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