Stage 2 SEM1

Semiconductors

The ionic depletion region in a pn junction forms when dopants in the n region donate electrons to dopants in the p region


A built-in potential VBI forms across the region

Thermodynamics

Optics

Quantum

Vector calculus

This current generates an electric field, which in turn creates an equal and opposite drift current.

Fundamentals

A quasistatic process is carried out in an infinite number of infinitesimal steps - the system can relax to equilibrium after each step

First law of thermodynamics


\(\Delta U=Q+W\) where \(U\) is the internal energy, \(Q\) is the heat added to the system and \(W\) is the work done on the system

\(\textbf{v}=\frac{\mathrm{d}\textbf{r}}{\mathrm{d}t}\)

Relations

\(S(U,V,N)\) can be considered a fundamental relation if the three postulates are met

\(S\) is continuous and differentiable wrt \(U\), \(V\) and \(N\)

\(S\) increases monotonically wrt \(U\)

The arc length of a curve \(C\) is given by \(s(t)=\int_{t_1}^{t}v(t)\;\mathrm{d}t\) - its full length is then simply \(s(t_2)\)

If a parameter \(x\) of \(f(x)\) is extensive, \(f(\lambda x)=\lambda f(x)\)


For intensive variables, generally \(f(\lambda x)\neq \lambda f(x)\)

Thermodynamic variables (AKA 'equations of state')

Intensive

\((\frac{\partial U}{\partial N})_{S,V}=\mu\)

Diffusion currents occur when there is a concentration gradient in carriers

Diffusion currents are dominated by the minority carrier concentrations, which are at their equilibrium values at either terminal

Bias

Forward

Reverse

Junction breakdown

Tunnelling between neighbouring valence and conduction bands under high can occur reverse bias

In a strong electric field, carriers gain enough energy to make ionising collisions, creating hole-electron pairs - this leads to a spike in \(I_p\)

High reverse bias leads to heat dissipation: this further increases the current, leading to thermal runaway

For large \(l\Rightarrow l = \sqrt{D\tau}\) where \(\tau\) is the recombination lifetime

Fermat's theorem: light travels between two points along path \(C\) such that \(\tau = \int_Cn(s)\,\mathrm{d}s\) is stationary under infinitesimal variations in the path

A path can be determined by solving \(\frac{\mathrm{d}\tau}{\mathrm{d}t}=0\)

Snell's law: \(n_i\sin\theta_i=n_r\sin\theta_r\)

An optical system is stigmatic under 3 conditions

P is a perfect image of S

P and S are conjugate points

System is reversible

A spherical lens has 2 focal lengths: \(f_o=\lim_{s_i\rightarrow\infty}s_o=\frac{n_m}{n_l - n_m}R\)
\(f_i=\lim_{s_o\rightarrow\infty}s_i=\frac{n_l}{n_l-n_m}R\)


Concave lens generally have negative focal lengths

For a thin lens, \(\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}\)


For multiple lens in contact, \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\ldots\)

Multiple lens can be used in achromatic doublets, systems which use a biconvex and phono-concave lens to reduce chromatic abberation

The p junction is connected to the positive terminal: \(n_p\) and \(p_n\) are reduced. Carriers are pushed towards and hence shrink the depletion region

The p junction is connected to the negative terminal: \(n_p\) and \(p_n\) are increased. Carriers are pulled away from and hence widen the depletion region

\(\phi_b\Rightarrow \phi_b-V_a\)

\(\phi_b\Rightarrow \phi_b+V_a\)

Heat engines

Extracting heat to do work reduces the entropy, so it has to be dumped into another system


In heat engines \(W=-Q_h-Q_c\)

\(\mathrm{d}S=\frac{\mathrm{d}Q}{T}\)

As \(T\) is variable, we have to use the infinitesimal \(\mathrm{d}Q\)

Notation: \(\mathrm{d}Q\) is positive when heat is added to a system

The maximum work is delivered if the process is reversible


\(\mathrm{d}S=\mathrm{d}S_h + \mathrm{d}S_c=0\)

\(\mathrm{d}Q_c=-\frac{T_c}{T_h}\mathrm{d}Q_h\)
\(\mathrm{d}W=-[1-\frac{T_c}{T_h}]\mathrm{d}Q_h\) where \(\eta_{max}=-\frac{\mathrm{d}W}{\mathrm{d}Q_h}=[1-\frac{T_c}{T_h}]\)

\(S\) is extensive (i.e. additive over its constituent subsystems)

\(\mathrm{d}Q=c(T)\mathrm{d}T\)

The line integral over a field is described by \(\int_C\textbf{F}\,\mathrm{d}r\), or \(\mathrm{d}\textbf{r}\) for a vector field


This can be performed by using the line element \(\mathrm{d}\textbf{r}=\textbf{v}\;\mathrm{d}t\) and taking the dot product

dropped image link

The natural parameterisation of a curve is described as a mapping from \(s\) (i.e. \(s\) is the curve parameter)

i) Find \(s\)

ii) Invert \(s=s(t)\Rightarrow t=t(s)\)

iii) Substitute \(t(s)\) into \(x(t)\), \(y(t)\) and find new limits

Double integrals

Integrating \(f(x,y)=1\) over a 2D region \(R\) gives the area of \(R\):


\(I=\int\int_R\mathrm{d}x\mathrm{d}y\)

ADD MORE

Operations

Normals

The normal to a surface mapped by \(u\), \(v\) is given by \(\textbf{n}=\textbf{v}_u\times\textbf{v}_v\) where


\(\textbf{v}_u=(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}), \textbf{v}_v=(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v})\)

The normal to \(f(x,y,z)=0\) is given by \(\textbf{n}=\pmb{\nabla}f\)

The normal to \(z=g(x,y)\) is given by \(\textbf{n}=(-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1)\)

The directional derivative \[\frac{\partial f}{\partial b}=\hat{\mathbf{b}}\cdot\pmb{\nabla}f\]

The curl of a vector field characterises its rotation and is given by
\(\mathrm{curl}\;\mathbf{v}=\pmb{\nabla}\times\mathbf{v}=\begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_x & v_y & v_z \end{vmatrix}\)
A vector field with all zero curl, i.e \(\mathrm{curl}\;\mathbf{v}=\mathbf{0}\), is irrotational

A wave function \(\Psi=\sum_n c(t)_n\psi_n\) can be described as a superposition of eigenstates, \(\psi_n\), with \(P_n=|\int_{-\infty}^{\infty}\Psi_n^*\Psi|^2=|c_n|^2\), where \(\sum_n P_n=1\)

Second law of thermodynamics


\(\Delta S>0\) - using \(\Delta S=\frac{Q}{T}\), this can be used to prove that heat can't be fully converted into work

If \(\mathrm{d}f\) is a perfect differential, \(\Delta f\) does not depend on the route, only the end points.


For \(\mathrm{d}f=M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y\), this requires \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\)

Potentials

For a system in contact with a heat reservoir with constant \(T\), \(V\) and \(N\), the equilibrium state is such that the Helmholtz Free Energy \(F(T,V,N)=U-TS\) is minimised

For a system in mechanical contact with a reservoir (i.e. variable \(V\)) with constant \(p\), \(S\) and \(N\), the equilibrium state is such that the Enthalpy \(H(S,p,N)=U+pV\) is minimised

For a system in mechanical and thermal contact with constant \(T\), \(p\) and \(N\), the equilibrium state is such that the Gibbs Free Energy \(G(T,p,N)=U-TS+pV\) is minimised

\((\frac{\partial U}{\partial V})_{S,N}=-p\)

The Euler relation: \(U=TS-pV+\mu N\;\Rightarrow\;\mathrm{d}U=T\mathrm{d}S-p\mathrm{d}V+\mu\mathrm{d}N\)

The Gibbs-Duhem relation: \(S\mathrm{d}T-V\mathrm{d}p+\mu\mathrm{d}N\)

By the relation \(\mathrm{d}f(x,y)=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y\), each differential can be matched with new derivatives


e.g. \(\mathrm{d}F=-S\mathrm{d}T-p\mathrm{d}V+\mu\mathrm{d}N\Rightarrow \frac{\partial F}{\partial T}=-S\)

Derivatives

click to edit

Another way to find the work done: \(W=\int\eta\mathrm{d}Q_h\)

Two curves intersect where \(\mathbf{r}(a)=\mathbf{r}'(b)\), i.e. \(t=a\), \(t'=b\); \(x\), \(y\) and \(z\) values can be compared to find the intersects

Operators

The divergence \(\mathrm{div}\;\textbf{v}=\pmb{\nabla}\cdot\textbf{v}=\frac{\partial\textbf{v}_x}{\partial x}+\frac{\partial\textbf{v}_y}{\partial y}+\frac{\partial\textbf{v}_z}{\partial z}\) of a vector field \(\textbf{v}\) is a scalar field, a measure of expansion from the origin


A vector field with \(\mathrm{div}\;\mathbf{v}=0\) is solenoidal

dropped image link

\[\pmb{\nabla}=(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})\]

The Laplacian \[\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\]

Subscript notation

dropped image link

Dummy subscripts are summed over, and generally appear twice in any expression


Free subscripts represent

\(\frac{\partial}{\partial x_i}\rightarrow \partial_i\)

click to edit

Two eigenstates are orthogonal if they correspond to different eigenvalues


\(\int_{-\infty}^{\infty}=\psi_m^*\psi_n=\delta_{m,n}\) where \(m=n\Rightarrow\delta_{m,n}=1\) and \(m\neq n\Rightarrow\delta_{m,n}=0\)


An eigenstate is normalised if \(\delta_{n,n}=1\)

\(\int_{-\infty}^{\infty}|\Psi(x,t)|^2=\int_{-\infty}^{\infty}|\psi(x)|^2=1\)

For \(t=0\), \(\Psi(x,0)=\psi(x)\) (?)

Time dependent SE


\(i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V(x)\Psi(x,t)\)

Generally for a finite box, \(\psi(x)_n=\sqrt{\frac{2}{a}}\sin{\frac{n\pi}{a}x}\)

Time independent SE


\(\frac{\mathrm{d}^2\varphi}{\mathrm{d}x^2}+\frac{2m}{\hbar^2}(E-V)\varphi(x)=0\)

Infinite wells

\(E_n=n^2(\frac{\hbar^2\pi^2}{2ma^2})\)

\(V=0\) for \(0\leq x\leq a\), else \(V=\infty\)

Finite wells

Find a general solution \(\varphi\) in each region, and find \(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\)

General ODE solutions

\(\Delta>0\Rightarrow \varphi=Ae^{\lambda_1x}+Be^{\lambda_2x}\)

\(\Delta<0\Rightarrow\varphi=e^{\alpha x}(A\cos\beta x + B\sin\beta x)\)

\(\Delta = 0 \Rightarrow\varphi=Ae^{\lambda x}+Bxe^{\lambda x}\)

Compare and eliminate boundary equations of \(\varphi\) and \(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\)

Break up into 3 regions with \(V_1>0\), \(V_2=0\) and \(V_3>0\) with \(E< V_1\), \(E< V_2\)

Physical

Fresnel diffraction dominates in the near field, whereas Fraunhofer diffraction dominates in the far field where \(R>>\frac{a^2}{\lambda}\)

Fourier transforms

\(\mathcal{F}[f(x)]=\bar{f}(u)=\int_{-\infty}^{\infty}f(x)e^{-i2\pi ux}\)


\(\mathcal{F}^{-1}[f(x)]=f(x)=\int_{-\infty}^{\infty}f(x)e^{i2\pi ux}\)

Standard results

\(\mathrm{rect}(\frac{x}{a}) \Rightarrow a\text{sinc}(au)=\frac{1}{\pi au}\sin{\pi au}\)

click to edit

\(u_p\) is a wave's amplitude with \(u=\frac{\theta}{\lambda}\)


For a single slit, \(u_p\propto a\mathrm{sinc}(au)\)


For a double slit, \(u_p\propto 2a\mathrm{sinc}(au)\cos(\pi ud)\)

Polarization

Linearly polarised light is a superposition of left-circularly and right-circularly polarized light.


It is rotated in an optically active medium; this is because left-circularly and right-circularly polarised light have different refractive indices


\(\theta(z)=\beta z\;\;\;\;\;\;\;\;\;\;\beta = \frac{\pi}{\lambda}(n_R-n_L)\)


Right handed rotation is dextrorotary; \(n_R>n_L\). Left handed motion is levorotary.

\(I(\theta)\propto |u_p|^2\)

\(E=h\nu\Rightarrow p=\hbar k\)

Harmonic oscillators

\(V(x)=\frac{1}{2}m\omega^2x^2\)

The thin lens equation:
\(\frac{n_m}{s_0}+\frac{n_m}{s_i}=(n_l-n_m)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{f}\)


For air (\(n_m=1\)), this shortens to:
\(\frac{1}{s_0}+\frac{1}{s_i}=(n_l-1)(\frac{1}{R_1}-\frac{1}{R_2})=\frac{1}{f}\)

For biconvex lens, generally \(R_1>0\) and \(R_1=-R_2\)

\(s_o=x_o+f\Rightarrow x_ox_i=f^2\)

Magnification \(M_T=\frac{y_i}{y_o}=-\frac{x_i}{f}\)

Capture

There are \(N-2\) subsidiary maxima between each main maximum


As \(N\rightarrow\infty\), the main maxima become narrower, more closely resembling a comb graph, and there are more subsidiary maxima between them

Convolution


\(f(x)*g(x)=\int_{-\infty}^{\infty}f(x')g(x-x')\mathrm{d}x'\)


\(\Rightarrow \mathcal{F}[f(x)*g(x)] = \mathcal{F}[f(x)]\cdot\mathcal{F}[g(x)]\)

Properties of Dirac function


\(\int_{-\infty}^{\infty}f(x)\cdot\delta(x-a)\mathrm{d}x=f(a)\)


\(\int_{-\infty}^{\infty}e^{i2\pi (x-a)u}\mathrm{d}u=\delta(x-a)\)

General solutions of electromagnetic waves \(u=u_0e^{i(kz-\omega t)}\Rightarrow u=u_0\cos(kz-\omega t)\)


with a phase velocity \(v=\frac{\omega}{k}\), \(k=\frac{2\pi}{\lambda}\)

Plane waves: \(u=u_0\exp{i(\mathbf{k}\cdot\mathbf{z}-\omega t)}\) with \(\mathbf{k}=k_x\mathbf{\hat{i}}+k_y\mathbf{\hat{j}}+k_z\mathbf{\hat{k}}\)

The principle of superposition: if \(u_1(z,t)\) and \(u_2(z,t)\) are solutions of the scalar wave equation, then so is \(u_1+u_2\)

\(|F_E| = q|E|\)

The Huggens Fresnel principle states that each element of a wavefront acts as a source of secondary wavelets, which diverge spherically. The resultant field is a superposition of these wavelets.

Malus' Law \(I(\theta)=I(0)\cos^2{\theta}\) for light passing through two linear polarizers, where \(\theta=\theta_2-\theta_1\neq 0\)

\(\)

Transmitted light is polarized in the plane of the polarizer

If there is a phase shift in the left or right-handed circular component, in general the direction of the linear polarisation is changed.

Fabry-Perot etalons


Coefficient of finesse \(F=\frac{4R}{(1-R)^2}\)


Finesse \(\mathcal{F}=\frac{\Delta\delta}{\bar{\delta}}=\frac{\pi\sqrt{F}}{2}\) is the ratio of peak separation to peak width; a measure of how clearly they are separated

Capture3

Finite barriers

\(\varphi(x)=A_1e^{ikx}+A_{1R}e^{-ikx}\) is a general solution for \(E>V\) (the sum of the incident and reflected wave) with \(k=\sqrt{\frac{2m}{\hbar^2}(E-V)}=\frac{2\pi}{\lambda}\)


For \(E < V\), \(\varphi\propto Ce^{-kx}\) decays exponentially


For \(E=V\), \(\varphi\propto Cx+D\)

\(T=\frac{k_T}{k_I}|\frac{A_T}{A_I}|^2\), \(R=|\frac{A_R}{A_I}|^2\)


\(T+R=1\)

Extensive

\((\frac{\partial S}{\partial V})_{U,N}=\frac{p}{T}\)

\((\frac{\partial S}{\partial N})_{U,V}=-\frac{\mu}{T}\)

\((\frac{\partial S}{\partial U})_{V,N}=\frac{1}{T}\)

\((\frac{\partial U}{\partial S})_{V,N}=T\)

If \(T(s,v)\) and \(p(s,v)\) are known, the fundamenal relation can be found by integrating


\(\mathrm{d}u=T\mathrm{d}s - P\mathrm{d}v\)

\(U=cNRT\)

\(W=\int_{V_0}^V p\mathrm{d}V\)

Carnot cycle

\(\frac{pV}{T}\) is constant for any ideal gas


\(pV^{\gamma}\) and \(TV^{\gamma -1}\) are constant for any adiabatic process

Second derivatives

\(c_p=\frac{T}{N}(\frac{\partial S}{\partial T})_p\)


\(c_v=\frac{T}{N}(\frac{\partial S}{\partial T})_V\)


\(\alpha=\frac{1}{V}(\frac{\partial V}{\partial T})_p\)


\(\kappa_T=-\frac{1}{V}(\frac{\partial V}{\partial p})_T\)

Electrons in the n region diffuse across the junction, leaving positive ions and creating negative ions in the p type region.


This results in an electric field from n to p