Philips Semiconductors
SA5211
Transimpedance amplifier (180 MHz)
∴I = η × -h---Pλ----c-- × e Amps (Coulombs/sec.)
where e = electron charge = 1.6 × 10-19 Coulombs
Responsivity R = -η-------h---λ-×--c-------e--- Amp/watt
I = P×R
Assuming a data rate of 400 Mbaud (Bandwidth, B = 200 MHz), the noise parameter
Zn may be calculated as:1
Z
=
I---E--Q--
qB
=
-(--1---.-6-----×-----1--4-0--1-–--1--×-9--)--1-(--02---–0--9-0-----×-----1---0---6---)
=
1281
(8)
where Z is the ratio of RMS noise output to the peak response to a single hole-electron
pair. Assuming 100% photodetector quantum efficiency, half mark/half space digital
transmission, 850nm lightwave and using Gaussian approximation, the minimum
required optical power to achieve 10-9 BER is:
PavMIN = 12h--λ-c--BZ = 12 × 2.3 × 10–19
200 × 106(1281) = 719 nW = –31.5 dBm = 1139 nW = –29.4 dBm
(9)
where h is Planck’s Constant, c is the speed of light, λ is the wavelength. The
minimum input current to the SA5211, at this input power is:
IavMIN
=
qPavMIN
--λ---
hc
J---o--1-u---l--e-
×
J---os--e-u--c-l--e-
×
q
=
l
=
7---0---7-----×-----1-2--0-.-3-–--9--×--×--1---10---.-–6--1--9-×-----1---0---–--1---9
=
500 nA
(10)
Choosing the maximum peak overload current of IavMAX = 60 µA, the maximum mean
optical power is:
PavMAX = h----c--l--aλ--v--qM----A----X- = 21----..-36-----××----11---00----––--11--99-60 × 10 µA = 86 µW or – 10.6 dBm (optical)
(11)
Thus the optical dynamic range, DO is:
DO = PavMAX – PavMIN= – 4.6 – (–29.4) = 24.8 dB
DO = PavMAX – PavMIN= – 31.5 – (–10.6)
(12)
1. S.D. Personick, Optical Fiber Transmission Systems, Plenum Press, NY, 1981, Chapter 3.
9397 750 07427
Product specification
Rev. 03 — 07 October 1998
© Philips Electronics N.V. 2001. All rights reserved.
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