Quiescent output of a JFET amplifier

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SUMMARY

The forum discussion centers on the discrepancies between calculated and simulated quiescent output of a JFET amplifier using PSpice (SIMetrix). The user calculated a drain current (ID) of 6.86 mA and a gate-source voltage (VGS) of 1.024 V, while the simulation yielded ID of 3.9189 mA and VGS of 6.12165 V. The 2N3819 JFET is not in saturation at a gate voltage of 6V, necessitating a lower gate voltage, such as 4V, to achieve accurate results. The user seeks clarification on the AC and DC components of VGS to resolve the issue.

PREREQUISITES
  • Understanding of JFET amplifier operation
  • Familiarity with PSpice (SIMetrix) simulation software
  • Knowledge of key JFET parameters such as IDSS and VP
  • Basic electronics principles, including AC and DC analysis
NEXT STEPS
  • Review JFET amplifier design principles and quiescent point analysis
  • Learn how to set up and interpret PSpice simulations for JFET circuits
  • Investigate the effects of gate voltage on JFET operation and saturation
  • Study the AC and DC components of gate-source voltage (VGS) in JFET amplifiers
USEFUL FOR

Electronics students, circuit designers, and engineers working with JFET amplifiers and simulation tools like PSpice, particularly those seeking to understand quiescent output and its implications in circuit design.

Callum Plunkett
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Homework Statement


I am to construct a JFET amplifier with Pspice (SIMetrix) to determine the quiescent output and to compare it with my own calculations. However my calculations don’t match the simulated output by quite a margin. Can anyone point me in the right direction as I’ve had very little feedback from the uni. Electronics is really (evidently) not my strong point and its the first time I've used Pspice.

Homework Equations


ID = IDSS(1-ID/VP)2
IDSS = 10
VP = 3

VGS = -IDRS

The Attempt at a Solution



ID = 6.86mA
VGS = 1.024V

upload_2019-3-9_2-1-34.png

from the simulation:
ID = 3.9189ma
VGS = 6.12165V
 

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The 2N3819 is not completely in saturation mode at gate voltage 6V. Therefore, channel current will be reduced. You need to operate at lower gate voltage (4V or such) for simplified current equation to work
 
It's not clear what the gate voltage is. What are the ac and dc components of Vgs?
 

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