Semiconductors: VT =
kT
= .02585V at T = 300K
q
for Si (default)
ni = 1.5 × 1010 # / cm 3
q/εS = 1.6 x 10-3 V.μm
ECE4853
εs = 0.1fF/μm
4kT = 1.6 x 10-20J
Exercises in photodiodes
q = 1.6 x 10-4fC
= 1.6 x 10-5fJ
version 1.8
7.1 The width of the intrinsic region of a silicon PIN (pπn) photodiode is WI = 30μm. For purposes of
response it is desired that the E-field at one end (the n+ end) be 4.0 V/μm and 1.0 V/μm at the other end
(a) What is the impurity concentration in the intrinsic region (the π region)?
(b) If the doping impurity concentrations at the p+ and n+ ends are equal and are 50 times greater than
the intrinsic layer what is the total width of the depletion (space-charge) region, including WI and the end
regions?
(c) What is the reach-through voltage for this PIN photodiode?
Answers: Ni = 6.25 x 1013 #/cm3, W = 31μm, VRx = 45.3V
7-2. A pπn silicon photodiode with WI = 20μm has ND+ = 1016 #/cm3, NA = 5 x 1013 #/cm3 in the intrinsic
(π) region and NA+ = 1016 #/cm3. The E-field at the center of the photodiode is 2.0 V/μm. Determine (a)
EMAX (b) the total width of the depletion region and (c) the applied voltage necessary to meet this field
requirement. (d) Estimate the bandwidth of this photodiode. Assume ve = vh = 105 m/s for silicon.
Answers: EMAX = 2.80V/μm, W =20.25um, VR = 39.6V
7-3. The quantum efficiency of the silicon pin photodiode shown is 0.5 and the absorption coefficient of
the semiconductor material is α = 400 cm-1. What is the quantum efficiency of a pin photodiode that has
the same construction but with an intrinsic region that is twice as thick?
Answer: η = 0.684
Hint: for doubling of thickness
(1- e-2x) = (1- e-x) (1+ e-x)
7-4. Inside a pin photodiode a free electron is accelerated by an electric field of 4.5V/μm. What distance
does the electron travel before reaching a velocity of 105 m/s?
Answer: 6.32nm
7-5. The triple-aperture photometer defines the brightness of an optical signal by adjusting the
opening of each aperture so that all three electrical signals are equal. Each aperture is equipped with
an optical filter that defines the wavelength that reaches its detector. Assume that the three
wavelengths are λ1 = 0.75μm, λ2 = 0.6μm and λ3 = 0.75μm, respectively. For calibration we would
set all openings be same and desire that the intensities of all wavelengths be equal. Determine the
thicknesses of the active regions of the three pin photodetectors that will be necessary, neglecting
energy losses, such that the minimum acceptable quantum efficiency η = 0.4. Assume the absorption
coefficient of the optical material is 500 cm-1 and the thickness of the thinnest detector W1 = 12μm.
Hint: In order that I1 = I2 = I3, all of the responsivities must be equal. And the longest wavelength
would have to accept the lowest quantum efficiency and therefore the smallest thickness.
Answers: W1 = 12μm, W2 = 16.6μm, W3 = 27.9μm
7-6. The system equivalent shown makes use
of a Germanium pπn photodetector which has
a responsivity R = 0.9 at λ = 1255 nm. Its
dark current is 20nA at this wavelength. The
equivalent network (shown) is of the form of
R1 = 0.25kΩ and C1 = 16pF. This RC time
constant also determines the bandwidth for the
system. Assume input signal Pin = 0.5μW.
Find: (a) quantum noise contribution due to dark current iDB2
(b) thermal noise contribution due to resistance iT2
(c) quantum noise contribution due to photon current iPH 2
(d) signal/noise ratio S/N.
Answers: (a) 0.4(nA)2 (b) 4000(nA)2 (c) 9.0(nA)2 (d)50.5
7-7. The system equivalent shown makes use of an
InGaAs RAPD pπpn photodetector with responsivity
R = 1.0 at λ = 1550 nm. It is biased for avalanche
multiplying factor M = 50 corresponding to ratio of
ionization coefficients of electrons vs holes to be k =
0.25. The dark current for this detector is 25nA
when operated at this wavelength. The equivalent
network (shown) is of the form of R1 = 0.5kΩ and C1 = 10pF and also determines the bandwidth for
this system. Assume input signal Pin = 0.5μW.
Find (a) quantum noise contribution due to dark current iDB2
(b) thermal noise contribution due to resistance iT2
(c) quantum noise contribution due to photon current iPH 2
(d) signal/noise ratio S/N.
Answers: (a) .014(μA)2 (b) 1600(nA)2 (c) 0.28(μA)2 (d) 2117 (same as 33.26dB)