Question

An 'n' channel JFET has \(I_{DSS}=8\text{mA}\), \(V_P=-5\text{V}\). Determine the minimum value of \(V_{DS}\) for \(V_{GS}=-2\text{V}\) in the pinch off region.

• A. --7 V
• B. 3 V
• C. 5 V
• D. 7 V

Hint:

For n-channel JFET, \(V_P\) is negative and \(V_{GS}\) is typically \(\le 0\).
Pinch-off condition: \(V_{DS} \ge V_{GS} - V_P\).
\(V_{DS(sat)} = V_{GS} - V_P\) is the drain-source voltage at the boundary between ohmic and saturation regions.

Answer 1

The Correct Option is B

For an n-channel JFET, the condition for operation in the pinch-off (saturation) region is: \[ V_{DS} \ge V_{GS} - V_P \] The minimum value of \(V_{DS}\) for pinch-off is \(V_{DS(sat)} = V_{GS} - V_P\). Given: Gate-to-source voltage, \(V_{GS} = -2\text{V}\). Pinch-off voltage, \(V_P = -5\text{V}\). \(I_{DSS} = 8\text{mA}\) (This is the drain current when \(V_{GS}=0\) and the JFET is in pinch-off; it is not needed for this specific calculation). Minimum \(V_{DS}\) for pinch-off: \[ V_{DS(min)} = V_{GS} - V_P = (-2\text{V}) - (-5\text{V}) = -2\text{V} + 5\text{V} = 3\text{V} \] So, the JFET enters the pinch-off region when \(V_{DS}\) reaches 3V (for \(V_{GS}=-2V\)). \[ \boxed{3 \text{ V}} \]