E = 1. 24 入(0)u0m eV

入0 =  ≅  =  = 0. 689 um = 689 nm

4.2)     An LED is formed by diffusing Zn into n-type GaAs and the junction is forward biased.

Will the voltage applied to the zinc side be positive or negative with respect to the n-side? On which side do you expect the majority of the light to be generated?

Zinc is a p-type dopant in GaAs. A pn junction is forward biased when a more positive voltage is applied to the p-region than is applied to the n-region.

→ A positive voltage is applied to the zinc-doped side of the LED.

Since the junction is formed by inwards diffusion of the p-dopant (zinc), a p+n junction is    formed. Since p > n, more minority holes are injected into the n-side than minority electrons injected into the p-side. Thus, most of the recombination (including the spontaneous emission that generates the light) occurs on the n-side.

4.3)     A p-n junction is formed so that the p-doping on one side is exactly equal to the n-doping on

the other.  (This can be achieved by growing so-called epitaxial films with controlled doping levels, one p-type, one n-type.)  The recombination lifetime of holes in the n-type film is 10-6 s, whereas the lifetime of the electrons in the p-type material is 10-9 s.  Which side of the junction will generate the most light?  What will be the ratio of the intensities?  (Assume the carrier mobilities to be approximately equal and neglect non-radiative recombination.)

Note: There is some missing information in this question about the length of the bases. Thus, since this is an LED (and we need recombination to generate light), my solution below assumes that both the n- and p-bases of the diode are “long” .

From the question, we know the following:

Na = Nd ; Tp = 10-6 s; Tn = 10-9 s;

Ln = √DnCn = √unCn and Ln = √DpCp = √upCp

Since n ≈  p, we see that

Lp ⁄Ln = √Cp ⁄Cn = √1000 ≅ 31.6

Next, we note that the concentration of minority holes and electrons at the edges of the deplection region are equal, since Na = Nd :

For a long base diode, we know that:V

Jn (0) =  (exp [] − 1) and Jp (0) =  (exp [] − 1)

Thus, since Dn = Dp, Na = Nd and Tp = 103 Tn, we see that:

Jn(0)       Lp         √upTp            

Jp(0)      Ln        √unTn

Thus, most of the light will be generated by the recombination of minority electrons on the p-side of the junction.  The ratio of the intensity of the light generated on the p-side of the junction to that generated on the n-side of the junction will be ~ 32.

4.4)     An LED is made in the form of an n+p junction.  The p-side is doped at a level of 1017 cm-3,

the electron mobility is 1000 cm2/Vs, and the recombination lifetime of electrons on the p-  side is 10-8 s.  If ni = 1012 cm-3, calculate the current density when a forward bias of 1 volt is applied.

This is an LED (and we need recombination to generate light), so we assume that the p-base of the diode is “long” . Since the junction is n+p (i.e. n >> p), then J  Jn(0) and the hole        injection into the n-region can be neglected.

J ≅ Jn (0) =          (exp [    ] − 1) =      √ (exp [    ] − 1)

J ≅ 1.60x10−19C √ (exp [] − 1)

J ≅ 8.14x10−8(exp[38.6] − 1)A cm −2 = 4.7x109A cm −2 !

Note: eV/q =  e V/q ≡ V

4.5)     The absorption coefficient of a semiconductor is 3000 cm- 1 .  Calculate the fraction of the

optical energy absorbed in a thickness of 2m.  Repeat for a thickness of 10m.

For d = 2 m:

 = 1 − exp(−ax) = 1 − exp(−3000 cm−1 2x10−4cm) = 0.45

For d = 10 m:

 = 0.95

4.6)     A p-i-n photodiode used to detect a signal with a wavelength of 1.55m employs a

heterostructure to avoid light absorption in the top (p-type) region.  The absorption coefficient of the intrinsic or “i” region (below the p-window) is 1x104 cm- 1 and its         thickness is 1 m.  If the device has an optical coating to reduce the surface reflection to 5%, calculate the responsivity of the photodiode in amps/watt.

The responsivity is given by: R =  = T[1 − exp(−ad)] () , where T is the

transmission coefficient through the surface, the term [1-exp(-d)] is the fraction of the      transmitted light which is absorbed in the active region, and Ephoton is the energy per photon (needed to determine the photon flux for an optical beam with a given power).

T = 1 − R = 1 − 0.05 = 0.95

[1 − exp(−ad)] = 1 − exp(−104 cm −110−4cm) = 0.63

 = ℎc(q入0) = 1.24qe(入)V(0) um =  = 1.25 V = 1.25 A w −1

Thus, R = 0.95(0.63)1.25 = 0.75 A W- 1

4.7)     If the above Si photodiode has a load resistor of RL = 104 , calculate the RC time constant

of the circuit and find the overall system bandwidth.  (Take cs = 11.7 for Si and assume the diode is 1mm in diameter – neglect the effect of the pre-amplifier.)

We know cs = 11.7; Wi = 1 m;

A = 冗r2 = 冗(5x10−4m)2 = 7.9x10−7m2

es e0A      11.7(8.85x10−12F m−1)7.9x10−7m2

Cdepl  =   wi     =                      1x10−6m                      = 82 pF

TRC = RCdepl = 1x104 Ω (8.2x10−11F) = 8.2x10−7s = 820ns