ECE110 Transparencies: Bipolar Junction Transistors and Logic Gates - Prof. Lippold Haken, Study notes of Electrical and Electronics Engineering

This document consists of transparencies from a university lecture on electrical engineering and computer science (ece) 110, covering topics such as bipolar junction transistors, binary logic gates, cmos logic, digital logic circuits, and boolean algebra identities. The transparencies also include examples, quiz questions, and exercises.

Typology: Study notes

Pre 2010

Uploaded on 02/24/2010

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L Haken ECE110 transparencies – set 2 page 1
ECE 110 Transparencies
Set 2 for Noon Lecture
Lippold Haken
Bipolar Junction Transistor.......................................................... 2
Binary Logic Gates ...................................................................... 15
CMOS Logic ................................................................................. 16
Digital Logic Circuits................................................................... 22
Boolean Algebra Identities.......................................................... 28
Minterms and Sum of Products.................................................. 31
Binary Numbers ........................................................................... 40
Binary Addition............................................................................ 43
Seven Segment Display................................................................ 46
Hexadecimal ................................................................................. 48
Digital Comparator...................................................................... 49
Multiplexer - Demultiplexer........................................................ 51
These transparencies include material by
Prof. Tim Trick and Prof. Marie-Christine Brunet
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ECE 110 Transparencies

Set 2 for Noon Lecture

Lippold Haken

Bipolar Junction Transistor.......................................................... 2

Binary Logic Gates ...................................................................... 15

CMOS Logic ................................................................................. 16

Digital Logic Circuits................................................................... 22

Boolean Algebra Identities .......................................................... 28

Minterms and Sum of Products.................................................. 31

Binary Numbers ........................................................................... 40

Binary Addition............................................................................ 43

Seven Segment Display ................................................................ 46

Hexadecimal ................................................................................. 48

Digital Comparator...................................................................... 49

Multiplexer - Demultiplexer........................................................ 51

These transparencies include material by

Prof. Tim Trick and Prof. Marie-Christine Brunet

Bipolar Junction Transistor

The BJT is a three-terminal device, so the IV graph will have a

family of curves.

In the active region, a BJT is an I b amplifier: I C = β * I B

When solving transistor circuits, we will deal with three BJT

parameters:

β

VBEON

VCESAT

Transistor Inverter Circuit

Transistor Parameters: β, V (^) BEON, VCESAT

If V i < VBEON, then transistor is operating in off region:

• I B = 0, IC = 0, VO = VCE = V CC

  • On Transfer Characteristic (V (^) o vs V (^) i Graph): V (^) OH = VCC V (^) IL = VBEON

If the transistor is operating in active region:

• V BE = V BEON

  • I (^) B = (Vi – V (^) BEON) / RB
  • I (^) C = β I (^) B
  • V (^) O = V (^) CE = V (^) CC – IC * RC
  • V (^) O = V (^) CE = V (^) CC – β I (^) B * RC
  • V (^) O = V (^) CE = V (^) CC – β (Vi – VBEON) * RC / RB
  • You must compute VCE to verify: V (^) CE > VCESAT
  • Slope of line is voltage gain (negative voltage gain)

If the transistor is operating in saturated region:

• V O = V CE = V CESAT

• V OL = V CESAT

  • One way to find VIH is to solve for V (^) IH in this equation: V (^) CESAT = VCC – β (V (^) IH – VBEON) * RC / RB
  • Another way to find VIH: ICSAT = (VCC -V (^) CESAT )/RC I (^) B = ICSAT /β, V (^) IH = VBEON+I (^) B*RB

Quiz LH5 (First quiz on Transistor Inverter)

Transistor Parameters: β =100, V BEON=0.7V, V CESAT=0.2V

a) If V i = 0.3V, what region is the transistor operating in?

What is I B, I C , and V O?

b) If V i = 1.9V, what region is the transistor operating in?

What is I B, I C , and V O?

c) If Vi = 3.7V, what region is he transistor operating in?

What is I B, I C , and V O?

d) What is V OH, V OL, VIL, V IH?

Transistor Inverter Example

Circuit has R b and R c.

We are interested in Vce in terms of Vin.

Quiz LH6 (Second quiz on Transistor Inverter)

Quiz LH6, continued

CMOS Logic