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For an n-channel JFET, the pinch off voltage Vp is - 4V, VDD is 10 V and the drain saturation current at zero gate bias IDD is 2 mA. The value of the saturated drain current for a gate voltage of -2 V is
1. 0.5 mA
2. 2 mA
3. 4.5 mA
4. 18 mA

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Correct Answer - Option 1 : 0.5 mA


Junction Field Effect Transistor (JFET):

  • It is one of the types of FET transistors.
  • It is used as controlled switches, voltage-controlled resistors, and as amplifiers.
  • BJT transistors are constructed with the PN-junctions but the JFET transistors have a channel instead of the PN-junctions. This channel is formed due to either P-type or N-type semiconductor materials.
  • The N-channel JFET has more current conduction than P-channel JFET because the mobility of electrons is greater than the mobility of holes. So the N-channel JFETs are widely used than P-channel JFETs.


Region of operation:

  • Ohmic Region: When VGS = 0 the depletion layer of the channel is very small and the JFET acts like a voltage-controlled resistor.
  • Cut-off Region: This is also known as the pinch-off region where the gate voltage, VGS is sufficient to cause the JFET to act as an open circuit as the channel resistance is at maximum.
  • Saturation or Active Region: The JFET becomes a good conductor and is controlled by the gate-source voltage, (VGS) while the drain-source voltage, (VDS) has little or no effect.
  • Breakdown Region: The voltage between the drain and the source, (VDS) is high enough to cause the JFET’s resistive channel to break down and pass uncontrolled maximum current.


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

Where ID = Drain current

IDSS = Maximum saturation current

VGS = Gate to source voltage

VP = Pinched-off voltage




Vp = -4 V

ID = 2 mA = 2 × 10-3 A

Gate voltage (VGS) = -2 V

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

\({I_D} = 2 \times {10^{ - 3}}{\left[ {1 - \frac{-2}{-4}} \right]^2} = 0.5 \times {10^{ - 3}} = 0.5\;mA\)

∴ Value of drain current is 0.5 mA.

The drain-source resistance is equal to the ratio of the rate of change in drain-source voltage and rate of change in drain current.

\({R_{DS}} = \frac{{{\rm{\Delta }}{V_{DS}}}}{{{\rm{\Delta }}{I_D}}} = \frac{1}{{{g_m}}}\)

Where RDS = Drain-source resistance

VDS = Drain to source voltage

ID = Drain current

gm= Trans-conductance gain

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