To find drain voltage for ntype JFET.

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In summary, the Ckt diagram for a ntype JFET shows that there are two potential sources of drain current, one coming from the gate source loop and the other from the drain source loop. The gate source loop has a negative voltage, while the drain source loop has a positive voltage.
  • #1
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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  • #2
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.
 
  • #3
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::rofl::tongue2: got it!

thanks...unplebeian
 

1. What is the drain voltage for an n-type JFET?

The drain voltage for an n-type JFET can be found using the equation VD = VDD - IDRD, where VDD is the supply voltage, ID is the drain current, and RD is the drain resistor.

2. How do you determine the supply voltage for an n-type JFET?

The supply voltage for an n-type JFET is typically specified by the manufacturer and can be found in the datasheet. It is important to use the correct supply voltage to avoid damaging the JFET.

3. What is the role of the drain resistor in finding the drain voltage for an n-type JFET?

The drain resistor is used to limit the drain current and prevent the JFET from being damaged. It also affects the drain voltage by creating a voltage drop across it according to Ohm's law.

4. Can the drain voltage be negative for an n-type JFET?

Yes, the drain voltage can be negative for an n-type JFET. This can occur when the drain current is high and the drain resistor value is low, resulting in a large voltage drop across the resistor.

5. How does the drain voltage affect the overall performance of an n-type JFET?

The drain voltage plays a crucial role in determining the operating point of the JFET and affects its gain, output impedance, and linearity. It is important to choose the appropriate drain voltage for the desired performance of the JFET circuit.

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