MTLE-6120: Advanced Electronic Properties of Materials Semiconductor - - PowerPoint PPT Presentation

mtle 6120 advanced electronic properties of materials
SMART_READER_LITE
LIVE PREVIEW

MTLE-6120: Advanced Electronic Properties of Materials Semiconductor - - PowerPoint PPT Presentation

1 MTLE-6120: Advanced Electronic Properties of Materials Semiconductor p-n junction diodes Reading: Kasap 6.1 - 6.5, 6.9 - 6.12 2 Metal-semiconductor contact potential p -type n -type p -type n -type Same semiconductor on both sides,


slide-1
SLIDE 1

MTLE-6120: Advanced Electronic Properties of Materials Semiconductor p-n junction diodes

Reading:

◮ Kasap 6.1 - 6.5, 6.9 - 6.12

1

slide-2
SLIDE 2

Metal-semiconductor contact potential

p-type n-type p-type n-type

◮ Same semiconductor on both sides, different doping ◮ Bands line up perfectly, but Fermi level does not ◮ Bands bend to line up Fermi level ◮ Equal doping Na = Nd ⇒ symmetric bending ◮ In general, contact potential shared between both sides ◮ Extreme limits: one side p+ or n+, ∼ Schottky junction

2

slide-3
SLIDE 3

Depletion region charge

p-type n-type p-type n-type

◮ Far from junction, EF = EF 0

⇒ ρ = 0

◮ Approaching junction, EF first deviates ∼ kBT from EF 0:

Small deviation from neutral ⇒ Debye screening regime

◮ Once EF more than few kBT away from EF 0 (towards center of gap):

n ≪ Nd (or p ≪ Na) ⇒ depletion; they were equal in neutral case

◮ Across junction, EF has to cross almost entire gap ≫ kBT ◮ Therefore, depleted width ≫ width where Debye screening applicable ◮ Assume ρ = +eNd for width wn on n-side,

−eNa for width wp on p-side, and 0 elsewhere

3

slide-4
SLIDE 4

Depletion region field and potential

◮ Neutrality of junction ⇒

Nawp = Ndwn

◮ Solve for electric filed using ∇ ·

E = E′(x) = ρ(x)/ǫ

◮ Solving from left, E(x) = −eNa(wp + x)/ǫ for x > −wp (0 otherwise) ◮ Solving from right, E(x) = −eNd(wn − x)/ǫ for x < wn (0 otherwise) ◮ Peak field E(0) = −eNawp/ǫ = −eNdwn/ǫ ◮ Solve for potential using ∇φ = φ′(x)ˆ

x = −E(x)ˆ x to get: φ(x) = eNa(x+wp)2

, −wp ≤ x ≤ 0

eNaw2

p

+ eNa(2xwn−x2)

, 0 ≤ x ≤ wn

p-type n-type p-type n-type p-type n-type

4

slide-5
SLIDE 5

Depletion region width

◮ Total width w0 ≡ wp + wn; Nawp = Ndwn ⇒ wp = w0Nd Na+Nd , wn = w0Na Na+Nd ◮ Total potential across region:

V0 = eNaw2

p

2ǫ + eNa(2w2

n)

2ǫ = eNaNdw2 2ǫ(Na + Nd)

◮ But total potential is the contact potential:

eV0 =

  • EF 0 + kBT ln Nd

ni

  • EF n

  • EF 0 − kBT ln Na

ni

  • EF p

= kBT ln NdNa n2

i ◮ Therefore depletion region width:

w0 =

  • 2ǫ(Na + Nd)V0

eNaNd =

  • 2ǫ(Na + Nd)kBT

e2NaNd ln NdNa n2

i ◮ When will depletion region width be set by λD?

5

slide-6
SLIDE 6

Applied bias

p-type n-type

◮ Barriers for e− from n → p and holes from p → n: e(V0 − V ) ◮ Corresponding ‘diffusion’ current: j2 exp −e(V0−V ) kBT ◮ Junction field drives ‘drift current’: j1 ◮ Must be balanced at V = 0 ⇒ j1 = j2 exp −eV0 kBT ≡ j0 ◮ Therefore j = j0

  • exp

eV kBT − 1

  • 6
slide-7
SLIDE 7

Magnitude of current

◮ So far IV -characteristics exactly like Schottky diode ◮ For Schottky diode, we found j0 ∝ exp −ΦB kBT

i.e. significant current when eV ΦB

◮ For pn-junction diode at zero bias: equal drift and diffusion currents = j0 ◮ Minority carrier diffusion current driven by concentration gradient ◮ In uniform n-semiconductor ˙

p = −(p − p0)/τh (minority carrier lifetime τh)

◮ In non-uniform semiconductor: ˙

p = −Dh∇2p (diffusion)

◮ Therefore in steady-state: Dh∇2p = (p − p0)/τ ⇒

p decays exponentially towards p0 with length scale Lh = √Dhτh

◮ Diffusion current jh = −eDhp′(x) ∼ eDhp0 Lh

= eDhn2

i

LhNd ◮ Total minority diffusion current

j0 = eDhn2

i

LhNd + eDen2

i

LeNa = eNcNv Dh LhNd + De LeNa

  • exp −Eg

kBT

◮ Therefore current significant for V > Eg/e

7

slide-8
SLIDE 8

Minority carrier concentration in depletion region

p-type n-type ◮ In n-depletion region, potential changes from neutral value by

(V0 − V )Na/(Na + Nd) (maximum at junction)

◮ Assume symmetric doping, potential changes by (V0 − V )/2 ◮ Hole concentration at junction

pM ∼ Na exp −e(V0 − V ) 2kBT ∼ ni exp eV 2kBT

◮ Hole recombination rate ∼ pM τh · wn 2 (averaged over region)

8

slide-9
SLIDE 9

Recombination current

◮ Current due to recombination of both e and h:

jr = eni wn τh + wp τe

  • exp

eV 2kBT

◮ Previous IV accounted for e and h transport separately, but not this

recombination (except that it is needed for equilibrium)

◮ Net current density therefore:

j(V ) = jdd0 eV kBT − 1

  • + jr0

eV 2kBT − 1

  • ◮ Frequently approximated as

j(V ) = j0 eV ηkBT − 1

  • with ideality factor η expected to be between 1 and 2

◮ Ideal diode: no recombination ⇒ η = 1

9

slide-10
SLIDE 10

Additional effects in reverse bias

◮ Space charge layer generation

◮ Reverse bias increases junction potential ◮ Higher field in space charge layer (depletion region) ◮ e and h in equilibrium: thermal generation vs recombination ◮ Field sweeps carriers away before they recombine ⇒ current ◮ Linearly increasing reverse current instead of saturated −j0

◮ Avalanche breakdown

◮ Depletion region: large field, few carriers ◮ If eEλ > Eg, carriers can excite additional e-h pairs ◮ Cascade process leading to sudden increase in current

◮ Zener breakdown

◮ Highly doped junctions ⇒ narrow depletion regions ◮ Potential larger than Eg: direct band-to-band tunneling ◮ Design sharp breakdown at specific potential (Zener diodes)

10

slide-11
SLIDE 11

Depletion layer capacitance

◮ Charge stored = eNawp = eNdwn = ew0NaNd/(Na + Nd) (per unit area) ◮ Substituting for width of depletion region:

q A = eNaNd Na + Nd ·

  • 2ǫ(Na + Nd)(V0 − V )

eNaNd =

  • 2ǫeNaNd(V0 − V )

(Na + Nd)

◮ Therefore differential capacitance:

Cd A ≡ ∂q A∂V =

  • ǫeNaNd

2(V0 − V )(Na + Nd)

◮ Typical value for Si, Na = Nd = 1017 cm3, C ∼ 0.1 µF/cm2 p-type n-type p-type n-type p-type n-type

11

slide-12
SLIDE 12

Light-emitting diode

◮ Basic design: just a p-n junction, but in direct band-gap material ◮ Current density → recombination near junction ◮ Fraction of recombination is radiative ⇒ light ◮ Spontaneous emission: light is incoherent and in random direction ◮ Efficiency: η = Plight/(IV ) (can be > 10% for direct semiconductors) ◮ Typically use asymmetric junctions: pn+ or np+: why?

12

slide-13
SLIDE 13

LED: light spectrum

◮ Minimum photon energy: Eg ◮ Peak photon energy ∼ Eg + kBT ◮ Spectral width ∼ 3kBT ◮ Due to distribution of both hole and electron energies ◮ For GaAs, Eg = 1.42 eV, λ = 870 nm (IR) ◮ ∆(hc/λ) ∼ 0.08 eV ⇒ ∆λ ∼ 50 nm ◮ Light emitted by LED can be absorbed by semiconductor ◮ Circumvent in hetero-junction LEDs

13

slide-14
SLIDE 14

Semiconductor laser

◮ Very similar to LED at the junction level ◮ Key difference: optical cavity using reflecting surfaces ◮ Start with spontaneous emission, sharpened by cavity resonance ◮ Stimulated emission builds up intensity in specific mode ◮ Require population inversion: e-h pairs waiting to recombine ◮ Population inversion achieved by bias: electrical pumping! ◮ Cavity resonance + laser amplification, ∆λ ∼ 0.1 nm ≪ kBT

14

slide-15
SLIDE 15

Photodiodes and solar cells

◮ LED in reverse? ◮ Not quite, can use indirect band gap materials ◮ Absorption still occurs right at band gap (how?) ◮ eh-pairs created in depletion region swept across by field ◮ Loss: recombination (radiative or non-radiative) ◮ Modified device characteristic

j(V ) = −jph + j0

  • exp

eV ηkBT − 1

  • ◮ Why non-zero current at V = 0 if it is in equilibrium?

◮ Photodiode: operated in reverse bias: why?

15

slide-16
SLIDE 16

Photovoltaic efficiencies

Single junction efficiency limited by band-gap; circumvent using multi-junction devices

  • S. Kurtz and D. Levi, NREL

16