[Ronja] BER plot ?

[J.D. Bakker]

[This is mostly off the top of my head; I might have missed a few 
factors here and there. Corrections are welcome]

>Hehe now I realize I can't even imagine how a signal to noise ratio
>in an optical system should be calculated.

Mostly the same as any other communications system -- with a link 
budget analysis.

Let's say you have a 10Mbps Manchester signal into a shot noise 
limited receiver, and you want to know the minimum input power for 
1e-9 BER.

Assuming ideal filtering, http://www.sss-mag.com/cber.html (and 
others) say this takes an Eb/No of 12.5dB. Now,

   C/N = Eb/No * (R/B)

where

   C/N is the Carrier (or signal) to Noise ratio
   R is the bit/symbol rate of the transmitted signal
   B is the receiver bandwidth

Manchester encoding with rectangular pulses requires a minimal 
bandwidth B of 2R, so C/N = 12.5 + 10*log(10e6/20e6) = 12.5 - 3 = 
9.5dB

How much noise does the receiver, um, receive ? Assuming that:

- the *only* noise in the receiver is shot noise (not realistic)
- the receiver uses an unmodified SFH203 photodiode
- daylight is 1000 lux

The incident daylight produces 80uA of photocurrent (source: SFH203 
datasheet). According to The Art of Electronics (and others, 
including 
http://optoelectronics.perkinelmer.com/content/whitepapers/NoiseandStabilityPINDetectors.pdf), 
the shot noise current is

   in = sqrt(2*q*B*id)

(I've ignored dark current, since it's way << the photocurrent)

   in is the shot noise
   q is the charge of an electron (1.6e-19 C)
   B is the measurement (receiver) bandwidth
   id is the DC photocurrent

so,

   in = sqrt(2*(1.6e-19)*(20e6)*(80e-6)) = 23e-9 A = 23nA

The daylight sensitivity of any Si PIN detector is ~0.5 A/W, so this 
corresponds to ~11nW, or 10log(11e-9/1e-3) = -49.5dBm. This is a LOT 
of base noise for a communication system; the noise floor of an 
802.11 system lies between -90...-100dBm.

All this means that the absolute minimum received power in our ideal 
system would have to be -49.5 + 9.5 = -40dBm, or 100nW. Now add 
receiver noise, non-ideal filtering and atmospherical attenuation and 
you'll end up nearer 1..10uW.

More information can be found at http://www.sss-mag.com/ebn0.html , 
or by googling for "link budget".

Comms engineers always add a margin on top of the link budget; this 
margin usually lies between 10 and 30dB depending on a gazillion 
factors. It's often funny to see project reports where an engineer 
has spent a month getting front-end noise from 1.5dB to 1.4dB, and 
then finishes by saying "...and we've added 35dB margin" ;-) (not 
that I don't see the value in optimizing front-ends, mind you).

Questions ? Suggestions ? Flames ?

JDB.
-- 
LART. 250 MIPS under one Watt. Free hardware design files.
http://www.lartmaker.nl/