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An N-channe! JFET has IDSS = 8 mA and Vp = -5 V. The minimum values of VDS for pinch- off region and the drain current IDS for VGS= - 2V in the pinch-off region are respectively.
- a)2 V and 1.8 mA
- b)3 V and 2.9 mA
- c)2 V and 2.9 mA
- d)3 V and 1.8 mA
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of ...
The minimum value of VDS for pinch-off to occur for VGS = -2V is
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An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of ...
Given parameters:
IDSS = 8 mA
Vp = -5 V
VGS = -2 V
To determine:
Minimum values of VDS for pinch-off region and the drain current IDS for VGS = -2 V in the pinch-off region.
Solution:
1. Pinch-off region:
In the pinch-off region, the JFET is in saturation mode and the drain current ID is constant. Therefore, the gate-source voltage VGS is equal to the pinch-off voltage Vp.
VGS = Vp = -5 V
2. Minimum VDS for pinch-off region:
The drain-source voltage VDS in the pinch-off region is the minimum voltage required to maintain the drain current ID at its maximum value IDSS. This voltage is given by:
VDS(off) = |Vp|
VDS(off) = |-5 V| = 5 V
Therefore, the minimum VDS for pinch-off region is 5 V.
3. Drain current IDS for VGS = -2 V in the pinch-off region:
The drain current ID for VGS = -2 V can be calculated using the following equation:
ID = IDSS[1 - (VGS/Vp)]^2
ID = 8 mA [1 - (-2 V/-5 V)]^2
ID = 8 mA [1 - 0.4]^2
ID = 2.88 mA
Therefore, the drain current IDS for VGS = -2 V in the pinch-off region is 2.9 mA (approx).
Hence, the correct answer is option B.
IDSS = 8 mA
Vp = -5 V
VGS = -2 V
To determine:
Minimum values of VDS for pinch-off region and the drain current IDS for VGS = -2 V in the pinch-off region.
Solution:
1. Pinch-off region:
In the pinch-off region, the JFET is in saturation mode and the drain current ID is constant. Therefore, the gate-source voltage VGS is equal to the pinch-off voltage Vp.
VGS = Vp = -5 V
2. Minimum VDS for pinch-off region:
The drain-source voltage VDS in the pinch-off region is the minimum voltage required to maintain the drain current ID at its maximum value IDSS. This voltage is given by:
VDS(off) = |Vp|
VDS(off) = |-5 V| = 5 V
Therefore, the minimum VDS for pinch-off region is 5 V.
3. Drain current IDS for VGS = -2 V in the pinch-off region:
The drain current ID for VGS = -2 V can be calculated using the following equation:
ID = IDSS[1 - (VGS/Vp)]^2
ID = 8 mA [1 - (-2 V/-5 V)]^2
ID = 8 mA [1 - 0.4]^2
ID = 2.88 mA
Therefore, the drain current IDS for VGS = -2 V in the pinch-off region is 2.9 mA (approx).
Hence, the correct answer is option B.
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An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer?.
An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer?.
Solutions for An N-channe! JFET has IDSS = 8 mA andVp = -5 V. The minimum values of VDS for pinch-off region and the drain current IDSfor VGS= - 2Vin the pinch-off region are respectively.a)2 V and 1.8 mAb)3 V and 2.9 mAc)2 V and 2.9 mAd)3 V and 1.8 mACorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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