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.