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Photodetectors

Solved Problems

Silvano Donati University of Pavia

Prentice Hall PTR Upper Saddle River, New Jersey 07458

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Library of Congress Cataloging in Publication Data Donati, Silvano Photodetectors; solved problems ISBN 0-13-020337-8A 1. Optoelectronic devices 2. Optical detectors 3. Photoelectric cells 4. Title Acquisitions editor: Bernard Goodwin. Editorial/production supervision: Vnbessa Moore Cover design director: Jerry Votta Cover design: Talar Agasyan Manufacturing manager: Alan Fischer Editorial Assistant: Lisa Kanzerlmann Marketing Manager: Anne Trowbridge

© 2000 by Prentice Hall PTR Prentice-Hall, Inc. A Simon & Schuster Company Upper Saddle River, New Jersey 07458

Prentice Hall books are widely used by corporations and government agencies for training, marketing, and resale. The publisher offers discounts on this book when ordered in bulk quantities. For more information, contact Corporate Sales Department, Phone: 800-382-3419; FAX: 201-236-7141; E-mail: corpsales@prenhall.com Prentice Hall PTR, One Lake Street, Upper Saddle River, NJ 07458. All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN 0-13-020337-8A Prentice-Hall International (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall Canada Inc., Toronto Prentice-Hall Hispanoamericana, S.A., Mexico Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, Inc., Tokyo Simon & Schuster Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro

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INDEX

Problems, Chapter 1 3 Problems, Chapter 2 5

Problems, Chapter 3 11 Problems, Chapter 4 15 Problems, Chapter 5 24 Problems, Chapter 6 44 Problems, Chapter 7 49 Problems, Chapter 8 58 Problems, Appendixes 62

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Foreword

A book is not complete without a solved-problem booklet, they are used to say, and I actually didn’t realize how true is this saying until trying to do one. When you try to make problems, the ups and downs of your book come out merciless and you can realize, much better than from the comments of a yet competent reviewer of your book, which are the strengths and the weaknesses of the work and, after all, of the knowledge you are trying to convey with your book. Well, I am happy that I had to suffer not so much from compiling a solved-problem booklet like this (in a few points, however, I did). Compared to the international standard, maybe this collected set of problem is classified as rather difficult. Actually, many first-level easy problems are already contained in the text under the form of examples, and I didn’t try to make duplicates of them in this booklet. I hope this piece of work may be useful to young scientist and students. For any comment and suggestion, as well as to point out errors, please e-mail to me at: donati@ele.unipv.it. For instructors, I have prepared a 380-slide set of Powerpoint presentation, taking all the Figures of the text and adding some more, along with the hand-outs. The set is available for free to those using the book in a course, contact me at donati@ele.unipv.it.

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CHAPTER 1 - PROBLEMS and QUESTIONS

Question 1-1 Is it correct to say that an image photodetector is just made by a number of individual single-point photodetectors, one for each the pixels to be resolved ? Answ: No, it is not. An image-photodetector requires two more mechanisms: the charge storage during non-reading time, and an adequate arrangement to sort out the pixel information in sequence. Question 1-2 Is the distinction between internal photoelectric effect devices and external photoelectric effect detectors a fundamental one or is purely nominal ? Answ: From the viewpoint of the user it is nominal, as any distinction (thermal, etc.) because the user is only interested to the signal out from the photodetector. But, from the viewpoint of the manufacturer and of the careful designer, it is substantial because each family determines a range of achievable performances. Question 1-3 Why the S/N ratio improvement offered by a good photodetector design should be so important ? Wouldn’t be just sufficient to increase the source power to get the same result ? Answ: Indeed, improving the S/N ratio by decreasing photodetection noise is equivalent to increase of the same amount the input light power that is made available. But, this is the key point: in photodetection we whish to use the least available power to achieve the desired performance. Rather, we can use the available extra power, once the photodetection noise has been properly minimized, to improve the quality of our setup, whatever it is.

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Problem 1-4 A distant galaxy (see Fig.P2) has emitted a photon of energy hν that, after a long time, has traveled up to being absorbed by a photodetector on the earth. Because of the Hubble recession, the photon reaching the earth has a red-shifted wavelength, or, its energy is hν’<hν, that is, considerably less than the initial energy hν. How can we reconcile this with the principle of energy conservation ? Answ: The received energy is not only hν’, because, if the emitting and receiving bodies are not at rest, but move by a relative velocity v, energy is also imparted to the receiving system by photon momentum hν/c. If the photodetector has a mass m and recedes with speed v, momentum conservation requires: mv + hν’/c = m (v+Δv) whence we get for the photodetector momentum variation: mΔv= hν’/c where ν’=ν/(1+v/c) is the Doppler shifted frequency. The energy actually imparted by the photon is then: hν’ + 1/2 m (v+Δv)2 - 1/2 m v2

= hν’ + m vΔv +1/2 m Δv2 (the last term being negligible)

= hν’ + v hν’/c = hν’(1+v/c) = hν [Actually, from general relativity it is ν’=ν √(1-v2/c2) /(1+v/c) and this second-order correction cancels out also the above term assumed as negligible]. ______________________________________________________________

CHAPTER 2

PROBLEMS and QUESTIONS

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Question 2-1 Comment on the advantages and disadvantages of reflection vs transmission photocathodes. Answ: Reflection photocathodes have the highest quantum efficiency η, but require an extra space in front of the sensitive surface to secure emitted photoelectrons collection. Transmission photocathodes are a little bit penalized in η but the release from the above constraint more than compensate from the application point of view in most cases. Question 2-2 A photocathode material has an absorption length Labs= 1 µm and a diffusion-length of hot carriers Ldif= 0.1 µm. A second photocathode has Labs= 0.1 µm and Ldif= 1 µm. Which is likely to have the largest η, other things being equal ? Answ: The second one. The quantum efficiency, or probability of emission is critically dependent on the escape probability, on its turn dependent on the free path of diffusion to reach the surface. Question 2-3 When the photoelectron is hot, hν>>Ep, it would appear (discussion of Section 2.3) that photoemission strongly decreases. Instead, diagrams of Figs. 2-8 and 2.9 show a substantially unaltered efficiency. How to explain this ? Answ: The hot photoelectron undergoes an energy loss of at least Eg. But, if the remaining energy is still sufficient to overcome the surface-barrier potential, the photoelectron can be emitted. Actually, as hν increases with ν (or with decreasing λ), an intermediate situation can be found, in which the photoelectron

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is first penalized but then recover. This is exemplified by the (modest) UV drop-off of the S-20 response in Fig.2-9, of the NaKSb in Fig.2-8, etc. Problem 2-4 A photocathodes has a quantum efficiency η=20% at λ=550 nm. It receives a luminous power of 10-3 lumen from a 550-nm laser. What is the photocurrent suuplied by the device? Answ: From App.A1 (page 360), the radiant power P corresponding to the luminous one Pl is P=Pl/Kmaxis 10-3 / 673 = 1.485 10-6 W. As the spectral sensitivity is σ=η λ /1.24 (µm) = 0.887 (A/W), we get I= σ P= 0.887. 1.485. 10-6= 1.32 µA. Problem 2-5 A S-20 photocathode has a spectral sensitivity σ=80mA/W at λ=450 nm and T= 20°C. It receives a radiant power of 10-9 W and is working at 80°C. What is the percentual error in our measurement if we neglect the σ-temperature dependance ? What if the wavelength would have been 800nm ? Answ: From Fig.2-10, the temperature coefficient of the spectral sensitivity is evaluated as α= 0.5%/°C. Therefore, the relative error from 20 to 80 °C is: αΔT= 0.5%/°C. 60= 30%, a large figure indeed. If wavelength is 800nm, then (Fig.2-10) α < 0.05%/°C and the error would have been negligible.

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Question 2-6 From Figures 2-7 through 2-9, reflection photocathodes appear to be always more efficient than their transmission counterparts. Also, there is a limit of η=50% in efficiency, and response goes not beyond 100nm on the UV side and 1.3 µm on the IR. Are these basic limitations of photocathodes ? Answ: A reflection photocathode has always a larger quantum efficiency than a transmission type, because photoelectrons do not have to cross the material thickness. As a guideline, transmission photocathode should be used with collimated (or, nearly collimated) light sources when a high η is the most important. Reflection photocathode are instead the preferred choice when using high NA objective lenses. The 50% limit of efficiency is a basic one because in the material, half of the photogenerated electrons point outwards and half inwards. In the UV, response could be extended well beyond 100nm with suitable materials (metals), but we should find the proper for them. Also, when hν>>1eV, it is better to try converting the energetic primary photon in a number of photons in the visible range (by means of a scintillator) rather than detecting it directly. In the IR, the limit is fundamental on the technology standpoint: there is no material with the appropriate bandgap and the appropriately low-energy surface barrier. Question 2-7 Why data about the dark current of photocathodes (Fig.2-11) are scattered on a factor of 5-10? Means that this parameter cannot be controlled properly? Answ: Actually, data in Fig.2-11 include of all the dark-current contribu-tions, that is, dark emission from the photocathode, emission from its edges and other electrodes, and emission due to residual and background radiation. When the

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fabrication process is reasonably well controlled, the spread of photocathode dark emission may be 10% only, but the other effects may mask this result. Problem 2-8 A 10-mm × 1-mm metal-wire (e.g., Tungsten) phototube is placed close (5-cm) to a 2-cm2 flame emitting as a blackbody at temperature T=6000 K (see the diagram of Fig.A1-5) in the range 200 to 300 nm, and assuming a spectral sensitivity of σ=20 mA/W, calculate the available photocurrent. Answ: From the diagram of Fig.A1-5, the radiance of the flame near λ=200nm is R≈ 500 (W/ cm2 . sr. µm). As the solid angle subtended by the receiver is: Ω€=0.1cm2/25cm2=0.004, we get for the power reaching the phototube: Pphd=500. 2(cm2) . 0.004. 0.1(µm)= 0.4 W. The photocurrent is therefore: Iph= σP=20m . 0.4= 8mA, a value large enough to actuate a relay or another flame-control circuit. Problem 2-9 Can the flame detector of Probl.2-8 be operated directly in a.c. from mains, i.e., without a rectifier section ? Answ: In a plain phototube with the photocathode sensitive to light and the anode in a different material (metal) not sensitive to light, operation of the circuit directly from mains would give a half-wave photocurrent response. Instead, if we have a tungsten-wire construction of the phototube, anode and photocathode interchange their role in positive and negative semiperiods, and the response in ac. So, a TRIAC can be switched-on on both semiperiods. Note also that, using a gas-filled phototube, we would add a gain G≈10 to the value calculated in Probl. 2-4, getting a current largely sufficient to operate directly a relay.

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Problem 2-10 A vacuum, fast phototube has a FWHM pulse response of 200ps. What is the pulse-delay we can expect from it ? Guess of three choices: 400, 600 and 1000 ps. Answ: From Fig.2-16, we find 600 ps as the typical delay time corresponding to a 200ps response (FWHM) phototube working at Vak=500 V. But, according to the good (or bad) design of the interelectrode flight structure, the figures of 1000 and 400 ps delay-time, respectively, might also be considered as possible values in a practical device. Problem 2-11 What is the maximum peak-power that can be detected with reasonable linearity from a vacuum, fast phototube working at a anode voltage 2000 V? Answ: From Eq.2.19, we have Jsat =2.3 Vak

3/2 (µA/cm2) as the saturation current density. Taking typical values of A= 1 cm2 and a spectral sensitivity σ=50 mA/W, we have: Isat= AJsat = 1. 2.3 10-6(2000)3/2 = 0.21 A, whence Isat= Psat/σ =0.21/ 50.10-3 = 4.2 W. _______________________________________________________________

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CHAPTER 3 PROBLEMS AND QUESTIONS

Problem 3-1. What is the S/N ratio of an ideal quantum detector receiving a digital signal containing N photons per bit ? What if the quantum efficiency is η<1 ? Answ: As the photodetected current is I= Ne/T, where T=1/F is the bit period (f=transmission rate), and the noise is i2

n= 2e I B= (since B=1/2T by Nyquist’s sampling theorem) or i2

n = 2e (Ne/T) (1/2T), we get: (S/N)2= I2 / i2

n= (Ne/T) 2 /2e (Ne/T) (1/2T) = N. If we consider a real quantum detector with an efficiency η detecting N photons, the result becomes (S/N)2= η N. Problem 3-2. A pin photodiode works at λ=1240 nm with a bandwidth B=80MHz, has a dark current 100 nA and is closed on a 500-Ω load. Its quantum efficiency is η=0.7. What is the noise-equivalent input power? What is the signal level at which the detector attains the quantum regime? What would change if the photodiode were an APD (avalanche photodiode) with α/β=5 and is working at a G=50 ? Answ: At λ=1240 nm and with η=0.7, the spectral sensitivity is σ=0.7 A/W. The load resistance is equivalent to a dc current 50mV/R=100µA, and therefore the total noise is: in

= [2eB(0.1+100)µA] 1/2= 50 nA which corresponds to a power Pn =72 nW.

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If we were using an APD photodiode with G=50 and α/β=5, we would have had a Fano factor F≈50/5=10 and the total noise would have been in

2= [2eB(0.1F +100/G2)µA] 1/2 = 0.00144 nA, or, at a input signal power: Psign=0.5 nW, i.e., the noise would be about 10 times smaller. Problem 3-3. A photodetector has a dark current of Idark, an internal gain G and an excess noise factor F. Find the noise figure NF in the quantum and thermal regimes. Answ: At the input, we receive a signal P accompanied by a noise √(2hνPB), so that (S/N)2

in= P/2hνB. The photodetector response is made up of a signal σPG accompanied by a Nois:e N2= 2e(σP+Idark) G

2FB+4kTB/R. Thus we have: (S/N)2

out= (σPG)2/[2e(σP+Idark)G2FB+4kTB/R] =

N2= 2e(σP+Idark) G2FB+4kTB/R. Thus we have:

(S/N)2out= (σPG)2/[2e(σP+Idark)G

2FB+4kTB/R] = =σP/[2e(1+Idark/σP)FB+2eIRB/σPG2] where IR= 2kT/eR indicates the current-equivalent of the termination resistance R. Combining, we get: NF=(S/N)2

i /(S/N)2out= (P/2hνB) [2e(1+Idark/σP)FB+2eIRB/σPG2] /σP

= (1/η) [1+FIdark/σP + IR/σPG2] (remember that σ= η e/hν). When the gain is large enough so as to make the termination-resistance contribution negligible, i.e., G2>>IR/σP FIdark, we get: NF = (1+FIdark/σP) /η Also, in terms of the photodetected current Ιph=σ P we can write the above as:

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NF = (1+FIdark/ Ιph) /η This is for the quantum regime. When G2<<IR/ΙphFIdark, the termination resistance noise is dominating and we find for the noise figure: NF =(1+ IR /ΙphG

2) /η = (1+2kT/eRΙphG2) /η.

Problem 3-4 Prove the following statement: a quantum-regime photodetector has a relative intensity noise floor given by –155 dB per unit frequency and unit current (that is, per Hz and mA). Answ: In the quantum regime we have (S/N)2= I/2eB. For I=10-3 and 2e=3.2. 10-

19 we get (N/S)2= 3.2. 10-16 equivalent to –160 +5=-155 dB. This figure is also the RIN (relative intensity noise) of an ideal laser, per Hz and dBm (=1mW), as it is: (S/N)2= P/2hνB which coincides (σP=I, σhν=e) with the above. Worth to note, –155 dB is also the of (S/N)2 at the break point of Fig.3-2. With this value, we can compute fast the (S/N)2 in various conditions, e.g., for B=1 GHz and I = 1µA (variations of 90 and 30 dB respectively) we get (S/N)2= 35 dB, etc. Problem 3-5 A pin photodiode works at λ=1240 nm with a bandwidth B=800kHz, has a dark current Idark=0.1 µA and is terminated on a R=250 kΩ load. At which input power PO does the detector reach the quantum regime of detection? What is the S/N ratio at this value of PO? Answ: The load-resistance noise is equivalent to a dc-current IR= 50 mV/R= 0.1 µA, so the total noise (added quadratically) is √2 Idark=0.14µA= I0. At λ=1240 nm and η=0.7 the spectral sensitivity is σ=0.7A/W, whence: P0= 0.2 µW.

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The total noise is: in= [2eIdark B +2eIR B]1/2 = √2. [2eIdark B]1/2 = 1.4. [3.2. 10-19 . 10-7. 0.8. 106]1/2= 0.225 nA and thus S/N= I0/ in= 0.1 µA /0.225 nA= 440 (=26 dB). Question 3-6. Define the noise-equivalent power of a detector, and the detecti-vities D, D* and D**. What are the BLIP and Dark-Current limits ? Answ: Follow the description of the text, pp.43-46. Question 3-7 If the responsivity is not a really meaningful figure of merit describing the ‘sensitivity’ of a detector, why should we consider it? Answ: The responsivity is actually an important parameter for the design of a photodetector-based instrument, as it gives the ratio of out-put electrical signal to input power. Though this parameter shall not be be taken as the ‘sensitivity’ of the system, it is necessary to evaluate the signal levels throughout the system. Question 3-8 Can the BLIP limit (Sect.3.3.2) be overcome by any practical detector ? Answ: This limit is for the operation of a detector against the thermal background at a given ambient temperature. Actually, a detector can also have a D* larger than than the BLIP limit, but when it is aimed to that kind of background (or, is looking to a thermal scene) its performance will be determined not ny its own intrinsic D*, but by the D*BLIP.

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Question 3-9 When we move from the thermal IR to the near-IR and to the VIS, the background limit loses its importance and the detectivity is limited by other factors. Specify which ones. Answ: It is the dark-current limit to dominate the D* performance in this case. As shown in Fig.3-3, at 300 K the performance of a detector is likely to be dominated by the dark current rather than by background noise below about λ=1.5 µm. _______________________________________________________________

PROBLEMS CHAPTER 4

Problem 4-1 A photomultiplier has 6 (respectively: 9, 12) stages with dynode gains all equal to g=3.5. If the single-electron response has a duration of Δτ=3 ns, what is the amplitude of the SER current pulse ISER? Can it be seen with a normal oscilloscope? Answ: Assuming the simple equation ISER=eG/Δτ=egn/Δτ, we get: ISER=1.6.10-19.(3.5)6 /3.10-9 =0.1µA for n=6 stages, = 4.3 µA for n=9 stages, = 0.18 mA for n=12 stages. In the first two cases, the sensitivity of a normal oscilloscope (typ. 1 mV/div) falls shorts and the SER will not be visible. In the last case, it will be barely visible. However, if we move the dynode gain from 3.5 to 4.5, the three peak currents become: 0.45µA, 41 µA and 3.6 mA and in the last case we can see it provided the oscilloscope bandwidth is adequate (min B=0.35/Δτ≈ 120 MHz).

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Problem 4-2 Of a dynode working with a gain g=5, we wish to double the gain. Shall we double the voltage difference at which it works? Answ: For most dynode materials, the g/V slope (see Fig.4.4) is slightly less than unity (typ. 0.78-0.82). So, to double the gain we need a 21/0.8= 2.38 times larger voltage. Problem 4-3 From the point of view of charge-measurement accuracy, would you prefer to use a PMT#1 with 1st-dynode gain=5, followed by 11-dynodes with gains 3.5, or another PMT#2 with 1st-dynode gain=10 and 11-dynodes with gains 2.5? Answ: It is the multiplier variance ε2

A=(1/g1)g/(g-1) that shall be considered. For the PMT#1 we have: ε2

A=0.2. 3.5/2.5= 0.28, while for PMT#2 it is: ε2A=0.1.

2.5/1.5= 0.17. Note that the total gain is G©=g1g

11= 5. 3.511=4.8. 106 and G2=10. 2.511=2.4. 105 in the two cases. Despite the smaller gain, PM#2 is better because it has a smaller charge variance. Problem 4-4 A photomultiplier has dynodes gain g=4 and a photocathode quantum efficiency η=10%. What is the noise figure for the detection of a charge pulse? And what is the noise figure for the detection of a photon packet containing F photons? Answ: The integral-response noise figure for the detection of R photoelectrons (Poisson-distributed) is: NF2

charge = 1+ε2

A where ε2

A=(1/g1)g/(g-1), so that:

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NF2 charge

= g/(g-1) = 4/3 = 1.33 That for the detection of photons is: NF2

photon = (1+ε2

A)/ η so that: NF2

photon= g/(g-1)η = 4/3 . 0.1= 13.3 Problem 4-5 Estimate the gain of the dynodes for the 12-stage PMT whose SER is reported in Fig.4-3 of the text. Answ: The SER displayed in Fig.4-3 exhibits a peak amplitude of S=2.5 mA, a duration (FWHM, full-width-half-maximum) of τFWHM=2.6ns; the amplitude spread read on the SER peak from the diagram is about ΔS =0.75 mA(p-p). From the SER area, we have a total gain G=S τFWHM/e=2.5 10-3 2.610-9/1.6 10-

19= 4 107. As we have 12 stages, if the dynode-gains were all equal, we should have: g=(4 107)1/12= 4.3. Now, we can compute the multiplier variance εA and compare it to the relative SER spread σSER/S. For all-equal dynodes it would be [see Eq.4-10]: εA=(g-1)-1/2=1/√3.3=0.55. As we may take (from statistics) ΔS≈3σSER, the rms deviation is evaluated as: σSER=0.25 mA, or: σSER/S=0.25/2.5=0.1, a value sensibly less than εA. Therefore, we shall imagine that the first dynode is working at a higher gain, and then the applicable expression for the multiplier variance is εA=[g1(1-1/g)]-

1/2 . In this case, we can solve for g1 as: g1 =1/[εA

2(1-1/g)]=1/(0.1)20.76= 13. This is not an unrealistic value, if the first dynode is a Ga-P type (Fig.4-4).

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Problem 4-6 At the sampling oscilloscope, I have observed the SER of a PMT and it had a very wide trace of fluctuation around the mean.What is all about it? Answ: The SER relative variance is given by ε2

A, i.e, the multiplier variance (typically ε2

A=(1/g1)g/(g-1)=0.2, or, εA=0.447). Thus, A 44.7% amplitude spread around the mean value is a very common deviation to be expected. Problem 4-7 The response of correlation-measurement accuracy requires two functions, namely ρDSPR and ρ SPR· to characterize the PMT (Sect.4.4.3). Is it a basic result? Or, we could have just one function describing the PMT ? Answ: Actually, we need both functions to take into account the extra randomness of the photocathode-to-first-dynode time of flight. Problem 4-8What is the time resolution we can get using a PMT when we have a short light pulse containing R=100 photons? Answ: If the light pulse is shorter than the SER, Eq.4.61 holds, and we get a time-accuracy localization of the pulse given by: ε2

t= τr2(1+ε2

A)/R For a typical PMT, we may take τr=2 ns and ε2

A=0.2 from which it comes: εt=2 ns . √(1+0.2) /√100= 220 ps.

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Problem 4-9 In wiring a photomultiplier voltage divider, two dynodes have been inadvertently interchanged. When the voltage has been switched-on, the PMT appears to be dead. Any reason for that ? Answ: The interdynode maximum voltage is not much more than the working voltage specified by the manufacturer. Thus, the lowest-order dynode misplaced at a higher voltage is likely to have worked at a voltage higher than the maximum permissible, with consequent destruction of it by power dissipation.

Problem 4-10 A PMT is closed on an anode load R=10kΩ. What is the bandwidth we can expect for the PMT ? Can the SER be resolved ? Answ: Assuming a anode-to-earth capacitance of C=5 pF, we get as a high-frequency cutoff: f2 = 1/2πCR= 0.16/ 5 10-12 104 = 3.2 MHz. The PMT is thus used in the moderate-bandwidth mode (as per Sect.4.7.1 of the text).Since the SER response is dominated by the pole corresponding to f2, single electrons give as a response an exponential-decaying pulse with time constant τ=RC=50 ns. The SER current peak-amplitude at the anode is ISER=Ge/τ =(typ.)=107 1.6 10-19/50 10-9= 32 µA, and gives a voltage output ISERR= 320 mV. So, it is likely to be visible on a normal oscilloscope.

Problem 4-11 A PMT having a SER risetime of tr=1 ns is detecting fast optical pulses (duration 50ps). Estimate the time-localization accuracy τΦ for R=10, 100 and 1000 photodetected photons.

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Answ: With a threshold-crossing decision circuit, we can take as a first approximation (more precise details are given in Sect.4.7.3): τΦ = 0.3tr = 0.3ns for a single photoelectron. For R= 10,100, 1000 photoelectrons, we will have: τΦ = 0.3tr /√R= 100,30,10 ps. Problem 4-12 Let the SER of a PMT be a Gaussian waveform, with delay τd and variance σSER : SER(t) = Ge gauss(t, τd, σSER) where explicitly we let: gauss(t, τd, σSER) = (√2πσSER)-1 exp-(t-τd)

2/2σSER2.

the PMT receives a gaussian-light pulse of the type I(t)=R gauss(t,0, τI). Calculate the mean output waveform and its variance.

Answ: The output pulse has a mean value: Iout(t)= I(t)*SER(t), where * stands for the convolution operator. In the Laplace-transform domain, expression becomes: Iout(s)= I(s).SER(s). In view of the gaussian transform being a Gaussian, i.e.: SER(s)=Ge exp -s2σSER

2/2, and I(s) =Rexp -s2τI 2/2, we get:

Iout(s) = exp -s2(σSER2+ τI

2)/2. Returning to the time domain, we have: Iout(t) = GeR [√2π (σSER

2+ τI2)]-1 exp-(t-τd)

2/2(σSER2+ τI

2) About the variance, in the rigid SER approximation (Sect.4.4.2), we have: σIout

2=(1+ε2A) I(t)*SER2(t)

and, repeating the above calculations with SER2, the result is obtained as: σIout

2=(1+ε2A) (Ge)2R [√2π σSER]-1 exp-(t-τd)

2/2(σSER2/2+ τI

2) We can now distinguish two cases:

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1) σSER>> τI (let us say σSER>0.3 τI for clarity). Because of the quadratic composition, the term τI is negligible and Iout(t)≈GeR [√2π τI)]

-1 exp-(t-τd) 2/2 σSER

2

σIout2=(1+ε2

A) (Ge)2R [√2π σSER]-1 exp-(t-τd)2/2(σSER

2/2) and we have a S/N ratio: (S/N) 2= σIout

2/ Iout2=(1+ε2

A)/R. 2) σSER<< τI (let us say σSER<0.3 τI for clarity). Then, in view of the quadratic composition rule in Iout(t) and σIout

2, the term σSER is now negligible and Iout(t)≈GeR [√2π τI)]

-1 exp-(t-τd) 2/2 τI

2

σIout2=(1+ε2

A) (Ge)2R [√2π σSER]-1 exp-(t-τd)2/2 τI

2

and the S/N ratio is: (S/N) 2= σIout2/ Iout

2=(1+ε2A)/R exp+(t-τd)2/2 τI

2

as expected, in this case the S/N ratio is maximum at the peak of response (t=τd) Problem 4-13 What is the minimum photon-rate which is detectable by a PMT in the single-photon counting regime? What is the same photon-rate when we add dark-subtraction ? Answ: With the plain photon-counting scheme (p.94) we can assume as the practical minimum photon-rate which is detectable as the one making 〈Ns〉 =〈Nd〉 [see eq.4.66)]. Therefore: Fmin = (ηdId /e) /ηηp With dark suntraction, the result [eq.(4.72)] is: Fmin-S = √(2ηdId /eT) /ηηp Thus, the improvement factor M by adding dark subtraction is: M = Fmin-S /Fmin = √(2/ηdId

. eT)

23

Problem 4-14 A PMT is placed at the focal plane of a 5-m diameter telescope. The PMT is made working in the single photon-counting regime. Calculate the magnitude of the dimmest star which can be detected with a reasonable signal-to-noise ratio, say S/N=10, on a 8-hour integration time.

Answ: Let us take an S-24 photomultiplier, most suited for the visible. Using a device with a 1-cm2 area, we can expect η=35% at λ=400nm (Fig. 2-7), and a dark current rate Id/e=10 cm-2 (Fig.2-11). With a first dynode gain g1=3 (Fig.4-23), we can choose the thresholds Q1/Ge=0.8 and Q2/Ge=2.5 and have, from the circuit of Fig. 4-24, typical values of ηp=0.7, ηd=0.8. The minimum detectable rate at S/N=1 is found from Eq.(4.72) as Fmin=√(2.0.8.10.1/T)/0.35.0.7=16.3/√T photon/√s for S/N=1 For S/N=10 we shall multiply the result by S/N (see page 95), and have: Fmin=163/√T photon/s. In a 8-hour (=28800s) integration period, this yields: Fmin=163/√28800=0.96 photon/s Turning to power, and being hν=4.8 10-19 J at λ=400nm, we get: Pmin =hνFmin= 4.61.10-19 W. Finally, to calculate the corresponding star magnitude, let us remind that (see p.362, App. A1) a 0-m star gives an illuminance of E0=2.65.10-6 lux, whence a power collected by the telescope: P0=E0 A/Kmax=2.65.10-6.lux(π 52/4)m2 /673 lm/W = 7.46.10-8 W. The magnitude corresponding to Pmin follows from Pogson’s scale (p.363) as: m= -2.5 Log10 Pmin / P0 = -2.5 Log10 6.03.10-12 =2.5.11.22 = 28

24

Thus, a m=28 th star is in the good reach of a 20m2 terrestrial telescope and a common PMT (or, a photographic plate). Of course, the choice of a standard, commercially available, 1-cm2 area PMT is not the best one, as a diffraction limited star is imaged on the focal plane in a much smaller area [≈(λ/NA) 2≈ 10-4 cm2]. If the sensitive area from which the dark-current also comes were restricted to such a value, we would be able to gain an additional -2.5 Log10(√10-4) = 5 stellar magnitudes. Also, in outer space where we can go to integrate up to, say, 10 days =240 h we would gain another -2.5 Log10(240/8)= 3.7 magnitudes, going in total to a limit of 34.7. Note that this figure is close to what is achieved by the Hubble Space Telescope (Fig. P-2), working with very-low-noise CCDs.

Problem 4-15 Estimate the time-localization accuracy σT that can be obtained by a PMT whose SER is as in Fig.4-3 in a PMT with n=12 stages, when the light pulse to be measured is short respect to the SER.

Answ: From Fig.4-3, the full-width-half-maximum duration of the SER is read as τFWHM=2.6ns. This corresponds to a time-of-flight variance (see Eq.4.62) σt=2.6/2.36=1.1 ns. If all dynodes time-of flight-dispersion σfi were equal, we should deduce that σfi = σt/√n= 1.1 / 3.46= 318 ps. Then, the time-accuracy would be (page 91 and Probl.4-5): σT = (1+ε2

A) σfi = (1+0.55) 318 ps= 493 ps. This result scales as the inverse of the √R square-root of number of detected photoelectrons. More likely, if the dynode dispersion is σfi=318 ps, the photocathode-to-first dynode time-of-flight dispersion may be 2-4 times better, and accordingly a reduction of factor 2-4 over the figure of 493 ps is to be expected.

25

Question 4-16 Describe the working principle of the Channeltron and its areas of application. Answ: Follow the text at pages 98-99. Worth to remember, not only for last generation PMTs the Channeltron turns out to be very useful, but it is invaluable also for post-intensification of fast CRT-tubes and in image intensifiers (ICTs).

Problem 4-17 At equal internal gain, is an MCP-PMT equivalent to a discrete-dynode PMT as far as the statistics of charge detection, or of timing accuracy, are concerned ?

Answ: Due to the continuous multiplication statistics, the MCP is worse that a discrete dynode PMT by a factor log g . k(n), see Eq.4.86. Correspondingly, the multiplier variance can be as high as ε2

A= 5-20 instead of ε2A= 0.3-0.5 of a

discrete-dynode structure. Th disadvantage is however recovered as the number of detected photoelectron R increases, as the statistics behaves as (1+ε2

A)/R. _____________________________________________________________

PROBLEMS CHAPTER 5 Problem 5-1A silicon pn photodiode works with a depletion-layer width W=50 µm and has a p-region undepleted-region thickness w=4 µm (see Fig.5-2). Calculate the PD-efficiency and spectral sensitivity at λ=800 and at λ=1000

26

nm. [assume diffusion lengths Ln=3 µm, Lp=10 µm]. Find the above for: (i) an untreated Si-air entrance window, and: (ii) for a Si-SiO2 entrance window. Answ:. From Fig.5-2.2 the absorption length in Si is found as Labs=1/α=10 µm and 50µm, respectively, at λ=800 and at λ=1000 nm. From Eq.(5.2) we have for the internal quantum efficiency: ηi = exp -(w-Ln)/Labs – exp-(w+W+Lp) /Labs By inserting numbers, we have, at λ=800 nm: ηi = exp - 1/10 – exp - 64/10 = 0.905 - 0.002 = 0.903 (*) and, at λ= 1000 nm: ηi = exp - 1/50 – exp - 64/50 = 0.989 - 0.278 = 0.711 (*) Note that the complement to 1 of the addenda in (*) is the loss in the undepleted regions. From Fig. 5-2.3, we find the reflection of the Si-air interface RSi≈0.30 and also R≈0.18 for the Si-SiO2. Thus, the total efficiency is: i) = (1-R) ηi = = 0.70. 0.903 = 0.632 @800 nm, = 0.70. 0.711= 0.498 @1000 nm (Si-air) = 0.82. 0.903 = 0.740 @800 nm, = 0.82. 0.711= 0.583 @1000 nm (Si- SiO2) Problem 5-2 What is the relative proportion of diffusion- and drift-limited frequency response (see Fig.5-2.9) of the Si-PD of Probl.5-1 ? Answ: at λ=800 nm, power reaching the depth z= w-Ln is exp-1/10=0.905. Of this value, a fraction 1-exp-Ln/Labs = 1-0.741=0.259 is dissipated in the p undepleted region, while the power dissipated in the n undepleted region is negligible. Thus, the relative proportions of diffusion and drift are:

27

ηiDn = 0.259. 0.905= 0.234 and ηd = ηi-ηiDn= 0.905-0.234= 0.371. For λ=1000 nm, power at z= w-Ln is exp-1/50=0.980, the p-undepleted dissipated fraction is 1-0.942=0.058. Thus, we get ηiDn = 0.058. 0.980= 0.057 and ηd = ηi-ηiDn= 0.711-0.057= 0.654. Problem 5-3 For the Si-photodiode of Probl.5-1, find the required dopant concentration level, the maximum supply voltage, the drift time and the specific capacitance. Answ: We take advantage of the diagram in Fig.5-2.11, in which we have the choice of moving along a horizontal line at the ordinate valueW = 50µm. If we stop in correspondence to an abscissa value ND=1013 cm-3, we get a working voltage Vbb=20 V and τd=5 ns. The capacitance per unit area is 2 pF/mm2. Material with ND=1013 cm-3 may be demanding, while the p-level dopant is uncritical, as we need only one much larger than ND, say, NA =1015 cm-3. Actually, to obtain exactly w=4µm with W=50 µm, we need a p-doping level NA= ND(W-w)/w= 1.25. 1014 cm-3. We may prefer going farther along the W = 50µm horizontal line, e.g., to ND=3.

1014 cm-3. This choice requires NA=3. 1016 cm-3 and will yield a faster response, τd=200 ps, and the same specific capacitance 2 pF/mm2. However, a much larger operating voltage is now needed, Vbb=500 V, very close to the photodiode breakdown.

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Problem 5-4 At the same operating (or bias) voltage Vbb, and depletion width W, how large is the gain we can obtain from a pin photodiode respect to an abrupt-junction pn photodiode? What about the specific barrier capacitance ? Answ: From Eq.(5.27), the pn-diode has τd= W2/2 Vbb µ*. From Fig.5-2.10, we have: Vbb=(ε/2)NDD2 and EB= ε NDD; in addition, it is D≈W. In terms of the breakdown field EB we have W2=2 Vbb/ε ND=2 VbbW /EB. Solving for W= 2 Vbb/EB and inserting in τd we get: τd= 2Vbb/EB

2µ*.

Now, for the pin-diode, we start from Eq.(5.36), τd= W2/4Vbb µ* and note that here EB=Vbb /W, so that in terms of EB it is: τd= Vbb/4EB

2µ*

Thus, at equal Vbb and depletion width, the pin photodiode is thus 8 times faster than the its pn counterpart. About the differential barrier capacitance Cb, Eqs.5.34 and 37 state that they are equal. Thinking of the pn-smooth distribution of charges respect to the δ-like distribution of the pin this may be surprising. However, it is correct: by increasing of ΔV the voltage across the junction, the ΔQ increase is contributed by new depleted material at the boundaries of the pre-existing depletion layer , in both cases (the integral capacitance being different). Problem 5-5 About fast photodiodes, comment on which semiconductors may be good candidates for improvement respect to Silicon as the starting point for a short response time.

29

Answ: Letting the intrinsic-response be the dominant limitation, we may note from Eq.5.36 that the effective mobility µ* is the parameter of importance once Vbb and W have been chosen. From Si with µe,h = 1500, 400 cm2/V. s, µ*= 315 cm2/V. s we have the following for different materials: µe,h =10 000, 400 cm2/V. s for GaAs µe,h = 80 000, 1800 cm2/V. s for InSb µe,h = 20 000, 280 cm2/V. s for InAs µe,h = 180, 120 cm2/V. s for GaP µe,h = 300, 15 cm2/V. s for CdS The relative figure of merit for uniform dissipation of photons in the depletion layer is the effective mobility µ*=(1/µe +1/µh). However, at λ<<λs, photon mainly absorbed in the first region and the drift time τd will then depend on the mobility of the carrier which is minoritary in the first region (p in a pin, n in a nip). Thus, for a GaAs-pin, the relevant µ is µe and we can get an improvement of 5 times in speed; in a InSb-pin the factor can go up to a factor 240. (The same argument applies for the fast part of the response, when µe,>>µh). Capacitance affecting the extrinsic response: while it is εSi=11.9 we have for other semiconductors: εGaAs=13.1, εInSb=17.9, εInAs=14.6, εGaP=11.0, εCdS=8.42. Problem 5-6 Using the data of Probls.5-1 to 5-4, calculate the reverse saturation current of the photodiode, assuming an active area A=1mm2. Answ: From Eq.5.13’ we have for the diffusion (Shockley) term:

30

Id= A e ni2 [Dp/ LpND + Dn/ LnNA]

where the intrinsic carrier concentration in silicon is ni =1.45 .1010 cm-3, and the

diffusion constants are: Dn,p =µn,p(kT/e)= 37 and 10 cm2/s (for n and p). Inserting numbers for the specific structure (Lp=3µm, ND=1013 cm-3, Ln=10µm, NA=1.25. 1014 cm-3) we get:

Id= 10-2 cm2 .1.6.10-19 C. 2.1.1020 cm-6 [10/3.10-4.1013+37/10-3.1.25.1014

(cm4/s)]= 1.6.10-21 (C/s) 2.1.1020 [3.3.10-9 +2.9610-10]= 1.2 nA. {This value is fairly high; however, if we choose the second solution outlined in Prob.5-3, namely ND=3.1014 cm-3 and NA=3.1016 cm-3, the current would become Id= 1.6.10-21. 2.1.1020 [1.1.10-10 +1.2.10-12]= 0.11 nA} The second contribution to reverse current is the generation-recom-bination term given by Eq.5.17: Ig-r= A e niW / 2τ here τ, the lifetime of carriers recombination (assisted by g-r levels), is proportional to Ng-r, the defects concentration. Assuming Ng-r<<NA, ND, the lifetime τ is fairly long, e.g., we can safely assume τ=1µs (or longer). In this way, we get: Ig-r= 10-2 cm2 .1.6.10-19 C. 1.45.1010 cm-3 . 50.10-4 cm / 2. 10-6 s= 0.06nA In this case, the most common in pn-PDs, the diffusion contribution Id dominates and the ideality factor (pages123-124) is unity.

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Question 5-7. Compare the typical circuits of PD transimpedance-preamplifiers for high frequency operation (Fig.5-3.6) and number out advantages and disadvantages of each. of the schematics. Answ: The 3-BJT schematic (Fig.5-3.6a) is perhaps the best to combine low-noise and good frequency performance. It fully exploits the fT of the active device, BJT with a high transconductance, only it has a modest low-frequency noise (at high feedback R) because of the non-negligible base current (Fig.5-3.9). When using an input-FET follower (Fig.5-3.6b), we get a better noise-performance at high feedback R, but the maximum attainable frequency is somehow smaller (see Fig.5-3.7). The schematic of Fig.5-3.6c yields a theoretical better performance, but requires one extra active component and more wiring that can override, if not carefully implemented, the bandwidth benefit. Question 5.8 Which photodiodes can be used for detection in the 2ndand 3rdwindows of optical fibers ? Which are the advantages of APDs respect to common pin-photodiodes, and in which range of signal frequency? Answ: In photodiodes for 2nd and 3rd window (around λ=1300 and 1550 nm) we use materials such as the ternary compounds gallium-aluminum-arsenide, grown on InP (indium phosphide) substrate. This allows to fabricate a pin-heterostructure, which is advantageous because radiation is relesed in the deplation layer and not in the entrance, undepleted, layer beneath the input window. As the material has a low α/β ratio, it is not well suited for APD-photodiodes, because the attainable gain is to be traded with the Fano excess-noise factor. At

32

the highest frequencies (e.g. 20-40 Gbit/s), therefore, a pin-photodiode is still preferred to APD. Problem 5-9 A transimpedance PD-preamplifier uses a bipolar-input Op-Amp with GB=10 MHz, Gdc= 10 6, Vos=0.5 mV, Ip+,-=-100 nA, and a photodiode with I0=1 nA, Rp=10 MΩ. Find the quiescent output voltage at P=0 and the maximum radiant power handled without saturation for several reasonable choices of the feedback resistance R. Assuming Ci=5 pF, find the high-frequency cutoff. Assuming also a voltage noise 30nV/√Hz, and a current noise 1pA/√Hz, find the critical current Iph0 What will improve and what will worsen if we use an FET-input Op-Amp? Answ: From Eq.5.40, we get the output quiescent voltage as: Vu0= -(I0-Ip-) R+Vos(1+R/Rp) Taking the sequence of values R=10k, 100k, 1M, 10MΩ, we get: Vu0= 0.1µA. (10k,100k,1M,10M)+0.5mV. [1+(0.001,0.01,0.1,1)] = 1.5mV, 10.5mV,100mV, 1V (the voltage offset is significant for the smallest value R=10k, then, it is dominated by the current term) As the signal swing is σPR, the maximum power handled before saturation at the supply voltage ±Vbb (or at a Vbb

’ just 1-2 V less Vbb) is: Pmax= Vbb

’ / σPR. Taking Vbb

’=10V and σ=1A/W, we get:

33

Pmax= 10 /(10k,100k,1M,10M)= 1mW, 0.1mW,10µW,1µW. About the frequency cutoff, using Eq.5.45: f2 =√GB fR where fR =1/2πRCi = 2.9M, 0.29M, 29k, 2.9kHz is the RC cutoff, we get the sequence: f2 = 5.4M, 1.7M, 540k, 170 kHz. The damping factor is, from Eq.5: χ=1/2 [(CR/(Ci+CR) √GB/fR + √ fR/GB] and, assuming CR=0.5pF, it is computed as : χ = 0.35, 0.46, 0.87, 2.6. (only for R=10k we have a modest 3-dB overshoot). Finally, from Eq.5.50’ we get for the noise break-point Iph0: Iph0= 2kT/eR+IA(1+RA

2/R2) +I0 where RA= vA/iA= 30(nV/√Hz) /1(pA/√Hz) = 30 kΩ and IA is the dc equivalent of shot-noise iA, i.e., iA

2= 2eIAB, whence IA=1. 10-24 / 2.1.610-19= 3µA. Inserting in previous equation yields: Iph0= (50mV)/R + 3µA [1+(30kΩ/R)2] +1nA= = 50mV/(10k,100k,1M,10M)+ 3µA [1+(30k/10k,100k,1M,10M)2] +1nA= = 5+30 µA, 0.5+3.3 µA, 0.05+3 µA, 0.005+3 µA (the last term is negligible in all cases). Thus, the result is: Iph0= 35, 3.8, 3.05, 3.0 µA, or Pph0= 35, 3.8, 3.05, 3.0 µW. With an FET-input Op-Amp, the voltage offset might be larger, say Vos=3-5mV, but the bias current (and its noise) is two or three decade less. So we have: Vu0= Vos=3-5mV The cutoff frequency is, of course, unaffected if we keep the same GB for the op-amp.

34

Also, we may have RA= vA/iA= 15(nV/√Hz) /0.01(pA/√Hz) = 1.5 MΩ and the sequence of break-point currents becomes: Iph0= 12, 0.57µA, 51nA, 6.3nA, thus much better at the highest resistance values than in nthe bipolar-input. Correspondingly, Pph0= 12 µW, 0.57µW, 51nW, 6.3nW. Problem 5-10 What is the best preamplifier scheme for the detection of quasi-dc very minute optical powers? Draw it and assess its performance. Answ: It is the dark-cancellation circuit (Sect.5.3.1.2). This circuit yields a signal: Vu= 2σPR and a quiescent output voltage: Vu0 = 2R Io[Vos/(kT/e)] - RΔIp + Vos Let us design it around a typical FET input Op-Amp (356 series) with Vos=3mV and Ip=200pA. By external compensation of the offset to (nominally) zero, the residual offset error comes from the thermal drift of Vos, typically ≈ mV/°C. We may conservatively assume Vos=25µV (there are units available even with <5µV). In addition, by specimen selection, one can reach ΔIp=5pA. Using a 5-mm2 Si photodiode with I0=1nA, we get from Eq.(5.51) a dark current term Vos/(kT/e) I0 =(25µV/25mV)1nA=1pA. The second term in dominates the third term for R>25µV/5pA=5MΩ. Using a very large resistance R= 500MΩ, we have Vu0=500MΩ.(2.1pA+ 5pA)=3.5 mV. Taking this the zero input power voltage as the minimum detectable power in dc we have: P=3.5 mV/(2σ500MΩ)= 3.5pW at σ=1A/W

35

(and the responsivity of the preamplifier is vu/P=1V/pW). Bandwidth is still given by Eq.(5.45), with 2R in place of R. As the PD capacitance at V≈0 is very large (thousands of pF), the response is limited to the low values (that is, a few Hz or tens of Hz) of interest in instrumentation. Noise is given by Eq.(5.52): vnu2 = 4kTB(2R) + [2e(σP+2Io)B+ iA2](2R)2 +vA2 For a 1-mm2 Si photodiode with Io=200pA and σ=1A/W, and with the above reported data, we get in the small-signal regime and for B=1 Hz: vnu=13µV, whence pni=13 fW (or ≈-110 dBm), a value actually achieved in practice. Problem 5-11 For a PD with a dark current Io=500pA, determine the achievable improvement in bandwidth that we can get by using the equali-zation technique. Is the improvement incidentally dependent on the op-amp parameters or is it likely to be obtained at all times? Answ: From Eq.(5.66) with κ=0.5 we have for the load resistance: R =50mV/300pA.1.5=110 MΩ. A discrete FET amplifier, or a FET-input op-amp may have as voltage and current noises: iA/√B=10 fA/√Hz and vA/√B=15 nV/√Hz, from which we find RA=15nV/10fA =1.5 MΩ. The bandwidth improvement obtained with equalization is therefore: f2/ f1= √[2(1+1/κ)]R/RA= √6.110/1.5 = 180. Note that the noise spectral density (in A/√Hz) corresponding to the dark current I0 is √(2eI0) = 14 fA/√Hz, whence the justification for κ=0.5. In practice, with Ci=5pF we would have an input cutoff frequency:

36

f1=1/2πRCi = 21kHz, and an equalized cutoff frequency: f2= 1/2πRACi=3.8 MHz, while the total current noise in the frequency band 0-f2 is the low-value: iin/√B=25 fA/√Hz. The improvement is almost always obtained, because in practical devices it is R/RA>>1. Indeed, R=50mV/I0 is to be compared to RA=(vA/√B)/(iA/√B) which, in a FET can be written as: (4kT/gm)1/2 / (2eIg)1/2 = (50mV/gmIg)1/2, gm being the FET transconductance and Ig the gate current. Since I0 and Ig are both reverse currents, they have the same order of magnitude at equal areas, while gm>>I0/50mV, whence the assertion.

Problem 5-12 How to select a good PD for the logarithmic conversion preamplifier, that is, one with an ideality factor as close as possible to unity on a wide range of currents? Which dark current (high or low) should be better? Answ: From page 124, we need ni(Dp/LpND)>W/2τ to have the diffusion reverse current dominating over generation-recombination current and hence n=1. Using Eq.5.31, W=√(2εV/eND), we can also express the above condition as: ni(Dp/Lp√ND)> √(εV)/τ√(2e) and this show that the low reverse bias voltage helps satisfying the disequality. The g-r current is usually low (see Prob.5-6) if no defects are deliberately introduced in the material to shorten the minority current lifetime τ and associated cutoff (as instead it is done in diodes and transistors). Ig-r is smaller the more W is shortened or the donor doping is made light.

37

In several commercial photodiodes (e.g., Hamamatsu PD S780) the photocurrent range in which n=1 and linearity to the logarithmic conversion is better than ±1% can span more than 4 decades, from tens nA’s to mA’s. Problem 5-13 Design the width and voltage across the multiplication region of a Si-APD with maximum attainable gain. For a reach-through structure, complete the APD design adding a consideration on the other doped regions. Answ: From Fig.5-4.1, we choose the point at E=1.5.105V/cm, at which α=1.

103 cm-1 and β=1. 101 cm-1. As the ratio α/β allows an optimal gain Mo=100, we get from Fig.5-4.3 a number of average multiplications αL=4.1. From here, L is calculated as L=4.1/1. 103 cm-1=41µm. Therefore, the required voltage across L is V=41µm. 1.5.105V/cm= 615 V. This is a large value indeed, which would probably result in a 1000-V range PD when we add also a thick dissipation layer. Instead, it is better to move to a slightly higher field, e.g. E=1.8.105V/cm at which (Fig.5-4.1) α=3. 103 cm-1 and β=8. 101 cm-1. Then, α/β =40 and we get: αL=3.2, L=11µm, and V= 200 V, a reasonable value. In a reach-through structure, the doping level of the multiplying region does not affect the gain, and will be kept the least possible (for example, 1013 cm-3) so that the other regions are sized without criticality. To do that, we can use again the diagram of Fig.5-2.12 (still applicable except for Cb) and take e.g., a W=10µm dissipation width, that adds 100 V to the total working voltage. Problem 5-14 Calculate the minimum number of photons which can be detected in a short illuminating pulse at a S/N ratio =2.

38

Consider the case of (i) a pin photodiode; (ii) an APD photodiode. Answ: This question is an assessment of how far are PDs from the single-photon performance of PMTs (whose typical SER has σ2/N2=0.25). The best starting point for considerations is perhaps the data of Fig.5-3.8, where the PD front-end performance is shown to approach closely the Johnson noise limit of the feedback resistance R. Here it is: in/√B = 1 pA/√Hz for R =10kΩ, with a slope of half a decade per decade, or, equivalently: in√R/√B = 100 pA. √Ω /√Hz (1) To make the comparison meaningful, we take for the operating pin-PD the typical bandwidth attained by a PMT, B=200 MHz. This bandwidth corresponds to a SER duration FWHM of 0.31/200MHz= =1.55ns (as per Eqs.4.61); moreover, from Fig.5-3.7, the 200MHz requires a feedback resistance value of R=2 kΩ. Thus, we have from (1): in = 100 (pA√Ω /√Hz). √200. 106/√2.103 = 32 nA Worth to note, this is current noise associated to a light pulse detected as a current pulse with a time constant 1/B (single-pole cutoff), i.e.: τ= 1/B = 5 ns. The current value associated to the pulse is then: is = R e/τ =R 1.6.10-19/5.10-9 = R 0.08 nA. Now, letting S/N= is/in = R 0.08 nA/ 32 nA = 2 we can solve R (the number of photoelectrons detectable at S/N=1) as: RPD=2. 32/0.08 = 800 photoelectrons for a pin-PD. For an APD, assuming for example Si with G=100, and an excess factor F=2, the noise Johnson resistance is increased by G2 to a value R= 20 MΩ. Then, we get: in = √2. 100 (pA√Ω/√Hz). √200.106/√20.106 = 0.45 nA Accordingly, at S/N= R 0.08 nA/ 0.45 nA = 2 we get as the minimum detectable number of photoelectrons:

39

RAPD= 2 . 0.45/0.08= 11 photoelectrons (2) Last, if the dark current Id is not small respect to (2kT/e)/G2R=2.5nA, we shall multiply (2) by a factor √Id/2.5nA. Assuming the realistic value Id= 25 nA, this factor is √25/2.5=3.1, and we get the typical APD performance: RAPD, Id lmt= 35 photoelectrons. Problem 5-15 While I was testing electrically a PD, having connected it across the (+) and (-) terminals of a digital voltmeter in my laboratory, I noted that, at dark, the PD voltage reading was 10-15 mV, and not zero as I expected. What is the explanation for that ? Answ: In the normal laboratory environment, you have 60-Hz power lines running above (or close to) you and your PD. A current signal jωCsVmains is fed in your circuit by the stray capacitance Cs of power line. Positive half cycles of current give a positive contribution (with a voltage limited by V<Vcut-in) while negative half cycles find the PD in reverse bias and give a V<0 contribution. The net sum of the two contribution is negative, thus we observe a dc (negative) voltage VPD across a PD working at dark in a normal laboratory environment. Depending on how strong the ac-line perturbation is, values in the range VPD =10 to 100 mV are common in practice. __________________________________________________________

PROBLEMS - CHAPTER 6

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Problem 6-1 A thermistor bolometer employs as the sensing element a bead of metal oxides, assumed round, with a r=0.5-mm radius and 0.05-mm thickness, and with tungsten connection wires 0.1-mm dia., 30-mm long. All is packaged in a glass-frit cylinder, 1 mm dia., 30-mm long. Calculate the time constant τ of the bolometer and its responsivity when read by a Wheatstone bridge fed by a Vbb= 1 V supply. Answ: Thermal resistance by radiation is: 1/Kt,rad = 4σT3A= [4. 5.67 10-12 .(300)3](W/cm-2K-4)A(cm2)=

= 4σT3A= 6.12 . 10-4 A(cm2) or, Kt,rad = 1630 (°C/W) / A (cm2)

in our case, A=2πr2 = 6.28 0.052 = 0.0155 cm2, and Kt,rad = 105 160 °C/W. The conduction contribution through the wires is: Kt,con = (1/κ) L/S, where κ is the thermal conductivity (for tungsten κ =1.74

W/cm°C) and L=3 cm the length, S= πd2/4=0.785.10-4 cm2 the section of wires. Thus we get (for two wires) Kt,con = (1/κ) 2L/S = (1/1.74) . 2. 3 /0.785.10-4 = 44 000 °C/W. Last, the glass tube thermal conduction contribution is Kt,con, env = (1/κ) L/S,

where for glass it is: κ =0.01 W/cm °C, L=3 cm and S= πd2/4=0.785.10-2 cm2. Thus, Kt,env = (1/0.01) . 3 /0.785.10-2 =38 200 °C/W. This numerical example shows that the envelope and wire contributions shall indeed be checked to avoid they become the dominant terms.

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Last, convection thermal exchange can be neglected (the element will be mounted in a vacuum envelope). Making the parallel combination of the above terms gives the total thermal resistance as: Kt, = Kt,rad // Kt,con // Kt,env =105 160 // 44 000 // 38 200 = 17 100 °C/W. The thermal capacitance of the sensor element is given by: Ct,= cv V where the specific heat is cv=0.5 J/°C. cm3 for a typical

semiconductor oxide mixture, and V=πr2.t is the element volume. Then, Ct,= 0.5 . 3.14 0.052. 0.005= 1.96 10-5 J/°C. Combining, we find the thermal time constant as: τ = Kt Ct = 17 100 . 1.96 10-5= 0.33 s a rather large value indeed, that however may be acceptable in several measurement applications. The signal developed in a Weathstone bridge by the bolometer is: ΔV= ΔR / 2R Vbb = α/2 Vbb ΔT where α = ΔR/R ΔT is the temperature coefficient of the thermistor (typ.α= -6%/°C). With this, we get: ΔV= 0.03 ΔT (V/°C). Using the thermal resistance value Kt,= 17 100 °C/W, we obtain for the responsivity: R = ΔV/ P = 0.03. 17 100 = 500 V/W (or 0.5 mV/µW). Problem 6-2 Consider the Vanadium Oxide thermistor bolometer of Fig.9-23.Here, we take a side W=100 µm wide, a thickness T=5 µm, and Alumi-num electrodes be 150 µm long and have a section of 15x 2 µm.

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Find the time constant τ of the bolometer and its responsivity when read by a Wheatstone bridge fed by a Vbb= 1 V supply. Answ: Repeating the calculation of Prob.6-1 we have: Kt,rad = 1630 (°C/W) / A (cm2) , A=W2= 0.012 cm2

then: Kt,rad = 1.6 107 (°C/W) The conduction resistance through the wires is: Kt,con = (1/κ) L/S = (1/ 2.38(W/cm°C)) 0.015(cm)/(15.10-4 . 2.10-4

(cm2))

= 2.1 104 (°C/W) which is prevalent respect to Kt,rad. Moreover, if the VOx film is under vacuum, Kt,rad can be disregarded. Thermal capacitance is: Ct,= cv V = 0.7 (J/°C.cm3)

. 0.012(cm2)

. 0.0005(cm)= 3.5 10-8 J/°C. The time constant results as: τ= Kt Ct,= 2.1 104 3.5 10-8 = 0.735 ms. The signal developed in a bridge incorporating the bolometer is: ΔV= ΔR / 2R Vbb = α/2 Vbb ΔT= 0.04/2 Vbb ΔT = 0.02ΔT (V/°C) The corresponding responsivity, given the thermal resistance Kt,= 2.1 104 °C/W, is: R = ΔV/ P = 0.02 . 2.1 104 = 420 V/W or, 0.42 mV/µW.

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Question 6-3 Does it make sense to fabricate an infrared detector with a detectivity surpassing the DBLIP value? Can e detector operate at a D* larger than this limit? Answ: The DBLIP limit comes about because of the shot noise associated with the blackbody background at which the detector is aimed, at a given temperature – usually taken T=300 K as in Fig.6-5. If we make a detector with a Ddet larger than DBLIP, when aimed to a thermal scene at T=300 K, the detector effective detectivity will be D*= (Ddet

-2+ DBLIP-2) –1/2 because of the quadratic

composition rule of noise. Thus, it is useless to push the detectivity (of the detector alone) beyond the BLIP limit appropriate to the background temperature of the scene being surveyed. Problem 6-4 Find the minimum detectable temperature difference that can be resolved by: i) a HgCdTe detector with D*= 2.5 1010 and ii), by a thermal detector with D*=4 108 . In both cases, assume a bandwidth B=10 Hz and α=2 mrad as the angle presented by the target to the detector. Answ: We may start from Eq.6.8 to compute exactly the signal collected by the two detectors. However, using Eq.6.15 provides a good approximation as well in a more straightforward calculation. Rewriting Eq.6.15: NEDT = 2kT2 (Dblip

2/D**) (1/ηNA) √(B/A) where we can take Dblip=6 1010 (see Fig.6-5 for λ=8-12µm) and let η=1. As a realistic choice for the objective lens, we may take a focal length F=50

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mm and a NA=0.25 (corresponding to a diameter D= 2F. NA = 25 mm). The detector side is then w=F tanα= 100 µm, or A= 1. 10-4 cm. Thus we get: NEDT = 2. 1.38 10-23. 3002 (621010 /2.5) (1/0.25) √(10/1. 10-4) = 0.00054 °C (HgCdTe) and NEDT = 2. 1.38 10-23 . 3002 (621010 /0.04) (1/0.25) √(10/1. 10-4) = 0.034 °C (bolometer) Taking realistically the minimum detectable temperature difference as ΔTmin=3 NEDT, we have from the above results: ΔTmin(HgCdTe)= 0.0016 °C, ΔTmin(bolometer)= 0.10 °C. Problem 6-5 Consider a thermovision using as a detector: i) a scanned-HgCdTe detector; ii) a focal-plane-array bolometer. Assuming the data of Probl.6-4, find the minimum detectable temperature difference for a frame at 30 Hz frame-rate made by a500x500-pixel frame. Answ: Passing from the single-point measurement to the 500x500 frame measurement, we change the bandwidth from 10 Hz to: 30x500x500= 0.75 MHz [case i)] 30 Hz [case ii)] thus, the NEDT become: NEDT HgCdTe = 0.00054 °C √0.75 M/10) = 0.15 °C NEDT bolometer = 0.034 °C √30/10) = 0.059 °C This example points how the advantage in using FPA-detectors, even of low detectivity, respect to a better quality one single-point detector, which is forced to operate by scanning the field of view, thus losing the position because of the bandwidth increase.

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Accounting the electronic preamplifier noise by means of a x2 worsening factor, and a factor x3 for a fair S/N, the minimum detectable temperature difference is: ΔTmin(HgCdTe)= 0.9 °C, ΔTmin(bolometer)= 0.35 °C. Problem 6-6 With reference to the discussion on page 227, what happens if a blackbody-radiant ΔT measurement is carried out in a surrounding with a non-uniform temperature distribution? Answ: The total radiance rtot(λ,T) is coincident to the blackbody radiance irrespective of the target emissivity ε if the target is surrounded by an ambient at the same temperature T (see Eq.6.16). But, for outdoor operation, for example, this assumption is no more valid. The clear sky (and also the cloudy sky) has a blackbody temperature (in the IR) much less than 300 K, thus the term (1-ε) rtot(λ,T) that we expected to compensate for the decreased target radiance εrtot(λ,T), is much less than needed. Therefore, the thermal contrast is no more simply εΔT and, to calculate ΔT, a more accurate description of the surrounding temperature and emissivity is required. ________________________________________________________________

PROBLEMS - CHAPTER 7

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Problem 7-1 A silicon solar cell has an area A=100 cm2. Find the optimum working point for a solar illumination at AM1.5 and the electrical available power, assuming an ideality factor n=1.1 and an average spectral sensitivity σave = 0.9 A/W. Answ: As we need I 00, we assume a dark-current density J0 = 0.5 nA/mm2 and find (Eq.7.3): I 00 = J0 A exp Eg/VT = 0.5 nA/mm2 .104 mm2. exp 1.12/0.026 = 6. 1012 A. From the data in Fig.7-3, the short-circuit photogenerated current is: Iph= A σave E1.5AM= 10-2 m2 .0.9 (A/W) . 807 (W/m2)= 7.26 A. The open-circuit voltage is then (Eq.7.4’): Voc = 1.12 V+ 1.1 . 26 mV. (-27.4) = 0.346 V and the optimum working voltage and current are, from Eqs.7.6, 7.7: V(opt)= 0.346 V –0.0282 . ln(1+0.346/0.0282) = 0.273 V I(opt)= 7.26 /(1+0.0282/0.346) = 6.71 A. Thus, the electrical power supplied by the cell is Pe= V(opt)I(opt) = 0.273 . 6.71 = 1.83 W, and, considering that the incoming power is: PS= A E1.5AM= 8.07 W the (internal) conversion efficiency turns out to be: ηc = 1.83/8.07= 22.7 %

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Problem 7-2 If we divide the single 100-cm2 cell in 100 individual 1- cm2 cells and assemble them in series, what is the supplied current and voltage? Does the series combination offer any advantage? Answ: As I 00 and Iph of the individual cell scales down of the same factor (the number of cell, N=100), we have the same Voc, V(opt) and ηc while currents become N times smaller. Then, for the series of N cells we have: V(opt)= N.0.273 V= 27.3 V I(opt)= 6.71 /N = 67.1 mA, and again ηc = 1.83/8.07= 22.7 %. Note that, for the series assembly, current and voltage assume values much more reasonable to use than the single cell. Also, the contact resistance becomes easier to manage in the series connection. Problem 7-3 What would be changed if we use the cell of Probl.7-2 in a syatem giving a concentration factor C=100 ? Answ: Roughly, voltage is about the same while current increse by C. But, more accurately, we have to repeat the calculation of Probl.7-1 to unveil that the more favourable ratio Iph /I 00 results in an increased ηc . Considering the single 1-cm2

cell we have: I 00 = J0 A exp Eg/VT = 0.5 nA/mm2 .102 mm2. exp 1.12/0.026 = 6. 1010 A. The photogenerated short-circuit current is: Iph= C A σave E1.5AM= 102 10-4 m2 .0.9 (A/W) . 807 (W/m2)= 7.26 A,

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and the open-circuit voltage is: Voc = 1.12 V+ 1.1 . 26 mV. (-22.8) = 0.476 V and the optimum working voltage and current are then: V(opt)= 0.476 V –0.0282 . ln(1+0.476/0.0282) = 0.415 V I(opt)= 7.26 /(1+0.0282/0.476) = 6.85 A. So, the electrical power supplied by the single cell is Pe= V(opt)I(opt) = 0.415 . 6.85 = 2.84 W. Considering that the incoming power per cell is PS= C A E1.5AM= 8.07 W the (internal) conversion efficiency turns out to be: ηc = 2.84/8.07= 35.2 % The system would accordingly supply a 41.5-V voltage and a 6.85-A current, optimal for a 41.5/6.85= 6.05-Ω load. Of course, if the cells are not exactly equal in performance, the one supplying the lowest current would determine the current available from the cell-series. This can be accounted for by a reduction of 0.8-0.9 of efficiency (and current). Also, the interconnection series resistance may become of importance (see Fig.7-9). Problem 7-4 Is the concentration optics of Probl.7-3 realizable with a single lens or does it requires a multiple-element arrangement? Answ: To have a real C=100 concentration factor on the cell, we need a lens area AL= Acell C/ηlens, where ηlens is the optical efficiency of the lens (ηlens =0.75 typically). For the whole 100-1 cm2 cell assembly we then need AL = 100 cm2. 100/0.75 = 1.33 m2 =(π/4) D2

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or, a lens with a D= 1.3-m diameter. Moreover, the required focal length is: F = 10 cm / θsun= 0.1m/ 9 mrad = 11.1 m Thus, the lens assembly would be rather bulky to move (the solar disk shall be tracked in the concentration system). Trying a more reasonable focal length, i.e. F=1.1 m, the covered cell size becomes dcell = F θsun= 1 cm (the single cell). The required lens diameter is DL = 13 cm. In this case, however, we have to interconnect electrically a lot of cells in the system and the mechanical layout is more complicate. Alternatively, we may anyway use a D= 1.3-m diameter to cover the full 100-cm2 cell array, and use a focal length multiplier. Choosing, e.g., a Fresnel lens with F= 2m, we put a negative lens (Barlow-lens) at a 10-cm distance in front of the array. With the proper focal length (typ. –30 cm), the combination has the desired focal length F = 11.1 m. Still another possibility is to use a truncated pyramid in place of the Barlow-lens. This element works on total reflection on the lateral walls, and enlarges the sun image to the exit base format (and also adds a better radiant uniformity on the cells). Problem 7-5 In an ideal multispectral photovoltaic system, could the 100% efficiency be theoretically reached? Answ: As the sun is a blackbody at 6000 K-temperature, the theoretical thermodynamic efficiency shall obey the Carnot theorem, and accordingly it is: η = (T2-T1) /T2 = (6000-300)/6000 = 95% where T1 is the temperature (assumed 300 K) of the cell.

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Problem 7-6 A thermal collector panel has an efficiency ηt_=60% and is able to supply hot water at T=77°C. A silicon photovoltaic cell has a ηcell_=10%. Both work at 27°C (300 K). Which is best from the thermodynamical conversion point of view? Answ: Actually, the solar collector panel has the heat quantity as an output. This can be converted to free energy with an efficiency: η= ηt (T2-T1) /T2 = 0.6 (350-300)/300 = 10% and supplies a fraction Q=0.6E of the energy received by the sun. Equivalently, the photovoltaic cell supplies free energy W that can be used to operate a reverse-refrigeration cycle supplying a quantity of heat Q given by: Q = W T2/(T2-T1) = 300 /(350-300) = 6 W = 6 ηcell E = 0.6 E. The systems are therefore equally good despite the different η. __________________________________________________________

PROBLEMS CHAPTER 8 Problem 8-1 A weak signal, P=1-nW (or, respectively: 1-pW, 1-µW) is detected by a coherent homodyne scheme, using a silicon photodiode with σ=1A/W, Idark= 0.1 µA, terminated on a 1-kΩ load. Find: (i) the minimum local oscillator strength that is required; (ii) the signal amplitude that is obtained, as compared to direct detection; (iii) the S/N ratio and its improvement respect to direct detection. Assume µ=1 and B=1 GHz.

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Answ: The condition IOL >> Idark + I + 2kT/eR yields for the local oscillator detected current IOL: IOL >> 0.1µA + 1pA +50mV/1kΩ = 50 µA (note that it is the third term, associated to the load resistance, to dominate). By satisfying the disequality with a multiplicative factor 10 and as we have σ=1A/W, we get for the required local oscillator power: POL = 500 µW (and IOL = 500 µA). For P=1 fW and 1 nW, the local oscillator power remains unchanged. The signal amplitude is: RσP = 1 µV (resp.: 1 nV, 1mV) (direct detection), RσPG = RI (1+2√IOL /I) (coherent detection) = 1µV. [1+2√(5.105)] = 1.4 mV (resp: 44µV, 44 mV). The S/N ratio is: (S/N)2 = 4I/2eB = 4. 10-9/3.2. 10-19. 109 = 12.5 (resp: 0.0125, 1.25 103) (coherent detection) and (S/N)2= I2/2eB(0.1µA + 1pA +50mV/1kΩ) = 10-18/3.2 . 10-19. 109. 50 10-6= 0.6 10-4 (resp: 0.6.10-7, 60) (direct detection) Note that, with the values assumed in this example, neither the coherent nor the direct detection are in the reach of the 1-pW signal at the given bandwidth. About the 1-nW signal, coherent detection becomes adequate while direct detection is not yet. Problem 8-2 A phase-modulated signal, of the type: E0 exp i[ωt+φ+2ks(t)] is detected by a homodyne coherent scheme. Write the expression of the output signal and guess the likely regime of detection.

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Answ: The homodyne detection scheme gives: I = I0 + 2√I0IS cos[φ-φlo+2ks(t)] Apart from a dc term I0 and a constant phase φ-φlo, the coherent detection of a phase-modulated signal supplies as a result the cosine function of the driving term s(t) of the phase-modulation. This result describes what exactly happens in a conventional interfero-meter: a laser beam is splitted in a beam divider, one portion of it goes to a distant target and comes back with a phase 2ks (optical pathlength by k=2π/λ). Upon recombination with the fixed-phase, reference beam, a truly homodyne detection scheme is clearly recognized. As a consequence, interferometers are most likely to work in the quantum-limit regime of detection (unless the reference beam power is kept unusually weak). Problem 8-3 How large is the signal associated with a N=10-photon-per-bit rate, the one giving a BER=10-9? Consider bit rates R= 1Mbit/s, 1Gbit/s. Assume λ=1.24 µm, and a spectral sensitivity σ= 0.8 A/W. What is the quantum noise and the S/N ratio of this signal ? Answ: The photodetected signal is: I = e N R = 1.6 . 10-19. 10 .106 = 1.6 pA @ 1Mbit/s, and = 1.6 nA @ 1Gbit/s; the corresponding powers, being hν=1eV @ λ=1.24 µm are: P = hν N R = Ι / σ = 2pW and 2 nW, respectively. Quantum noise is (note that it is B=R/2 from Nyquist theorem): in = [2eI B]1/2= [2eI R/2]1/2= [1.6.10-19.1.6.10-12 .106]1/2

= 0.51 pA @ 1Mbit/s, and = 5.1 nW @ 1Gbit/s.

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From these numbers, the signal to noise is found back as: (S/N)2 =(I /in)

2= (1.6/0.51)2 = 10 (in both cases). Problem 8-4 Given the data of Probl.8-1, what is the theoretical minimum µ that would be tolerable to still have an advantage in using coherent versus direct detection? Answ: By comparing the S/N ratio of the two cases, we get the break-point coherence factor µ as: µ = (S/N)dir

/(S/N)coh = √(0.6 10-4 /12.5) = 0.22 10-2 (P=1-nW)

and √0.6.10-7/0.0125 = 0.22 10-2 (P= 1 pW) √60 /1.25 103 = 0.22 10-2 (P=1 µW) of course, these are always the same value because the noise is the same in all cases. Problem 8-5 A coherent detection uses a non-ideal local oscillator, that has: a circular polarization instead of the linear one of the signal, a mode size (Gaussian) twice as wide as that of the signal, and a phase rms deviation σΦ= 0.5 rad. Calculate the total coherence factor µ and the S/N penalty. Answ: We have: µpol = [1,0] × [1/√2, 1/√2] = 1/√2= 0.707; µsp = ∫ gauss (r,w0) gauss (r,2w0) 2πrdr = 1/√5= 0.48

µΦ = 1 - σΦ2/2 = 0.75

and collecting the terms µ = µΦ µsp µpol = 0.707 0.48 0.75 = 0.24 S/N penalty is µ = 0.24 = 6.2 dB

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Problem 8-6 Evaluate the number of photons per bit to achieve a BER=10-9 in a SOPSK (state-of-polarization-shift-keying) transmission with homodyne detection, where the ‘1’ is transmitted with a given state of polarization (e.g., a linear one) polarization (for example, same as that of the local oscillator) and the ‘0’ with the polarization state orthogonal to that of the ‘1’. Answ: The ‘1’ signal is N=2√N0NS , the ‘0’ signal is zero, so that the optimum threshold is S=N/2. Noise σN on both levels is √N0 (see also p.258). Thus we have: BER = erfc S/σN = erfc √NS and we obtain NS =36 photons per bit. Problem 8-7 Again consider a SOPSK coherent detection, and assume using a balanced detector with an input Glan-beamsplitter dividing the incoming polarization. What is the number of photons per bit to achieve a BER=10-9 ? Answ: In this case, each photodiode will detect a symbol, the ‘0’ and the ‘1’. The difference signal at the output of the balance detector swings from –N to +N and is twice as large the individual ones, ΔN=4√N0NS, while noise on the –N to +N levels is always √N0. Optimum threshold is S= ΔN/2. Thus we have: BER = erfc S/σN = erfc 2√NS and accordingly we get NS =9 photons per bit. Problem 8-8 Should optical receivers for a 100 Gbit/s transmission rate become developed and available for system use, which would be their expected sensitivity (in µW or dBm) ?

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Answ: Looking the practical results reported in the diagram of Fig.8-4 and by extrapolating the bit-rate to 100 Gbit/s, we may expect that the sensitivity of actual receivers should be: 3 - 5 µW (or -25 to –23 dBm) for direct detection 0.3–0.5 µW (or -35 to –33 dBm) for coherent detection ≈ 5 µW (or –23 dBm) for optical preamplified detection Problem 8-9 An optical amplifier with a gain G=103 is intended for use as a preamplifier at λ=1500 nm. Calculate: the output ASE, the input-equivalent ASE, the NEP (noise-equivalent- input) for the full bandwidth available (40 nm) and for a B=10 GHz electrical bandwidth. What is changed if we filter the amplifier output with a narrowband filter with Δλ=0.5 nm ? Answ: The ASE for a single polarization state is, from Eq.(8.25): ASEout = nsp hν (G-1) Δν where nsp,the inversion factor, that can be taken nsp =0.9 as a typical value. For both polarizations, or, when no polarization selection is performed, we have ASEout/tot=2 ASEout and accordingly ASEout/tot = 2 . 0.9 .1.6 10-19. (1.24/1.55) 103. (40/1500) 200 1012 (in the above expression, we compute hν as 1eV by the ratio of wavelength λ to 1.24 µm, and Δν as the fractional wavelength Δλ /λ by the optical frequency 200 THz that corresponds to λ =1500nm) So we have: ASEout/tot= 1.8 .1.28 10-19.103. 5.33 1012= 1.23 mW This is the dc optical power at the output of the amplifier. Referred to the input, we have ASEin/tot= ASEout/tot/G = 1.23 µW and the associated noise for a bandwidth B=10GHz is (Eq.8.26): NEPASE

2 = 2 hν ASEin/tot B = 1.28 10-19. 1.23 10-6 . 1010 = 1.57 10-15

or, NEPASE = 39.6 nW.

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To get the total noise, we shall add the excess term (Fig.8-11), F≈2 in a good amplifier and for a not too small signal PS>ASEin/tot. Thus, we may conclude that our optical amplifier has, for PS>1.23 µW either: (i) a quantum-noise performance with excess factor F, or: (ii) a noise 2.

39.6 nW = 80 nW (or equivalently –41 dBm) – whichever is larger. If an optical filter with Δλ=0.5 nm is used, the ASE decreases by a factor 0.5/40=1/80 [so that ASEin/tot=15 nW], and the NEP by 1/√80≈1/9 (reaching –50 dBm). This figures (compare with data in Fig.8-4) are very interesting for applications, albeit limited to a specific wavelength of operation. Problem 8-10 A signal from a narrow-line DBR laser around 1500 nmn is fed in another similar laser, detuned of Δf=20 GHz from the first..The laser gain is α=c 200 cm-1, loss is Γ=c10 cm-1, cavity length is L=200 µm, mirror facets reflectivity is R=0.3, power emitted is Pout= 1mW. Calculate the heterodyne injection gain and the signal amplitude for a detected power PS=1pW. Answ: The power in the laser cavity is P00= Pout/(1-R)=1.4mW; mirror field transmittance is T=√(1-R)=0.84. Using Eq.8.39, we can compute the injection gain as: G = 2√[P00/PS] [Tc/2L(α−Γ)] = 2 [1.4. 10-3/10-12] 1/2 0.84/(2 0.02 . 190) = 7.48. 104. 0.11= 8200 If a photodiode is place on the rear mirror of the laser where 1mW is emitted as well as from the front mirror, the photodetected current is (assuming σ=1A/W and after Eq.8.38): I0 = I00 + G IS cos (Ω00-ΩS)t = 1mA[dc] + 8200 1pA[20 GHz] = dc term + 8.2 nA at signal frequency.

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As a verification, we shall check that the signal frequency Δf is within the bandwidth of injection response. In the laser diode, the gain linewidth is much larger than Δf (usually THz’s), while the cavity linewidth is given by Δfline=f/Q, Q being the quality factor of the cavity. Calculating Q = 4πL/λ(1-R) = 12.56.0.02/0.00015.0.7= 2300 we get: Δfline= 200 THz/2300= 870 GHz, a value much larger than Δf. Problem 8-11 In a heterodyne or homodyne injection detection, what is the minimum signal that can be detected by a 1-mW semiconductor laser with the parameters of Probl.8-10 ? Evaluate it both for a small detuned signal Δf injected in the laser and for a weak echo at the same frequency of the laser. Answ: As injection detection, no matter if heterodyne or homodyne, is a coherent detection albeit with a penalty factor not much less than unity, the detectable signal is the same as for coherent detection. Typical values reported in experiments (see References on page 293-94), both with He-Ne and GaAlAs and InGaAsP diodes are, e.g.: for heterodyne injection: - pW’s for small/moderate bandwidth (<1MHz) - nW’s for high bandwidth (≈1GHz) for homodyne injection: - -90 dB of attenuation (respect to P00) at small bandwidth (≈1kHz) - -50 dB of attenuation, high bandwidth (≈1GHz)

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Problem 8-12 Could the QND scheme of detection be implemented with a pump at the same frequency of the signal? How shall the setup of Fig.8-16 be modified? Answ: Entering in the setup of Fig.8-16 with orthogonal pump and signal polarizations, and using a Glan polarization s in place of the IF filters, we can indeed superpose pump and signal with virtually no loss. However, we then need an index of refraction nonlinearity n2 associated with orthogonal axes, that is a material with non-vanishing nonlinear permettivity χ(3)

2221 along the signal (1) and pump(2) polarization axes. Problem 8-13 What is the squeezing factor we can obtain from a diode laser converting the injected electrons into photons with an efficiency η=0.95 ? What is the squeezing factor when this laser signal is launched in a fiber? Answ: Ideally, if all of the electrons were converted into photons, we would have F=0. Whatever the nature, a loss ε=1-η is accounted for by Eq.8-56: F’ = 1- ε (1- F) = 1- (1-η) (1- F) and, introducing in this equation F=0, η=0.05, we get F’= 0.05. If we attempt to launch in a fiber such a squeezed signal, as the launching efficiency laser-to-fiber is seldom ηlaunch>80%, we would immediately get a degradation, again from the above equation, to a squeezing factor Flaunch: Flaunch = 1- ηlaunch (1- F’) = 1 – 0.8 0.95 = 0.24. The same result would supply the observed squeezing factor if we attempt to detect the laser emission with a photodiode having a quantum efficiency η=80%.

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Problem 8-14 Why the squeezed radiation actually improves the interferometer readout (pages 286-287) while it does not help improving in data transmission and detection ? Answ: Because these are physically different situations. In transmission, information is carried by the signal, and a squeezed local oscillator can’t help [Eq.(8.57)] because of the cross-multiplication of signal amplitude and squeezing factor. As the information limit due to photon statistics is already bounded at the transmission side, it is quire expectable that we cannot overcome the quantum limit of detection by squeezed states. Only a squeezed source can improve, but with the severe limitation of propagation attenuation. In an interferometer, opposedly, information is internal to the optical path and thus a cleaner, squeezed signal can indeed help to read it with less noise. ______________________________________________________________

PROBLEMS CHAPTER 9 Problem 9-1 With a Sb2S3 vidicon tube and a F=50 mm, D= 22 mm objective lens, we pickup a scene illuminated at Esc=500 lux. What is the target current ? What is the S/N ratio of the image ? Assume an average scene diffusivity δ=0.3, a photoconductive gain G=10, a target load Rt=10 kΩ and an electron beam current ipe=1µA. Answ: The F-number of the lens is F/=22/50=0.44; the typical luminous sensitivity of the Sb2S3 target is σl = 20 nA/lux (see Fig.9-9; this value includes the G=10 gain). From Eq.(9.5), written in terms of luminous unit-area sensitivity σl, we have: it = σl δ Esc / 4(F/2)

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= 20 (nA/lux). 0.3 .500 (lux) / 4.(2.2)2 = 150 nA Assuming the typical value of dark current I0=10nA, and κ=3,the noise (Eq.9.4) is given by: σit

2 = 2e [1+2(M-1) 2] (it +I0)B + 2eκ ipeB +4kTB/Rt = 2e{[1+2(M-1) 2] (it +I0)+ κ ipe +50mV/Rt }B = 3.2 10-19 {[1+2.9 2] 160.10-9 + 3.10-6 +50mV/10kΩ}.5 106 = 3.2 10-19 {162 .0.15 10-6+3.10-6+ 5.10-6}.5 106 (incidentally, comparing the three terms reveals that shot noise is the dominant term). Going on the calculation, σit

2 = 3.2 10-19 {162 .0.15 10-6+3.10-6+ 5.10-6}.5 106 = 3.2 10-19 {34 .10-6} 5.106= 54 10-18 whence σit = 7.4 nA; the S/N is finally given by: S/N = 150 nA / 7.4 nA = 20 (or, 26 dB, a fair value for an image pickup).

Problem 9-2 Consider a 3-Φ Si-CCD with 5 µm cell size, with an acceptor doping level NA=1014 cm-3, and a dark current per pixel I0=0.01 pA. What is the clock driving voltage required? What is the saturation level and the dynamic range? What is the signal for a 100 lux illumination on the CCD?

Answ: Taking a field VG/wox =5.105 V/cm, a value close to the maximum set by oxide dielectric strength, Eq.(9.9) gives the stored charge at saturation as: Qmax = εox (VG/wox)= 3.9. 8.86.10-14 (F/cm).5.105 V/cm = 1.7.10-7 Ccm-2, or, in terms of the number of electrons, Nmax = Qmax / e = 1.7.10-7 Ccm-2/ 1.6.10-19 C = 1.1.1012 cm-2. For the elemental cell, of area A=5x5 (µm)2= 2.5 10-7 cm2, the saturation charge becomes: Nmax.A = 2.75 .105 electrons

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Compared to the dark electrons collected in a T=20ms frame, Ndark = I0T/e = 0.01 10-12 A 20 10-3 s /1.6.10-19 C = 1.6 .103 electrons this amounts to a dynamic range of the video signal Nmax /Ndark = 2.75 .105 / 1.6 .103 = 170 The required amplitude of clock voltages is (Eq.9.9): VG=Qmaxwox/εox, that, for a typical value wox=100 nm yields: VG = 2.10-7 Ccm-2 10-5 cm / 3.9 8.86.10-14 F/cm = 5.8 V Correspondingly, from Eq.(9.7) the well extension is found as: W = [ 2 εs Φs / e NA]1/2

= [ 2 .11.9 .5.8 .8.86 10-14 F/cm /1.6.10-19 C.1014 cm-3]1/2 = 8.7 µm. Last, the signal for a 10-lux illumination on the CCD is: iout = η σl A Esc where η =2/3 is the fill-factor, and σl can be estimated (Fig.9-15) assuming a 0.3A/W average radiant sensitivity and a spectral content of 100 lm/W. Thus, iout = 0.66 0.3A/W .2.5 10-11 m2 .10 lux /100 lm/W = 0.5.10-12 A This current yields, in a frame period, Nout = ioutT/e =0.5.10-12 .20.10-3 s /1.6.10-19 C = 6.104 electrons a value well within the dynamic range. Problem 9-3 Other things being equal (frame rate, number of pixels, MTF, etc.) would you prefer a smaller or a larger size CCD ? Answ: A smaller CCD is preferable, of course, because it requires a smaller objective lens. The difficulty in reducing the CCD size is first, of course, a technological problem.

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If size is scaled down by a factor, say K, we have that: (i) signal and dark current both decrease by K, and the dark-limited S/N ratio decreases by √K; (ii) transfer efficiency improves a little, as the MTF does; (iii) saturation charge increases by K. Problem 9-4What is the best theoretical MTF for a 2-on-3 phase integratiing CCD at Nyquist frequency ? Answ: Even if the cells are tightly spaced, when integration on the frame period is on 2 out of 3 cells, the relative filling is η=2/3=0.66. From Fig.9-17, accordingly, we have MTF≈0.82. Problem 9-5 An electro-optical system is made by the cascade of two subsystems, each with a gaussian MTF and a limiting resolution of 60 cycle/mm. What is the limiting resolution of the whole system ? Answ: Writing the subsystem MTF as: MTF (k) = exp – k2 /2k0

2 where k0 is the rms of the gaussian, we have: exp – k2 /2k0

2= 0.03 at k=60 cyc/deg, or k2 = -2k0

2 log 0.03 The cascade has: MTF (k) = MTF1(k) MTF2(k) = = exp – k2 /k0

2 To find the MTF, we equate this quantity to 0.03, thus obtaining: k2 = - k0

2 log 0.03 or k = √(-k0

2 log 0.03) =(1/√2) √(-2k02 log 0.03)

=(1/√2) 60 cyc/deg = 42.5 cyc/deg

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Problem 9-6What is the typical gain required to an image intensifier to enable seeing of a scene at the dimmest natural illumination? Answ: The smallest level of natural illuminance is 10-4 lux (see page 362). From the Blackwell’s diagram (Fig.A3-1), we can see that we need a backgroung brilliance of about 10 nit to get a good photopic vision. Thus, the required brilliance gain is: G = 10 nit /(1/π)10-4 lux = 3 105. Problem 9-7 A 1’’ streak-camera tube has a screen limiting resolution of 50 lp/mm. The deflection plates have a sweep conversion factor of 80 V/mm. What is the time resolution attainable? Answ: On the 1’’ output screen, we get as the total number of lines: N = 25.50 = 1250 lp. The voltage amplitude needed to get the full deflection on the screen is V = 80.25 = 2000V. If this signal is a linear ramp with a risetime τ=5 ns, the time resolution corresponding to the single resolved line is accordingly: Δτ = τ / N = 5 ns / 1250 = 4 ps. ________________________________________________________

PROBLEMS APPENDIXES

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PROBLEM A1.1 Given that a 0-m star yields an illumination of E0 = 2.65 10-6

lux, calculate the stellar magnitude of the sun and of the moon. Answ: From Table A1-1, we have that the illuminations supplied by the sun and of the moon are: Esun =130 000 lux, Emoon =0.1 lux, Its stellar magnitude, from Pogson scale, is accordingly: msun = -2.5 Log10 Esun / E0 + m0 = -2.5 Log10 1.3 105 / 2.65 10-6 = -2.5 (11 –0.3) = -26.7 and mmoon = -2.5 Log10 Emoon / E0 + m0 = -2.5 Log10 10-1 / 2.65 10-6 = -2.5 (5 –0.4) = -11.5 PROBLEM A1.2 Calculate the loss in dB of the glass of your window (in the visible). Answ: Assuming nglass=1.5, at each glass/air interface there is a reflection loss: R = [(nglass- nair)/ (nglass- nair)]

2= (0.5/2.5) 2= 4% Transmittance is thus 0.96 and, applying the dB formula (page 363) or reading directly on the nomogram of Fig.A1-4, we have: lossdB = 0.17 dB (1 surface), 0.34 (total) PROBLEM A1.3 For a Si-photodiode, the rated radiant spectral sensitivity is σ=0.7 A/W at λ=850 nm, while the luminous sensitivity is σlum= 5 mA/lm (for a 2850-K lamp). Considering that 1 lm = (1/673)W = 1.48 mW at λ=550 nm, how can the two data be reconciled?

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Answ: As the 2850-K lamp used to define the luminous sensitivity has a rather wide spectral range of emission, the two data are not in contrast . By interpolation of the data presented in Fig.A1-5, we can estimate that the 2850-K blackbody has a spectral emission (at 50% points) extending from λ=0.50 to 3.0 µm. As the Si-PD can be assumed to have a cutoff at λ= 1.0 µm, the spectral width of detection is Δλσ= 500 nm, while it is ΔλV= 100 nm (Fig.A1-3). Eq. A1.15 gives accordingly (as a first approximation): σlum= (σmax/Kmax)

. Δλσ/ ΔλV

= 0.7 (A/W) / 673 (lm/W) . 500 nm / 100 nm = 5.2 mA/lm PROBLEM A2.1 Going out to take a picture of the newly reported comet – or of the usual night sky - should you prefer to take with you your telescope (having a 20-cm diameter and a 200-cm focal length) or your camera (having an objective with 40-mm diameter and a 80-mm focal length) ? Explain why. Answ: The photographic film, irrespective of its speed or type (B/N, color) will have an exposition threshold dictated by the illuminance you are able to shine on it. Thus, what actually is important, is not just to have the largest diameter (or collecting power) of the optical system, but its numerical aperture. The telescope has NA≈ (20/2)/200= 0.05, while the camera has NA≈ (40/2)/80=0.25 and is therefore better for taking photographs. Of course, from the standpoint of being able to expose the faintest stars or light details. The telescope, if its numerical aperture is sufficient to expose the film, will yield a better angular resolution (stars or light details better separated).

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PROBLEM A2.2. By assuming that sun and planets are blackbodies, calculate the earth and planets temperature by applying the invariance of brilliance (it is:Tsun=6000 K). What happens if the emissivity is not unity ? Answ: For unitary emissivity, the balance of power collected and emitted requires that BsunΩsunAearth= Bsun(πθsun

2) πRearth2= Bearthπ(4πRearth

2). As they are blackbodies, it is Bsun=σ Tsun

4 and Bearth=σ Tearth4 . Thus, we get:

Tearth4=Tsun

4(θsun2/4). For Tsun= 6000 K, this yields Tearth= 285 K.

For planets other than earth, Tplanet= 285 K (Learth/Lplanet)1/2 because of the

θsun≈1/Lplanet and Tplanet

4≈θsun2 dependences.

If emissivity were not unity, this would leave the sun contribution unaffected, as its temperature is assumed as the visual one, while planets would collect less power (a fraction ε) from the sun, but also it would radiate less, just by the same factor ε. Therefore, provided ε does not change with wavelength of emission/ radiation, the temperature is the same as above. An exception: if εvisible<εinfrared, then the planet radiates less than that it receives and therefore the temperature would be considerably higher [by a factor (εvisible/εinfrared)1/4]: this is the case of planet Venus. PROBLEM A2.3 An engineer of road-lighting crossed his head with the idea of illuminating tunnels by optical fibers collecting sun light. Using the brilliance invariance, calculate for him: i) the power that can be collected from the sun at AM1.5 with a multimode fiber having a 50-µm diameter and a 30°-numerical aperture; and ii) the same, at the maximum of practicable concentration with a lens (specify which).

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Answ: As the fiber acceptance is AΩ= (π/4)D2 π (sin 30°)2= 1.5.10-3 mm2, and the sun gives an illumination E= 130.000 lm/m2 at AM1.5, the useable power is Plum= 130000. 1.5 10-3. 10-6= 0.2 10-3 lm, a very poor starting level indeed. But, with concentration, we may increase this figure by a factor C= (NA) 2/θsun

2= 50000 (NA) 2, where NA is the numerical aperture of the collecting lens (i.e., a Fresnel plastic lens with typ. NA=0.5). Also, using a bundle of N (typ. 1000) individual optical fibers packed together, we can have Plum= 0.2 10-3 lm . N C= 2500 lm, already a respectable value of some interest if we can now tolerate the system requirement of adding a mechanical movement to the lens to keep the solar disk tracked and focused on the fiber bundle. PROBLEM A2.4 Can it be true the old tale of Archimedes devising a mirror solar-burner of foe-ships sails? Assuming burning temperature of sail is 300°C, and mirrors held by hands, with 40 cm diameter and 30% reflectivity (for copper), calculate how many mirrors are required to burn sails at 200 m distance (of course, assume a sunny day, at noon, no clouds). Answ: By radiance invariance it is Rsail=Rsun. Writing a power balance of the received and radiated powers at the sail, we have: Psail= r Rsun (Amirror / d2) Asail = σTsail

4 2Asail, where the factor 2 is because both sail surfaces radiate, and r denotes mirror reflectance. Since Rsun= σTsun

4 4/π (Lambertian emitter) we get: Amirror=(2π d2 /r)(Tsail /Tsun) 4 and, inserting numbers, Amirror=(2π 40000 m2 / 0.3) (600/6000) 4 = 83.8 m2. As the mirror area is (π 0.42m2) /4 = 0.126 m2 we require N=665 mirrors.

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PROBLEM A2-5.Consider an Y branching device on a planar substrate, made of equal lossless waveguides up to the branching region, which on its turn is smooth and without any geometrical defect. What is the power out of the lower waveguide of the Y when two equal powers P enter the two upper waveguides of the Y? What is the power out of the two upper waveguides when a power P enters the lower waveguide? What is the phaseshift of the fields E1, E2 out of the upper waveguides when a field E enter the lower waveguide (assuming a perfectly symmetrical geometry?) Answ: By the invariance of radiance, the power out the lower waveguide cannot be larger than the input power P. Thus, the Y branch has an inherent and unavoidable 3dB attenuation and it is P = P1=P2. When exiting at the upper waveguides, the input power P is split evenly and, because of the structure symmetry, the relative optical phaseshift of the outputs must be zero. As field vectors E1 and E2 shall add to give E=√P, this means that they are both half the amplitude of E, or, in power: P1=P2=P/4 (once again, there is a 3-dB loss in power). PROBLEM A2-6.I am not satisfied of the result of Problem A2.5. Add a more convincing reasoning. Answ: The results are correct, and the 3-dB loss that undergoes the power balance can be understood as follows. Consider the mode spatial distribution in the lower leg arriving at the branching transition of the Y-coupler. We can take it as a Gaussian. When split into the two upper-leg waveguides, the Gaussian is spatially split in two halves, which are different from the mode (again a

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Gaussian) of the guides. Thus, a mode- conversion loss is entails, which accounts exactly for the 3-dB loss. In a X-coupler with 4 input/output ports, no intrinsic loss is found, because there is no necessary mode conversion. Or, stated differently, the Y-coupler is a sort of maimed X-coupler, where the missing leg accounts for the loss. PROBLEM A2.7 To make an endoscope, we want to use a SELFOC fiber (with parabolic index profile, it re-focalizes periodically the image) to guide an image of N=500x500 pixel (in visible, λ=0.5µm). If the numerical aperture is sinθ=0.3, what is the minimum diameter required? Answ: By the invariance of acceptance (or of the mode number), we have: N λ2 = 0.25.106 λ2 = AΩ= (π/4) D2 π(0.3)2 whence we find: D= 0.5 mm. PROBLEM A2.8 It is probably well known to the reader that, when laser light is shed on a diffusing surface, it re-radiated light exhibiting a grainy appearance called speckle pattern. This is because the diffuser has destroyed the spatial coherence of the light, at least locally. At a distance z from a diffuser, each speckle-grain has a longitudinal (parallel to the z-axis) dimension (say sl) and a transversal (perpendicular to the z-axis) dimension st . By considering that a speckle is an individual bright (or dark) spot inside which no structure can be resolved, or, it is a single-mode, calculate its dimensions applying the λ2-acceptance value of the single mode. Answ: Consider first the tranversal dimension st (taken as a diameter) as seen at a distance z from a diffuser of diameter D.

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For a speckle at a distance z from the diffuser, the solid angle within which rays are coming to the speckle is Ω=(π/4)D2/z2, and the receiving surface is A=(π/4) st

2. By letting their product equal to λ2 for a single mode we have: A.Ω=(π/4) st

2.(π/4)D2/z2= λ2 . By solving for st we get: st =(4/π) λ z /D. In addition, as the rays are angled of θ=D/2z around the speckle, they sort out a length sl =st/θ before diverging in size more than st. Thus, the longitudinal speckle dimension is: sl =(8/π) λ z2/D2. Worth to note, the near-to-unity multiplying factors of st and sl, namely 4/π and 8/π, are dependent on the choice of the illumination on the diffusing surface. It can be shown that, for a a real laser beam with a Gaussian spatial-distribution, these factors become exactly unity. As a last remark, the total volume occupied by the speckle grain, or, by the coherence region, V=(π/4)st

2 sl can be computed as V=(8/π) λ 3z4/D4 or, for a Gaussian beam, we have: V.Ω2= λ 3, This equation holds more generally for any single-mode geometry or situation we may possibly encounter. PROBLEM A2.9 Should the Greek astronomer Eratostene (200 BC) have had a photometer, ha could have succeeded in computing the earth-to-moon distance. Hint how. Answ: Eratostene had measured the earth diameter Dearth (looking at the sun elevation change between Alexandria and Assuan). He could have proceeded with the following reasoning.

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Pointing the photometer to the new moon, he could have determined the ratio Bm/e/ Bm/s of the brilliance of the moon illuminated by the earth Bm/e to the brilliance of the moon illuminated by the sun Bm/s . These quantiities are: Bm/s=αm BsunΩsun/π, and Bm/e=αm BeΩe/π, where Bsun, Be are the sun and earth (surface) brilliance, Ωsun, Ωe are the solid angle under which sun and earth are seen from the moon and αm is the moon albedo. Thus, we have: Bm/e/ Bm/s = BeΩe / BsunΩsun. Now, Be=αe BsunΩsun/π where the earth albedo αe can be measured as the ratio of the brilliances of a white paper sheet to the (average) earth surface. Inserting in previous expression yields: Bm/e/ Bm/s= αe BsΩsun Ωe / π BsunΩsun.= αe Ωe / π whence the earth-moon distance Le/m is solved as: Ωe = π D2

earth / L2

e/m = Bm/e/ Bm/s π./αe or L2

e/m = αe D2

earth Bm/s /Bm/e Problem A3.1 At a good illumination level, say ≥100 nit of background radiance, the eye contrast limit of 0.005 is approached. When you are under full-lighmoon illumination, what is the limit contrast you can get? Answ: From Table A1-1, we have that the moonlight illumination is Emoon =0.1 lux, or Bmin= 1/π Emoon = 0.3 nit. The Blackwell diagram (Fig.A3-1) gives for such a Bmin value: Clim = 0.015-0.03 (@ α= 200-600 mrad) or, for a fine detail with α= 1.2 mrad, we find the required contrast as: C1.2mrad = 0.8

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Problem A3.2 From data contained in Blackwell’s visual acuity diagram (Fig.A3-1), estimate the minimum number of photons perceivable to the eye in rel time (with an integration time of, say, 0.1 s) in photopic and scotopic conditions. Answ: At the boundary (Bmin=2.10-3 nit) of scotopic vision, the Blackwell diagram can be approximated by the following expression (look at the Bmin=2.10-3nit point of the second curve from the top): Bmin

. α2. C= const. = 2.10-3 . (1.2 mrad)2 . 20 = 57.6.10-9 lux. (1) For high contrast C it is (see definition in the y-scale of Fig.A3-1): Bmin C= Bmax- Bmin ≈ Bmax, the above expression (1) becomes: Bmax πα2/4= (π/4) 57.6.10-9 = 0.45.10-7 lux. On the human eye of normal aperture D=7 mm (in dim light conditions, i.e. for Bmin=2.10-3 nit), this figure corresponds to a collected power ending in the single-element sensitive area of the eye: Bmaxπα

2/4(πD2/4). The value of power is: P= Bmax πα2 (πD2/4)= 0.45.10-7 lux . 38.4 mm2= 17.3 10-13 lm Now, let us recall that, at λ=550 nm, it is: 1 lm = 1/673 W , so that P= (1/673) 17.3 10-13 lm = 25.7 10-16 W. In T=0.1s, the number of photons received by a power P is: N = (P/hν) T, and, at 550 nm the photon energy is: hν = 1.6 10-19. (1.24/0.55)= 3.6 10-19 W,

whence we get for the minimum number of photons perceived by the eye: Nphot = (P/hν) T = (25.7 10-16/3.6 10-19).0.1 = 710 photons. On the other hand, in scotopic conditions, which require (as mentioned in the text) the adaption to darkness for a 10-30 minute period, eq.(1) becomes:

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Bmin. α2. C= const. = 3.10-5 . (1.2 mrad)2 . 100 = 4.2.10-10 lux,

that is, we gain a factor 137 in sensitivity respect to photopic vision. Therefore, the minimum number of photons perceived by the eye is: Nscot = 710 /137= 5.2 photons Indeed, this is a remarkable result of performance. Problem A4.1 Find the variance of the electric field fluctuation for a field in thermal equilibrium with an ambient at absolute temperature T. Answ: From Eq.A4.13, we shall compute σ2E = 〈ΔE2〉 = k /⎥δ2S/δE2⎥E= 〈E〉

where k is the Boltzmann constant. As it is: δQ=TdS=dU-δL, we take dU=0 at equilibrium and express δL as: δL= E δD. V where V is the volume of the quantization space, V= A.L =A cτ = Ac /2B in terms of the observation bandwidth B. Thus entropy is: S = ∫ δQ/T= ∫ - δL/T = ∫ -E δD. V/T = - ε0E

2. V/T. By differentiating twice the above result, we have: δS/δE = -2 ε0EV/T, and δ2S/δT2= -2ε0V/T = -2 A cε0 /2BT Noting that c=1/√µ0ε0 whence ε0 c=√ε0/µ0 =1/Z0 , we get: δ2S/δT2= 2 A /2BT Z0 so that, by substituting, the field variance is: σ2E = k /(A /BT Z0)=

1/2 kT (2Z0/A) B

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Note the resemblance of this expression to Eq.8.60 derived under the second quantization condition, where kT and hν interchange. Problem A4.2 Generalize Eq. 8.60 and determine the wavelength of transition between the thermal- and quantum-dominated regimes of field fluctuations at absolute temperature T. Answ: From the result of Probl.A4.1 and Eq.8.60, we can state that, as a generalization, σ2E =1/2 (kT+hν) (2Z0/A) B Thermal fluctuations of the electric field dominates for kT> hν, quantum fluctuations in the opposite case. The break point is at: kT > hc/λ or at: λ = hc/ kT = 47.9 µm (see also Eq.6.10’). Problem A5.1 Your Β=60-MHz oscilloscope has a 1-mV/division vertical sensitivity. The input resistance is 1-MΩ. Because of Johnson noise vn= [4kTBR]1/2, you should then expect to see a trace on the scope with vn=[4 .40 .10-22 60 .10-6 1 .106]1/2= 0.98 mV rms noise, that should be clearly visible. However, no noisy trace is actually seen (trace is much thinner). How you explain this ? What is the correct expression of your noise voltage in this case? Answ: We shall take account of the non-negligible effect of the input capacitance C across the 1-MΩ resistance. we have also a parasitic capacitance C, typically C=10 pF. The parallel combination of R and C yields a high frequency cutoff much smaller than the oscilloscope measurement bandwidth B.

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Because of this, the spectral density of noise is vn2(ω), =(4kTR) (1+ω2/ωRC

2), where /ωRC=1/RC is the high frequency cutoff of the RC-group. By integrating vn

2(ω) from f=0 to f=∞ we get for the total voltage noise seen at the input: vn

2 =(4kTR) ∫0..∞ (1+ω2/ωRC2)df = 4kTRωRC ∫0..∞ (1+ω2/ωRC

2)d(ω/ωRC) = 4kTR/2πRC atan(ω/ωRC)⎜ 0..∞ = 4kTR/2πRC (π/2), viz. vn

2= kT/C. For example, using C= 10pF for the input capacitance we find: vn = [kT/C]1/2= [40 .10-22/ 10.10-12] = 0.02 mV which is well below the scope sensitivity.

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