Introduction / Context:
For a JFET, the Shockley model provides the transfer characteristic and a simple relation between the transconductance and the device parameters. From a single operating point (VGS, gm) together with IDSS, one can back-calculate the pinch-off voltage VP (negative for n-channel JFETs on typical substrates).
Given Data / Assumptions:
- IDSS = 2 mA.
- VGS = −2 V.
- gm(VGS) = 0.5 mS = 0.0005 S.
- Shockley relations: ID = IDSS(1 − VGS/VP)^2 and gm(VGS) = (2 IDSS / |VP|) * (1 − VGS/VP).
Concept / Approach:
We use the gm expression because it involves a single power of (1 − VGS/VP) and is algebraically simpler for solving VP. Remember that for an n-channel JFET, VP is negative; we often denote |VP| as the magnitude. Compute |VP|, then assign the sign.
Step-by-Step Solution:
Let x = |VP|. For an n-channel JFET, VP = −x.gm(VGS) = (2 IDSS / x) * (1 − VGS/VP) = (2 × 0.002 / x) * (1 − (−2)/(−x)) = (0.004/x) * (1 − 2/x).Set gm = 0.0005 S: 0.0005 = (0.004/x) * (1 − 2/x).Rearrange: 0.0005 x = 0.004 − 0.008/x ⇒ 0.0005 x^2 = 0.004 x − 0.008.Multiply by 2000: x^2 − 8x + 16 = 0 ⇒ (x − 4)^2 = 0 ⇒ x = 4 V.Thus VP = −4 V.
Verification / Alternative check:
Using gm0 = 2IDSS/|VP| = 4 mA/V / 4 V = 1 mS at VGS = 0; with VGS = −2 V, gm = gm0(1 − 2/4) = 1 mS × 0.5 = 0.5 mS, consistent.
Why Other Options Are Wrong:
−2 V or 2 V: Do not satisfy the gm relation with the given IDSS and VGS.+4 V: Sign is wrong for an n-channel device's pinch-off voltage.−1 V: Also inconsistent with the computed gm.
Common Pitfalls:
Dropping the negative sign of VP; mixing up mS (millisiemens) with mW; using the current equation instead of gm and creating algebraic complexity.
Final Answer:
-4 V
Discussion & Comments