To find drain voltage for ntype JFET.

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The forum discussion focuses on calculating the drain voltage (Vd) for an n-type JFET with given parameters: V2 = -10V, V1 = +10V, R1 = 1M ohm, and R2 = 10k ohm. The user initially calculated a drain current (Id) of 22.5 mA, exceeding the maximum drain current (Idss) of 10 mA, indicating an incorrect assumption of the JFET's operating region. The solution involved recognizing that the JFET was biased in the ohmic region rather than the active region, leading to the application of a voltage divider formula to find Vd.

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sphyics
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Hi all,
Ckt diagram for ntype JFET
The values are as follows
V2 = -10 V , V1 = +10V
R1 = 1M ohm
R2 = 10 k ohm

IDss (drain current with shorted gate condition [tex]^{}_{}Vgs[/tex]= 10mA
Vp = 4v {Vp= pinch off voltage}
To find drain voltage Vd =?

http://img233.imageshack.us/img233/4007/19716664vu3.png

https://www.physicsforums.com/newattachment.php?do=manageattach&p=1600867

///////// SOLVED //////

For gate source loop
Vgg = V2 = -10 V
Vdd = V1 = +10V

From kirchhoffs law

(-Vgg ) – (Ig)* (Rg) – Vgs = 0 since (Ig ~= 0)

Vgs = (-Vgg) ------------------------------------------------------- ***********

By Shockley’s equation

Id = Idss( 1-(Vgs /Vp)^2

Id = 22.5 mA

For drain source loop

Vdd – (Id)*(Rd) – Vds = 0

Vds = Vdd – (Id)*(Rd)
Getting an absurd answer for Vds (= -215V)

aim was to obtain Vds then Vds = Vd – Vs; as Vs = 0, Vds = Vd


Can anyone help me out by pointing my fault in analysis. Or is there any other method to solve for Vd .
 

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Actually your answer if correct in one sense.

1. First notice that 22.5mA is greater than the Idss. So you have a problem.
2. Vgs= -10V. So what does this tell you about the state of the JFET. ON/OFF? The negative value of VDS is also a hint here.


Directly solving by equations by brute force won't help.
 
unplebeian said:
Actually your answer if correct in one sense.

1. First notice that 22.5mA is greater than the Idss. So you have a problem.
2. Vgs= -10V. So what does this tell you about the state of the JFET. ON/OFF? The negative value of VDS is also a hint here.


Directly solving by equations by brute force won't help.


brute force method really worked ;) and thanks u gave me a wonderful idea

as Id > Idss my assumption that its biased in active region is false... it must be in ohmic region(biased) same for Vgs although i could have inferred from vgs itself :(

the ohmic resistance is Rds =_____
replacing JFET by Rds and applying Voltage divider formulae to find Vd______________:cool::smile::-p got it!

thanks...unplebeian
 

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