By Mingyang Li. [TOC] # Derivations ## PREREQ.: 4 types of charged species in a semiconductor Their population densities are: - Electrons: n - Holes: p - Positive donor ions: N_d - Negative acceptor ions: N_a ## Starting with Microscopic Expression of Current... Start with: $$j=nqv_d$$ ### … to Hall Effect we have: $$v_d=\frac {j_x}{nq}$$ The total force a charge carrier feels under Hall experiment is: $$F_y=qE_y\pm q(v_d\times B_z )$$ where subscripts y and z indicates direcitons. In steady state, $$F_y=0\\ \Rightarrow qE_y = q(v_d\times B_z )\\ \Rightarrow E_y = v_d B_z$$ Insert the expression for v_d here: $$E_y = \frac {j_x}{nq} B_z$$ Now we have one variable for each directions. Nice. By denoting the constant factor **as Hall coefficient**: $$R_H\equiv \frac1{nq}$$ we can write: $$E_y=R_H j_x B_z$$ ### … to Ohm's Law Since[^3]: $$v_d=\mu_dE$$ [^3]: This equation, therefore also Ohm's Law, is only valid under low electric fields. We can write: $$j=nq\mu_dE$$ By writing the constants collectively as a "conductivity": $$\sigma=nq\mu_d$$ we can simplify it to: $$j=\sigma E$$ Which is the Ohm's Law. Instead of writing w.r.t. conductivity, we often use the "resistivity": $$j=\sigma E=\frac E \rho \Leftrightarrow E=\rho j \text{ where } \rho\sigma=1$$ ## In A Depletion Region ### Depletion Approximation #### Construction Using n-type semiconductor as example, we use the **Possion's Equation**: $$\frac{d^2V}{dx^2}=-\frac{\rho(x)}\epsilon$$ where \rho is the total charge concentration. In semiconductor region (i.e. NOT bulk region[^2]): $$\rho(x)\equiv \begin{cases} e\cdot(p+N_d^+-n+N_a^-) & \text{in semiconductor region} \\ 0 & \text{in bulk region} \end{cases}$$ [^2]: This equation also infers that V=0 across the whole bulk region, because there is no charge to generate the field. which can be reduced by: - Assuming a doped semiconductor: $$\rho(x)=e\cdot(p+N_d^+-n+N_a^-) \\ \approx e\cdot(0+N_d^+-0+N_a^-) \\ = e\cdot(N_d^++N_a^-)$$ - Assuming a n-type: $$\rho(x)\approx e\cdot(N_d^++0)\\ = e\cdot N_d^+$$ where + suggests "concentration of ionized donor atoms". - Assuming all donor atoms have their e-'s ionized: $$\rho(x)\approx e\cdot N_d$$ Thus, this equation can be written as: $$\frac{d^2V}{dx^2}=-\frac{e\cdot N_d}\epsilon$$ Central idea of this approximation: The carrier density n and p should stay 0 throughout the depletion region 0<=x<=d at the N-side. This gives the boundary conditions: $$V=0 \text{ @ } x=0\\ V=V_d+V_a\text{ @ }x\geq d$$ where: - ε: permittivity, a constant. - N_d: population of donor ions (positively charged) - V_d: depletion voltage - V_a: applied voltage - e: electron charge #### Electric Field Integrate for once, we have the **electric field along the depletion region on the n-side**: $$E(x)=-\frac{dV}{dx}=-\frac{eN_d}\epsilon (x-d)$$ #### Voltage Potential Integrate again, the solution is the **voltage potential along the depletion region on the n-side**: $$V(x)=-\frac{eN_d}\epsilon(\frac12x^2-dx)$$ #### Depletion Length Set V(x) = V_d+V_a (treat as constant) to find d: $$V(x=d)=-\frac{eN_d}\epsilon(\frac12d^2-d\cdot d)=V_d+V_a\\ \Rightarrow d=\sqrt{\frac{2\epsilon(V_d+V_a)}{e\cdot N_d}}$$ or simply, by representing V = V_d + V_a: $$d=\sqrt{\frac{2\epsilon V}{e N_d}}$$ ### EXTRA: Use depletion region as capacitor Parallel-plate capacitance: $$C=\frac{\epsilon A}{d}$$ We consider its per-area variation: $$C=\frac{\epsilon}{d}=\frac{\epsilon}{\sqrt{\frac{2\epsilon V}{e N_d}}}\\ =\epsilon\sqrt{\frac{e N_d}{2\epsilon V}}=\sqrt{\frac{e\epsilon N_d}{2 V}}$$ or you can derive this via: $$C=\frac{dQ}{dV}=...=\sqrt{\frac{e\epsilon N_d}{2 V}}$$ A *Matt-schottky Plot* is a C^-2 - V_a plot: $$\frac1{c^2}=\frac{2(V_d+V_a)}{e \epsilon N_d}$$ Its slop is: $$\frac{d\frac1{c^2}}{dV_a}=\frac2{e\epsilon N_d}$$ ## In a semiconductor ### Effective Mass General formula: $$m^*=\frac{\hbar^2}{\frac{\partial E}{\partial k^2}}$$ People usually write: - m^*_e as the effective mass of an **electron**. - m^*_p as the effective mass of a **hole**. ### Effective density of states Effective DoS at the conduction band edge: $$N_c\equiv 2(\frac{2\pi m^*_e k_BT}{h^2})^\frac32$$ Effective DoS at the valence band edge: $$N_v\equiv 2(\frac{2\pi m^*_p k_BT}{h^2})^\frac32$$ ### Number of e-'s per unit volume Electron population density: $$n_e=n=N_ce^{-\frac{E_C-E_F}{k_BT}}$$ Hole population density: $$n_p=p=N_ve^{-\frac{E_F-E_V}{k_BT}}$$ ### Using them on semiconductors #### For intrinsic semiconductor Apparently, $$n=p\\ \Rightarrow N_ce^{-\frac{E_C-E_F}{k_BT}}=N_ve^{-\frac{E_F-E_V}{k_BT}}\\ \Rightarrow \frac {N_c} {N_v} e^{-\frac{E_C-E_F}{k_BT}}=e^{-\frac{E_F-E_V}{k_BT}}$$ Take ln on both sides: $$\Rightarrow \ln\frac {N_c} {N_v} =\frac 1{k_BT}(E_C-E_F-E_F+E_V)\\ \Rightarrow \ln\frac {N_c} {N_v} =\frac 1{k_BT}(E_C+E_V-2E_F)\\ \Rightarrow E_F = \frac12[ (E_C+E_V)-k_BT \ln\frac {N_c} {N_v}]$$ Typically, when measuring "band energies", we take the valent band as reference: $$E_V\equiv 0$$ Thus, $$E_F = \frac12[ (E_C+0)-k_BT \ln\frac {N_c} {N_v}]\\ = \frac12[ E_g-k_BT \ln\frac {N_c} {N_v}]$$ Since N_c ~= N_v for a intrinsic semiconductor, we have: $$E_F\approx \frac12 E_g$$ Meaning: the Fermi level of an intrinsic semiconductor lies at the mid-gap. #### For any semiconductor For any semicondoctor, these 2 relations give the concentrations of all 4 kinds of charged species within a homogeneous region: ##### charge balance $$n+N_a=p+N_d$$ - For an intrinsic semiconductor, approximately: $$n=p$$ - For a n-type semiconductor, approximately: $$n=N_d$$ - For an p-type semiconductor, approximately: $$N_a=p$$ ##### the Law of Mass Action According to the *Law of Mass Action*, the product of the concentrations of both types of carriers should equal to the intrinsic carrier population squared: $$n_e\cdot n_p=n_i^2$$ Expand it: $$\Rightarrow N_C\cdot N_V e^{-\frac{E_C-E_V}{k_BT}}$$ Remember that the Band Gap is just the difference between the Conduction Band and the Valence Band: $$\because E_g\equiv E_C-E_V \\ \therefore n_i^2 = N_C\cdot N_V e^{-\frac{E_g}{k_BT}}$$ This is useful.[^1] [^1]: See Notability Notes. You can express the charge carrier density w.r.t. intrinsic levels. Omitted here. #### Use it to derive the Built-in Potential in a PN junction For N-region, the electron population density is given bt Charge Balance: $$n=N_d=N_ce^{-\frac{E_C-E_F}{k_BT}} \text{, labeled as}= N_ce^{-\frac{qA}{k_BT}}\\ \Rightarrow A=\frac{k_BT}q \ln\frac{N_c}{N_d}$$ For P-region, these two laws give: $$\text{Charge Balance}\Rightarrow p=N_a\\ \text{Law of Mass Action}\Rightarrow n\cdot p=n_i^2$$ Combined, we have the n density in P-region: $$n=\frac{n^2_i}{N_a}$$ The n - N_c relation is still applicable: $$n=N_ce^{-\frac{E_C-E_F}{k_BT}} \text{, labeled as}= N_ce^{-\frac{qB}{k_BT}}\\ \Rightarrow \frac{n^2_i}{N_a} = N_ce^{-\frac{qB}{k_BT}}\\ \Rightarrow B=\frac{k_BT}q \ln\frac{N_cN_a}{N_i^2}$$ Thus, the Built-in Potential is: $$\phi_{bi}=B-A=\frac{k_BT}q \ln\frac{N_cN_a}{N_i^2}-\frac{k_BT}q \ln\frac{N_c}{N_d}\\ =\frac{k_BT}q [\ln\frac{N_cN_a}{N_i^2}-\frac{k_BT}q \ln\frac{N_c}{N_d}]\\ =\frac{k_BT}q \ln\frac{N_dN_a}{N_i^2}$$ # Formulae ## Current densities in junctions ### in a PN junction $$j=(j_\text{0,e- in n} + j_\text{0, e- in p})(e^{-\frac{eV_a}{k_BT}}-1)$$ where: - j_\text{0,e- in n}  is the electron current in N-region when no voltage is applied. - j_\text{0,e- in p}  is the electron current in P-region when no voltage is applied. - V_a is the applied voltage. ### in a Schokky junction (semicon.->metal) $$j=j_0 (e^{-\frac{eV_a}{k_BT}}-1)$$ This includes: - Current from metal->semiconductor: $$j_{m\rightarrow s}=j_0$$ - Current from semiconductor->metal: $$j_{s\rightarrow m}=j_0 e^{-\frac{eV_a}{k_BT}}$$ ## In the depletion region (using N-type as example) ### Net charge density: Following above, we have: $$\rho (x)=e(N_d-n)$$ where: - e: e- charge ### Electric field: $$E(x)=-\frac{eN_d}{\epsilon}(x-d)$$ Where: - \epsilon: permittivity - d: length of depletion region - x: distance from the point of interest to the junction interface ### Electric potential / voltage: $$V(x)=\int E(x) dx\\ = -\frac{eN_d}{\epsilon}(\frac{x^2}{2}-dx)$$