bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx 1 Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering 43 Chp12 Prob 12.24 Solution Common-Source pFET Amplifier

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Engineering 43. Chp12 Prob 12.24 Solution Common-Source p FET Amplifier. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. ENGR43 P12.24. Use LOAD-LINE Analysis to Find v o ( t ) for the Ckt below Note: This a p FET Circuit. - PowerPoint PPT Presentation

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Page 1: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx1

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp12Prob 12.24

SolutionCommon-Source pFET

Amplifier

Page 2: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx2

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

ENGR43 P12.24

Use LOAD-LINE Analysis to Find vo(t) for the Ckt below

• Note: This a pFET Circuit

Page 3: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx3

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

The pFET Characteristic

Page 4: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx4

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

P12.24 GamePlan

Draw LoadLine for the 1 kΩ Resistor load, then find the operating (vDS, iD) Values corresponding to This Set of vGS values • vGS = [7v + 1V•sin(0)] −10V

– This is VGSQ = −3 V

• vGS = [7v + 1V•sin(π)] −10V– This is vGS,max = −4 V

• vGS = [7v + 1V•sin(3π/2)] −10V– This is vGS,min = −2 V

Page 5: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx5

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

Page 6: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx6

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

Page 7: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx7

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

Page 8: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx8

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

Page 9: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx9

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices

Page 10: Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

© [email protected] • ENGR-43_Prob_12-24_pFET_Csource_Amp_Soln.pptx10

Bruce Mayer, PE Engineering-43 • Electrical Circuits & Devices