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.

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
  - $A_{vo}$ $A_{v o}$ / $A_{is}$ $A_{i s}$ / $G_m$ $G_{m}$ / $R_m$ $R_{m}$ - for the voltage/current source (no loads)
    - intrinsic gain
  - $A_v$ $A_{v}$ / $A_i$ $A_{i}$ - also considers $R_i$ $R_{i}$ and $R_o$ $R_{o}$
    - loaded gain
  - $G$ $G$ - everything, considers source resistance and multiple stages if applicable
    - overall system gain
- CD Amplifier
  - Source follower because $v_S$ follows $v_I$
- CS Amplifiers
  - - $r_o$ $r_{o}$ defines how leaky the transistor is due to channel-length modulation (increasing $v_{DS}$ $v_{D S}$ also increases $i_D$ $i_{D}$ ). See [[#How $r_o$ $r_{o}$ works]]. Without channel-length modulation ( $\lambda=0$ $λ = 0$ ), $r_o$ $r_{o}$ is $\infty$ $\infty$
      - $r_o=\frac1{\lambda I_D'}$
    - $R$ 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 $A_v=\frac{v_o}{v_i}=-g_m(r_o||r_D||r_L)$ $A_{v} = \frac{v _{o}}{v _{i}} = - g_{m} (r_{o} ∣∣ r_{D} ∣∣ r_{L})$ - 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 $g_m$ $g_{m}$ indicates how much $i_D$ $i_{D}$ changes for small changes in $v_I$ $v_{I}$ AKA $v_{GS}$ $v_{GS}$ . How much is transferred over.
    - $g_m\stackrel\Delta=\frac{i_D}{v_{GS}}|_Q$
  - Drain resistance $r_o\stackrel\Delta=\frac{v_{DS}}{i_D}$
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 - $\frac{I_{OUT}}{V_{IN}}$ $\frac{I _{O U T}}{V _{I N}}$ , how much the current output changes based on how the input voltage is modified
    - process transconductance $k'=\mu_n C_{ox}$ $k^{'} = μ_{n} C_{o x}$
      - 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, $V(t)=V(\infty)+[V(0^+)-V(\infty)]e^{-t/\tau}$ $V (t) = V (\infty) + [V (0^{+}) - V (\infty)] e^{- t / τ}$
  - 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 $P=IV$ $P = I V$
  - 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
- 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)
- $\epsilon_r=\frac\epsilon{\epsilon_0}$ $ϵ_{r} = \frac{ϵ}{ϵ _{0}}$
  - $\epsilon_r$ - relative dielectric constant
  - $\epsilon$ - dielectric constant
  - $\epsilon_0$ - permittivity of free space (9E-12 F/m)
- $C=\frac{\epsilon A}{d}$ $C = \frac{ϵ A}{d}$
  - 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]])
- $I_{R_1}=I_\text{tot}\frac{R_\text{tot}}{R_1}$ $I_{R_{1}} = I_{tot} \frac{R _{tot}}{R _{1}}$ . 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

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 ( $i_D$ ). $r_o=\frac1{\lambda I_D'}$