14. For the network of Fig. 6.78, V₁ = 9 V. Determine: (a) ID- (b) Vs and Vps- (c) VG and VGS. (d) Vp. 18 V • 750 ΚΩ VG VGS '91 ΚΩ ID 2 ΚΩ VDS Vs 0.68 ΚΩ Figure 6.78 Problem 14 V=9V IDSS=8 mA

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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(P4) JFET (Course: Electronic Devices and Circuit Theory).

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§ 6.4 Voltage-Divider Biasing
14. For the network of Fig. 6.78, VD = 9 V. Determine:
(a) ID.
(b) Vs and Vps-
(c) VG and VGS.
(d) Vp.
• 750 ΚΩ
m
VG
18 V
+
VGS
91 ΚΩ
ID
2 ΚΩ
Vps
o V₂ = 9 V
0.68 kQ2
Figure 6.78 Problem 14
IDSS = 8 mA
Transcribed Image Text:§ 6.4 Voltage-Divider Biasing 14. For the network of Fig. 6.78, VD = 9 V. Determine: (a) ID. (b) Vs and Vps- (c) VG and VGS. (d) Vp. • 750 ΚΩ m VG 18 V + VGS 91 ΚΩ ID 2 ΚΩ Vps o V₂ = 9 V 0.68 kQ2 Figure 6.78 Problem 14 IDSS = 8 mA
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