# JFET biasing problem

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 $V_{DS}$ 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 $V_S$). – Null Apr 28 '15 at 20:21

## 1 Answer

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.