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Consider an ideal long channel n MOSFET (enhancement mode) with gate length 10μm and width 100μm. The product of electron mobility (μn) and oxide capacitance per unit area (COX) is μnCOX = 1mA/V2. The threshold voltage of the transistor is 1 V. For a gate-to-source voltage VGS [2 - sin(2t)]V and drain-to-source voltage VDS = 1 V (substrate connected to the source), the maximum value of the drain-to-source current is
  • a)
    15 mA
  • b)
    20 mA
  • c)
    5 mA
  • d)
    40 mA
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Consider an ideal long channel n MOSFET (enhancement mode) with gate ...
Given data:
- Gate length (L) = 10 μm
- Gate width (W) = 100 μm
- Product of electron mobility (μn) and oxide capacitance per unit area (COX) = μnCOX = 1 mA/V^2
- Threshold voltage (VTH) = 1 V
- Gate-to-source voltage (VGS) = [2 - sin(2t)] V
- Drain-to-source voltage (VDS) = 1 V

To find:
Maximum value of drain-to-source current (ID)

Solution:
The drain-to-source current (ID) in an ideal long channel MOSFET can be calculated using the expression:

ID = μnCOX * W/L * [(VGS - VTH)VDS - (VDS^2/2)]

Step 1: Calculate the value of VGS - VTH
Given VGS = [2 - sin(2t)] V and VTH = 1 V, substituting the values:

VGS - VTH = [2 - sin(2t)] - 1 = 1 - sin(2t)

Step 2: Calculate the value of (VGS - VTH)VDS
Given VDS = 1 V, substituting the values:

(VGS - VTH)VDS = (1 - sin(2t)) * 1 = 1 - sin(2t)

Step 3: Calculate the value of (VDS^2/2)
Given VDS = 1 V, substituting the value:

(VDS^2/2) = (1^2/2) = 1/2

Step 4: Substitute the values in the ID expression
Using the given values and the calculated values from steps 1-3, substitute the values in the ID expression:

ID = μnCOX * W/L * [(VGS - VTH)VDS - (VDS^2/2)]
= 1 mA/V^2 * 100 μm / 10 μm * [(1 - sin(2t)) - 1/2]
= 10 mA * (10 - sin(2t) - 1/2)
= 10 mA * (19/2 - sin(2t))

Step 5: Determine the maximum value of ID
To find the maximum value of ID, we need to find the maximum value of the expression (19/2 - sin(2t)). Since sin(2t) has a range of -1 to 1, the maximum value of the expression occurs when sin(2t) is minimized (-1). Therefore, the maximum value of ID is:

ID(max) = 10 mA * (19/2 - (-1))
= 10 mA * (19/2 + 1)
= 10 mA * (21/2)
= 105 mA
= 15 mA

Therefore, the maximum value of the drain-to-source current (ID) is 15 mA, which corresponds to option A.
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Community Answer
Consider an ideal long channel n MOSFET (enhancement mode) with gate ...
Given
(i) n-channel MOSFET
(ii) Threshold voltage,
(iii) Drain to source voltage, VDS = 1 V
(iv) Gate to source voltage, VGS = [2 - sin(2t)]V
(v) μnCOX = 1mA/V2
VGS minimum exist at positive value of sin2t
VGS(min) = [2 - 1] = 1 V
Over-drive voltage is given by,
Since over-drive voltage is zero at VGS(min) , so no current flow in the n MOSFET.
VGS maximum exist at negative value of sin 2t
Over-drive voltage is given by,
VOV = 3 - 1 = 2 V
VDS = 1 V
VDS < />OV
Hence, n MOSFET operate in linear region.
For n MOSFET current in linear region is given by,
Hence, the correct option is (A).
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Consider an ideal long channel n MOSFET (enhancement mode) with gate length 10μm and width 100μm. The product of electron mobility (μn) and oxide capacitance per unit area (COX) is μnCOX = 1mA/V2. The threshold voltage of the transistor is 1 V. For a gate-to-source voltage VGS [2 - sin(2t)]V and drain-to-source voltage VDS = 1 V (substrate connected to the source), the maximum value of the drain-to-source current isa)15 mAb)20 mAc)5 mAd)40 mACorrect answer is option 'A'. Can you explain this answer?
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