2
\$\begingroup\$

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

\$\endgroup\$
  • \$\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 Apr 28 '15 at 20:21
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.