A JFET has the following parameters: IDSS = 30 mA, VGS(off) = - 5V, VGS = -4.5 V. Find the value of drain current.

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SSC JE Electrical 10 Oct 2023 Shift 2 Official Paper-I
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  1. 30 mA
  2. 0.5 mA
  3. 15 mA
  4. 0.3 mA

Answer (Detailed Solution Below)

Option 4 : 0.3 mA
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Detailed Solution

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Concept:

In pinch off or saturation region in JFET, drain current is given by Shockley equation and is written as

\({I_D} = {I_{DSS}}{\left( {1 - \frac{{{V_{GS}}}}{{{V_{GS\left( {off} \right)}}}}} \right)^2}\)

Calculation:

Given IDSS = 30 mA

VGS = -4.5 volt

VGS(off) = -5 volt

\({I_D} = 30{\left( {1 - \frac{{ - 4.5}}{{ - 5}}} \right)^2} = 0.3\;mA\)

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