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Ch11Sec2

# Ch11Sec2 - THE FIBER FORUM Fiber Optic Communications...

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Dr. Joseph C. Palais 11.2 1 THE FIBER FORUM Fiber Optic Communications Dr. JOSEPH C. PALAIS PRESENTED BY

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Dr. Joseph C. Palais 11.2 2 Section 11.2 Signal-to-Noise Ratio
Dr. Joseph C. Palais 11.2 3 Consider the equivalent circuit of a photodiode receiver. R d v i S C d C d = diode’s junction capacitance (small) R d = diode’s junction resistance (large) R s = diode’s bulk series (n and p) resistance (small) R s

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Dr. Joseph C. Palais 11.2 4 i S is the photocurrent. As before, it is given by i s = ( η eP/hf) = ρ P For simplicity, assume R s = 0 and R d = infinite. Also neglect C d for purposes of noise calculations, since it does not affect the noise in the circuit. The simplified receiving circuit, including all sources of thermal and shot noise is now:
Dr. Joseph C. Palais 11.2 5 R L i S i 2 NT i 2 NS We will use this circuit to compute SNR.

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Dr. Joseph C. Palais 11.2 6 11.2.1 Constant Power SNR Let the incident optical power P be a constant. This corresponds to a binary 1 in a digital system. Compute the SNR. SNR = average signal power / average noise power These are the electrical powers. From the equivalent circuit, we see that SNR = (R L i 2 S ) / (R L i 2 NS + R L i 2 NT ) SNR = P ES / (P NS + P NT ) SNR = i 2 S / ( i 2 NS + i 2 NT )
Dr. Joseph C. Palais 11.2 7 These equations are general. For the special case where P = a constant: i S = i S = ( η e/hf)P = ρ P P ES = R L i 2 S = ( η e P/hf) 2 R L P NT = R L i 2 NT = (4kT f /R L ) R L = 4kT f P NS = R L i 2 NS = 2e[ I D + ( η e P/ hf)] f R L Then [( η e P/ hf) 2 R L ] {2e[ I D + ( η e P/ hf)] f R L }+ 4kT f SNR = (11-8)

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Dr. Joseph C. Palais 11.2 8 Special cases: Case 1 : signal current >> dark current i s >> I D and shot noise >> thermal noise 2e f ( η e P/hf) >> (4kT f /R L ) Then, the SNR equation simplifies to: SNR = [( η e P/hf) 2 R L ] / [2e f R L ( η e P/hf)] SNR = η P/2hf f (11-9) This is called the shot-noise limited SNR or the quantum-noise limited SNR .
Dr. Joseph C. Palais 11.2 9 This is a very good result (high SNR). It is usually not the actual result, because P is not usually large enough to make the assumptions leading to it valid. If P is large, we have no SNR problems. Since i S = ( η eP/hf) the shot-noise limited SNR (11-9) can be rewritten as SNR = i S /2e f (11-10)

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Dr. Joseph C. Palais 11.2 10 Case 2: thermal noise >> shot noise In this case the SNR (11-8) becomes: SNR = [( η eP/hf) 2 R L ] /4kT f (11-11) This is the thermal-noise limited SNR. It is valid when the received power is low, which is normally the situation.
Dr. Joseph C. Palais 11.2 11 Example: Light source is an LED, 10 mW output power, λ = 0.85 μ m. The system losses are: coupling loss = 14 dB fiber loss = 20 dB connector losses = 10 dB Total loss = 44 dB

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Dr. Joseph C. Palais 11.2 12 Compute the received power. dB = log (P R /P T ) - 44 = log (P R /P T ) (P R / P T ) = 10 - 4.4 P R = 10 x 10 - 4.4 = 10(3.98 x 10 - 5 ) P R = 4 x 10 - 4 mW Alternative calculation for the received power: P T = 10 mW = 10 dBm Loss = - 44 dB P R = - 34 dBm
Dr. Joseph C. Palais 11.2 13 Check : dBm = 10 log P R -34 = 10 log P R P R = 10 - 3.4 = 4 x 10 - 4 mW This result checks.

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Ch11Sec2 - THE FIBER FORUM Fiber Optic Communications...

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