Electronic- Semiconductor Properties and PN junctions

Question 1

Graph of log(carrier concentration) Vs 1/temperature

Answer 1

Steep straight line down from top left for intrinsic behaviour. Then flat line for extrinsic behaviour. Then straight shallow line down in freeze out region. Slopes give activation/ionisation energies

Question 2

Formula for electrical conductivity

Answer 2

σ=neμ(electrons)+neμ(holes)

Question 3

What is electron and hole mobility due to?

Answer 3

Response of charge carriers to and applied field. Scattering of charge carriers

Question 4

What are electrons scattered by?

Answer 4

Disorder including phonons (thermal) and impurities and defects (structural)

Question 5

Formula for mobility due to thermal disorder

Answer 5

μth=1/((mc*^5/2)T^3/2)

Where mc* is conductivity effective mass

Question 6

Formula for mobility due to structural disorder

Answer 6

μim=(T^3/2)/((Nim)m^1/2)
Where m is effective mass
Nim is ionised impurity density

Question 7

Formula for total mobility

Answer 7

μ=(1/μth + 1/μim)^-1

Question 8

How do electrons move in an applied field?

Answer 8

They lose the extra energy they have gained resulting in an average drift velocity and not continuous acceleration. There are abrupt losses in velocity when collisions occur

Question 9

Graph of carrier drift velocity Vs electric field

Answer 9

Log Vs log scale. For low electric fields is linear region where vd=μE (drift velocity and electric field). But then starts to curve and reaches a peak before decreasing due to phonon and impurity (mainly impurity) scattering decreasing mobility

Question 10

Formula for mean free time between scattering events

Answer 10

1/τAv=1/τth + 1/τim

Subscript Av means average

Question 11

Formula for mobility involving mean free time between scattering events

Answer 11

μ=(e/m*)τAv

Question 12

Formula for total resistivity

Answer 12

ρtot=ρth+ρim

Question 13

How does mobility change with impurity concentration?

Answer 13

It decreases initially very gradually but then later a lot. So decreases with increasing number of carriers

Question 14

Log(σ) Vs 1/T for extrinsic semiconductors

Answer 14

Same as for log of carrier concentration but horizontal bit actually decreases as temperature increases (going left) due to reduction in mobility caused by increased scattering of fixed number of carriers

Question 15

4 point probe method

Answer 15

4 probes at equal distance apart across the semiconductor. Current sent in and out through the outer most probes. Voltage measured between two inner most probes.
Rs=(V/I)CF (correction factor).
ρ=RsW (sample thickness)
Rs is resistance across separation of probes S
Used for calculating resistivity

Question 16

How to avoid temperature sensitivity

Answer 16

Work in plateau region. Use semiconductor with large bandgap. Extrinsic semiconductor with donor/acceptor levels close to band energies

Question 17

What is a thermistor?

Answer 17

An electronic device that makes use of deliberate temperature sensitivity. Has a temperature-dependent resistivity

Question 18

Two methods of exiting electrons from valence to conduction band

Answer 18

Absorption of photon with energy greater than that of band gap. Thermal excitation from phonons

Question 19

What happens when a CB electron recombines with a VB hole when they are close together?

Answer 19

Either emission of photon of greater energy than Eg or non-radiative relaxation, i.e emission of phonons (heating)

Question 20

Direct bandgap materials

Answer 20

In the E-k diagram the CB min and VB max are directly above each other. There is no change in k (momentum) required for electron excitation/relaxation. Optical transition need not involve phonons. Means transition probabilities high, high absorption coefficients. Examples are III-V and II-VI materials. Used extensively in optical applications

Question 21

Absorption coefficient

Answer 21

I=I0e^-αx
α=4πke/λ
I and I0 are final and initial light intensities
x is semiconductor thickness
ke is extinction coefficient

Question 22

Indirect band gap materials

Answer 22

Have a shift in relative position of top of VB and bottom of CB. Means electrons at bottom of CB have very different momentum (k) to those at top of CB. Phonons must be involved in transition because photons have very little momentum. Requirement of phonons means absorption/emission is strongly temperature dependent and unlikely (photon absorption must coincide with phonon absorption). Means absorption coefficients much lower than 10^3 but still high compared to coloured glass. Examples include Si and Ge

Question 23

How does absorption coefficient for indirect band gap materials vary with wavelength?

Answer 23

α increases with photon energy (smaller λ) much more slowly than for direct

Question 24

How does photoconductivity work?

Answer 24

Use light to excite electronic transitions. Photon with energy greater than Eg generates electron-hole pair. Excited electrons and holes can conduct electricity. Conductivity increases depending on α, light intensity, lifetime of excited state (τ). VB and CB are sloped due to electric field

Question 25

What happens when light is switched off for photoconductivity?

Answer 25

Current decays exponentially due to loss of photocarriers from recombination. May be characterised by more than one time constant. Ideally decays so 0

Question 26

What is dark current?

Answer 26

The current due to the small number or charge carriers due to thermal excitation. Means current from photoconductivity doesn’t decay to 0

Question 27

Light dependent resistor

Answer 27

Use photoconductivity to create light sensitive resistor. Generally made from intrinsic direct band gap semiconductor. Gives high sensitivity to light and high resistivity (intrinsic) so low dark current

Question 28

How are P-N junctions created?

Answer 28

Bring together p-type and n-type semiconductors. n has lots of electrons in CB and p has lots of holes in VB. A depletion region forms around the contact point where the n-type CB electrons and p-type VB holes recombine due to diffusion of carriers. This occurs until the two Ef are equal (n-type band diagram lowers). This determines the depletion width.

Question 29

Energy effects for electron-hole recombination at P-N junction

Answer 29

Loss of electrons from n reduces n Fermi level. Resulting new positive charge (of donor ions) decreases energy of n VB and CB. Electron gain in p increases p Fermi level. Resulting net negative charge (of acceptor ions) increases energy of p VB and CB. Occurs until Fermi levels equal

Question 30

What is true of depletion region with no applied field?

Answer 30

It contains no mobile charge carriers so has a high resistance and no current flow

Question 31

What happens when a voltage is applied to a pn junction in forwards bias?

Answer 31

Electrons added to n-type region reduce the positive ion charge at the depletion layer. Holes added to the p-type region reduce the negative ion charge at the depletion layer. Narrows depletion layer and lower potential until current starts to flow

Question 32

What happens when a voltage is applied to a pn junction in reverse bias?

Answer 32

More electrons pushed to p-type region. More holes pushed to n-type region. This increased charge across depletion layer and increases depletion voltage. The current is prevented from flowing

Question 33

Current in forwards and reverse bias

Answer 33

Forwards: If=I0(exp(eVapp/kBT)-1)
Reverse: If=I0(1-exp(-eVapp/kBT))
Think I0 is dark current

Question 34

How do photodiodes work?

Answer 34

Potential difference across depletion region means any new electrons or hole swept out of the junction in opposite directions. A photon is absorbed in depletion region to create e-h pair. e goes to n-type and h goes to p-type. Current produced constitutes the signal of photodiodes which indicate number of incident photons

Question 35

How do LEDs work?

Answer 35

Use effect of e-h pair recombination. Electron from n-type side and hole from p-type side brought together to generate a photon of energy hv>Eg. Direct band gap material gives highest efficiency. Emission wavelength range determined by semiconducting band gap