Calculate Gate Source, Drain Current, and Source Voltage of a JFET

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Discussion Overview

The discussion revolves around calculating the gate-source voltage, drain current, and source voltage of a Junction Field Effect Transistor (JFET). It includes elements of homework problem-solving and theoretical analysis.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the gate voltage (Vg) using a voltage divider formula but initially arrives at an incorrect value of -7V instead of -4V.
  • Another participant suggests that after determining Vg, they would find the transconductance parameter (Kn) and drain current (Id), and then set up an equation to find the gate-source voltage (Vgs).
  • A different participant calculates the voltage across a potential divider and concludes that the gate voltage is -4V, which aligns with the corrected value mentioned earlier.
  • One participant requests clarification on specific terms (ka', W/L, and Ka) that are not standard in JFET models, indicating a need for additional context or equations.
  • A later reply indicates that one participant has resolved their issues and expresses hope for a good outcome on their test.

Areas of Agreement / Disagreement

Participants generally agree on the corrected gate voltage of -4V, but there are unresolved questions regarding the specific equations and terms used in the analysis.

Contextual Notes

Some assumptions regarding the definitions of terms and the equations used in the calculations are not fully clarified, which may affect the understanding of the discussion.

DODGEVIPER13
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Homework Statement


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Homework Equations


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The Attempt at a Solution


So what I started with is Vg=(R2/(R1+R2))Vdd - 10 = -7 Volts this is incorrect though I was told it is -4 V what did I do
 

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Theoretically after I solve for this I would get Kn and Id. Then I would set up an equation to find Vgs.
 
The voltage across the potential divider R1+R2 is 20V. So the voltage across R2 is..

20 * R2/(R1+R2) = 6V

So the gate is 6V above the -10V rail which is -4V.
 
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Thanks man
 
What equation were you given that includes ka', W/L and Ka? These are not standard terms in a JFET model so you need to give us your equation(s).
 
Oh yah I got it all figured out man thanks just had the test too hope I did well
 

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