ECE492 Section 1 Notes: Difference between revisions

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Let's try Phosphorous for Penta-Valent example:
Let's try Phosphorous for Penta-Valent example:
* 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. :)
'''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





Latest revision as of 02:03, 27 January 2022

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 * 10-19J).
  • 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.

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 D2 and D3 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 D2 and D3 open? Since anode voltage is higher than cathode, then diode acts like a closed switch [in ideal case].
    • Therefore, both D2 and D3 are open.

Insert example 1B

Now let's consider this next example.

  • Step 1: Find I1, I2, and I3 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 and plug it into to find x2: