Short notes

JFET Short Notes

In a common-source amplifier built around a 2N5457 N-channel JFET with a 10 kΩ drain resistor and V_DD = 15 V, the drain current settles at a value set entirely by the gate-source voltage — there is no gate current at all, because the gate is a reverse-biased PN junction. With V_GS = 0 V, current equals I_DSS; for the 2N5457, that is typically 1 mA to 5 mA. Apply −1 V to the gate and the channel pinches down; reach V_P (pinch-off voltage, around −5 V for this device) and I_D drops to nearly zero.

EEE, ECE, EI

How it works

The JFET channel is a bar of N-type (or P-type) silicon with ohmic contacts at drain and source. The gate is a P-type region diffused on both sides, forming two PN junctions that are always reverse biased during normal operation. V_GS controls the depletion region width — more negative V_GS widens the depletion regions, narrows the channel, and reduces I_D. The Shockley equation governs this: I_D = I_DSS · (1 − V_GS/V_P)². Transconductance g_m = 2I_DSS/|V_P| · (1 − V_GS/V_P) peaks at V_GS = 0 and reaches a typical value of 2–4 mA/V for small-signal JFETs.

Key points to remember

JFETs are depletion-mode only — they are ON at V_GS = 0 and turned off by increasing the gate voltage magnitude. N-channel JFETs require V_GS between 0 and V_P (negative); P-channel JFETs require positive V_GS. The drain resistance r_d in the saturation region is typically 10 kΩ to 1 MΩ. Because gate current is essentially zero (typically < 1 nA), JFET amplifiers present extremely high input impedance — up to 10⁸ Ω — making them ideal for pH meters and electrometer circuits. The 2N5484 and BFW10 are commonly cited in Indian textbook examples and exam problems.

Exam tip

The examiner always asks you to derive the transconductance expression g_m from the Shockley equation by differentiation — show the step from I_D = I_DSS(1 − V_GS/V_P)² to g_m = dI_D/dV_GS clearly.

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