Page 42 - ebook PdP FET
P. 42
EXAMPLE TABLE OF CONTENTS
ANSWER:
ii) CUT-OFF POINT, VDS (cut-off)
VDD = IDRL + VDS ; ID = 0A
VDD = VDS
VDS(cut-off) = VDD CHAPTER 4
= 20V
iii. Q POINT
VDSQ = VDD
= 20V
2
= 10V CHAPTER 3
iv. DC load line of the JFET amplifier
CHAPTER 2
CHAPTER 1
35
