# To find drain voltage for ntype JFET.

1. Feb 7, 2008

### sphyics

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 $$^{}_{}Vgs$$= 10mA
Vp = 4v {Vp= pinch off voltage}
To find drain voltage Vd =?????

http://img233.imageshack.us/img233/4007/19716664vu3.png [Broken]

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

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

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

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

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2. Feb 21, 2008

### unplebeian

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.

3. Feb 24, 2008

### sphyics

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______________:rofl::tongue2: got it!!!!!!!

thanks...unplebeian