Basic Amplifier Configurations

Q1: What are the basic amplifier configurations? $\frac {g_mr_\pi R_E}{1+(\frac{R'_L}{r_o})(1+\frac{R_E}{r_\pi})}$
A1: There are 8 basic amplifier configurations: 4 basic BJT amplifier configurations and 4 basic MOS amplifier configurations. Note that that taking the limit as $r_\pi \to \infty$ of the BJT equations produces the MOS equations.

Common Base:

$A_v=g_m(r_o || R'_L)$

$R_i=r_\pi || \frac {r_o+R'_L}{1+g_mr_o}$ 

Common Gate:

$A_v=g_m(r_o || R'_L)$

$R_i= \frac {r_o+R'_L}{1+g_mr_o}$  

Common Source:

$A_v=-g_m(r_o || R'_L)$

$R_i=\infty$ 

Common Emitter:
$A_v=-g_m(r_o || R'_L)$
$R_i=r_\pi$ 

Common Emitter with $R_E$:

$A_v=-\frac{g_mR'_L}{1+g_mR_E+(R'_L/r_o)(1+R_E/r_\pi)}$

$R_i=r_\pi + \frac {g_mr_\pi R_E}{1+(R'_L/r_o)(1+R_E/r_\pi)}$

Common Source with $R_S$:

$A_v=-\frac{g_mR'_L}{1+g_mR_E+(R'_L/r_o)}$

$R_i=\infty$

Common Collector/Emitter Follower:

$A_v=\frac{g_m(r_o || R'_L)}{1+g_m(r_o || R'_L)}$

$R_i=r_\pi + \beta (r_o || R'_L)$ 

Common Drain/Source Follower:

$A_v=\frac{g_m(r_o || R'_L)}{1+g_m(r_o || R'_L)}$

$R_i=\infty$ 

Introduction to Small Signal Model

Q1: How can I analyze real amplifier circuits?
A1: You can divide the amplifier circuit into its signal circuit and bias circuit. For now, we will consider the signal circuit.

Q2: How do I map real circuit elements to their signal circuit counterparts?
A2: By using their small signal models (SSM):

Resistors and capacitors behave the same in signal circuits

Independent voltage sources mapped to short circuits while
independent current sources are mapped to open circuits

Things change a bit with nonlinear elements.
The diode maps to a resistor of $r_d=\frac{nV_T}{I_D}$

$r_\pi=\frac{V_T}{I_B}$
$g_m=\frac{I_C}{V_T}=\frac{\beta}{r_\pi}$
$r_o\approx \frac{V_A}{I_C}$

Both the NPN and PNP map to the same SSM

$g_m=\frac{2I_D}{V_{OV}}$
$r_o\approx \frac{1}{\lambda I_D}$
Both the NMOS and PMO map to the same SSM

Mirror Circuit

Q1: What's a mirror circuit?
A1:

$Q_{ref}$ and $Q_1$ are identical NPN transistors.
This is why current splits evenly to each of their bases.

Q2: What state is $Q_{ref}$ in?
A2: Since $i_{C,ref}>0$, we know that $Q_1$ must be on. Hence, $V_{BE,ref}=V_{D0}$
Note that the base and collector share the same node in $Q_{ref}$.
Hence, $V_{BE,ref}=V_{CE,ref}=VD_{0}$
Therefore, $Q_{ref}$ is always active.

Q3: How is $I_1$ related to $I_{ref}$?
A3: First, let's find an expression for $I_{ref}$:
By KCL: $I_{ref}=2i_B+i_{c,ref}$
Since $Q_{ref}$ active: $I_{ref}=2\frac{i_{C,ref}}{\beta}+i_{C,ref}$
$I_{ref}=i_{C,ref}(\frac{2}{\beta}+1)$

Next, let's examine $I_1$:
$I_1$=$i_C$
$I_1=\beta i_B$
$I_1=\beta \frac{i_{C,ref}}{\beta}$
$I_1=i_{C,ref}$
$I_1=\frac {I_{ref}}{\frac{2}{\beta}+1}$

For $\beta \gg 1$:
$I_1\approx I_{ref}$

By KVL through $Q_{ref}$ (either the BE or CE branch):
$I_1\approx I_{ref}=\frac{V_{CC}-V{D0}+V{EE}}{R}$

Q2: Why use a mirror circuit?
A2: You can use a mirror circuit to behave as a current source in order to bias a BJT.

Q3: Why not use a mirror circuit?
A3: The mirror circuit produces integer multiples of the reference current. For a more flexible option, you might want to look into current-steering circuits.

MOS Circuits

Q1: What states can a MOS be in?
A1: A NMOS may be in:

1. Cutoff
• $v_{OV}\leq 0$
• $i_D=0$
2. Saturation
• $v_{OV}\geq 0$ and $v_{DS}\geq v_{OV}$
• $i_D=0.5\mu_nC_{ox}\frac{W}{L}v_{OV}^2(1+\lambda v_{DS})$
3. Triode
• $v_{OV}\geq 0$ and $v_{DS}\leq v_{OV}$
• $i_D=0.5\mu_nC_{ox}\frac{W}{L}(2v_{OV}v_{DS}- v_{DS}^2)$

A couple of notes:

• $v_{OV}=v_{GS}-v_{tn}$ where $v_{OV}$ is the overdrive voltage and $v_{tn}$ is the threshold voltage associated with the NMOS
• The expressions for $i_D$ in saturation and triode both contain the pre-factor $0.5\mu_nC_{ox}\frac{W}{L}$ 
• $\mu_n$ is the negative charge carrier effective mobility
• $C_{ox}$ is gate-oxide capacitance per unit area
• $W$ is the gate width
• $L$ is the gate length
• The expression for $i_D$ in saturation contains the channel-length modulation parameter $\lambda$. 
• If channel-length modulation can be neglected, then $\lambda=0$
• The channel-length modulation parameter is related to early voltage by: $\lambda=\frac{1}{v_A}$
• The phenomenon of channel-length modulation in MOSFETs can be compared to Early effect in BJTs

For a PMOS:

1. $v_{OV}=v_{GS}-v_{tn}\to v_{OV}=v_{SG}-|v_{tp}|$
2. $\mu_n \to \mu_p$
3. $v_{DS}\to v_{SD}$
4. $v_{GS}\to v_{SG}$

Q2: How do I solve MOS circuits?

A2: For an NMOS circuit:

1. Write the KVLs for $v_{GS}$ and $v_{DS}$
2. Assume cutoff by setting $i_D$=0. Check that you are in cutoff.
3. Assume saturation. Check that $v_{DS}\geq v_{OV}$
4. You're in triode mode. 

Note that the equations for $i_D$ in saturation and triode mode are quadratic. Although you may find 2 roots, only keep the 1 that's valid for that particular state.

MOS Device

Q1: What is a MOSFET?
A1: A MOSFET is a metal oxide semiconductor field-effect transistor. They're often abbreviated as MOS devices. A MOSFET has 4 terminals:

1. Gate
2. Body
3. Drain
4. Source

Q2: What are some types of MOS devices?

A2: There's n-channel metal oxide semiconductor (NMOS) and p-channel metal oxide semiconductor (PMOS).

Current direction is reversed in the PMOS

Note the 4 terminals of the NMOS

In practice, the body terminal is often connected
to the source terminal of the NMOS

Same goes for the PMOS

Combining the body and source terminals
simplifies the NMOS to a 3 terminal device

Same for the PMOS

Q3: What circuit variables do I need to look for in a MOSFET?
A3: For a MOSFET in which the body and source terminals are connected, there are only 3 currents and 3 voltages to deal with. By KVL, you know that 1 voltage may be written in terms of the other 2. Same goes for current. So it looks like we need to account for 4 circuit variables, as was the case for the BJTs. However, MOSFETs differ from BJTs in that one current is already specified: $i_G$ is always 0. Hence, you only need to keep track of 3 circuit variables.

For an NMOS you might choose to keep track of:

1. $i_D$
2. $v_{GS}$
3. $v_{DS}$

For a PMOS, the current direction is reversed and the voltage subscripts are flipped. 

Universal and Empty Set

Q1: What is the complement of a set $A$?
A1: The complement of a set $A$ is the set that contains all elements that do not belong to A.

$A^c=\{ x: x\not\in A\}$

It follows that:
$(A^c)^c=A$

Q2: What is the universal set?
A2: The universal set $\Omega$ is the set that contains all conceivable elements.
Hence, $A \cup A^c = \Omega$

Q3: What is the complement of the universal set?
A3: The empty set {} is the complement of the universal set. The empty set contains no elements.
Hence, $A \cap A^c = \{\}$

Union and Intersection

Q1: What is a union of sets?
A1: A union of sets is a a set whose elements are belong to at least one of the sets in the union.
Given: For 2 sets $A$ and $B$:
$A\cup B=\{ x: x\in A\ or\ x\in B\}$
Given: For countable collection of infinite sets $C_1, C_2, C_3, ...$:
$\bigcup\limits_{i=1}^{\infty}C_i=C_1 \cup C_2 \cup C_3 \cup ... =\{ x: x\in C_i\ for\ some\ i\}$

Q2: What is an intersection of sets?
A2: An intersection of sets is a set that contains only those elements that are shared by all sets that form the intersection.
Given: For 2 sets $A$ and $B$:
$A\cap B=\{ x: x\in A\ and\ x\in B\}$
Given: For countable collection of infinite sets $C_1, C_2, C_3, ...$:
$\bigcap\limits_{i=1}^{\infty}C_i=C_1 \cap C_2 \cap C_3 \cap ... =\{ x: x\in C_i\ for\ all\ i\}$

Q3: What is a disjoint collection of infinite sets $C_1, C_2, C_3, ...$?
A3: A disjoint collection of infinite sets is a collection of sets whose intersection is the empty set.
$\bigcap\limits_{i=1}^{\infty}C_i=\{\}$

Q4: What does it mean if a family of sets $P$ composed of cells $P_1, P_2, P_3, ...$ partition a set $A$?
A5: If $P$ partitions a set $A$, the following 3 conditions hold:

1. The partition does not contain the empty set
• $\{\}\not\in P$
2. The union of the cells is the set $A$. The partition covers $A$.
• $\bigcup\limits_{i=1}^{\infty}P_i=A$
3. No 2 cells share any elements. The partition is pairwise disjoint.
• $P_i\cap P_j=\{\}\ for\ all\ i\not =j$

Subsets

Q1: How do I indicate that an element belongs to a particular set?
A1: Going back to the coin example, suppose you wanted to indicate that heads was an element of the subset of coin tosses $A$.
You could write: $H\epsilon A$

Q2: What is a subset?
A2: A subset is a set whose every element belongs to another set. For example, consider the set of outcomes denoting the toss of a single die:
$D=\{1, 2, 3, 4, 5, 6\}$
Then consider the set $C=\{1, 2, 3\}$
Clearly, each element of C is also an element of D. Hence, C is a subset of D:
$B\subset D$

Also, D is a subset of itself:
$D\subset D$

Q3: What is set equality?
A3: Suppose you have 2 sets, $A$ and $B$.
$A=B$ if and only if $A\subset B $ and $B\subset A$

Introduction to Set Theory

Q1: What is a set?
A1: A set is a collection of elements.
For example, the set of outcomes for a coin toss may be represented as:
$A=\{H, T\}$
where A is the set of outcomes including heads and tails.

Q2: What is the size of a set?
A2: The size of a set is the number of elements it contains.
For the coin toss example, it was clear that $A$ represented a finite set of outcomes.
Moreover, |A|=2. That is the size of A is 2.

However, you can consider other sets, like the set of integers $\Bbb Z={..., -2, -1, 0, 1, 2, ...}$.
Not only is the set of integers an infinite set, but it is countably infinite. Like the finite set of coin tosses, you could go about counting these elements in a systematic way, even if there are an infinite amount of elements.

On the other hand, you could look at the unit set interval $S=[0,1]$, which includes all real numbers greater than or equal to 0 and less than or equal to 1. Like the set of integers, the unit set interval has an infinite number of elements. But there is no conceivable by way by which you could attempt to count the elements in the unit set interval. Hence, it is an uncountable infinite set. Yet we can still associate a size to it: $|S|=1$. In this context, the size of the unit set interval refers to its length, and not the number of elements it contains.

Box-Tukey Paradigm

Q1: What is the Box-Tukey paradigm?
A1: The Box-Tukey paradigm is the conceptual framework for statistical reasoning named after George Box and John Tukey:

Power Reduction Formula

Q1: How can I simplify $cos^2(x)$?
A1: Recall that Euler's formula implies that:
$cos(x)=\frac{1}{2}(e^{jx}+e^{-jx})$

Hence,
$cos^2(x)=\frac{1}{4}(e^{jx}+e^{-jx})^2$
$cos^2(x)=\frac{1}{4}(e^{2jx}+2+e^{-2jx})$
$cos^2(x)=\frac{1}{2}(\frac{e^{2jx}+e^{-2jx}}{2}+1)$
$cos^2(x)=\frac{1}{2}(cos(2x)+1)$
$cos^2(x)=\frac{1+cos(2x)}{2}$

Q2: How can I simplify $sin^2(x)$?
A2: Using similar logic as in A1:
$sin(x)=\frac{1}{2j}(e^{jx}-e^{-jx})$

Hence,

$sin^2(x)=-\frac{1}{4}(e^{jx}-e^{-jx})^2$
$sin^2(x)=-\frac{1}{4}(e^{2jx}-2+e^{-2jx})$
$sin^2(x)=-\frac{1}{2}(\frac{e^{2jx}+e^{-2jx}}{2}-1)$
$sin^2(x)=-\frac{1}{2}(cos(2x)-1)$
$sin^2(x)=\frac{1-cos(2x)}{2}$

Q3: How can I simplify $tan^2(x)$?
A3: Using the results of A1 and A2:
$tan(x)=\frac{sin(x)}{cos(x)}$
$tan^2(x)=\frac{sin^2(x)}{cos^2(x)}$
$tan^2(x)=\frac{1-cos(2x)}{1+cos(2x)}$

Bode Plot of Magnitude

Q: How can I graphically depict ratio of output to input power for a LTI system?

A: Suppose the following:

We could examine $\frac{P_o}{P_i}$

Alternatively, we could examine $log(\frac{P_o}{P_i})$ which has dimensionless units of bels

More commonly, we use $10log(\frac{P_o}{P_i})$ which gives us decibels (dB)

Let's take a closer look at this $10log(\frac{P_o}{P_i})$:

$P_i=\frac{1}{2}Re\left\{V_iI_i^*\right\}$

$P_i=\frac{1}{2}Re\left\{\frac{V_iV_i^*}{R}\right\}$

$P_i=\frac{1}{2}Re\left\{\frac{|V_i|^2}{R}\right\}$

$P_i=\frac{|V_i|^2}{2R}$

Similarly,

$P_o=\frac{|V_o|^2}{2R}$

Hence, $10log\frac{P_o}{P_i}$ =

=$10log\frac{|V_o|^2}{|V_i|^2}$ 

=$20log\frac{|V_o|}{|V_i|}$ 

Recall that $|H(j\omega)|=\frac{|V_o|}{|V_i|}$

So $10log(\frac{P_o}{P_i})=20log|H(j\omega)|$

To visualize the ratio of power with respect to angular frequency, $\omega$ sits on the horizontal axis using a logarithmic scale while $20log|H(j\omega)|$ sits on the vertical axis. From these 2 axes, a Bode plot may be constructed. For a rough and quick idea of the circuits behavior, straight line approximations are commonly used.

BJT Circuits

Q1: What are the states a BJT may be in?
A1: An NPN may have 3 states, which can be linearly modeled as:

1. Cutoff: BE junction is reverse biased
• $i_B=0$, $i_C=0$
• $v_{BE}<v_{D0}$
2. Active: BE junction is forward biased, BC junction is reverse biased
• $v_{BE}=v_{D0}$, $i_B \geq 0$ 
• $i_C=\beta i_B$, $v_{CE}\geq v_{D0}$ 
3. Saturation*: BE junction is forward biased, BC junction is forward biased
• $v_{BE}=v_{D0}$, $i_B \geq 0$ 
• $v_{CE}=v_{sat}$, $i_C < \beta i_B$ 

*There are actually 3 subcategories of saturation: soft saturation, near cutoff saturation, and deep saturation. Here, we'll use saturation to mean deep saturation.

For a PNP, flip the order of the subscript letters for voltages and remember that current directions reverse (relative to the NPN)

For silicon, $v_{sat}=0.2V$

Q2: How can I solve BJT circuits?

A2: To specify the 6 BJT parameters, recall that you only need to find 2 currents and 2 voltages. For example, you might choose to calculate:

1. $v_{BE}$
2. $v_{CE}$
3. $i_{B}$
4. $i_{C}$

To calculate the 4 BJT parameters above, you may use the following algorithm: 

1. Write down the KVL equation for the BE and CE voltages.
2. Assume BJT is off: set $i_B=0$ and $i_C=0$. Check this assumption using the BE-KVL.
• If it follows that $v_BE<v_{D0}$, your assumption is correct. You can then use the CE-KVL to solve for $v_{CE}$
• Otherwise, the BJT has a forward biased BE junction. Set $v_BE=0.7$ to compute $i_B$
3. Assume the BJT is active: set $i_C=\beta i_B$. Check this assumption using the CE-KVL.
• If it follows that $v_{CE} \geq v_{D0}$, your assumption is correct
• Otherwise, the BJT must be in saturation
4. The BJT is in saturation: set $v_{CE}=v_{sat}$. 
• Use the CE-KVL to find $i_C$.

Bipolar Junction Transistor

Q1: What is a bipolar junction transistor (BJT)?
A1: A bipolar junction transistor is a two-junction, three-terminal semiconductor device. "Bipolar" refers to the presence of oppositely-charged charge carriers (electrons and holes). The device is divided into three regions:

1. Emitter
2. Base
3. Collector

A lead is connected to each region.

Notice the junctions separating the base from the emitter and collector

Q2: What are some types of BJTs?
A2:

1. NPN
   
2. PNP

Q3: Why doesn't the NPN transistor act like two PN junction diodes connected in series?
A3: Although the models above suggest that a symmetry shared by the emitter, base, and collector regions, the actual devices are designed to avoid this symmetry.

Q4: How do I label the circuit variables associated with the NPN and PNP ?

A4: A single letter subscript indicates the region to which the current is associated with. A two letter subscript indicates the voltage at the first letter with respect to the second letter.

NPN Circuit Variables
Even though there's are 6 circuit variables,
2 can be written in terms of the others
using KCL and KVL. Same goes for the PNP.

PNP Circuit Variables
Notice that current directions and
letter order in the subscripts for voltage
 are reversed with respect to the NPN

Q5: How might I see the NPN and PNP in a schematic?

A5: 

NPN Symbol
Notice the arrow pointing away
from the base and towards the emitter

PNP Symbol
Notice the arrow pointing towards
 the base and away from the emiter

iv Characteristic of a Diode

Q1: How can I understand the iv characteristic of a diode?
A1: You could use the Shockley diode equation to relate $i_D$ to $v_D$:
$i_D=I_S(e^\frac{v_D}{nv_T}-1)$

• $I_S$ should be familiar to you. For a rough idea, $I_S$ can range from $10^{-9}$ to $10^{18}$A
• n is a constant ideality factor that ranges from 1 to 2. We'll largely assume the ideal case where n=1.
• $v_T$ is a constant thermal voltage. At room temperature, $v_T\approx 26mV$

Let's confirm that the Shockley diode equation is consistent with what you already know:

• When the diode is forward-biased, the current through the diode increases dramatically.

$v_D$ is sufficiently large $\rightarrow$ The exponential dominates the expression inside the parentheses 

$i_D=I_Se^\frac{v_D}{nv_T}$

• When the diode is reverse-biased, a small current flows in reverse.

$v_D$ is sufficiently small $\rightarrow$ The -1 dominates the expression inside the parentheses 

$i_D=-I_S$

But note that the Shockley diode equation does not account for a diode's breakdown region.

Q2: Why should I be concerned that the Shockley diode equation describes a nonlinear relationship between voltage across and current through a resistor?
A2: Because the presence of a nonlinear element in the circuit means that you can't solve the circuit analytically. You could use numerical or graphical methods of solution. But the method we'll look at is the piecewise linear model (PWL) of the diode:

1. Replace the exponential that takes off past the knee voltage $v_{D0}$ by a vertical line.

• For silicon, $v_{D0}$ has an approximate range of 0.6-0.7V

1. Replace $I_S$ with 0 (assumes the reverse saturation voltage is negligible).

We can define 2 states: 

1. On: $v_D=v_{D0}$
2. Off: $v_D<v_{D0}$

Again, like the Shockley diode equation, the PWL model doesn't account for the diode's breakdown region. 

But unlike the Shockley diode equation, the PWL model gives us a straightforward algorithm by which diode circuits may be solved:

Basic Method of Solving Diode Circuits

1. Write down the circuit equations and simplify as much as possible
2. Assume the diode is on or off to get a diode equation. Use the diode equation to solve the circuit and find $i_D$ and $v_D$
3. If $i_D$ or $v_D$ satisfy the inequality associated with the state, then you have solved the diode circuit. Otherwise, repeat Step 2 assuming the diode is in the other state.

Q3: How is the basic method of solving diode circuits limited?

A3: The basic method of solving a diode circuit only gives a solution that assumes that the diode is in a particular state. To give a more general solution, you could solve the diode circuit parametrically (i.e. treating the diode as a parameter).

Q4: In the PWL model, a vertical line replaces the exponential at knee voltage. Why not replace the exponential with a non-vertical line that has a positive slope?

A4: There are two reasons for why diodes are not usually modeled that way:

1. The model introduces another parameter (slope), whose value is somewhat arbitrary
2. A positively-sloped line is not that much of a better fit for the exponential compared to a vertical line

Q5: So the PWL model seems pretty handy. But can I account for the diode's breakdown region?

A5: You can use a Zener piecewise linear model of the diode. The Zener PWL is the same as the PWL except that:

• At the breakdown voltage or Zener voltage $v_Z$, the exponential rise in negative current is replaced by a vertical line.

Hence, we may define 3 states (the first 2 of which are familiar):

1. On: $v_D=v_{D0}$
2. Off: $-v_Z<v_D<v_{D0}$
3. Zener: $v_D=-v_Z$

Again, if you want to know why a sloped line isn't used in place of a vertical line at $-v_Z$ refer back to the explanation in A4.

Usually, hitting the Zener voltage is catastrophic. But there are diodes that are specially made to operate in this breakdown region: the Zener diodes. Zener diodes are useful to serve as reference voltages.

Current through a PN Junction Diode

Q1: How can I describe relative amounts of charge carriers in semiconductors?
A1: Recall that the p-type semiconductor is characterized by a high concentration of holes while the n-type semiconductor is characterized by a high concentration of free electrons.
Hence, holes and electrons are the majority carriers in the p-type and n-type semiconductor (respectively).
In turn, electrons and holes are the minority carriers in the p-type and n-type semiconductor (respectively). Theses minority carriers are generated by thermal excitation.

Q2: Why introduce majority and minority carriers?
A2: You can use majority and minority carriers to examine describe the 2 types of current in a junction diode.
1. Drift Current: Current that flows due to an applied electric field
2. Diffusion Current: Current that flows due to charge carriers moving from a high to low concentration
Recall that holes and electrons diffuse across the junction, which sets up an internal electric field responsible for the junction voltage. The diffusion of these majority carriers also sets up a diffusion current:
$I_{diff}$
Recall that this junction voltage prevents further diffusion of:
1. Holes from the p-type to n-type semiconductor
2. Free electrons from the n-type to p-type semiconductor
In other words, the junction voltage prevents further diffusion of both majority charge carriers.

However, recall that the minority carriers are still present in the diode. When they reach to the depletion layer, the electric field associated with the junction voltage pushes:
1. Free electrons from the p-type semiconductor to the n-type semiconductor
2. Holes from the n-type semiconductor to the p-type semiconductor.
This movement of minority carriers establishes a drift current that flows to from the n-type to p-type semiconductor.
$I_S$
We call this drift current the reverse saturation current of the diode.

Q3: How are are the diffusion current and the reverse saturation current related?
A3: Suppose that no external electric field is applied to the diode. The diode has a diffusion and reverse saturation current, but the diode must remain neutral overall, so:
$I_{diff}=-I_S$

Q4: How can I distinguish between the diffusion current and the reverse saturation current? Besides looking at whether the carriers they specify are majority/minority carriers?
A4: As mentioned at the end of A1, minority carriers are thermally generated. Because of this, the magnitude of the reverse saturation current increases with temperature. Roughly speaking, the reverse saturation current doubles for every 7°C rise in temperature.

On the other hand, the diffusion current of majority carriers depends on the externally applied voltage reaching a certain threshold, after which point the diffusion current scales exponentially.

Q5: How does diffusion current and reverse saturation current relate to forward and reverse biasing?

A5:

Let $i_d$ be the current through the diode

In forward biasing, the externally applied voltage has overcome a certain threshold. So the diffusion current dominates.

$i_d\approx I_{diff}$

In reverse biasing, the externally applied voltage is below that threshold. So the reverse saturation current dominates.

$i_d\approx -I_{S}$

LTI and Transfer Functions

Q1: What's the easiest way to find the zero-state solution for a dissipative, LTI system with a sinusoidal input?
A1: By using a transfer function (that makes use of phasors).

Suppose you know that $x(t)=Kcos(\omega t)$ and the linear differential equation with constant coefficients that relates your input to the output that you want to find.
First, we manipulate the differential equation:
$a_n\frac{d^m}{{dt}^m}y(t)+a_{m-1}\frac{d^{m-1}}{{dt}^{m-1}}y(t)+...+a_1\frac{d}{dt}y(t)+a_0y(t)=$
$b_n\frac{d^n}{{dt}^n}x(t)+b_{n-1}\frac{d^{n-1}}{{dt}^{n-1}}x(t)+...+b_1\frac{d}{dt}x(t)+a_0y(t)$
In a more compact form:
$\sum\limits_{i=0}^ma_i\frac{d^i}{dt^i}y(t)=\sum\limits_{k=0}^nb_k\frac{d^k}{dt^k}x(t)$

Now, you can use the fact that $x(t)=Kcos(\omega t)$ by plugging in the corresponding phasor $B$ multiplied by $e^{j\omega t}$. As you'd expect, taking the real part of the phasor gives you back the original sinusoidal input.

After substitution:
$\sum\limits_{i=0}^ma_i\frac{d^i}{dt^i}y(t)=\sum\limits_{k=0}^nb_k\frac{d^k}{dt^k}Be^{j\omega t}$

Even though there are complex coefficients, the equation is still linear. Hence, you can use the guess and check method to determine a solution.

Suppose that the real part of $Ae^{j\omega t}$ corresponds to the output $y(t)$.
After substitution:
$\sum\limits_{i=0}^ma_i\frac{d^i}{dt^i}Ae^{j\omega t}=\sum\limits_{k=0}^nb_k\frac{d^k}{dt^k}Be^{j\omega t}$
$\sum\limits_{i=0}^ma_i(j\omega)^iAe^{j\omega t}=\sum\limits_{k=0}^nb_k(j\omega)^nBe^{j\omega t}$

$A\sum\limits_{i=0}^ma_i(j\omega)^i=B\sum\limits_{k=0}^nb_k(j\omega)^k$

For neatness, you can define to polynomials:
1. $P_D(j\omega)\equiv\sum\limits_{i=0}^ma_i(j\omega)^i$
2. $P_N(j\omega)\equiv\sum\limits_{k=0}^nb_k(j\omega)^k$
$AP_D(j\omega)=BP_N(j\omega)$

We now define the transfer function, which captures the relationship between input and output:
$H(j\omega)\equiv\frac{P_N}{P_D}$
$A=BH(j\omega)$

Converting from the complex phasor corresponding to the output to the real output:
$y(t)=Re\left\{Ae^{j\omega t}\right\}$
$y(t)=Re\left\{BH(j\omega)e^{j\omega t}\right\}$

Q2: I thought the transfer function was ratio of the output to input. Is this new definition in A1 just the same way of saying this?

A2: No, the transfer function as defined in A1 is the ratio of the complex polynomial associated with the output to the complex polynomial associated to the input.

Q3: I noticed that A1 assumes the sinusoid input is a cosine. What about a sine input? Or a combination of the two?
A3: The following table relates sinusoid inputs to output of a dissipative, LTI system:

Input: x(t)	Output: y(t)
$Acos(\omega t) =Re\left\{Ae^{j\omega t}\right\}$	$Re\left\{AH(j\omega)e^{j\omega t}\right\}$
$Bsin(\omega t) =Re\left\{-jBe^{j\omega t}\right\}$	$Re\left\{-jBH(j\omega)e^{j\omega t}\right\}$
$Acos(\omega t)+Bsin(\omega t) =Re\left\{(A-jB)e^{j\omega t}\right\}$	$Re\left\{(A-jB)H(j\omega)e^{j\omega t}\right\}$

Q4: Couldn't I write $Bsin(\omega t)$ as $Im\left\{Be^{j\omega t}\right\}$?

A4: Yes, you could. And then $y(t)$ would be $Im\left\{BH(j\omega)e^{j\omega t}\right\}$.

But for the sake of consistency, we will adopt the convention that uses the real part of the complex number. 

Band Theory for Conductors

Q1: What's the difference between conduction and valence electrons?
A1: Conduction electrons are not associated with particular atoms, and hence move freely and conduct current. Valence electrons are associated with particular atoms, and do not necessarily move as freely.

Q2: What are some of the key ideas of band theory?
A2: Band theory introduces the concepts of the conduction band, valence band, and energy gap.
The conduction band arises from the cumulative effect of having enough conduction electrons at different energy levels.
The valence band arises from the cumulative effect of having enough valence electrons at different energy levels.
For both the conduction and valence bands, the energy levels are so close together that they can be treated as a continuum (and hence, the term "band").
The energy gap is the difference in energy from the bottom of the conduction band to the top of the valence band.

Q3: How does band theory explain conductors, insulators, and semiconductors?
A3: Conductors have no energy gap; in fact there is overlap between the valence and conduction band, meaning that some of the valence electrons are free to conduct.
Insulators have a large energy gap, so their valence electrons are not free to conduct.
Semiconductors have a smaller energy gap, which would suggest that they too would be unable to conduct. In fact, this is the case at 0K. But at room temperature, there's usually enough thermal excitation to bump some of the valence electrons in to conduction band.

Q4: Why is this smaller energy gap important for semiconductors?
A4: As mentioned in A3, the smaller energy gap enables some of the valence electrons to feasibly reach the conduction band at room temperature. But the smaller threshold also explains helps to explain doping. Consider the energy gap of a PN junction. The p-type semiconductor puts holes at the bottom of the energy gap, which makes it easier for valence electrons to enter these extra energy levels. The n-type semiconductor puts extra electrons at the top of the energy gap, which makes it easier for electrons at these energy levels to enter the conduction band.

An/Cations & An/Cathodes

Q1: What is an ion?
A1: An ion is an atom that is not electrically neutral. You can categorize an ion in 2 ways based on its charge parity:
1. If the ion is positively charged (due to a deficiency in electrons), it's a cation.
2. If the ion is negatively charged (due to a excess of electrons), it's an anion.

Q2: What is an electrode?
A2: An electrode is a conducting contact that interfaces with some nonmetallic region. It may serve as an exit or entry point for current.

Q3: How can I categorize electrodes?
A3: You can categorize an electrode in 2 ways based on the type of ion it attracts:
1. If the electrode attracts cations, then it is a cathode
2. If the electrode attracts anions, then it is an anode

Conductivity

Q1: What is conductivity?
A1: Conductivity is how well a material conducts current.

Q2: Why should I care about conductivity?
A2: Many electronic devices owe their properties to the conductivities of the different materials from which they are made up of.

Q3: How do I categorize these materials of different conductivities?
A3: Starting at the two extremes, a conductor conducts well while an insulator conducts poorly. A semiconductor, as the name suggests, has a conductivity between an conductor and insulator.

Diode Biasing

Q1: What is biasing?
A1: Biasing is when you establish a potential difference across an element so that it operates in a specific way.

Q2: Okay, so I have a general sense of what biasing is. But how does it apply to a diode that makes use of a PN junction?
A2: Well, you can apply a voltage drop across the diode in 2 ways: in the same direction as the drop in internal junction voltage or in the opposite direction.

Q3: What happens with the first case?
A3: When the applied voltage drop is in the same direction as the drop in junction voltage, the diode is said to be reverse biased. Since the applied voltage drop reinforces the internal junction voltage, the insulating depletion layer widens. Hence, a reverse-biased PN junction diode conducts negligible current. However, once that voltage drop reaches a high enough magnitude, the electric field is strong enough to upset the covalent bonds at the depletion zone, generating electron-hole pairs. When this threshold voltage or breakdown voltage is exceeded, there is a massive surge in conduction. Exceeding breakdown voltage unintentionally could be catastrophic if it causes the diode to melt. On the other hand, special diodes called Zener diodes are designed to make use of this breakdown voltage.

Q4: What about the other case?
A4: When the applied voltage drop is in the opposite direction as the drop in junction voltage, the diode is said to be forward biased. Negligible current flows through the diode, until the externally applied voltage overcomes the internal junction voltage. This threshold value is called knee voltage. Once applied voltage exceeds voltage, the diode is able to conduct much more current.

PN Junction

Q1: How do I make a PN junction?
A1: You can physically interface a p-type and n-type semiconductor together. The plane along which they meet is a PN junction.

Q2: Why would I want to make a PN junction?
A2: At least 1 PN junction is used in nearly every semiconductor device.

Q3: What's so special about the PN junction?
A3: In the neighborhood of where p-type and n-type semiconductors are in contact with each other, some of the holes from the p-type semiconductor combine with the free electrons from the n-type semiconductor. Where this neutralization of charge carriers occurs is called the depletion layer.

Q4: Why should I care about the depletion layer?
A4: The depletion layer is where the ionized acceptor and donor atoms come into play. The separation of negatively-charged acceptor atoms and positively-charged donor atoms establishes an electric field in the N to P direction across junction. This internal electric field prevents further diffusion of charge carriers across the junction. The potential difference across the junction is called the junction voltage.

P-Type and N-Type Semiconductors

Q1: What's an application of doping?
A1: You can use doping to make p-type and n-type semiconductors.

Q2: What semiconductor can I use to make these p-type and n-type semiconductors?
A2: You could use silicon, a Group IV element. A silicon atom has 4 valence electrons, which may be paired up with another 4 electrons from 4 neighboring silicon atoms through 4 covalent bonds (for the sake of stability). At room temperature, not all of the electrons are held in covalent bonds in pure silicon. Some of these electrons can move around.

Q3: What's the difference between p-type and n-type semiconductors?
A3: The difference lies in the parity of their charge carriers, charged particles that are free to move around. P-type semiconductors have positive charge carriers while n-type semiconductors have negative charge carriers.

Q4: How can I dope silicon to make a p-type semiconductor?
A4: You can introduce boron, a Group III element, as an impurity. Neighboring silicon atoms see boron's lack of a fourth valence electron as a hole. For stability, boron may take the electron of a nearby silicon atom. Hence, boron is referred to as an acceptor atom. After accepting the electron, the boron atom becomes a negative ion and the hole is relocated from boron to a silicon atom. In turn, this silicon atom may fill its hole with an electron from another silicon atom. And so on. The takeaway is that this hole moves freely and serves as an positive charge carrier.

Q5: How can I dope silicon to make n-type semiconductors?
A5: You can introduce phosphorus, a Group V element, as an impurity. Neighboring silicon atoms see phosphorus's fifth electron as excessive. For stability, phosphorus may give its extra electron to a nearby silicon atom. Hence, phosphorus is referred to as a donor atom. After donating its electron, phosphorus becomes a positive ion and the extra electron is relocated from phosphorus to a silicon atom. In turn, this silicon atom may rid itself of the extra electron by donating it to another silicon atom. And so on. The takeaway is that this extra electron moves freely and serves as a negative charge carrier.

Introduction to Semiconductor Doping

Q1: What's a semiconductor?
A1: A semiconductor is a substance whose electrical conductance falls between a conductor and an insulator.

Q2: Why should I care about semiconductors?
A2: Semiconductor devices are ubiquitous in modern electronics.

Q3: What's doping?
A3: Doping is the technique by which impurities are introduced to a pure semiconductor.

Q4: Why should I care about doping?
A4: You can use doping to alter the electrical properties of (what were originally pure) semiconductors.

Zero-State Solution using Complex Numbers

Recall:
$\tau \frac{di_f(t)}{dt}+i(t)=Kcos(\omega t)$
We can rewrite this as:
$\tau \frac{di_f(t)}{dt}+i(t)=Re\left\{Ke^{j\omega t}\right\}$

We will ignore that we are taking the real part of the polar form of the complex number momentarily.
Instead, we will solve the problem as if we had a complex input voltage.
$\tau \frac{di_f(t)}{dt}+i(t)=Ke^{j\omega t}$

Then, once we find $i_f(t)$ that solves the complex input voltage, we'll remember to take the real part of this answer to find the real zero-state solution for current.

Suppose $i_f(t)=Be^{j\omega t}$
$Bj\omega \tau e{j\omega t} + Be{j\omega t} = Ke{j\omega t}$
Since $e{j\omega t} \neq 0$:
$Bj\omega \tau + B = K$
$B=\frac {K} {1+j\omega \tau}$
$i_f(t)=\frac {K} {1+j\omega \tau}e^{j\omega t}$

Now, we take the real part of this solution to find our real answer:
$i_f(t)=Re\left\{\frac {K} {1+j\omega \tau}e^{j\omega t}\right\}$
$i_f(t)=Re\left\{\frac {v_{in}\omega C} {1+j\omega \tau}e^{j\omega t}\right\}$

RC Circuit Example

Q: Suppose you have a circuit consisting of a voltage source $v_{in}(t)=v_{in}sin(\omega t)$ in a series RC circuit.
What is the complete solution for the current $i(t)$ through the circuit?

A: By KVL:
$v_{in}(t)=v_R+v_C$
$v_{in}(t)=Ri(t)+v_C$
Differentiating both sides:
$\frac{dv_{in}(t)}{dt}=R\frac{di(t)}{dt}+\frac{dv_C}{dt}$
From the definition of a capacitor:
$\frac{dv_{in}(t)}{dt}=R\frac{di(t)}{dt}+\frac{1}{C}i(t)$
$C\frac{dv_{in}(t)}{dt}=RC\frac{di(t)}{dt}+i(t)$
Defining a time constant $\tau \equiv RC$:
$C\frac{dv_{in}(t)}{dt}=\tau \frac{di(t)}{dt}+i(t)$
$\tau \frac{di(t)}{dt}+i(t)=C\frac{dv_{in}(t)}{dt}$

$i(t)=?$
$i(t)=i_h(t)+i_f(t)$

$i_h(t)=?$
$\tau \frac{di_h(t)}{dt}+i_h(t)=0$
Suppose $i_h(t)=Ae^{st}$:
$\tau sAe^{st}+Ae^{st}=0$
Since $e^{st} \neq 0$:
$\tau sA+A=0$
Since we are looking for a non-trivial solution - i.e. $A \neq 0$:
$\tau s + 1 =0$
$s = -\frac{1}{\tau}$
$i_h(t)=Ae^{-\frac{t}{\tau}}$

$i_f(t)=?$

In this post, a basic method will be used to solve the zero-state solution for current. But a method using complex numbers is examined in another post.
$\tau \frac{di_f(t)}{dt}+i(t)=C\frac{dv_{in}(t)}{dt}$
$\tau \frac{di_f(t)}{dt}+i(t)=v_{in}\omega Ccos(\omega t)$
For cleaner notation, let $K \equiv v_{in}\omega C$
$\tau \frac{di_f(t)}{dt}+i(t)=Kcos(\omega t)$

Suppose $i_f(t)=Dcos(\omega t)+Esin(\omega t)$
 $-D \omega \tau sin(\omega t)+E \omega \tau cos(\omega t)+Dcos(\omega t)+Esin(\omega t)=Kcos(\omega t) $
 $(E-D \omega \tau) sin(\omega t)+(D+E \omega \tau) cos(\omega t)=Kcos(\omega t) $

$\begin{cases} E-D \omega \tau = 0 \\ D+E \omega \tau=K \end{cases}$

$E=D \omega \tau$
$D+D \omega^2 \tau^2=K$
$D=\frac{K}{\omega^2 \tau^2}$

$i_f(t)=\frac{K}{\omega^2 \tau^2}(cos(\omega t)+\omega\tau sin(\omega t))$
$i_f(t)=\frac{v_{in}\omega C}{\omega^2 \tau^2}(cos(\omega t)+\omega\tau sin(\omega t))$

Loads

Q1: What is an electrical load?

A1: A load is a 2-terminal network containing no independent sources.

Q2: Why is important that a load contains no independent sources?

A2: By Thevenin's theorem, a load may be modeled as a resistor (from the perspective of the source).

Introduction to Networks

Q1: What defines a 2-terminal network?
A1: A 2-terminal network is defined by its $iv$ characteristic. Note that this is similar to how a 2-terminal element is defined by its$iv$ characteristic.

Q2: What defines a 2-port network?
A2: A 2-port network is defined by its transfer function that describes how it modifies an input. Since 2 terminals form its input port while the other 2 form its output port, a 2-port network is also called a 4-terminal network.

Q3: Why do we care about networks?
A3: Networks allow us chunk circuits into functional blocks. Circuits may be reduced to more basic networks in analysis. Circuits may also be built up from basic networks in design.

Circuit Theory

Q1: What is circuit theory?
A1: Circuit theory is an approximation of Maxwell's equations.

Q2: What function do models serve?
A2: A model approximately represents of some aspect of the physical world.
There is often trade-off between the accuracy of the model and its ease of use.
Idealizations are often quick and user-friendly, and can give a rough idea about how a circuit behaves.

Q3: What are the idealizations made by circuit theory?
A3: Circuit theory:
1. Treats the speed of light as infinite
$c \to \infty$
2. Assumes that electrical and magnetic fields associated with individual elements do not interact with any other elements
That is, elements can only "talk" to each other through wires

Q4: Why learn circuit theory if they don't perfectly capture real circuits?
A4: For design and analysis, perfectly accurate models are not needed (or even conceivable) to still yield useful information about particular real-world circuits.

Circuit Response

Q: How can I analyze the response of a circuit?
A: Consider a 2nd-order circuit. By superposition, the complete response is the sum of the solution to the homogeneous equation and the inhomogeneous equation.
$x(t)=x_h(t)+x_f(t)$

Q2: What is $x_h$ called?
A2: $x_h$ is called the zero-input response (ZIR), natural response, or transient response.
This is the response when there is no input to the system.

Q3: What is $x_f$ called?
A3: $x_f$ is called the zero-state response (ZSR), forced response, or steady-state response.
This is the response when there is no energy initially stored by the system.

Second-Order Circuit

Q: What is a 2nd-order circuit?

A: A 2nd-order circuit is a circuit in which any voltage/current inside the circuit may be described by a 2nd-order differential equation:

$\frac{d^2x(t)}{dt^2}+c_1\frac{dx(t)}{dt}+c_0x(t)=f(t)$

where $c_1$ and $c_0$ are constants

Linear Time-Invariant Systems

Q1: What is a linear time-invariant system (LTI)?
A2: A linear time-invariant system is a system that is linear and time-invariant.

Q2: Why do we care about LTI systems?
A2: We care about LTI systems because we can analyze them using the tools of LTI theory.

Time-Invariant Systems

Q: What is a time-invariant system (TIV)?

A: A time-invariant system is a system in which a time shift in the input produces an equivalent time shift in the output.

Given: $x(t) \to y(t)$

Then: $x(t+\delta) \to y(t+\delta)$

Dissipative Systems

Q1: What is a dissipative system?
A1: A dissipative system is a system for which the zero-input response $\to 0$ as $t \to \infty$
where $t$ is time

Q2: Why shoud I be on the lookout for dissipative systems?
A2: By definition, the zero-input response goes zero as $t \to \infty$
Hence, we know that the complete solution as $t \to \infty$ is the same as the zero-state response.

Q3: How do I determine if a system is dissipative?
A3: Suppose the system in question is a LTI system that relates an input $x(t)$ to an output $y(t)$ by a linear differential equation with constant coefficients. Then the homogeneous counterpart to that differential equation is:
$\sum\limits_{i=0}^na_i\frac{d^i}{dt^i}y_h(t)=0$
Since the equation is linear, you can guess a form of $y_h(t)$ and if the guess works, you've found the solution since it's unique.
Guess: $y_h(t)=Ae^{st}$
$\sum\limits_{i=0}^na_i\frac{d^i}{dt^i}Ae^{st}=0$
$\sum\limits_{i=0}^na_iAs^ne^{st}=0$
$\sum\limits_{i=0}^na_is^n=0$
Solving this equation gives the roots of the homogeneous equation (i.e. $s_1, s_2, ..., s_n$) and is called the characteristic equation or characteristic polynomial.

Suppose you've solved for the n roots for the nth order characteristic polynomial.
Then, $y_h(t)=\sum\limits_{i=0}^nA_ie^{s_it}$
In general, $s_i$ is complex (i.e. $s_i=a_i+jb_i$).
For sake of ease, let's look at the contribution of a specific $A_ie^{s_it}$ to $y(t)$:
$A_ie^{s_it}=A_ie^{(a_i+jb_i)t}$
$A_ie^{s_it}=A_ie^{a_it}e^{jb_it}$
So the specific contribution is the product of three parts:

1. $A_i$
2. $e^{a_it}$
3. $e^{jb_it}$

Since $A_i$ is a constant, it doesn't diminish as $t\to\infty$. Since $e^{jb_it}$ is a complex exponential, it doesn't diminish as $t\to\infty$. So $A_i$ and $e^{jb_it}$ are not dissipative.

This leaves us with $e^{a_it}$. This is the term that you care about: the only way you can diminish the effects of $A_i$ and $e^{jb_it}$ is if $e^{a_it}\to 0$ as $t\to\infty$. And $e^{a_it}$ is dissipative only if $a_i$ is negative. This in turn makes the whole contribution dissipative. Note that if $a_i$ is positive, then $e^{a_it}$ shoots up to infinity as $t\to\infty$. On the other hand, if $a_i=0$, then $e^{a_it}=1$. This outcome does nothing to counter the effects of $A_i$ or $e^{jb_it}$.

As a summary: 

$\begin{cases} a_i <0,&\mbox{Contribution is dissipative} \\ a_i\geq0,&\mbox{Contribution is not dissipative} \end{cases}$

This gives us an easy method for determining whether $y_h(t)$ is dissipative:

1. Solve for each of the roots of the characteristic polynomial corresponding to the original homogeneous equation. 
2. Examine the real part of each root. If all real parts are positive, then you know that each contribution is dissipative. The sum of dissipative terms is also dissipative, so you can then conclude that $y_h(t)$ is dissipative.

Linear Systems

Q1: What is a linear system?
A1: Consider a system that maps:
$x_1 \to y_1$
$x_2 \to y_2$
The system is linear if it satisfies:
1. Additivity: The output of a sum of inputs is the sum of each input's individual outputs
$x_1 + x_2 \to y_1 + y_2$
2. Homogeneity: The output of a scaled input is the output of that original input scaled by the same factor
$ax_1 \to ay_1$
$bx_2 \to by_2$
where a and b are constants

Q2: Can the two conditions of A1 be encapsulated more succinctly?
A2: Yes. The more succinct restatement is as follows:
Consider a system that maps:
$x_1 \to y_1$
$x_2 \to y_2$
The system is linear if it satisfies additivity and homogeneity:
$ax_1 + bx_2 \to ay_1 + by_2$

Q3: Why care about linear systems?
A3: Systems that are linear allow us to use the technique of superposition.

Introduction to Logarithms

Q1: What is a logarithm?
A1: A logarithm is a convenient way of expressing an exponential function.
Given: $y=b^x$
Equivalent: $\log_b(y)=x$

Q2: What's a common logarithm?
A2: A common logarithm is a logarithm that uses base 10.
It's written as: $\log(y)=x$

Q3: What's a natural logarithm?
A3: A natural logarithm is a logarithm that uses base $e$.
It's written as: $\ln(y)=x$

Logarithms of Complex Numbers

Q: How do I take the logarithm of a complex number?
A: Convert the complex number to polar form before taking the logarithm:
$z=|z|e^{j\theta}$
$ln(z)=ln|z|+ln(e^{j\theta})$
$ln(z)=ln|z|+j\theta$

Roots of Unity

Q: What are the roots for the case $1^{\frac{1}{3}}$?
A: Since 1 is a real number and therefore complex, we can apply our method of solving for the roots of a complex number.
Rewriting 1 in polar form:
$1=e^{j\theta}$ for $\theta=0, 2\pi, 4\pi$
$1^{\frac{1}{3}}=e^{\frac{j\theta}{3}}$
$1^{\frac{1}{3}}=1,e^{j\frac{2\pi}{3}},e^{j\frac{4\pi}{3}}$

Kirchoff's Laws

Q1: What are Kirchoff's circuit laws?

A1:

1. Kirchoff's Current Law (KCL): The sum of all currents into a node is zero

$\sum\limits_{j=1}^ni_j=0$

2. Kirchoff's Voltage Law (KVL): The sum of all voltages in a loop is zero

$\sum\limits_{j=1}^nv_j=0$

Q2: Why are Kirchoff's circuit laws useful?

A2: You can use Kirchoff's circuit laws to analyze circuits. KCL is used in nodal analysis while KVL is used in mesh analysis. Nodal or mesh analysis may be used to solve circuits.

Q3: Could I get away with just learning nodal analysis? Or just mesh analysis?

A3: You could, but it's handy to be comfortable with both. For example, if you're solving for an unknown voltage, then nodal analysis is more intuitive. On the other hand, if you're solving for an unknown current, then you'd probably turn to mesh analysis.

Q4: Is there another reason (besides convenience) for learning nodal and mesh analysis?

A4: With both methods of analysis, you can check your work. For example, you might first use nodal analysis to solve a circuit. You can then resolve the same circuit using mesh analysis. Disagreement between the results constitutes a failed sanity check (i.e. at least one of your methods of analysis was implemented incorrectly). Agreement between the results means that you passed a sanity check (i.e. your results are not wrong in an obvious way). Note that passing a sanity check does not guarantee correctness of your results. Still, sanity checks are good practice for avoiding blatantly wrong answers.

Maclaurin Series

Q: What is a Maclaurin series?
A: A Maclaurin series a special case of the Taylor series expansion about 0.
$f(x)=\sum\limits_{n=0}^n \frac {{f(0)}^{(n)}x^n}{n!}$

Maclaurin Series for Sine and Cosine

Q1: What is the Maclaurin series for cosx?
A1: $cosx=1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$
$cosx=\sum\limits_{n=0}^\infty \frac{(-1)^nx^n}{2n!}$

Q2: What is the Maclaurin series for sinx?
A2: $sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$
$sinx=\sum\limits_{n=0}^\infty \frac{(-1)^nx^n}{(2n+1)!}$

Euler's Formula

Q1: How can $e^{j\theta}$ be expressed in terms of trigonometric functions?
A1: $e^{j\theta}$ can be simplified using a Maclaurin series:
Let $\theta \to x$:
$\Rightarrow e^{j\theta} \to e^{jx}$
$e^{jx}=\sum\limits_{n=0}^n \frac {j^nx^n}{n!}$
Expanding the Maclaurin series:
 $e^{jx}=1+jx-\frac{x^2}{2}-j\frac{x^3}{3!}+\frac{x^4}{4!}+j\frac{x^5}{5!}-\frac{x^6}{6!}-j\frac{x^7}{7!}+...$
Collecting real and imaginary coefficients:
 $e^{jx}=[1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+...]+[j(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...)]$
 Note that the expressions in the square brackets are the Maclaurin expansions of $cosx$ and $sinx$

$e^{jx}=cosx+jsinx$
$e^{j\theta}=cos\theta+jsin\theta$

Q2: What's the expression for $e^{-j\theta}$?
A2: Let $j \to -j$:
$e^{-j\theta}=cos\theta+jsin\theta$

Q3: Is there one equation that encapsulates the results from A1 and A2?
A3: $e^{\pm j\theta}=cos\theta\pm jsin\theta$

Powers of the Imaginary Unit

Q: What are the integer powers of the imaginary unit?
A:

$j^0=1$
$j^1=j$
$j^2=-1$
$j^3=-j$

This basic four-part pattern continues for consecutive integer powers above and below this arbitrarily selected sequence

Complex Conjugates

Given: $z = a + jb = |z|e^{j\theta}$

Q1: What is the complex conjugate of $z$ in Cartesian form?

A1: $z^* = a - jb$

Q2: What is the complex conjugate of $z$ in polar form?

A2: Note that $z^*$ is simply the reflection of $z$ over the real axis of the complex plane. It then follows that:

1.The magnitude of $z^*$ is the same as the magnitude of $z$

2. The angle that $z^*$ makes with respect to the positive real axis is the negative value of the angle that $z$ makes with the positive real axis. 

Hence, $z^*= |z|e^{-j\theta}$

Q3: How can you simplify the expression $zz^*$?
A3: Although the multiplication suggests that you use polar form to simplify the product, Cartesian form yields an interesting result.
$zz^*=(a+jb)(a-jb)$
Notice that the cross-terms (i.e. the O and I products of FOIL) negate each other.
$zz^*=a^2+b^2$
$zz^*={|z|}^2$

Complex Number Exponentiation

Given: $z$ a complex number

Q1: What is $z^n$? Assume $n$ is an integer.

A1: Since exponentiation is repeated multiplication, it's convenient to use the complex polar form of $z$:

$z=\left|z\right|e^{j\theta}$

$z^n={\left|z\right|}^ne^{jn\theta}$

This result is known as De Moivre's formula.

Q2: Suppose you don't have $z$ raised to an integer power. Instead, you have $z^{\frac{1}{n}}$. How do you find the $n$ roots of this expression?
Q3: To answer Q2, let's look at the specific case where $n$ is 3. What are the 3 roots of $z^{\frac{1}{3}}$?
A3: Following the logic from A1,
Root 1: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
But this is only 1 root. We know we need to find 2 more. Note that $2\pi$ can be added to the angle $\theta$ without changing the result. We can write 2 more cases:
Root 2: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+2\pi)}{3}}$.
Root 3: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+4\pi)}{3}}$.
Q4: This gives us 3 roots - just what we want. But why not continue adding 2$\pi$ to the angle again? Couldn't this produce a 4th root?
A4: Suppose you added another 2$\pi$. Then you'd get the fourth case:
Root 4: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+6\pi)}{3}}$.
But note that this reduces to  $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}+2\pi}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
which is the same as Root 1. So there's no need to continue adding 2$\pi$ past the Root 3. Any other roots found by adding 2$\pi$ past this point are redundant.
A2: To answer the general case of Q2, we substitute $n$ for 3 to find that the $n$ roots of $z^{\frac{1}{n}}$ are:
 $z^{\frac{1}{n}}={\left|z\right|}^{\frac{1}{n}}e^{\frac{j(\theta+2k\pi)}{n}}$ where $k$ is an integer in the range 0, 1, ... , $n-1$

Complex Multiplication/Division

Given:
$z_1=a_1+jb_1=\left|z_1\right|e^j\theta_1$
$z_2=a_2+jb_2=\left|z_2\right|e^j\theta_2$

Q1: What is $z_1z_2$ in Cartesian form?
A1: Using the FOIL method and collecting the real and imaginary parts:
$z_1z_2=(a_1a_2-b_1b_2)+j(a_1b_2+a_2b_1)$

Q2: What is $z_1z_2$ in polar form?
A2: $z_1z_2=\left|z_1\right|\left|z_2\right|e^{j(\theta_1+\theta_2)}$

Q3: What is $\frac{z_1}{z_2}$ in Cartesian form?
A3: Multiplying both the numerator and denominator by $z_2^*$:
$\frac{z_1}{z_2}=\frac{(a_1a_2+b_1b_2)+j(a_2b_1-a_1b_2)}{a_2^2+b2_2}$

Q4: What is $\frac{z_1}{z_2}$ in polar form?
A4: $\frac{z_1}{z_2}=\frac{\left|z_1\right|}{\left|z_2\right|}e^{j(\theta_1-\theta_2)}$

Complex Addition/Subtraction

Given:
$z_1=a_1+jb_1$
$z_2=a_2+jb_2$

Q1: How do I add complex numbers in Cartesian form?
A1: Complex numbers are added component-wise:
$z_1+z_2=(a_1+a_2)+j(b_1+b_2)$

Q2: How do I subtract complex numbers in Cartesian form?
A2: Complex numbers are subtracted component-wise:
$z_1-z_2=(a_1-a_2)+j(b_1-b_2)$

Cartesian and Polar Forms of Complex Numbers

Q1: What is a complex number?
A1: A complex number $z$ is a number with:
1. A real component $a$
2. An imaginary component $b$
where $z=a+jb$ where $j=\sqrt{-1}$

This is the Cartesian form of the complex number.

Q2: How else may a complex number be expressed?
A2: Alternatively, $z$ may be expressed in polar form with:
1. A magnitude $\left|z\right|$
2. An angle $\theta$
where $z=\left|z\right|e^j\theta$

Q3: Why should I bother learning Cartesian and polar form of complex numbers?
A3: Cartesian form is convenient for complex addition/subtraction while polar form is convenient for complex multiplication/division.

Q4: How do I switch from Cartesian to polar form?
A4: Use the relations:
1. $\left|z\right|=\sqrt{a^2+b^2}$
2. $\tan{\theta}=\frac {b}{a}$

Q5: How do I switch from polar to Cartesian form?
A5: Use the relations:
1. $a=\left|z\right|\cos{\theta}$
2. $b=\left|z\right|\sin{\theta}$

Ideal Sources

Q1: What is an ideal voltage source?
A1: An ideal voltage source is a two-terminal element that guarantees a fixed voltage across its terminals, regardless of the current through it.

Q2: What is an independent current source?
A2: An ideal current source is a two-terminal element that guarantees a fixed current through its terminals, regardless of the voltage across it.

Q3: What are independent sources?
A3: Independent sources are sources whose guranteed voltage/current does not depend on any parameter of the circuit.

Q4: What are dependent sources?
A4: Dependent sources are sources whose guranteed voltage/current depends on a separate circuit parameter. The four flavors are:
1. Voltage-Controlled Voltage Source (VCVS)
2. Current-Controlled Voltage Source (CCVS)
3. Voltage-Controlled Current Source (VCCS)
4. Current-Controlled Current Source (CCCS)

Introduction to Resistors

Q1: What is a resistor?
A1: A resistor is a two-terminal element whose voltage across its terminals is proportional to the current through it. The proportionality constant is called resistance (R).
$v=iR$

Q2: What determines resistance?
A2: Resistance is determined by the composition and geometry of the resistor. Reaistance is influenced by temperature and strain. Some resistors may also react to light levels.

Q3: How do I simplify resistors in series?
A3: The equivalent resistor has a resistance equal to the sum of the individual resistances.

Q4: How do I simplify resistors in parallel?
A3: The equivalent resistor has a resistance equal to the inverse of the sum of the inverted individual resistances.

Introduction to Inductors

Q1: What is an inductor?
A1: A capacitor is a two-terminal element whose voltage across its terminals is proportional to the time rate of change of current through it. The proportionality constant is called inductance (L).
$v=L\frac{di}{dt}$

Q2: What determines inductance?
A2: Inductance is determined by the composition and geometry of the inductor.

Q3: How do I simplify inductors in series?
A3: Inductors in series add like resistors in series.

Q4: How do I simplify inductors in parallel?
A3: Inductors in parallel add like resistors in parallel.

Introduction to Capacitors

Q1: What is a capacitor?
A1: A capacitor is a two-terminal element whose current through its terminals is proportional to the time rate of change of voltage across it. The proportionality constant is called capacitance (C).
$i=C\frac{dv}{dt}$

Q2: What determines capacitance?
A2: Capacitance is determined by the composition and geometry of the capacitor.

Q3: How do I simplify capacitors in series?
A3: Capacitors in series add like resistors in parallel.

Q4: How do I simplify capacitors in parallel?
A3: Capacitors in parallel add like resistors in series.

Introduction to Current

Q1: What is current?
A1: Current is the amount of positive charge that passes through a given area per unit of time.

Q2: What are the units for current?
A2: The basic unit for current is defined as the ampere (A). It is one of the seven fundamental SI units. One ampere is equal to one unit of one coulomb per second. Amperes are commonly called amps.

Q3: I took another look at the definition of current in A1. Wouldn't this mean that if electrons are flowing in a particular direction in my circuit, then the current would be flowing in the opposite direction?
A3: You are absolutely correct: positive current would be in the direction opposite to electron flow. If current were redefined as the amount of negative charge that passes through a given area per unit  time, then positive current would be in the same direction of electron flow. The fact that current is defined in reference to the flow of positive charge is a product of historical convention.

Q4: Why do should I keep track of currents over time?
A4: Changes in current usually tell us that something interesting is happening in a circuit.

Introduction to Voltage

Q1: What is voltage?

A1: Voltage is the amount of energy per unit of charge needed in moving from a lower electric potential to a higher electric potential. Alternatively, voltage is the amount of energy per unit charge released in moving from a higher to lower electrical potential.

Q2: What are the units for voltage?

A2: Voltage is measured in joules per coulomb (following from the definition given in A1). The volt (V) is defined as one joule per coulomb.

Q3: What is the voltage at a point?

A3: Recall from definition A1 that voltage is a relative physical quantity, concerned with differences in electric potential. Since there is no difference in electric potential at a point with respect to itself (at a particular instance in time), there is zero voltage. 

Q4: So the voltage at any point with respect to itself is zero. But why do I see circuit diagrams with points labeled with nonzero voltages?

A4: It is true that the voltage at any point with respect to itself is zero. And it is perfectly valid to label points or nodes in a circuit with nonzero voltages. The solution to this apparent paradix: not all voltages are measured at a node with respect to that same node. That wouldn't be very useful - you'd have a mess of zeroes cluttering your diagram. Instead, one node is usually picked as a reference node, the node to which the voltages at all other nodes are measured with respect to. You can designate this reference node arbitrarily, but some selections may be intuitive than others. Because this reference node is widely used, we give it a name ground.

Q4: Why should we keep track of voltage with respect to time?

A4: Changes in voltage indicate that something interesting is happening in the circuit.