• To calculate optical receiver sensitivity, total noise in the receiver is

Substituting these values and solving equation (5.2.3) gives

Substituting these values and solving equation (5.2.3) gives

… (5.3.1)

… (5.3.2)

… ( 5.3.3)

Where,

This equation is known as thermal noise characteristic of an optical receiver.

• The optimum gain to achieve desired BER for receiver is given by –

… (5.3.4)

Assuming no ISI i.e. γ =1 Where,

Q is parameter related so S/N ratio to achieve desired BER.

W is thermal noise characteristic of receiver. X is photodiode factor. I2 is normalized BW.

Mean Square Input Noise Current

• The mean square input noise current is gives as –

… (5.3.5)

1. Shot noise Current :

1. Thermal Noise :

• Shunt Noise :

1. Series Noise :

1. Total Noise:

… (5.3.6)

… (5.3.7)

Example 5.3.1 : An InGaAs PIN photodiode has the following parameters at a wavelength of 1300 nm : ID = 4 nA, η = 0.9, RL = 1000 Ω and the surface leakage current is negligible. The

incident optical power is 300 nW (- 35 dBm) an the receiver bandwidth is 20 MHz. Find the various nosie terms of the receiver.

Solution : Given :

λ = 1300 nm ID = 4 nA

η = 0.9

RL = 1000Ω

Pincident = 300nW

B = 20 MHz

Mean square quantum noise current

Ans.

Mean spark dark current

= 2 (1.6 x 10-19) (20 x 106) (4 x 10-9)

= 0.256 x 10-19 Amp                                                                      … Ans.

Mean square thermal noise current

Where B is Boltzman constant = 1.38 x 10-23 J/K

T = (25o+273o) = 298 K

= 3.28 x 10-16 Amp                                                                       … Ans.