SEMICONDUCTOR DEVICE PHYSICS AND DESIGN

202 CHAPTER 4. JUNCTIONS IN SEMICONDUCTORS:P-NDIODES

The radiative recombination depends upon the radiative lifetimeτrand the non-radiative lifetime
τnr. To improve the efficiency of photon emission one needs a value ofτras small as possible
andτnras large as possible. To increaseτnrone must reduce the material defect density. This
includes improving surface and interface qualities.
The LED current is then given by

J=eRspontQW+J 0 exp

[

e

(

Vbi−Vturnon)

)

kBT

]

+JSNS

The parasitic currents are the second and third terms in the expression. The second term repre-
sents current injected over the barrier and the third term the current recombining at the maximum
recombination plane.

Example 4.5Calculate thee-hrecombination time when an excess electron and hole

density of 1015 cm−^3 is injected into a GaAs sample at room temperature.

Since 1015 cm−^3 or 1021 m−^3 is a very low level of injection, the recombination time is

given by equation 4.9.8 as

1

τr

=

1

2 τo

(

2 π^2 m∗r

kBTm∗em∗h

) 3 / 2

p

=

1

2 τo

(

2 π^2

kBTm∗e+m∗h

) 3 / 2

p

Usingτo=0.6nsandkBT= 0.026 eV, we get form∗e= 0.067mo,m∗h=0. 45 mo,

1

τr

=

1021 m−^3

2 ×(0. 6 × 10 −^9 s)

[

2 × 3. 1416 ×(1. 05 × 10 −^34 Js)^2

(0. 026 × 1. 6 × 10 −^19 J)×(0. 517 × 9. 1 × 10 −^31 kg)

] 3 / 2

τr =5. 7 × 10 −^6 s∼= 9. 5 × 103 τo

We see from this example that at low injection levels, the carrier lifetime can be very long.

Physically, this occurs because at such a low injection level, the electron has a very small

probability of finding a hole to recombine with.

Example 4.6In twon+pGaAs LEDs,n+pso that the electron injection efficiency is

100% for both diodes. If the nonradiative recombination time is 10 −^7 s, calculate the

300 K internal radiative efficiency for the diodes when the doping in thep-region for the

two diodes is 1016 cm−^3 and 5 × 1017 cm−^3.

When thep-type doping is 1016 cm−^3 , the hole density is low and thee-hrecombination

time for the injected electrons is given by equation 4.9.8 as

1

τr

=

1

2 τo

(

2 π^2 m∗r

kBTm∗em∗h

) 3 / 2

p