# Need help finding the drain current of a jfet

I have this simple dc jfet circuit (homework):

First of all I had to find the values of the drain and source resistors given the data above, this was pretty simple and I got: $$R_D=4k\Omega$$ $$R_S=11k\Omega$$

Next, $$\I_{DSS}\$$ get's doubled to $$\8mA\$$, now I'm asked to find $$\I_D\$$.

Using: $$I_D = I_{DSS}\left [ 1-\frac{V_{GS}}{V_P} \right ]^{2}$$

I've found out that: $$I_{D1} =1.16mA$$ $$I_{D2} =1.0258mA$$

During the calculations I've assumed the transistor is saturated thus: $$V_{DG} > V_P$$ It seems like both $$\I_D\$$'s fulfill the above requirement.

How can I tell which one is correct ?

• Have u already checked if both currents fulfill your saturated supposition? Mar 27, 2015 at 14:04
• I have, I get $V_{DG} = 5.36v$ and $V_{DG} = 5.9v$ respectivly, they are both larger than $V_P = -2$
– Mike
Mar 27, 2015 at 14:19
• Well if you see mathematically , Id is directly proportional to Idss. For Idss = 4 mA , Id = 1 mA . So for Idss = 8 mA, Id = ?? (I think I'm right, just a bit confused why no one else has pointed this out. Feel free to correct me) Mar 27, 2015 at 15:23
• but $V_{GS}$ is also a function of $I_D$
– Mike
Mar 27, 2015 at 16:18
• @Mike But $V_S = 3V$ will also satisfy the current equation. See my answer. Mar 28, 2015 at 9:03

The image (taken from wikipedia) below shows the JFET chara. It can be seen that the drain current reduces to zero as $V_{GS}$ approaches pinch off ($V_{P}$). And the channel is off for $|V_{GS}|>|V_{P}|$ and no current flow happens.

So the transistor is in saturation and the Shockley's equation

$$I_D = I_{DSS}\left [ 1-\frac{V_{GS}}{V_P} \right ]^{2}$$

is valid only if $|V_{GS}| < |V_P|$ and $V_{DG} > V_P$.

Now calculating $V_{GS}$ in your case,

case1: $I_{D1} =1.16mA$ $$V_{GS} = -2.76V$$ But $V_{P}=-2V$ so $|V_{GS}| > |V_P|$ and hence transistor is in cut-off.

case2: $I_{D2} =1.0258mA$ $$V_{GS} = -1.2838V$$ Here, $|V_{GS}| < |V_P|$ and hence transistor is in saturation.

So $I_{D} =1.0258mA$ is the correct answer.

PS: You should have faced this problem while calculating the value of $R_S$ also.

$$1mA = 4mA\left( 1 + \dfrac{10-1mA\times R_S}{2}\right)^2$$

$R_S = 11k\Omega$ and $R_S = 13k\Omega$ will satisfy this equation. The value $R_S = 13k\Omega$ can not be used because of the same reason discussed above.