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445.664 445.664
LightLight--Emitting DiodesEmitting Diodes
Chapter 4.Chapter 4.LED Basics : Electrical PropertiesLED Basics : Electrical Properties
Euijoon YoonEuijoon Yoon
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pp--n junction band diagramn junction band diagram
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Schockley EquationSchockley Equation
• VD (diffusion voltage)
NA, ND : acceptor and donor concentrationni : intrinsic carrier concentration
The diffusion voltage represents the barrier that free carriers must overcome in order to reach the neutral region of opposite conductivity type.
• Schockley equation (the I-V characteristic of a p-n junction)
2i
DAD
nNNln
ekTV =
}1/)({ −−⎟⎟⎠
⎞⎜⎜⎝
⎛= +
kTVVeeNDNDeAI DD
n
nA
p
p
ττDn, p : electron and hole diffusion constants
p,nτ : electron and hole minority carrier lifetimesA : cross-sectional area, V : diode bias voltage
• Threshold voltage (Vth)-the voltage at which the current strongly increase- Vth ~ VD
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Diode currentDiode current--voltage characteristicsvoltage characteristics
• Forward voltage is approximately equal to Eg/e.
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Diode forward voltageDiode forward voltage
• Most semiconductor LEDs follow the solid line, except for LEDsbased on III-V nitrides.
1. Large bandgap discontinuities -> additional voltage drop2. Poor contact technology in the nitrides3. Low p-type conductivity in bulk GaN4. Parastic voltage drop in the n-type buffer layer
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Deviations from the ideal IDeviations from the ideal I--V characteristicV characteristic
•)/( kTneVeII ideals=
( ) )/()( kTnIRVeeIR
IRVI idealss
p
s −=−
−•
nideal : ideality factor
Rs : parasitic series resistance
Rp : parasitic parallel resistance
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NonNon--ideal Iideal I--V characteristicsV characteristics
• Problems area of diodes can be identified from I-V characteristic.- A series resistance can be caused by contact resistance or by the resistanceof the neutral region.
- A parallel resistance can be caused by any channel that bypasses the p-n junction. The bypass can be caused by damaged regions of the p-n junction orby surface imperfections.
Parallelresistance
Rp=dV/dI
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Methods to determine series resistanceMethods to determine series resistance
• Method shown above suitable for series resistance measurement.
• At high currents, diode I-V becomes linear due to dominance of series resistance. Diode series resistance can be extracted in linear regime.
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Carrier distribution in pCarrier distribution in p--n homon homo-- and heterojunctionsand heterojunctions
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Double heterostructure (DH)Double heterostructure (DH)
• Einstein relation- Dn = kTμn / e - Dp = kTμp / e
• In typical semiconductors, the diffusion length is of the order ofa several micrometers.ex) in p-type GaAs, Ln = (220 cm2/s x 10-8 s)1/2 = 15μm
• Minority carriers are distributed over quite a large distance. The large recombination region in homojunctions is not beneficial forefficient recombination.
• Introduction of double heterostructure (DH) makes it possible toconfine the carriers in active region. The thickness of the regionin which carriers recombine is given by the thickness of the activeregion rather than the diffusion length.
• Diffusion length (L)- Ln = (Dnτn)1/2
- Lp = (Dpτp)1/2
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The effect of heterojunctions on device resistanceThe effect of heterojunctions on device resistance
• One of the problem introduced by heterostructures is the resistancecaused by the heterointerface and it can have a strong deleteriouseffect on device performance, especially in high-power devices.
• The resistance introduced by abrupt heterostructures can be completelyeliminated by parabolic grading.
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Grading of heterostructuresGrading of heterostructures
• In order to compensate for the parabolic shape of the depletion potential, the composition of the semiconductor is varied parabolically as well, so that an overall flat potential results.
• If the potential created in the depletionregion is equal to Eg/e, then electronswill no longer transfer to the small-bandgap material. The thickness of thedepletion region is,,,
• The “spikes” are a result of the lineargrading assumed, and would not resultfor parabolic grading.
2D x2
eNε
=φ
Δ
D2CD
NeE2W Δε
=
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Carrier loss in double heterostructuresCarrier loss in double heterostructures
• The concentration of electrons with energy higher than the barrier
∫∞ρ=
BFDDOSB E dE)E(fn kT/)EEeNn B
CBFn( −
=
(Fermi-Dirac distribution) (Boltzmann distribution)
• The diffusion current density of electrons leaking over the barrier
nnnn
L)0(neD
dx)x(dneDJ B
0xB
0x −=−= ==
nFnn L/xekT/)EE(eNL/xe)0(n)x(n BCBB
−−−=−=
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Carrier overflow in double heterostructuresCarrier overflow in double heterostructures
• The current density at which the active region overflows,,,
DH
3C
2C
overflow eBWkTE
3N4J ⎟
⎠⎞
⎜⎝⎛ Δ⎟
⎠
⎞⎜⎝
⎛π
= (for DH structure LED)
QW
2
oC2overflowWeB)EE(
)2/h(*mJ ⎥
⎦
⎤⎢⎣
⎡−Δ
ππ= (for QW structure LED)
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Carrier overflow Carrier overflow -- For DH structure LEDFor DH structure LED
BnpeW
Jdtdn
DH−=
• The rate equation of carrier supply to (by injection) and removal from the active region (by recombination) is given by
(B: the bimolecular recombination coefficient)
• At high injection levels, the Fermi energy rises and will eventually reach thetop of the barrier. At that point, . Using this value, the currentdensity at which the active region overflows is given by
CCF EEE Δ=−
• For high injection densities: n=p• Under steady-state conditions (dn/dt=0), DHeBW
Jn =
DH
3C
2C
overflow eBWkTE
3N4J ⎟
⎠⎞
⎜⎝⎛ Δ⎟
⎠
⎞⎜⎝
⎛π
=
3/2
C
CF
Nn
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kTEE
⎟⎟⎠
⎞⎜⎜⎝
⎛ π=
−
• In the high-density approximation, the Fermi level is given by
(Nc: effective density of states)
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Carrier overflow Carrier overflow -- For QW structure LEDFor QW structure LED
• For QW structures, we must employ the 2D density of states, rather than the3D density of states. The Fermi level in a QW with one quantized state withenergy E0 is given by
⎥⎥⎦
⎤
⎢⎢⎣
⎡−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
− 1Nnexpln
kTEE
D2c
D20F ( n2D : the 2D carrier density per cm2,
Nc2D : the effective 2D density of states)
kT)2/h(
*mN 2D2
Cππ
=
• Nc2D is given by
• The high-degeneracy approximation (at high carrier densities) is employedand one obtains
D22
0F n*m
)2/h(EE ππ=−
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Carrier overflow Carrier overflow -- For QW structure LEDFor QW structure LED
• The rate equation for the QW of carrier supply to (by injection) and removalfrom the active region (by recombination) is given by
D2D2D2D2
pnBeJ
dtdn
−=
(B2D B/WQW : the bimolecular recombination coefficient for a 2D structure)≈
• For high injection densities: n2D=p2D
• Under steady-state conditions (dn2D/dt=0), eBJW
eBJn QW
D2D2 ==
• At high injection levels, the Fermi energy will reach the top of the barrier. At that point, . Using this value, the current density at which the active region overflows is given by
0C0F EEEE −Δ=−
QW
2
oC2overflowWeB)EE(
)2/h(*mJ ⎥
⎦
⎤⎢⎣
⎡−Δ
ππ=
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Saturation of output power due to leakageSaturation of output power due to leakage
• In order to avoid the carrier overflow, high current LEDs mustemploy thick DH active regions, or many QWs of multiple QW (MQW) active regions, or a large injection (contact) area.
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Electron blocking layersElectron blocking layers
• The barrier in the valence band is screened by free carriers so thatthere is no barrier to the flow of holes in the p-type confinement layer.
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Diode forward voltageDiode forward voltage
• The total voltage drop across a forward-biased LED is given by
• One finds experimentally that the diode voltage can be slightly lowerthan the minimum value predicted by above Eq’n, i.e. ~ Eg/e
eEE
eEEIR
eEV oVoC
Sg −Δ
+−Δ
++=
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Temperature dependence of diode voltageTemperature dependence of diode voltage
• Diode voltage decreases with increasing temperature due to decreasein energy gap and increase in saturation current density.
• The temperature dependenceof the diode voltage
• first term : due to the changes of the Fermi level with temperature• The contribution of the first term is generally smaller than the contribution of the second term.
e)T(E
*I
Ilne
kT)T(V g
S+=
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Temperature dependence of diode voltageTemperature dependence of diode voltage
• Diode forward voltage can be used to assess junction temperaturewith high accuracy.
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Drive circuitsDrive circuits
• Constant current-drive circuit
• Constant voltage-drive circuit
• What are the advantages and disadvantages?