Electronic- Semiconductor Properties and PN junctions Flashcards
Graph of log(carrier concentration) Vs 1/temperature
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
Formula for electrical conductivity
σ=neμ(electrons)+neμ(holes)
What is electron and hole mobility due to?
Response of charge carriers to and applied field. Scattering of charge carriers
What are electrons scattered by?
Disorder including phonons (thermal) and impurities and defects (structural)
Formula for mobility due to thermal disorder
μth=1/((mc*^5/2)T^3/2)
Where mc* is conductivity effective mass
Formula for mobility due to structural disorder
μim=(T^3/2)/((Nim)m^1/2)
Where m is effective mass
Nim is ionised impurity density
Formula for total mobility
μ=(1/μth + 1/μim)^-1
How do electrons move in an applied field?
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
Graph of carrier drift velocity Vs electric field
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
Formula for mean free time between scattering events
1/τAv=1/τth + 1/τim
Subscript Av means average
Formula for mobility involving mean free time between scattering events
μ=(e/m*)τAv
Formula for total resistivity
ρtot=ρth+ρim
How does mobility change with impurity concentration?
It decreases initially very gradually but then later a lot. So decreases with increasing number of carriers
Log(σ) Vs 1/T for extrinsic semiconductors
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
4 point probe method
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
How to avoid temperature sensitivity
Work in plateau region. Use semiconductor with large bandgap. Extrinsic semiconductor with donor/acceptor levels close to band energies
What is a thermistor?
An electronic device that makes use of deliberate temperature sensitivity. Has a temperature-dependent resistivity
Two methods of exiting electrons from valence to conduction band
Absorption of photon with energy greater than that of band gap. Thermal excitation from phonons
What happens when a CB electron recombines with a VB hole when they are close together?
Either emission of photon of greater energy than Eg or non-radiative relaxation, i.e emission of phonons (heating)
Direct bandgap materials
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
Absorption coefficient
I=I0e^-αx α=4πke/λ I and I0 are final and initial light intensities x is semiconductor thickness ke is extinction coefficient
Indirect band gap materials
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
How does absorption coefficient for indirect band gap materials vary with wavelength?
α increases with photon energy (smaller λ) much more slowly than for direct
How does photoconductivity work?
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
What happens when light is switched off for photoconductivity?
Current decays exponentially due to loss of photocarriers from recombination. May be characterised by more than one time constant. Ideally decays so 0
What is dark current?
The current due to the small number or charge carriers due to thermal excitation. Means current from photoconductivity doesn’t decay to 0
Light dependent resistor
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
How are P-N junctions created?
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.
Energy effects for electron-hole recombination at P-N junction
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
What is true of depletion region with no applied field?
It contains no mobile charge carriers so has a high resistance and no current flow
What happens when a voltage is applied to a pn junction in forwards bias?
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
What happens when a voltage is applied to a pn junction in reverse bias?
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
Current in forwards and reverse bias
Forwards: If=I0(exp(eVapp/kBT)-1)
Reverse: If=I0(1-exp(-eVapp/kBT))
Think I0 is dark current
How do photodiodes work?
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
How do LEDs work?
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
