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Unformatted text preview: Dr. Joseph C. Palais 11.2 1 THE FIBER FORUM Fiber Optic Communications Dr. JOSEPH C. PALAIS PRESENTED BY Dr. Joseph C. Palais 11.2 2 Section 11.2 SignaltoNoise 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 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. 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 = (118) 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 (119) This is called the shotnoise limited SNR or the quantumnoise 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 shotnoise limited SNR (119) can be rewritten as SNR = i S /2e ∆ f (1110) Dr. Joseph C. Palais 11.2 10 Case 2: thermal noise >> shot noise In this case the SNR (118) becomes: SNR = [( η eP/hf) 2 R L ] /4kT ∆ f (1111) This is the thermalnoise 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 Dr. Joseph C. Palais 11.2 12 Compute the received power....
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This note was uploaded on 01/26/2011 for the course EEE 546 taught by Professor Palais during the Spring '10 term at ASU.
 Spring '10
 PALAIS

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