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Physical electronics devices and ics miscellaneous

Physical Electronics Devices and ICs

  1. Consider the JFET circuit given below—
    Current ID is given by
    Assume ID = 12 1 +
    VGS
    2mA—
    4


    1. 2·26 mA
    2. 3·39 mA
    3. 1·48 mA
    4. 2·78 mA
Correct Option: A

The given circuit

From given circuit VGS = VG – VS = 0 – VS
(Since VG = 0V and VS = IDRS)
or VGS = – IDRS = – ID.1 kΩ


Now, ID = 121 + VGS2 mA
4


Now, ID = 121 - ID2
4

or ID = 12 1 + ID2 – 2.1. ID
164

or ID = 12
16 + ID2 – 8ID
16

or 4 ID = 48 + 3 ID2 – 24 ID
or 3 ID2 – 28 ID + 48 = 0
or ID =
28 ± √282 – 4 × 3 × 48
2 x 3

ID = 2·26 mA
Hence alternative (A) is the correct choice.



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