Two n channel MOSFETs are fabricated and biased in saturation region in such a way that the first one has width as well as VGS-VTH double as those of the second one. All other parameters remain the same. What is the ratio of drain currents of the transistors?

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  1. 2 ∶ 1
  2. 4 ∶ 1
  3. 8 ∶ 1
  4. 16 ∶ 1

Answer (Detailed Solution Below)

Option 3 : 8 ∶ 1
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Detailed Solution

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Explanation:

 

  • The drain current (ID) of a MOSFET in the saturation region is given by the equation:

    ID = (1/2) × μn × Cox × (W/L) × (VGS - VTH,

    where:
    • μn: Electron mobility
    • Cox: Gate-oxide capacitance per unit area
    • W: Width of the MOSFET channel
    • L: Length of the MOSFET channel
    • VGS: Gate-to-source voltage
    • VTH: Threshold voltage
  • All other parameters (μn, Cox, and L) are constant in this problem.
  • Therefore, the drain current is directly proportional to both the width of the MOSFET (W) and the square of the overdrive voltage (VGS - VTH).

Solution:

Let the parameters of the second MOSFET (MOSFET 2) be:

  • Width: W
  • Overdrive voltage: (VGS - VTH)
  • Drain current: ID2

For the first MOSFET (MOSFET 1), the parameters are:

  • Width: 2W (double the width of MOSFET 2)
  • Overdrive voltage: 2(VGS - VTH) (double the overdrive voltage of MOSFET 2)
  • Drain current: ID1

The drain current for MOSFET 2 is:

ID2 = (1/2) × μn × Cox × (W/L) × (VGS - VTH

The drain current for MOSFET 1 is:

ID1 = (1/2) × μn × Cox × (2W/L) × [2(VGS - VTH)]²

Simplifying ID1:

ID1 = (1/2) × μn × Cox × (2W/L) × 4(VGS - VTH

ID1 = 4 × (1/2) × μn × Cox × (W/L) × (VGS - VTH

ID1 = 8 × ID2

Conclusion:

The ratio of the drain currents of the two MOSFETs is:

ID1 : ID2 = 8 : 1

The correct answer is Option 3.

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