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SSCS - San Fernando Valley Chapter

IC Design Cheat Sheet

α→alpha β→beta γ→gamma δ→delta ϵ→epsilon ε→varepsilon ζ→zeta η→eta θ→theta ι→iota κ→kappa λ→lambda μ→mu ν→nu ξ→xi ο→omicron π→pi ρ→rho σ→sigma τ→tau υ→upsilon ϕ→phi χ→chi ψ→psi ω→omega  \begin{aligned} &\alpha \to \text{alpha } \newline &\beta \to \text{beta } \newline &\gamma \to \text{gamma } \newline &\delta \to \text{delta } \newline &\epsilon \to \text{epsilon } \newline &\varepsilon \to \text{varepsilon } \newline &\zeta \to \text{zeta } \newline &\eta \to \text{eta } \newline &\theta \to \text{theta } \newline &\iota \to \text{iota } \newline &\kappa \to \text{kappa } \newline &\lambda \to \text{lambda } \newline &\mu \to \text{mu } \newline &\nu \to \text{nu } \newline &\xi \to \text{xi } \newline &\omicron \to \text{omicron } \newline &\pi \to \text{pi } \newline &\rho \to \text{rho } \newline &\sigma \to \text{sigma } \newline &\tau \to \text{tau } \newline &\upsilon \to \text{upsilon } \newline &\phi \to \text{phi } \newline &\chi \to \text{chi } \newline &\psi \to \text{psi } \newline &\omega \to \text{omega } \newline \end{aligned}

Vds≤(Vgs−VT):Id=12KpWL[2(Vgs−VT)Vds−Vds2]Vds>(Vgs−VT):Id=12KpWL[(Vgs−VT)2] \begin{aligned} &V_{ds} \leq (V_{gs} - V_T): &I_d &= \frac{1}{2} K_p \frac{W}{L} [2(V_{gs}-V_{T})V_{ds} - V_{ds}^{2}] \newline &V_{ds} > (V_{gs} - V_T): &I_d &= \frac{1}{2} K_p \frac{W}{L} [(V_{gs}-V_{T})^{2}] \end{aligned}

Id=12KpWL[2(Vgs−VT)Vds−Vds2]I_d = \frac{1}{2} K_p \frac{W}{L} [2(V_{gs}-V_{T})V_{ds} - V_{ds}^{2}]

Id=12KpWL[(Vgs−VT)2]I_d = \frac{1}{2} K_p \frac{W}{L} [(V_{gs}-V_{T})^{2}]

Vds≤(Vgs−VT):Id=12KpWL[2(Vgs−VT)Vds−Vds2] V_{ds} \leq (V_{gs} - V_T): I_d = \frac{1}{2} K_p \frac{W}{L} [2(V_{gs}-V_{T})V_{ds} - V_{ds}^{2}]

c=±a2+b2 c = \pm\sqrt{a^2 + b^2}

f(x)=∫−∞∞f^(ξ)e2πiξxdξ f(x)=\int_{-\infty}^{\infty} \hat{f}(\xi) e^{2 \pi i \xi x} d \xi

ρDvDt=∇⋅P+ρf\begin{equation*} \rho \frac{\mathrm{D} \mathbf{v}}{\mathrm{D} t}=\nabla \cdot \mathbb{P}+\rho \mathbf{f} \end{equation*}

E=∑iEi=E1+E2+E3+⋯\begin{equation} \mathbf{E}=\sum_{i} \mathbf{E}_{i}=\mathbf{E}_{1}+\mathbf{E}_{2}+\mathbf{E}_{3}+\cdots \end{equation}

a=b+c\begin{equation} a = b+c \end{equation}

((100∥50∥3600)(100∥50∥3600)+100)⋅13⋅(0.75)≈0.062\left( \frac{(100 \parallel 50 \parallel 3600)}{(100 \parallel 50 \parallel 3600) + 100 } \right) \cdot \frac{1}{3} \cdot \left( 0.75 \right) \approx 0.062