ECE492 Section 1 Notes - Revision history

Eddynetweb: Jan 26 class publishing.

2022-01-27T02:03:52Z

Jan 26 class publishing.

@@ Line 103: / Line 103: @@
 Let's break the problem down:
 * Step 1: Fill me in. :)
 ===Assignment 1.1===

Eddynetweb: Additional lecture content.

2022-01-24T19:47:35Z

Additional lecture content.

← Older revision		Revision as of 19:47, 24 January 2022
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	* Pair Phosphorous with Silicon atoms, knowing that Phosphorous has an outer electron count of 5.		* 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. :)

	===Assignment 1.1===		===Assignment 1.1===

Eddynetweb at 19:17, 24 January 2022

2022-01-24T19:17:06Z

@@ Line 34: / Line 34: @@
 * 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).
 ===Assignment 1.1===

Eddynetweb: Update lecture work with content.

2022-01-24T18:56:51Z

Update lecture work with content.

@@ Line 9: / Line 9: @@
 == Section 1.1 ==
-Insert lecture notes here!
+We first need to consider the following materials:
 ===Assignment 1.1===

Eddynetweb: Create Electronic Circuits wiki for note taking.

2022-01-24T18:28:54Z

Create Electronic Circuits wiki for note taking.

New page

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 ==

Insert lecture notes here!

===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>