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JFET VGS for given IDS and gm

Short answer

VGS=VP(1−√(ID/IDSS))=−4.5(1−√(3/9))=−1.90V. gm=gm0√(ID/IDSS) with gm0=2IDSS/|VP|=4mS, giving gm=2.31mS.

An n-channel JFET has a pinch-off voltage of -4.5V and IDSS = 9 mA. At what value of VGS will IDS be equal to 3 mA? What is its gm at this IDS?

Given: VP = −4.5 V, IDSS = 9 mA, ID = 3 mA.

(a) VGS for ID = 3 mA

ID = IDSS (1 − VGS/VP)²

1 − VGS/VP = √(ID/IDSS) = √(3/9) = 0.5774

VGS = VP (1 − 0.5774) = −4.5 × 0.4226 = −1.90 V

(The other root, 1 − VGS/VP = −0.5774, gives VGS = 1.5774 VP = −7.10 V. This is beyond the pinch-off voltage, where the JFET is cut off and the equation does not apply, so it is not valid.)

(b) Transconductance at this ID

gm0 = 2 IDSS/|VP| = 2 × 9 / 4.5 = 4 mS

gm = gm0 (1 − VGS/VP) = 4 × 0.5774 = 2.31 mS

Check: gm = gm0 √(ID/IDSS) = 4 × √(1/3) = 2.31 mS.

Answer: VGS = −1.90 V, gm = 2.31 mS (2.31 mA/V).

✓ Answer checked by Claude Opus 5.5 (high effort)

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