View source for ECE492 Section 1 Notes

Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. 

My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. 

== Prologue ==

Hey how are you. I will insert something here later. :)

== Section 1.1 == 

We first need to consider the following materials: 
* Conductors
* Insulators
* Semiconductors

Consider the concept of the charge for Oxygen (O):
* 8 protons around the nucleus
* 8 electrons (two on the inner orbit, 6 on the outer orbit). 

Let's look at the outer most orbit in context with a conductor:
* There are two bands: valency and conduction. 
* Electrons are always in the valency band. 
* In between, there is an energy gap, which is approximately 0 (meaning electrons can move freely between valency band and conduction band). 
* That is the reason why conductive materials act as they do. 

Let's look at the outer most orbit in the context of insulators: 
* There are two bands: valency and conduction. 
* Electrons are always in the valency band. 
* In between, there is an energy gap, which is very high (meaning electrons have a very hard time moving between the valency band and conduction band). 
* That is the reason insulators, are well, insulators (or their property is insulative). 

Let's look at the outer most orbit in the context of semi-conductors: 
* There are two bands: valency and conduction. 
* Electrons are always in the valency band. 
* In between, there is an energy gap, which is exactly 1.12eV (1eV = 1.602 * 10<sup>-19</sup>J). 
* Once electrons are in the conduction band, the semi-conductor will act as a conductor. 

This entire class is based on semi-conductor materials. 

=== What is a semiconductor? ===

It sometimes allow current, and sometimes it doesn't.

Semi-conductors only have '''4''' electrons in its outermost orbit. 
* Carbon
* Silicon 
* Germanium
* Lead

What's the best two?
* Silicon and Germanium

Consideration of Silicon: 
* Silicon will acquire other silicon atoms to form a covalent bond. 

Now let's take a Silicon atom: 
* Take a pure silicon atoms and put energy to them: go from insulator to conductive properties, and electrons from the furthest orbit detach and jump to another position.  
* When electrons detach, it leaves a hole where it once was, that is viewed as a positive charge for clarity. 

Doping: Adding impurity to a pure semiconductor material. 
* Tri-Valent Impurity: majority of "holes" (positive charge), also know as P-type semiconductor. 
* Penta-Valent Impurity: majority of electrons, also known as N-type semiconductor. 

Let's take Aluminum for Tri-Valent example:
* Pair the Aluminum with Silicon atoms, knowing that Aluminum has an outer electron count of 3. 
* Silicon will be bounded to the three outer Aluminum electrons, but 1 Silicon will be left.

Let's try Phosphorous for Penta-Valent example:
* Pair Phosphorous with Silicon atoms, knowing that Phosphorous has an outer electron count of 5. 

Now what is P-N Junction? It's a diode. 
* A P-type semiconductor connected to an N-type semiconductor.
* Recall that P-type just has a lot of "holes" (positive). 
* Recall that N-type has a lot of electrons (negative).
* Shouldn't these holes and electrons combine? Electric field is present due to electrons on the N-type side of the P-N junction. This electric field does not allow for combining. 
* There is an anode and a cathode on the P-type and N-type side respectively. A diode can be represented with A -> C. 
* Now let's connect a battery on each side of the junction, and we supplement 0.5V.
* The electric field barrier becomes thinner and thinner, but not completely gone. 
* Now apply 1.0V, the electrons and holes will push against the junction. 

Now consider a reverse biased diode: 
* A diode where the energy source charge is reverse from the P-N junction. 
* Holes to negative and electrons to positive. 

Important consideration: A P-N Junction (diode, or biased/reversed biased diode) allows current in only one direction. 

What are the V-I characteristics of the P-N Junction/Diode:
* What is PIV? Peak Inverse Voltage. 
* Consider the graph below for a typical V-I characteristic of a diode. 

https://media.geeksforgeeks.org/wp-content/uploads/20211025120251/VIcharacteristics.jpg

Therefore, important considerations added for P-N Junction diode:
* A P-N Junction (diode, or biased/reversed biased diode) allows current in only one direction.
* Acts like a closed switch in forward biased condition. 
* Acts like an open switch in reversed biased condition.  

Now consider the following example: Find the current passing through each diode. 
* We'll need to break the problem apart. 
* Current always flows from higher potential to lower potential.
* Assume ideal diode. 

Let's break the problem down: 
* Step 1: Fill me in. :)

'''Insert example 1A'''

Now let's consider the following example:

* Step 1: Find all possible current flow directions (current always flows from high potential to low potential) and make a determination of whether a diode is open/closed/don't know. 
* Step 2: We'll take the assumption that D<sub>2</sub> and D<sub>3</sub> are closed because we don't know what they are. 
** Note: We can't have 3 different voltages at the same node, so our assumption is wrong. 
* Step 3: Now we'll verify that our assumption is correct.
** Is D<sub>2</sub> and D<sub>3</sub> open? Since anode voltage is higher than cathode, then diode acts like a closed switch [in ideal case]. 
** Therefore, both D<sub>2</sub> and D<sub>3</sub> are open. 

'''Insert example 1B'''

Now let's consider this next example.

* Step 1: Find I<sub>1</sub>, I<sub>2</sub>, and I<sub>3</sub> assuming ideal diodes. 
** Find all possible current flow directions (current always flows from high potential to low potential). 
* Step 2: We'll assume that 


===Assignment 1.1===

We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: 

<math> 
\begin{align}
2x_{2}+x_{3}=5 \\
x_{3}=\frac{9}{3} \\
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\
2x_{2}+\frac{9}{3}=5 \\
2x_{2}=2 \\
x_{2}=1
\end{align}
</math>