Question

The drain current in a JFET is given by:

• A. $I_D = I_{DSS} \left(1 - \frac{V_{GS}}{V_P} \right)^2$
• B. $I_D = I_{DSS} \left(1 + \frac{V_{GS}}{V_P} \right)^2$
• C. $I_D = I_{DSS} \left(1 - \frac{V_P}{V_{GS}} \right)^2$
• D. $I_D = I_{DSS} \left(1 + \frac{V_P}{V_{GS}} \right)^{1/2}$

Hint:

Remember: In JFETs, $I_D$ decreases with more negative $V_{GS}$ and becomes zero at pinch-off.

Answer 1

The Correct Option is A

Step 1: Understanding JFET characteristics
In a JFET, the current $I_D$ depends on the gate-source voltage $V_{GS}$ and the pinch-off voltage $V_P$. 
Step 2: Shockley’s Equation
The standard equation for drain current in the saturation region is: 
$I_D = I_{DSS} \left(1 - \frac{V_{GS}}{V_P} \right)^2$ 
Step 3: Validity
This formula is valid only for $V_{GS}<0$ for n-channel and $V_{GS}>0$ for p-channel. 
Conclusion: The drain current in JFET is governed by Shockley’s quadratic law.