Solve: Answers should be expressed using 2 decimal places a. k (transfer constant) = mA/v2 b. Ip = mA c. VG = V d. Vs = V e. VGS = V

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Solve: Answers should be expressed using 2 decimal places
a. k (transfer constant) =
mA/v2
b. Ip =
mA
c. VG =
V
d. Vs =
e. VGs =
V
f. Vps =
V
g. Rin(tot) =
kohms
h. Voltage gain, Av =
>
Transcribed Image Text:Solve: Answers should be expressed using 2 decimal places a. k (transfer constant) = mA/v2 b. Ip = mA c. VG = V d. Vs = e. VGs = V f. Vps = V g. Rin(tot) = kohms h. Voltage gain, Av = >
+Vpp
Rp
C3
Vout
R1
RL
Vin
Rs
Given:
- VDD = 26 V
R1 =17 kohms
· R2 = 5 kohms
· RD = 750 ohms
• Rs = 0 kohms
RL = 14 kohms
• ID(on) = 16 mA at VGS(on) = 6 V
VGS(th) = 3 V
• gm = 27 ms
Transcribed Image Text:+Vpp Rp C3 Vout R1 RL Vin Rs Given: - VDD = 26 V R1 =17 kohms · R2 = 5 kohms · RD = 750 ohms • Rs = 0 kohms RL = 14 kohms • ID(on) = 16 mA at VGS(on) = 6 V VGS(th) = 3 V • gm = 27 ms
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