chapter 10

# chapter 10 - THE FIBER FORUM Fiber Optic PRESENTED BY Click...

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Click to edit Master subtitle style 5/13/10 Dr. Joseph C. 10.1 11 THE FIBER FORUM Fiber Optic Dr. JOSEPH C. PALAIS PRESENTED BY

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5/13/10 Dr. Joseph C. 10.1 22 Chapter 10 Modulatio n
5/13/10 Dr. Joseph C. 10.1 33 Section 10.1 LED Circuits

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5/13/10 Dr. Joseph C. 10.1 44 10.1.1 Analog Modulation t PSP (dP/di) = a1 Pdc Id c ISP t i is Power P
5/13/10 Dr. Joseph C. 10.1 55 i Id c is i = Idc + is i = Idc + ISP sin ϖ t ϖ = modulation frequency P = Pdc + PSP sin ϖ t modulation factor: m’ = ISP / Idc Analog modulation circuit:

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5/13/10 Dr. Joseph C. 10.1 66 For 100% modulation: Isp = Idc and thus m’ = 1 t Idc Is p m’ = 1 0 i In general 0 ' 1 m S
5/13/10 Dr. Joseph C. 10.1 77 The optic modulation factor is m = PSP / Pdc Thus P = Pdc + mPdc sin ϖ t P = Pdc( 1 + m sin ϖ t ) We earlier noted the frequency response of an LED. It was: ( 29 ϖτ = + 1 SP SP 2 a I P 1

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5/13/10 Dr. Joseph C. 10.1 88 τ = carrier lifetime (average recombination time) a1 = slope of the power vs. current curve Thus m = PSP / Pdc = m = ( 29 1 ϖτ + 1 SP 2 1 dc a I a I 1 ( 29 + 2 m' 1 m / m’ ϖ 1
5/13/10 Dr. Joseph C. 10.1 99 Transistor amplifier LED driver: Ra Rin 50 Rb 5 k Re 60 ib ic LED (LD) b c e 2k Signal vin iE Vdc = 5 V C

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5/13/10 Dr. Joseph C. 10.1 1010 50 30 10 0 1250 750 250 1 2 3 4 vce (volts) ic(mA) ib ( μ A) Load Line = Q point (operating point)
5/13/10 Dr. Joseph C. 10.1 1111 Current amplification: β = ( ic / ib ) = [(30 - 10)mA / (750 - 250) μ A] = 40 When there is no signal (vin = 0), the transistor operates at the Q-point. In this example, the base current is IB = 763 μ A and the collector current is IC = β IB = 40(763) μ A IC = 31 mA (see text for full analysis)

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5/13/10 Dr. Joseph C. 10.1 1212 An input signal (vin ) varies ib and thus varies ic about the 31 mA dc level. Note that the collector current saturates at about ic = 55 mA. Thus the total swing in collector is limited to ISP = 55 - 31 = 24 mA The maximum modulation factor is then m’ = (ISP / Idc) = (24 / 31) = 0.774 Follow the circuit design equations in the text. These calculate the load line and the Q-point.
5/13/10 Dr. Joseph C. 10.1 1313 Linearity We have been assuming a linear characteristic for the LED output. Slope = a1 Pdc Id c is i P P

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5/13/10 Dr. Joseph C. 10.1 1414 The output power is : P = Pdc + a1 is where a1 = slope of the characteristic curve is = signal current In practice, LEDs vary from linearity . A suitable mathematical model is : P = Pdc + a1 is + a2 is2 (10-9) This is a power series expansion about the bias point (Pdc, Idc )
5/13/10 Dr. Joseph C. 10.1 1515 Let the signal current be: is = I sin ϖ t Then P = Pdc + a1 I sin( ϖ t) + a2 I2 sin2 ( ϖ t) P = Pdc + a1 I sin ( ϖ t) + a2 I2 [(1 - cos(2 ϖ t) / 2] P = Pdc + 0.5 a2 I2 + a1 I sin( ϖ t) - 0.5 a2 I2 cos(2 ϖ t) The last term is the second harmonic distortion. Pdc + 0.5 a2 I2 a1 I 2 ϖ 0.5 a2 I2 0 ϖ Signal spectrum

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5/13/10 Dr. Joseph C. 10.1 1616 The total harmonic distortion (THD) is: THD = ( electric power in harmonics/electric power in fundamental) as measured at the receiver The received current is iR = ρ P iR = ρ [Pdc + 0.5 a2 I2 + a1 I sin( ϖ t) - 0.5 a2 I2 cos(2 ϖ t)]
5/13/10 Dr. Joseph C. 10.1 1717 Electrical power in the receiver is proportional to the current squared . Thus THD = [(0.5 a2 I2)2 / (a1 I)2 ] THD = 0.25 [ (a2/ a1) I ] 2 (THD)dB = - 10 log { 0.25 [ (a2/ a1) I ]2 } Distortion due to nonlinearity in LEDs and LDs is the biggest problem in analog modulation. Cable TV often uses optical analog modulation to transmit television signals.

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chapter 10 - THE FIBER FORUM Fiber Optic PRESENTED BY Click...

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