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Solve: Answers should be expressed using 2 decimal places a. k (transfer constant) = MAN2 b. Ip = mA c. VG = V d. Vs = V e. VGs = V

Solve: Answers should be expressed using 2 decimal places a. k (transfer constant) = MAN2 b. Ip = mA c. VG = V d. Vs = V e. VGs = V

Introductory Circuit Analysis (13th Edition)
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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Given:

  • VDD = 21 V
  • R1 =12 kohms
  • R2 = 5 kohms
  • RD = 580 ohms
  • RS = 0 kohms
  • RL = 15 kohms
  • ID(on) = 17 mA  at VGS(on) = 5 V
  • VGS(th) = 2 V
  • gm = 25  mS

Solve:  Answers should be expressed using 2 decimal places

  1. k (transfer constant) = Blank 1 mA/V2  
  2. ID = Blank 2 mA 
  3. VG = Blank 3 V 
  4. VS = Blank 4 V  
  5. VGS = Blank 5 V 
  6. VDS = Blank 6 V
  7. Rin(tot) = Blank 7 kohms
  8. Voltage gain, Av = Blank 8
QUESTION 4 +VpD Rp C3 RL Vin R2 Rs Given: • VDD = 21 V • R1=12 kohms • R2 = 5 kohms • RD = 580 ohms • Rs = 0 kohms • RL = 15 kohms • ID(on) = 17 mA at VGS(on) = 5 V • VGS(th) = 2V • gm = 25 ms
Transcribed Image Text:QUESTION 4 +VpD Rp C3 RL Vin R2 Rs Given: • VDD = 21 V • R1=12 kohms • R2 = 5 kohms • RD = 580 ohms • Rs = 0 kohms • RL = 15 kohms • ID(on) = 17 mA at VGS(on) = 5 V • VGS(th) = 2V • gm = 25 ms
Solve: Answers should be expressed using 2 decimal places a. k (transfer constant) = mAN2 b. Ip = mA c. VG = V d. Vs = V e. VGs = V f. Vps = V g. Rin(tot) = kohms h. Voltage gain, Ay =
Transcribed Image Text:Solve: Answers should be expressed using 2 decimal places a. k (transfer constant) = mAN2 b. Ip = mA c. VG = V d. Vs = V e. VGs = V f. Vps = V g. Rin(tot) = kohms h. Voltage gain, Ay =
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