spring 2007ee130 lecture 28, slide 1 lecture #28 announcements for problem 2 of hw#9, use a = 10 -7...

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EE130 Lecture 28, Slide 1 Spring 2007 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today Due on Friday, April 27 You may work with one partner OUTLINE • The PNPN thyristor Reading: Chapter 13

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Page 1: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 1Spring 2007

Lecture #28

ANNOUNCEMENTS

• For Problem 2 of HW#9, use A = 10-7 cm2

• The Term Project will be posted online today– Due on Friday, April 27– You may work with one partner

OUTLINE

• The PNPN thyristor

Reading: Chapter 13

Page 2: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 2Spring 2007

The Silicon Controlled Rectifier (Thyristor)

• The SCR is a PNPN device– Used in high-power switching applications

Page 3: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 3Spring 2007

• The SCR can be modeled as 2 interconnected BJTs

SCR Operation

Page 4: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 4Spring 2007

Forward blocking state

• Few holes injected into N region

• Thermal R-G current only

Page 5: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 5Spring 2007

Conducting State

1. Holes injected into N

2. Due to wide N-P depletion region, holes are swept into P, turning on PNP device

3. This turns on NPN device

• This regenerative effect maintains conduction even at low voltage, down to a minimum “holding current”.

Page 6: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 6Spring 2007

Quantitative Analysis• We can write:

• Normally, (1+2) <1

• However, as VAK , base width narrowing causes (1+2) 1, IAK

21

0201

022011

1

RRAK

RAKRAKAK

III

IIIII

Page 7: Spring 2007EE130 Lecture 28, Slide 1 Lecture #28 ANNOUNCEMENTS For Problem 2 of HW#9, use A = 10 -7 cm 2 The Term Project will be posted online today –Due

EE130 Lecture 28, Slide 7Spring 2007