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 ?
Best Answer
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.