Design the JFET circuit for the largest in swing. Use the self-bias circuit shown in Fig. 6. Assume that VGS (0) = -1.3 and Vcc = 15 volts. Furthermore, assume that ipss = 20 mA. Using Fig. 4b, draw the load line and identify the Q point. Explain why this will allow the largest swing. Use ip = ipss (1- VGS VGS(0) to show what happens to i, and vps when you have a swing of 0.2 volts in vcs form its operating point (that is, change vas by ±0.2 volts and compute the corresponding iD and VDs). RD RG Rs 0 20nA GS=-1.3 VGS 12 10nA -0- Fig. 6. Circuit for Examples 2 &3. BA-C 50 100 150 200 □ ID(J1) UDS Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of 0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3, respectively.
Design the JFET circuit for the largest in swing. Use the self-bias circuit shown in Fig. 6. Assume that VGS (0) = -1.3 and Vcc = 15 volts. Furthermore, assume that ipss = 20 mA. Using Fig. 4b, draw the load line and identify the Q point. Explain why this will allow the largest swing. Use ip = ipss (1- VGS VGS(0) to show what happens to i, and vps when you have a swing of 0.2 volts in vcs form its operating point (that is, change vas by ±0.2 volts and compute the corresponding iD and VDs). RD RG Rs 0 20nA GS=-1.3 VGS 12 10nA -0- Fig. 6. Circuit for Examples 2 &3. BA-C 50 100 150 200 □ ID(J1) UDS Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of 0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3, respectively.
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Transcribed Image Text:Design the JFET circuit for the largest in swing. Use the self-bias circuit shown in
Fig. 6. Assume that VGS (0) = -1.3 and Vcc = 15 volts. Furthermore, assume that
ipss = 20 mA. Using Fig. 4b, draw the load line and identify the Q point. Explain
why this will allow the largest swing. Use ip = ipss (1-
VGS
VGS(0)
to show what
happens to i, and vps when you have a swing of 0.2 volts in vcs form its operating
point (that is, change vas by ±0.2 volts and compute the corresponding
iD and VDs).
RD
RG
Rs
0
20nA
GS=-1.3 VGS
12
10nA
-0-
Fig. 6. Circuit for Examples 2 &3.
BA-C
50
100
150
200
□ ID(J1)
UDS
Fig. 4b. The IV characteristics of an n-channel JFET (J113). The plots are for VGs increments of
0.05 volts. VGS(0) -1.3. The yellow and blue load lines are for examples 2 &3,
respectively.
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