HW 8 Solution

1) d = 400 mu m, lambda_o = 1.33 mu m, n = 4, eta_{slope} = 0.35 W/A, I_{th} = 35 mA, Delta lambda = 3 nm, alpha = 10 cm^{-1} = 10^3 m^{-1}
a) P_o = 2mW - eta_{slope} ( I - I_{th} ) => I = 2 / {I_{slope}} + I_{th} = 2/ {0.3} +35 = 41.67 (mA)
b) d = {m lambda_o} / 2n => m = round ({2nd} / lambda_o) = round ({8 times 400 times 10^{-6}} / {1.33 times 10^{-6}} ) = 2406
c) Delta f_{FSR} = c / {2nd} = {3 times 10^8} / { 2 times 4 times 400 times 10^{-6}} = 9.375 times 10^{10} (Hz)
d) Delta lambda_{FSR} = {Delta f_{FSR}} / nu lambda_o = (Delta f_{FSR}) lambda_o^2 / c = 9.375 times 10^{10} times (1.33 times 10^(-6))^2 / {3 times 10^8} = 5
e) R = ( {4 -1 } / {4 + 1 } )^2 = 0.36
f) t_{th} = alpha + 1 / {2d} ln ( 1/ {R^2} ) = 10^3 + 1 / {800 times 10^{-6}} ln ( 1/ 0.36^2) = 3554 (m^{-1})

2) From problem 4 of HW4, we know the fiber operating under single mode. From problem 1 of HW5, | D_{"intra"} | = 5.785 {ps} / {km-nm}. Delta lambda = 2nm, lambda = 1.3 mu m, and Delta tau = 11.57L (ps) (L in km), R_b = 0.3 {Gb}/s, and Delta tau_{sys} = 1 / {4 R_b} = 1 / {4 times 0.3} = 0.833 (ns).
a) Delta f_t = Delta f_r = 2 GHz
tau_t = tau_r = {0.35} / {2 times 10^9} = 0.175 (ns)
tau_f = sqrt{ Delta tau_{sys}^2 - tau_t^2 - tau_r^2} = 0.7957 (ns)
Since t_{f} > 0 and real, this system is feasible.
L = {tau_{f} (ps)} / {11.57} = {795.7} / {11.57} = 68.78 (km).
b) Delta f_t = 800 MHz => tau_t ={0.35} / {8 times 10^8} = 0.4375 (ns)
tau_f = sqrt{ 0.833^2 - 0.175^2 - 0.4375^2} = 0.6873 (ns)
Again the system is feasible.
L = {tau_{f} (ps)} / {11.57} = {687.3} / {11.57} = 59.48 (km).
c) Delta f_t = Delta f_r = 0.5 GHz, tau_t = tau_r = {0.35} / {0.5 times 10^9} = 0.7 (ns)
We can estimate that tau_{f}^2 = -2.856 (ns) which means tau_f is imaginary, i.e. This system is not feasible.

3) 13DSH001: tau_r = 0.35 ns, C_d = 1 pF, R_L = 1.8 K Omega
OKI OL303A: tau_t = 0.3 ns, Delta lambda = 2 nm (use typical values).
Fiber: | D_{"intra"} | = 5.785 {ps} / {km-nm}, Delta lambda = 2 nm.
a) tau_{CR} = 2.19 C_d R_L = 2.19 times 10^{-12} times 1.8 times 10^3 = 3.942 (ns)  > 4 tau_r. Circuit time determine the effective rise time => rise time = 3.942 (ns) and Delta f_r = {0.35} / {3.942} = 0.08898 (GHz).
b) R_b = 40 Mb/s, Delta tau_{sys} = 1 / { 4 R_b} = 6.25 (ns)
L = {tau_{f} (ps)} / {11.57} = {4841} / {11.57} = 518.7 (km).
c) There are many ways: decrease Delta lambda, decrease R_b, decrease tau_r and decrease tau_t etc. Consider changing R_b
2 times 4.841 = sqrt { Delta tau_{sys}^2 - 0.3^2 - 3.942^2}
Delta tau_{sys} = 8.838 (ns) and R_b = 1 / { 4 R_b} = 28.29 (Mb/s).

4) ccR = 0.3 A/W,  I_d = 2 nA, R_L = 1000 Omega, Delta f = 60 MHz, T = 273+15 = 288 (k).
a) {4 k T} / R_L Delta f = 2 q (I_d + I_{ph} ) Delta f => I_d + I_{ph} = {2 k T } / {q R_L} = 2 times 26 times 10^{-3} times 288 / 300 times 1 / 1000 = 4.992 times 10^{-5} (A)
Since I_d = 2nA << I_{ph} => I_{ph} = 4.992 times 10^{-5} (A)
P_{"in"} = I_{ph} / {ccR} = 1.66 times 10^{-4} (W)
b) \bar {i_T^2} = \bar {i_s^2} = 2 q ( I_{ph} + I_d ) Delta f = 2 times 1.6 times 10^{-19} times 4.992 times 10^{-5} times 60 times 10^6 = 9.585 times 10^{-16} (A^2)
SNR = I_{ph}^2 / { \bar {i_T^2} + \bar {i_s^2}} = I_{ph}^2 / { 2 \bar {i_T^2} } = (4.992 times 10^{-5})^2 / {2 times 9.585 times 10^{-16}} = 1.3 times 10^6 = 61.14 (dB)
c) P_{shot} = \bar {i_s^2} R_L = 9,585 times 10^{-16} times 1000 = 9.585 times 10^{-13} (W)

5) Use minimum ccR = 0.35 at 830 nm to get minimum I_{ph}.
13DSH001 area A = 0.04 mm^2 => P_{"in"} = I A = 200 times 0.04 times 10^{-2} = 0.08 (W), I_{ph} = ccR P_{"in"} = 0.028 (A)
13DSH003 area A = 0.2 mm^2 => P_{"in"} = I A = 200 times 0.2 times 10^{-2} = 0.4 (W), I_{ph} = ccR P_{"in"} = 0.14 (A)
13DSH005 area A = 0.8 mm^2 => P_{"in"} = I A = 200 times 0.8 times 10^{-2} = 1.6 (W), I_{ph} = ccR P_{"in"} = 0.56 (A)
b) On the graph of the spec. sheet, relative responsivity at 500nm is ccR / ccR_830 = 0.63, i.e. ccR = 0.63 times 0.35 = 0.2205
13DSH001 I_{"ph"} = 0.01764 (A)
13DSH003 I_{"ph"} = 0.0882 (A)
13DSH005 I_{"ph"} = 0.3538 (A)
c) eta = ccR {hf} / q. At 830 nm, eta = 0.35 times 1.24 / {0.83} = 0.5229
At 500nm, eta = 0.2205 times 1.24 / {0.5} = 0.5468
d) Delta f = 0.35 / tau_r and use min. tau_r to achieve max. Delta f:
13DSH001, Delta f = 0.35 / {0.35} = 1 (GHz)
13DSH003, Delta f = 0.35 / {1} = 0.35 (GHz)
13DSH005, Delta f = 0.35 / {2.5} = 0.14 (GHz)
e) tau_r / 4 >= 2.19 C_d R_L => R_L <= tau_r / { 4 times 2.19 C_d}
13DSH001: R_L <= {0.35 times 10^{-9}} / {4 times 2.19 times 10^{-12}} = 39.95 (Omega)
13DSH003: R_L <= {10^{-9}} / {4 times 2.19 times 2.5 times 10^{-12}} = 45.66 (Omega)
13DSH005: R_L <= {2.5 times 10^{-9}} / {4 times 2.19 times 5.5 times 10^{-12}} = 51.89 (Omega)

6) Delta f = 10 MHz, ccR = 0.3 A/W, I_d = 1 nA, SNR = 30dB = 10^4, P_{"in"} = P_o ( 1 + cos omega t ), T = 40 + 273 = 313 K, R_L = 50 Omega.
SNR = { 1/2 P_o^2 ccR^2} / {[ 2 q ( ccR P_o + I_d) + {4 kT}/ {R_L} ] Delta f} =>P_o^2 approx 2 {SNR}/ {ccR^2} q [ 2 ccR P_o + {4 k T(ev)} / R_L ] Delta f (drop I_d since ccR P_o ">>" I_d)
P_o^2 = 3.556 times 10^{-8} ( 0.6 P_o + {4 times 26 times 10^{-3} times 313 / {300}} / {50} ) => P_o^2 = 3.556 times 10^{-8} ( 0.6 P_o + 2.17 times 10^{-3})
P_o = { 3.556 times 10^{-8} times 0.6 +- sqrt{ ( 3.556 times 10^{-8} times 0.8)^2 + 4 times 3.556 times 10^{-8} times 2.17 times 10^{-3}}} / 2
= 8.785 (mu W) (pick the positive solution)
Check if ccR P_o ">>" I_d, \mathbbR P_o = 2.639 ( mu A) ">>" 1 nA. Hence assumption is correct.