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Question

# An ideal n−channel MOSFET has the following parameters,threshold voltage VT=0.65V and oxide thickness is 0.4nm.It is found that the MOSFET operating in saturationregion with VDS=6V. Then the power dissipation in theMOSFET with VGS=4 V isμnCaxWL=1.5×10−3A/V2,

A
50.50 mW
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B
37.65 mW
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C
25.25 mW
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D
75.50 mW
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Solution

## The correct option is A 50.50 mWGiven,μnCoxWL=1.5×10−3A/V2VT=0.65VVGs=4VVDS=6Vpower diipation in the MOSFET is,P=VDS×IDSWhere, IDS is drain to source saturation current.Since the MOSFET is operating in saturation region,IDS=12μn CoxWL(VGS−VT)2IDS=8.42×10−3Apower dissipation, P=6×8.42×10−3∴P=50.50 mW

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