204 CHAPTER 4. JUNCTIONS IN SEMICONDUCTORS:P-NDIODES
The injection efficiency is now (assuming no recombination via traps)
γinj=
eDnnpo
Ln
eDnnpo
Ln +
eDppno
Lp
=0. 98
Example 4.8Consider thep-n+diode of the previous example. The diode is forward
biased with a forward-bias potential of 1 V. If the radiative recombination efficiencyηQr=
0.5, calculate the photon flux and optical power generated by the LED. The diode area is
1mm^2.
The electron current injected into thep-region will be responsible for the photon
generation. This current is
In =
AeDnnpo
Ln
[
exp
(
eV
kBT
)
− 1
]
=
(10−^2 cm^2 )(1. 6 × 10 −^19 C)(30 cm^2 /s)(6. 8 × 10 −^5 cm−^3 )
5. 47 × 10 −^4 cm
[
exp
(
1
0. 026
)
− 1
]
=0.30 mA
The photons generated per second are
Iph=
In
e
·ηQr =
(0. 30 × 10 −^3 A)(0.5)
1. 6 × 10 −^19 C
=9. 38 × 1014 s−^1
Each photon has an energy of 1.41 eV (= bandgap of GaAs). The optical power is thus
Power =(9. 38 × 1014 s−^1 )(1.41)(1. 6 × 10 −^19 J)
=0.21 mW
4.10 PROBLEMS ....................................
Problem 4.1Why does the potential in ap-ndiode fall mainly across the depletion region
and not across the neutral region?
Problem 4.2An abrupt GaAsp-ndiode hasNa=10^17 cm−^3 andNd=10^15 cm−^3.
(a) Calculate the Fermi level positions at 300 K in thepandnregions.
(b) Draw the equilibrium band diagram and determine the contact potentialVbi.
Problem 4.3Consider an Sip-ndiode doped atNa=10^17 cm−^3 ;Nd=5× 1017 cm−^3
at 300 K. Plot the band profile in the neutral and depletion region. Also, plot the electron
and hole concentration from thep-tothen-sides of equilibrium. How good is the
depletion approximation?
