse more easily to the other side (minority carrier injection as they are minority carriers on the other side). The drift component of current is unaffected however the diffusion component becomes larger.
121
What assumptions are made for a p-n junction under forward bias?
That carrier transport across the junction is sufficiently rapid that electron conc in the p-type material at the edge of the junction (minority carriers) is at equilibrium with respect to the n-type. Similarly for holes.
The change in depletion region width when voltage is applied does not change the electron conc in the n-type of the hole conc in the p-type (majority carriers)
122
Derive expressions for the amount of minority carrier injection in each side of the p-n junction under forward bias.
Check slide 138
123
Sketch a graph of minority carrier injection on both slides of a p-n junction under forward bias.
Check slide 139
124
Find expressions for the diffusion current of minority carriers from minority carrier injection that results from a p-n junction under forward bias.
Check slide 140
125
Derive the ideal diode equation.
Check slide 141
126
Explain the shape of the IV curve of a p-n junction.
Conducts under forward bias as lower potential barrier across the depletion region and diffusion current increases due to minority carrier injection.
Non-conducting in reverse bias mode as potential barrier across the junction increases and diffusion current lowers to zero leaving only a small drift current.
127
What happens when a p-n junction is placed under reverse bias?
The increased height of the potential barrier in the junction means it is harder for majority carriers to diffuse across the junction so diffusion currents reduced.
Minority carriers still drift across the junction as it is unaffected by bias.
128
What happens to the diffusion current in a p-n junction wrt the drift current under different bias conditions?
Zero bias: equal to the drift current
Forward bias: Much greater than the drift current
Reverse bias: Much less than the drift current
129
Sketch and explain the current distribution across a p-n junction under forward bias where N(A) ≈ N(D)
Check slide 145
130
Derive an expression for the total current across an asymmetric p-n junction where N(D) >> N(A) stating assumptions made.
Check slide 146
131
By what methods does a p-n junction breakdown under reverse bias?
Avalanche
| Tunnelling (aka Zener)
132
When does a p-n junction breakdown under reverse bias?
If the maximum electric field exceeds a certain value for the material. Breakdown occurs and excessive current flows.
133
Derive an expression for the breakdown voltage of a one sided p-n junction (N(D) >> N(A)).
Check slide 148
134
Sketch a plot of breakdown voltage against one sided carrier conc.
Check slide 149
135
What relationship does the breakdown voltage follow for a one sided p-n junction where N(D) >> N(A)?
V(BD) proportional to 1/N(A)
136
How does avalanche breakdown occur?
Minority carriers are accelerated in depletion region of the reverse bias junction. Their speed depends on the field and time between scattering events. When the KE of carriers exceeds the width of the gap, then the a collision can create another e-/h+ pair (impact ionisation). The new pair can cause further impact ionisation this the current increases uncontrollably.
137
What changes to factors cause E(crit) to increase for avalanche scattering? (E(crit) is critical field strength for breakdown)
The band gap increases
The scattering time decreases
The temperature increases (dV(BD)/dT is +ve)
138
How does tunnelling breakdown occur?
Breakdown occurs when electrons tunnel from full valence band into adjacent empty conduction band.
139
How does the probability of tunnelling change with certain factors?
It decreases as the band gap increases (height of barrier) and the electric field decreases (width of barrier increases).
140
Breakdown usually occurs by what method when?
V(BD) ≥ 6∆E/e avalanche
V(BD) ≤ 4∆E/e tunnelling
Mixture for 4∆E/e ≤ V(BD) ≤ 6∆E/e
All where dV(BD)/dT ≈ 0 (this produces good voltage regulator diodes
141
Sketch how depletion region width varies with reverse bias for varying N(A) for a one sided N(D) >> N(A) p-n junction.
Check slide 153
142
Give expressions for depletion region width for reverse bias applied to p-n junction at low and high voltages.
Check slide 153
143
Sketch and label a metal-semiconductor junction.
Check slide 155
144
What forms when a metal and semiconductor come into contact?
Schottky contact with a barrier height, built in potential and depletion region width.
145
Give an expression for the Schottky barrier height between a metal and semiconductor under different bias conditions for an n-type semiconductor.
Check slide 158
146
The Schottky barrier between a metal and p-type semiconductor is a barrier to what?
Hole transport.
147
What are surface states of a semiconductor?
Deep levels found at the surface due to:
Broken bonds
Impurity atoms
Oxide layers etc
148
What is the issue with surface states on a semiconductor?
The surfaces become charged producing bent bands even before the metal is deposited.
149
In theory what should the sum of the energy barriers between a metal and semiconductor be for an n-type and p-type of the same material?
Should sum to the energy gap.
150
Why doesn't the sum of the energy barriers for an n-type and p-type paired with a metal equal the energy gap?
When surface states are present (on most real semiconductors) they affect the band structure at the surface of the semiconductor before metal is deposited on it(states below the Fermi level become occupied).
151
Give an expression for the density of surface states in a real semiconductor due to charge neutrality.
Check slide 161
152
What is needed for ohmic contacts between metal and semiconductors?
Low resistance - symmetrical characteristics.
153
How to achieve ohmic contacts between metals and semiconductors?
Use metal with low barrier height (often not good enough)
| Form a highly doped surface layer to semiconductor meaning transport is by tunnelling through the very narrow barrier.
154
What is a rectifier diode used for?
To convert an AC voltage to a DC voltage.
155
How does a single rectifier diode respond across the cycle of AC?
On the forward half cycle, R(F) << R(L) so V(out) ≈ V(in).
| On the reverse half cycle, R(R) >> R(L) so V(out) ≈ zero.
156
How can we use rectifier diodes to change AC to DC?
Use a diode bridge connected to a capacitor. Sketch on slide 10
157
What diode is needed when you need to place a switching signal , between 0V (OFF) and 5V (ON) on a part of a circuit when the signal is very fast, and need to retain electrical isolation when in the OFF state.
Fast switching diode.
158
What happens to depletion region width when a diode is switched?
It changes and so does the charge contributing to the junction capacitance.
The shortest time associated with this charge is tau ≈ RC
R is the total series resistance and C is the junction capacitance.
159
When a diode switches, what happens to the current levels in the diode?
The current level changes as minority carriers are injected across the depletion region.. The shortest time associated with this change is minority carrier lifetime, tau(min)
160
How to optimise performance of a fast switching diode.
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Lower C (junction capacitance)
Lower R (total series resistance)
Lower tau(min)
```
161
The problem with fast switching diodes is that for small C, the depletion region width is large so low doping is needed and for small R, high conductivity is needed so high doping is required in the remainder of the device. What is the solution?
```
Low doping is used for the p-n junction which is attached to a highly doped bulk. The series resistance in the minimally doped p-n junction is lowers it is very thin (≈5µm)
Low tau(min) is obtained by doing a local diffusion of Au into the region of the junction.
```
162
What problem is a voltage regulating diode a solution to?
The need to provide a stabilised voltage independent of output current and small variations in supply of voltage.
163
How does a voltage regulating diode work?
Simply use a reverse biased diode at it's breakdown voltage as slope means basically independent of current.
As known as reference diode or Zener diode.
164
What problem is a tunnel diode a solution to?
The need to create a divide with negative differential resistance for high frequency (microwave) oscillator circuits.
165
What is the doping of a tunnel diode?
Both the n and p sides are degenerately doped.
166
Describe the shape of the depletion region across a tunnel diode.
Very bent so the conduction band in the n side is below the intrinsic Fermi energy and the valance band in the p type is above the intrinsic Fermi energy. This means that the n and p sides are very conductive of electrons and holes respectively.
167
Sketch the effect of forward and reverse bias on a tunnel diode.
Check slide 16
168
What does a tunnel diode do?
Uses the property of negative resistance to stop oscillations from decaying wrt a time constant (tau = 2RC/L (L is inductance))
When the effective resistance is negative the oscillations grow.
The voltage is held around the bias point for this to happen.
169
Sketch a tunnel diode oscillator.
Check slide 17
170
What is another solution to producing and maintaining microwave oscillations other than a tunnel diode?
A transfer electron oscillator.
171
What property does the transfer electron oscillator use?
The negative differential mobility that appears in GaAs. Above a threshold field there exists negative differential mobility. Since V(D) = µE(field) as the field increases, the velocity of the carriers decreases.
It results from the spillover of electrons from the full initial and direct conduction band into an indirect one of greater energy.
172
What happens in a transferred electron oscillator?
A field is applied to GaAs. If sufficiently high, a uniform distribution of free carriers is unstable and instead a bunch of carriers will form (normally at contact).
These bunched electrons move across the device and when it reaches other contacts gives rise to a pulse in the external circuit.
The pulse frequency is given by the drift velocity and active length of the device.
173
What input optical devices are there?
Photon detectors
| Solar cells
174
What output optical devices are there?
LEDs
| Lasers
175
What happens to photons in photon detectors?
They are absorbed and photons with energy greater than the band gap generate electron-hole pairs
176
Give an expression for the current in a photon detector.
Check slide 22
177
What is absorption coefficient in a photon detector?
A coefficient in the exponent that relates to the amount proportion of photons that are absorbed at a certain energy.
178
Which semiconductors generally have higher absorption coefficients, direct or indirect?
Direct as no change in k, so all energy into band gap and only one quantum process.
179
How can we detect the presence or intensity of light?
Use a diode that adsorbs light of required wavelength. e-/h+ pairs are generated near the junction, drift in the electric field of the depletion region or diffuse then drift across: they create an electric current. The junction is held under reverse bias so only minority carriers cause current.
180
How can reverse bias current in a standard photodiode be used to find intensity of light?
The reverse bias current is proportional with intensity so it can be calibrated.
181
Give equations for the reverse bias current in a standard photodiode.
Check slide 26
182
What factors are considered when designing a standard photodiode?
Selecting the right material to match the wavelength of light to be detected.
Selecting the right geometry to match position of e-/h+ pairs.
183
What problem does a PIN photodiode solve?
The need for increased speed of detection to allow timing of very fast signals.
184
Sketch the structure of a PIN photodiode.
Check slide 30.
185
How does a PIN diode work?
It collects electron hole pairs from the depletion region that has an increased width (as it is the width of the intrinsic material)
The time required to collect pairs is the sum of diffusion and drift times (diffusion is slow). Diffusion is prevented by masking all but the depletion region from incident light.
In a standard photodiode, masking reduces the depletion region width and sensitivity, thus by increasing the depletion region width with the intrinsic layer, sensitivity is recovered.
186
What problem does the avalanche photodiode (APD) solve?
Need increased sensitivity for detection of very small signals?
187
How does an avalanche photodiode (APD) work?
Uses the photodiode in breakdown mode.
Photons produces e-/h+ pairs in depletion region which are accelerated by strong E field and undergo impact ionisation to generate more electrons (amplification).
Intrinsic amplification can take values up to 100 times.
188
State equations for the short and open circuit currents of solar cells along with the open circuit potential.
Check slide 34
189
What is the limit of the open circuit potential for a solar cell?
The potential of the band gap.
190
Sketch I against V for a solar cell.
Check slide 34.
191
Derive an expression for the maximum power of a solar cell in terms of potentials and currents.
Check slide 35
192
Give equations for the EQE and IQE of a solar cell.
Check slide 37
193
What material are most solar cells made from?
Silicon.
194
Why are heterojunction solar cells used for rich applications?
They have a higher efficiency but are more expensive to produce.
195
Sketch the structure of a typical silicon solar cell.
Check slide 38
196
What how do tandem solar cells provide greater efficiency?
Cells are connected in series to collect photons from different parts of the spectrum.
197
Basically, how do photon emitters work?
Under forward bias, excess of minority carriers in p-n junction will cause e-/h+ pairs to recombine at the edge of the depletion region. Some of this recombination will result in the emission of photons.
198
What is the difference between radiative recombination and non-radiative recombination.
Radiative is dominant in direct band gap semiconductors and is direct recombination with the emission of a photon. Non-radiative recombination is dominant in indirect band gap semiconductors and is generally a bad emitter of light as recombination happens as phonon-photon emission.
199
What impact do defects have on the quality of photon emitting products?
Reduce quality as they tend to promote non-radiative recombination.
200
How do LEDs provide a continuous choice of wavelengths in the visible spectrum?
Use of ternary compounds and change in composition to tune band gap. Modern LEDs use AlN-GaN-InN system.
201
What solution is used to increase the efficiency of electron-hole pair/photon conversion (internal quantum efficiency) in indirect band gap materials to reduce power dissipation?
Introduce a deep electron trap to increase the rate of radiative recombination. See sketch of band gap on slide 49.
202
What problem is the double heterojunction LED a solution to?
The need to reduce power dissipation by increasing internal quantum efficiency.
203
How does a double heterojunction LED work?
Use of a very thin material of narrower band gap between the p and n layers.
Potential barriers at interfaces confine the injected carriers in this layer thus increasing the minority charge carrier conc and therefore increasing the ratio of radiative/non-radiative recombination.
(Non-radiative recombination through defect states begins to saturate).
The spectral output is very narrow thus can be turned for specific requirement.
204
How can white LEDs be produced?
Add phosphor powder next to diode to down convert a fraction of the light emitted to a second colour.
Down convert is absorption of first photon to emit a second of lower energy.
Typical phosphor is Yttrium Aluminium Garnet (YAG)
White balance is typically poor from YAG even though broad spectrum as lack of red. This can be fixed by a second phosphor powder.
205
Why were lasers needed?
Need a bright coherent source of light.
206
What are lasers used for?
Data transfer, DVD/CD, printing.
207
What is stimulated photon emission?
Where the electric field of a photon may interact with an electron in a higher energy level and stimulate it to recombine producing a new photon with the same energy, direction and phase.
208
What condition to do with emission is true for lasers?
The stimulated photon emission must be much greater than the spontaneous emission in order for coherent waves to be emitted.
209
What is population inversion for a laser diode?
Stimulated emission must be greater than photon absorption so there must be a high carrier conc present in the higher energy level. In thermal eqm, the carrier conc in the higher level is smaller than the lower so energy must be input in order to keep the higher level filled.
210
Sketch the structure of a laser diode.
Check slide 60
211
What problem does the double heterojunction laser diode solve?
The need for reduced power consumption by increasing the fraction of stimulated emission.
212
How does a double heterojunction laser diode work?
Potential barriers form that concentrate the electrons and holes near the p-n junction and thus achieve local population inversion at relatively small currents.
Force total internal optical reflection that confines photons to p-n junction, thus achieving large photon energy density at the same position as the population inversion.
213
How can focused high intensity light emission for micro machining and surgical applications be achieve in high power laser diodes?
Current density can further be improved bu confining the charge carriers in a narrow strip in the plane of the semiconductor.
This allows for continuous wave operation at RT.
214
What problem do surface emitting laser diodes solve?
Reduced cost of laser diodes.
215
How do surface emitting laser diodes work?
Use a vertical cavity reducing amount of material and cost
In plane mirrors are used using upper and lower distributed Bragg reflects.
Selective proton implantation can also be used to confine current flow and improve localisation.
216
How do Bragg reflectors work in a surface emitting laser diode?
Made by growing alternate layers of material with different refractive index where each layer reflects a fraction of incoming light.
217
Sketch a pnp and npn bipolar transistor.
Check slide 67
218
Sketch the band structure of the non bipolar transistor.
Check slide 68
219
What bias is the emitter/base junction under in a bipolar junction transistor with common base connection?
Forward bias so the input circuit resistance is low.
Small changes in input voltage gives substantial rise to the current flowing from the emitter (mostly straight through the base) into the collector.
220
What bias is the base/collector junction under in a bipolar junction transistor with common base connection?
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Reverse bias so has large resistance (R(out)) and can withstand high voltages.
Since V(out) = I(C)R(out), small changes in I(C) give large changes in the output voltage, so voltage amplification occurs.
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221
Sketch the common base connection version of a bipolar junction transistor.
Check slide 69
222
Give basic equations for the common base configuration of the bipolar junction transistor for the steady state current, small changes in current, current transfer ratio, current amplification ratio and the relationship between amplification and transfer.
Check slide 70
223
What conditions are needed for a high current transfer ratio in a bipolar junction transistor with common base configuration?
High ratio of emitter to base doping.
Narrow width of base region
Long majority carrier lifetime in the base
High minority carrier mobility in the base
224
Sketch the common emitter configuration of the bipolar junction transistor.
Check slide 71
225
Give expressions for voltage amplification and current amplification in a bipolar junction transistor in common emitter configuration.
Check slide 71
226
What is design considerations must be made in designing a bipolar junction transistor at each junction.
1. E/B high alpha so N(D)/N(A)>>1
2. Base high alpha so narrow width
3. B/C reasonable V(BD) so large W(BC)
4. B/C low capacitance for fast response so large W(BC)
5. B/C depletion region must not extend too far into the base (punching through), so collector doping << base doping.
227
How can a bipolar junction transistor be adapted to increase the speed and efficiency to increase amplification bandwidth or switching frequency?
Use a small base width.
| Reduce the transit time across the base region by replacing diffusion with drift using a graded base transistor.
228
What is a graded base bipolar junction transistor?
A transistor where the doping conc varies exponentially with distance which in turn means the band edges slope linearly with distance which provides the built in field.
This happens naturally when bipolar transistors are made by diffusing dopants into the substrate to make the emitter and base regions.
229
How can the high frequency limitation of a bipolar transistor be improved?
It is limited by the transit time of carriers across the base so reducing this time will increase the frequency.
The RC time constant of the device should be as short as possible. The important resistances are the E/B junction and the resistance of the collector.
230
What does FET stand for?
Field effect transistor
231
What happens in a MOSFET when the channel is activated?
The semiconductor layer beneath the gate is inverted allowing for conduction between the drain and the source.
232
Sketch the IV curve for a MOSFET between source and drain, considering different points in its activation.
Check slide 83
233
Sketch a plot of current through the drain against the V(DS) for a MOSFET.
Check slide 84
234
In order for the field effect transistor to work, what must be true of the SiO2 layer?
It must be thin and free of impurities.
235
What applications other than computer chips are MOSFETs used for?
Amplification and other analogue processing applications.
236
What limitations are there with MOSFETs?
Temperature limitation
| High frequency limitation
237
How does the temperature sensitivity of a MOSFET compare to that of a BJT?
MOSFETs are majority carrier devices, less so sensitive to temperature than BJT.
238
What is the high frequency limitation of a MOSFET?
The speed depends on RC as the charge in the channel needs to be redistributed, so C(SG), C(DG) and R(of channel and wiring) need to be minimised.
Also depends on source/drain transit time which can be reduced by reducing the length of the channel.
239
How does a metal semiconductor field (MESFET) effect transistor work?
Metal in contact with conducting channel of same material as source and drain just less heavily doped.
At no bias on the gate, the transistor is on.
For sufficiently high gate voltage, the whole of the region between gate and substrate is inverted, turning the switch off.
240
Describe how a CCD works.
In each pixel the CCD usually has 4 electrodes that are part of a the MOS capacitor. Bias is applied to the middle two electrodes and charge can be collected underneath from the e-/h+ pairs generated by incident photons. When collection time is complete, the stored charge can be transported across to the row end by switching on and off electrodes and the charge can be converted into intensity for an image to be formed.
