Question

For an n-channel JFET, the pinch off voltage V_p is - 4V, V_DD is 10 V and the drain saturation current at zero gate bias I_DD 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

Answer 1

Correct Answer - Option 1 : 0.5 mA

Concept:

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 V_GS = 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, V_GS 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, (V_GS) while the drain-source voltage, (V_DS) has little or no effect.
• Breakdown Region: The voltage between the drain and the source, (V_DS) 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 I_D = Drain current

I_DSS = Maximum saturation current

V_GS = Gate to source voltage

V_P = Pinched-off voltage

 

Calculation:

Given:

V_p = -4 V

I_D = 2 mA = 2 × 10^-3 A

Gate voltage (V_GS) = -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 R_DS = Drain-source resistance

V_DS = Drain to source voltage

I_D = Drain current

g_m= Trans-conductance gain