ECE492 Section 2 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 2.1

We'll begin by covering a Bi-polar Junction Transistor, also known as BJT.

What is it exactly? Well, it's a transistor.

• It is a 3 terminal devices.
• It is a unsymmetrical device.
• It amplifies the signal.

There are two types of BJTs:

• NPN, in which...
• PNP, in which...

Now we'll cover (B) Common Emitter Current gain:

• $B={\frac {I_{c}}{I_{b}}}$
• 50 < B < 200

...and (Alpha) Common Base Current Gain:

• $Alpha={\frac {B}{B+1}}={\frac {I_{c}}{I_{e}}}$

V-I Characteristics of Transistors:

• AAAA

NPN Transistor Example:

• Cut Off: Both junction are reverse biased.
• Saturation: Both junction are forward biased.
• Active: One junction is forward biased, another one is reverse biased.

Understanding cut off regions for NPN transistors:

• Cut off region (open switch - both junctions one reverse biased)
• Saturation region (closed switch)
• Active region (amplification - one junction forward biased another is reverse biased)
• Breakdown region (Damaged - acts like open circuit essentially)

How do you know when one of these three (or four counting breakdown) apply?

• Voltage
  • V_E < V_B < V_c
  • V_E < V_B > V_c
  • V_E > V_B < V_c
  • V_E > V_B > V_c
• NPN
  • Active
  • Saturation
  • Cut off
  • Reverse active
• PNP
  • Reverse active
  • Cut off
  • Saturation
  • Active

Understanding common emitter (CE) configuration

• Consider the following configurations...

Consider the following example:

• We'll assume that V_B is equal to 0 (since it is grounded).
• Start solving the example by assuming active mode of operation.
• In active mode of operation
  • B_BE = 0.7 (NPN), B_CE >= 0.3V
  • B_EB = 0.7 (PNP), B_EB >= 0.3V

Therefore, V_EB = 0.7V = V_E - V_B such that V_B = 0

Header 2

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

Assignment 1.1

We'll take this $x_{3}={\frac {9}{3}}$ and plug it into $2x_{2}+x_{3}=5$ to find x₂:

${\begin{aligned}2x_{2}+x_{3}=5\\x_{3}={\frac {9}{3}}\\{\text{Plug in }}x_{3}={\frac {9}{3}}{\text{ into }}2x_{2}+x_{3}=5{\text{:}}\\2x_{2}+{\frac {9}{3}}=5\\2x_{2}=2\\x_{2}=1\end{aligned}}$