I tried solving this problem, however I feel like there's something missing, I am asked to bias a N-channel JFET with using this provided data:

  • Idss= 12 [mA]
  • VDD= 12 [V]
  • VP = -5 [V]
  • VDS=VDD/2

using a voltage divider circuit, however I can't seem to get an equation for the resistance values.

Here's what I've done so far: The circuit

What I've done so far

  • \$\begingroup\$ The \$V_{DS}\$ spec tells you how much voltage can be across \$R_S\$ and \$R_D\$, but it seems like you need to know the desired output voltage (whether the output is \$V_D\$ or \$V_S\$). \$\endgroup\$
    – Null
    Commented Apr 28, 2015 at 20:21

1 Answer 1


You "feel like there's something missing" in the task... and you are right, there are missing two pieces of information to determine the required bias (\$ V_{GS} \$).

From the given information you can write:

$$ I_D = \frac{V_{DD}}{2 \cdot (R_S + R_D)} = \frac{k \cdot V_{DD} - V_{GS}}{R_S} = I_{DSS} \cdot \left(1 - \frac{V_{GS}}{V_P}\right)^2 $$ where \$k\$ is the \$ R_1, R_2 \$ divider ratio \$ \frac{R_2}{R_1+R_2} \$.

So you have to know either:

  1. \$ R_S, k \$ and you can calculate the required \$ V_{GS} \$, (then \$ I_D \$ and \$ R_D \$), or
  2. \$ R_S, R_D \$ and you can calculate \$ I_D \$, then the required \$ V_{GS} \$ (and finally \$k\$), or
  3. \$ R_D, k \$ and again calculate the rest

Then, to solve the \$ R_1, R_2 \$ divider out of the \$k\$ value, you have to know (choose) value of one of the resistors (or their sum), of course.


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