Answer to Kondo's Si problem in EDIT2013
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EDIT2013_Kondo_ans/Makefile
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EDIT2013_Kondo_ans/Makefile
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all:
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pdflatex kondo_ex.tex
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clean:
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rm -rf *.aux kondo_ex.pdf *.log
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168
EDIT2013_Kondo_ans/kondo_ex.tex
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EDIT2013_Kondo_ans/kondo_ex.tex
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\documentclass[a4paper]{article}
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\usepackage[utf8x]{inputenc}
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\usepackage{ucs}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{epsf}
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\usepackage{float}
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\usepackage{url}
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\usepackage{fancyhdr}
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\usepackage{verbatim}
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\usepackage{color,listings}
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\usepackage{booktabs}
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\usepackage{tabularx}
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\usepackage{natbib}
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\usepackage[all]{xy}
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\usepackage{graphicx}
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\usepackage{graphics}
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\usepackage{color}
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\usepackage{listings}
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%\makeatletter
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%\def\@xobeysp{}
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%\makeatother
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\usepackage{color,listings}
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\author{Tran Hoai Nam}
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\title{Answer to quizzes in \\Detectors of LHC Experiments lecture}
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\begin{document}
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\maketitle
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\section*{Quiz 1: Number of protons for $\mathcal{L}=10^{34}$
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cm$^{-2}$s$^{-1}$}
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Total luminosity of LHC:
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\[
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\mathcal{L} = f N_b L_{bunch\_cross} = f N_b \frac{N_p \times N_p}{4\pi\sigma_x\sigma_y}
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\]
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\[
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\rightarrow N_p = \sqrt{ 4\pi \sigma_x\sigma_y \frac{\mathcal{L}}{fN_b}}
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\]
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where: $f$ is revolution frequency;
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\[
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f \approx \frac{c}{l_{LHC}}
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= \frac{3\times 10^8 [ms^{-1}]}{27 \times 10^3 [m]} \approx 10^4 [s^{-1}]
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\]
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Substitute the numbers, number of proton in each bunch is:
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\[
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N_p =
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\sqrt{4\times 3.14 \times 16 \times 10^{-6} [m]\times 16 \times 10^{-6} [m]
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\times
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\frac{10^{34} \times 10^{4}[m^{-2}s^{-1}]}{2808 \times 10^4[s^{-1}]}}
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\approx 1.07 \times 10^{-11}
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\]
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\section*{Quiz 2: Characteristics of a silicon detector}
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\subsection*{1) Full depletion voltage}
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Let $V$ be the full depletion voltage for the 300 $\mu m$ Si detector:
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\[
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d = \sqrt{\frac{2\varepsilon^{'} \varepsilon_0}{q_e N_{eff}}V}
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\rightarrow V = \frac{q_e N_{eff}}{2 \varepsilon^{'} \varepsilon_0} d^2
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\]
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\[
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V = \frac{1.6 \times 10^{-19} [C] \times 2 \times 10^{12} [cm^{-3}]}
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{2\times11.6\times 8.854 \times 10^{-14} [F cm^{-1}]} (300 \times 10^{-4} [cm])^2
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\approx 140.2 [V]
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\]
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\subsection*{2) Noise}
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Capacitance of the detector when it is fully depleted:
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\begin{align*}
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C &= \varepsilon^{'} \varepsilon_0 \frac{S}{d} \\
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&= 11.6 \times 8.854 \times 10^{-14} [F cm^{-1}]
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\frac{1 [cm^2]}{300 \times 10^{-4} [cm]} \\
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&\approx 34.2 \times 10^{-12} [F] \\
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&= 34.2 [pF]
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\end{align*}
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Equivalent noise charge square:
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\[
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ENC^2 = \left(\frac{e^3}{36q_e}\right)^2
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\left[ \frac{3}{\tau} (4k_B T R_S) C^2 + \frac{5\tau}{3}(2q_eI_{leak})\right]
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\]
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We have:
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\begin{align*}
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k_B T &= 8.62 \times 10^{-5} [eV K^{-1}] \times 273.15 [K] \\
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&\approx 2.35 \times 10^{-2} [eV] \\
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&= 2.35 \times 10^{-2} \times 1.6 \times 10^{-19} [J]\\
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&= 2.35 \times 10^{-2} q_e [JC^{-1}]
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\end{align*}
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Thus:
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\begin{align*}
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ENC^2 =
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& \frac{2.72^6}{36^2 \times 1.6 \times 10^{-19}[C]} \times \\
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& \times \Biggl[
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\frac{3}{20 \times 10^{-9} [s]}
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(4 \times 2.35 \times 10^{-2} [JC^{-1}] \times 200 [\Omega])
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(34.2 \times 10^{-12} [F])^{2} + \\
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& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
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(2 \times
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10 \times 10^{-6} [A])
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\Biggr]
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\end{align*}
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\begin{align*}
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ENC^2 &=
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\frac{2.72^6}{36^2 \times 1.6 \times 10^{-19}[C]}
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\times \Biggl[
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3.3 \times 10^{-12}[Js^{-1}C^{-1}\Omega F^2] + 0.3 \times 10^{-12} [C]
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\Biggr] \\
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&\approx 7.0\times 10^6
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\end{align*}
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(dimensionally speaking, $Js^{-1}C^{-1}\Omega F^2
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= VA \times C^{-1}\times VA^{-1}\times C^2V^{-2} = C$)
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\[
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\rightarrow ENC \approx 2.65 \times 10^3
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\]
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The $ENC$ is $2.65 \times 10^3$, and it can be seen that the contribution from
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capacitance is dominant.
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%\begin{align*}
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%ENC^2 =
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%&\left(
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%\frac{2.72^3}{36\times 1.6\times 10^{-19}[C]}
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%\right)^2 \times \\
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%& \times \Biggl[
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%\frac{3}{20 \times 10^{-9} [s]}
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%(4 \times 8.62 \times 10^{-5} [eV K^{-1}]
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%\times 273.15 [K] \times 200 [\Omega])
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%(34.2 \times 10^{-12} [F])^{2} + \\
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%& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
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%(2 \times 1.6 \times 10^{-19}[C] \times
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%10 \times 10^{-6} [A])
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%\Biggr]
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%\end{align*}
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%\begin{align*}
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%ENC^2 =
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%&\left(
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%\frac{2.72^3}{36\times 1.6\times 10^{-19}[C]}
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%\right)^2 \times \\
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%& \times \Biggl[
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%\frac{3}{20 \times 10^{-9} [s]}
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%(4 \times 2.35 \times 10^{-2}[eV] \times 200 [\Omega])
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%(34.2 \times 10^{-12} [F])^{2} + \\
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%& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
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%(2 \times 1.6 \times 10^{-19} [C]\times
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%10 \times 10^{-6} [A])
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%\Biggr]
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%\end{align*}
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%\begin{align*}
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%ENC^2 =
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%&\left(
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%\frac{2.72^3}{36\times 1.6\times 10^{-19}[C]}
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%\right)^2 \times \\
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%& \times \Biggl[
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%\frac{3}{20 \times 10^{-9} [s]}
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%(4 \times 2.35 \times 10^{-2} \times 1.6 \times 10^{-19}[J] \times 200 [\Omega])
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%(34.2 \times 10^{-12} [F])^{2} + \\
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%& + \frac{5 \times 20 \times 10^{-9}[s]}{3}
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%(2 \times 1.6 \times 10^{-19} [C]\times
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%10 \times 10^{-6} [A])
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%\Biggr]
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%\end{align*}
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\end{document}
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