Calculating Current Through JFET Transistor System

[Hybird]

I have a problem of calculating the current through a system. It is a simple JFET transistor with the source grounded. The gate is reverse biased with a 3v potential, and the drain has a 2000 ohm resistor in series then a positive 20v potential in series which is of course grounded. How would I go about finding the current through the load resistor?

 

[Hybird]

...Assuming an n-type Si transistor

Also I don't have any other information.. I know that the more and more you apply a negative voltage on the gate, the smaller the channel gets which leads to more resistance across the source-drain channel. The answer I am lookin for is 2mA.. that would lead to a resistance of 8000 ohms, cause we still have a 2000 ohm resistor in series therefore 10000 ohm and with a voltage across the source-drain channel of 20 volts.. you get your 2mA answer, I don't see how you can get this bridge voltage without knowing something about a family of curves or the like?

 

[ravenprp]

To calculate the current through the load resistor in this JFET transistor system, we can use Ohm's Law and the current-voltage relationship of a JFET transistor.

First, we need to determine the drain-source voltage (Vds) of the JFET transistor. Since the gate is reverse biased with a 3v potential and the source is grounded, the Vds can be calculated as 20V - 3V = 17V.

Next, we can use the current-voltage relationship of a JFET transistor, which states that the drain current (Id) is equal to the transconductance (gm) multiplied by the square of the gate-source voltage (Vgs) minus the threshold voltage (Vt). In this case, we can assume that the threshold voltage is zero since the transistor is in saturation mode.

Therefore, Id = gm * (Vgs)^2

To find the transconductance, we can use the formula gm = 2 * √(Idss * (Vgs - Vt)), where Idss is the maximum drain current when Vgs = 0.

Since the source is grounded, Vgs = 3V. Assuming Idss = 10mA, we can calculate gm as 2 * √(10mA * (3V - 0V)) = 6mA/V.

Now, we can plug in our values to calculate the drain current (Id): Id = 6mA/V * (3V)^2 = 54mA.

Finally, we can use Ohm's Law to find the current through the load resistor (I = V/R). Since the load resistor has a value of 2000 ohms and the drain-source voltage is 17V, the current through the load resistor is 17V/2000 ohms = 8.5mA.

Therefore, the current through the load resistor in this JFET transistor system is 8.5mA.