Circuits

This is a combination of knowledge that I have gained from ENGR 40M and EE 101A, both of which I took during summer quarter in 2025, and self-learning.

• Phasors

  • sin(x)=cos(x-90°)

• Impedance $Z$

  • $Z_C=\frac1{j\omega C}$

  • $Z_L=j\omega L$

  • $Z=R+jX$

    • R - resistance

    • X - reactance

• Admittance $Y\equiv\frac1Z=G+jB$

  • G - conductance

  • B - susceptance

• Capacitors have negative reactance and inductors have positive reactance

• Four amplifier types

  • 

    These four are equivalent. It's like VCVS, CCCS, VCCS, and CCVS. To remember, think of transconductance as having the gain of conductance so output current/input voltage and for transresistance as having the gain of resistance so output voltage/input current.

    • When there is an output voltage (for transresistance and voltage amplifier), it should be measured when the output terminals are open for there to be no current flowing through $R_o$ which would decrease $v_o$.

    • When there is an output current, it should be measured when short-circuited so that none of the output current flows through $R_o$ (you get all the current flowing from the current source in $i_o$)

    • trans - input and output of different types/at different places (voltage → current, current → voltage)

    • m in $G_m$ and $R_m$ is for "mutual" like in "mutual inductance" because the current and voltage are at separate ports (similar to trans-)

  • Equations for swapping between them using transformations (see derivation at [[#Four Interchangeable Amplifier Types]]): $A_{vo}=A_{is}\frac{R_o}{R_i}=G_mR_o=\frac{R_m}{R_i}$

  • Three parameters necessary for forming an amplifier: $R_{\text{IN}}$, $R_{\rm OUT}$ and gain

  • Gains

    • $A_{vo}$/$A_{is}$/$G_m$/$R_m$ - for the voltage/current source (no loads)

      • intrinsic gain

    • $A_v$/$A_i$ - also considers $R_i$ and $R_o$

      • loaded gain

    • $G$ - everything, considers source resistance and multiple stages if applicable

      • overall system gain

  • CD Amplifier

    • Source follower because $v_S$ follows $v_I$

  • CS Amplifiers

    • 

      • $r_o$ defines how leaky the transistor is due to channel-length modulation (increasing $v_{DS}$ also increases $i_D$). See [[#How $r_o$ works]]. Without channel-length modulation ($\lambda=0$), $r_o$ is $\infty$

        • $r_o=\frac1{\lambda I_D'}$

      • $R$ models the voltage source's real-world imperfection (cannot supply infinite current)

      • Capacitors block out DC bias, just allowing the signal to pass through

    • Gain $A_v=\frac{v_o}{v_i}=-g_m(r_o||r_D||r_L)$ - is lower when there's a load resistance

      • When no load resistance, open circuit gain $A_{vo}=-g_m(r_o||r_D)$

    • $R_D$ allows the output current to be converted into an output voltage

    • Early voltage $V_A$ is $\frac1\lambda$ so $r_o=\frac{|V_A|}{I_D}$ meaning it's good if $V_A$ and $r_o$ are large because then the MOSFET is not very leaky

    • Bias circuit - the part of the circuit that sets the DC voltage so the transistor can operate in a small-signal linear region without distortion

    • Operating point - point at which load line (which describes all current-voltage pairs that the load constrains) matches the characteristic equation of the device

      • Q-point - point at which signal voltage is 0 (quiescent as in dormant/inactive)

    • A VCCS is a transconductance amplifier

    • transconductance $g_m$ indicates how much $i_D$ changes for small changes in $v_I$ AKA $v_{GS}$. How much is transferred over.

      • $g_m\stackrel\Delta=\frac{i_D}{v_{GS}}|_Q$

    • Drain resistance $r_o\stackrel\Delta=\frac{v_{DS}}{i_D}$

• Graphs

  • voltage transfer characteristic (VTC) $v_{out}$-$v_{in}$ graph

• MOSFETs

  • MOSFET acts as a VCCS with a transconductance $g_m$

  • Operate in saturation region for amplifiers and triode region for CMOS

  • CMOS - complementary MOS, manufacture nMOS and pMOS transistors on the same device

  • body is connected to source for both nMOS and pMOS

  • $r_o$ is for saturation

  • $r_{DS}$ is for triode (variable resistor)

  • nMOS in enhancement-mode

    • transconductance - $\frac{I_{OUT}}{V_{IN}}$, how much the current output changes based on how the input voltage is modified

      • process transconductance $k'=\mu_n C_{ox}$

        • mobility $\mu_n$

    • $g_{DS} = \frac{1}{r_{DS}}=\mu C_{ox} \frac W L v_{ov}=k_nv_{OV}$ when $v_{DS}$ is negligible

    • $I=g_{DS}v_{DS}$

    • called enhancement-type MOSFET because increasing $v_{GS}$ above $V_t$ enhances the region's conductivity. For depletion-mode MOSFET, increasing $v_{GS}$ above $V_t$ depletes the region (stops it from conducting).

    • $C_{ox}=\frac{\varepsilon_{ox}}{t_{ox}}$ where $\epsilon_{ox}=3.9\epsilon_0$

    • Equations for the three regions (cutoff, triode, and saturation):

    • $k'=\mu C_{ox}$

    • $k=\mu C_{ox}\frac W L$

    • channel-length modulation occurs because as $v_{DS}$ increases, the channel length changes, changing the current

• Drift velocity $v_D=\mu E$ where $\mu$ is the mobility

• RC circuit (derivation at [[#RC Circuit]])

  • In general, $V(t)=V(\infty)+[V(0^+)-V(\infty)]e^{-t/\tau}$

    • I interpret this as the first term being the permanent response and the second being the transient response due to the $e^{-t/\tau}$ multiplier. Since a-b means to a from b, the sum is V(∞) + a vector that disappears with time that says go to V(0) from V(∞). Here it is animated:

      

  • Charging

    • $Q(t)=Q_{\text{charged}}(1-e^{-t/\tau})$

    • $I(t)=\frac{V_b}Re^{-t/\tau}$

    • $V(t)=V_b(1-e^{-t/\tau})$

  • Discharging

    • $Q(t)=Q_0e^{-t/\tau}$

    • $I(t)=\frac{-V}Re^{-t/\tau}$

    • $V(t)=V_0e^{-t/\tau}$

• NOT, NAND, NOR Gates

• Tellegen's theorem - in an electrical network, sum of instantaneous powers in all branches is 0

• Diode

  • Assuming

    • Assume that it is ON ($V_f$ is reached). If we can show that there is a positive current through it, then there must be $V_f$ across the diode.

    • Assume that it is OFF. If we can show that there is a voltage across it greater than $V_f$, then it must be ON.

  • Assume all OFF. When in parallel and choosing which one to turn ON, choose the one with lower $V_f$

  • Models

    • ideal diode - V_f=0

    • constant voltage drop model

      • FKA piecewise linear model - V_f=0.7

    • Shockley model $I = I_s \left( e^{\frac{V_D}{n V_T}} - 1 \right)$

    • reverse bias model

  • Open circuit voltage, short circuit current

  • Ideal diode

    • Acts like an open circuit when voltage less than the forward voltage

    • Acts like a constant voltage source for voltages greater than the forward voltage

    • Ideal diode's forward voltage is 0V

  • p-n junction

    • Depletion region

  • Silicon diode forward voltage - 0.7 V

  • Doping

    • P-type doping - Boron, adds a hole

    • N-type doping - Phosphorus, adds a free electron

    

• High voltage is used in transmission because

  • High voltage means that the current goes down for the same power because $P=IV$

    • A lower current means that the voltage drop between the generator and substation, $V=IR$, is smaller since the $R$ of the transmission line is constant and the $I$ decreases. The voltage drop across the transmission lines modeled as resistors went down even though the voltage on the line is higher. With a constant R and a lower current with HV, $P_{\text{loss}}=I^2R$ goes down and $P_{\text{loss}}=\frac{V^2}R$ also goes down since V in the power equation measures the voltage drop, which went down.

• Inductor

  • $L=k\mu AN^2$

    • k - geometry

    • $\mu$ - permeability

    • A - cross-sectional area

    • N - # turns

  • Energy stored is $\frac12LI_{\rm final}^2$

  • constant-current device

  • $v=Li'$

  • equivalent combinations like that of a resistor

  • Time constant $\tau=\frac{L}R$

• Capacitors

  • constant-voltage device

  • Voltage cannot change instantaneously on a capacitor (unless we do dirac delta functions)

  • $\epsilon_r=\frac\epsilon{\epsilon_0}$

    • $\epsilon_r$ - relative dielectric constant

    • $\epsilon$ - dielectric constant

    • $\epsilon_0$ - permittivity of free space (9E-12 F/m)

  • $C=\frac{\epsilon A}{d}$

    • C ∝ A and C ∝ 1/d

  • $Q=CV$ and $I=CV'$

  • Types

    • Ceramic - cheap, low C

    • Polymer - HV

    • Electrolytic - polar, large C

    • Surface mount capacitors

    • IC capacitor

• Waves

  • $\frac1f=T=\frac\lambda{v}\to f=\frac{v}\lambda$

• Nodal analysis: write out KCL (for nodes with unknown voltages) in terms of voltages

  • Full steps

    1. Choose reference node and label the nodes' voltages. Use $V_1, V_2, ...$ for unknown. Use voltage divider to find out relevant voltages if possible.

    2. Label current directions

    3. Apply KCL at all non-reference nodes (using Ohm's law integrated in the eqns)

• Current divider ([[#Current Divider|derivation]])

  • $I_{R_1}=I_\text{tot}\frac{R_\text{tot}}{R_1}$. mn current ∝ inverse of %resistance in same component

    • $I_{R_1}=I_\text{tot}\frac{R_2}{R_1+R_2}$ (special case of two resistors)

• Voltage divider ([[#Voltage Divider|derivation]])

  • $V_1=V_\text{tot}\frac{R_1}{R_\text{tot}}$. mn voltage ∝ %resistance in same component

• Resistor Chart

  

• Double-subscript notation:

  

• Combination in series and parallel

  • Equivalent resistance for series resistors is sum because KVL means voltages add up and current is the same in series: $V_\text{eq}=V_1+V_2+V_3$ becomes $I(R_{\text{eq}})=I(R_1+R_2+R_3)$

  • Equivalent resistance for parallel resistors is the way it is because KCL means currents add up and voltages constant: $I_{\rm eq}=I_1+I_2+I_3$ becomes $\frac V{R_{\rm eq}}=\frac V{R_1}+\frac V{R_2}+\frac V{R_3}$

  • These all have the same behavior for equivalent resistance/inductor:

    $V=IR$ - resistance

    $V=LI'$ - inductance

  • Likewise, these all act the same (flipped, so in parallel they act like resistors would in series)

    $I=VG$ - conductance

    $I=CV'$ - capacitance

Derivations

Op-Amp

Four Interchangeable Amplifier Types

Amplifier $A_{vo}=-g_MR_D$

Also, $|A_{v}|=g_mR_L'\approx g_mr_o=(\frac{2I_D}{V_{OV}})(\frac{1}{\lambda I_D})=\frac2{\lambda V_{OV}}$

Transconductance $g_M$ Expressed in More Ways

In saturation, $g_M=k_nV_{OV}=\frac{2I_D}{V_{OV}}=\sqrt{2I_{DQ}k_n}$

Transconductance $g_M$ in the 3 MOSFET Modes

$g_M=k_nv_{OV}$ in saturation

$g_M=k_nv_{DS}$ in triode

Amplifier $i_D$

How $r_o$ works

o in $r_o$ is for "output" as in "output resistance" since it models the leakiness of a current source. For an ideal current source, increasing the voltage wouldn't affect how much the current source delivers. For a non-ideal leaky current source (which $r_o$ simulates), increasing voltage ($v_{DS}$) also increases current output ($i_D$).

$r_o=\frac1{\lambda I_D'}$