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An n-channel JFET has $${I_{{\text{DSS}}}} = 2\,{\text{mA}}$$   and Vp = -4 v. Its transconductance gm = in (mA/V) for an applied gate to source voltage VGS = -2 v is

Answer & Solution
Correct Answer: Option B
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7 Comments
Charan Kumar
Charan Kumar 3 years ago
gm = go(1-vgs/vp)

go = 2 idss/|vp|
= 2*0.002/4 = 0.001
gm = 0.001(1- (-2/-4))
=0.005 = 0.5mA/V
Jayvee Semetara
Jayvee Semetara 5 years ago
n channel JFET
𝑰𝑫𝑺 = 𝟐 𝒎𝑨
𝑽𝑫𝑺 = −𝟐 𝑽
The expression for gm is given as
𝒈𝒎 =
𝟐 𝑰𝑫𝑺𝑺
|𝑽𝑷|
[𝟏 −
𝑽𝑮𝑺
𝑽𝑷
]
𝒈𝒎 =
𝟐 × 𝟐 × 𝟏𝟎−𝟑
|−𝟒|
[𝟏 −
(−𝟐)
(−𝟒)
]
= 𝟎. 𝟓 𝒎𝑨/𝑽
Option (b)
Sonu Sharma
Sonu Sharma 6 years ago
I'd=Idss (1-Vgs/Vp)2
Id=1/2ma
Gm=I'd/Vgs
Gm=1/4
Sonu Sharma
Sonu Sharma 6 years ago
I'd=Idss (1-Vgs/Vp)2
Id=1/2ma
Gm=I'd/Vgs
Gm=1/4
Vijaya Durga
Vijaya Durga 6 years ago
Id= Idss(1 - Vgs/Vp)2 = 0.5mA
IT_25_PREETI_GUPTA
IT_25_PREETI_GUPTA 6 years ago
Ans should be 0.75.
Jora Arulious
Jora Arulious 8 years ago
Can you please provide the solution