# Magnetically Coupled Circuits - Turns ratio $n=N_2/N_1\to v_2=nv_1$ - $\tilde{v}=jwL _1\tilde{I}_1 +j\omega M \tilde{I_2}$ where $M=\frac{\mu N_1 N_2}{\ell}$ - If dots on same side, $\tilde{v}_{1}+\tilde{v}_{2}$, otherwise its $\tilde{v}_{1}-\tilde{v}_{2}$ # Ideal Transformers - Following rules apply if $\tilde{I}_{L}$ goes into transformer - If dots on opposite side, $\tilde{v}_{2}=n\tilde{v}_{1}, \tilde{I}_{2}=\frac{\tilde{I}_{1}}{n},z_{2}=n^2z$ - If dots on opposite side, $\tilde{v}_{2}=-n\tilde{v}_{1},\tilde{I}_{2}=-\frac{\tilde{I}_{1}}{n},z_{2}=n^2z_{1}$ # Semiconductors - Intrinsic carrier concentration $n_{i}$ - Total carrier concentration $n$ - Holes $p$ - For intrinsic semiconductor $n_{i}=n=p$ - For doped semiconductor, $np=n_{i}^2$ - Doping is useful if $N_{D},N_{A}\gg n_{i}$ # PN Junctions - $N_{D}x_{n}=N_{A}x_{p}$ where $x_{n}$ is depletion width on $n$ side and $x_{p}$ is depletion width on $p$ side, $w=x_{p}+x_{n}$ - Current density: movement of +ve and -ve charged particles create current. Current density $J$ is $\frac{I}{cm^3}$ - $J_{n,diff}=qD_{n} \frac{dn}{dx}$ - $J_{p,diff}=-qD_{p} \frac{dp}{dx}$ - $J_{p,drift}=q\mu_{p} \mathcal{E}_{x}$ - $J_{n,drift}=-q\mu_{n}n\mathcal{E}_{x}$ - $\mathcal{E}_{max}=\frac{2v_{bi}}{w}$ where $v_{bi}$ is "built-in voltage" - $v_{bi}=\frac{kT}{q}\ln\left( \frac{N_{a}N_{D}}{n_{i}^2} \right)$ - $v_{bi}=\frac{2kT}{q}\ln\left( \frac{N_{A}^2}{n_{i}} \right)$ for $N_{A}=N_{D}$ - $\mathcal{E}_{max}=\frac{v_{bi}}{x_{p}}$ $x_{n}=\sqrt{ \frac{2\varepsilon _{r}\varepsilon_{0}}{q} \left( \frac{N_{D}}{N_{A}(N_{A}+N_{D})} \right) v_{bi} },~x_{p}=\sqrt{ \frac{2\varepsilon _{r}\varepsilon_{0}}{q} \left( \frac{N_{A}}{N_{A}(N_{A}+N_{D})} \right)v_{bi} },~w=\sqrt{ \frac{2\varepsilon_{r}\varepsilon_{0}}{q} \left( \frac{N_{A}+N_{D}}{N_{A}N_{D}} \right)v_{bi} }$ >[!note] Bias in PN Junction >1. No bias: $v_{s}=0$ -> no flowing current thru PN junction b/c $e^-$ @ equilibrium >2. Forward bias: $v_{s}>0$ -> shrinks depletion zone; when $v_{s}>v_{bi}$ (typically 0.7V), current starts flowing from $p\to n$ or $e^-$ from $n\to p$ >3. Reverse bias: $v_{s}<0$ -> widens depletion zone since (-) from $v_{s}$ adds to $p$ and (+) adds to $n,$ no current flows, $I=0$. Tries to create current $I_{0}$ in opposite direction ($n\to p$), $I_{0}\approx E-12$ due to minority carriers ## Diodes $ \begin{cases} \text{On (FB)} & v_{d}=0,~I_{d}\geq0 \\ \text{Off (RB)} & I_{d}=0, ~v_{d}\leq0 \end{cases} $ - VI relationship given by $I_{D}=I_{O}\left( \exp\left( \frac{qv_{D}}{kT} \right)-1 \right)$ - $V_{s}=V_{D}+R_{D}\left[ I_{D}\left( \exp\left( \frac{qv_{D}}{kT} \right) -1\right) \right]$ (pg. 13) - **Graphical method** 1. Calculate $I_{D}=v_{\frac{s}{R}}$ 2. Draw a line between the points $(0,I_{D})$ and $(v_{S},0)$. 3. Intersection of this line and the function $I_{D}(v_{D})$ gives an approximation for $v_{D}$ and $I_{D}$, iterate until point converges - **Ideal Diode Model** - Assume $I_{D}=0$ for RB and $v_{bi}=0$ for FB. - Main condition for RB is $v_{D}<0$ and $I_{D}>0$ for FB 1. Guess diode bias - FB -> short - RB -> open 2. Solve circuit and check for consistency. If you get an inconsistent conclusion, retry with the other assumption ## MOSFETs - Ohmic region follows $v_{gs}-v_{T}>v_{ds}$ - $v_{g}-v_{T}>v_{d}$ for $v_{s}=0$ - Saturation region follows $v_{ds}>v_{gs}-v_{T}$ - $v_{d}>v_{g}-v_{T}$ for $v_{s}=0$ - $I_{ds,ohmic}=k\left[ (v_{gs}-v_{T})v_{ds}-\frac{1}{2} v_{ds}^2 \right]$ - $I_{ds,sat}=k\left[ \frac{1}{2}(v_{gs}-v_{T})^2 \right]$ - $P_{mosfet}=v_{ds}I_{d}=I_{d}(v_{d}-v_{s})$ - $v_d=v_{dd}-I_{d}R_{d}$ >[!note] Amplification >- Overdrive voltage $v_{ov}=v_{gs}-v_{T}$ >- $I_{D}=\frac{k}{2}V_{OV}^2$ >- Transconductance $g_{m}=kv_{OV}=\frac{2I_{D}}{v_{OV}}=\sqrt{ 2kI_{D} }$ >- Voltage gain $A_{v}=\frac{v_{o}}{v_{i}}\approx-g_{m}R_{D}$