Relation between V threshold and L

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aryajur

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Why does the threshold voltage of a MOS increase when we increase the L of the transistor?
 

As we increase the length of the transistor, the area under the gate increases and thus larger volyage is required to attract more charges to cause inversion in the channel region.
 

To maintain the same charge density we would not need any additional voltage since the Capacitance per unit area is constant, so I don't see why increasing the area would require more voltage to keep the charge constant.
Any other suggestions?
 

I think the question needs to be rephrased:

"Why does the threshold drop when the device gets shorter?"

This is because as the channel length decreases, the width of the drain and source depletions becomes comparable to the channel length. Hence, the fixed depleted dopants under the gate will help reduce the threshold voltage (smaller amount of charge would be needed due to the already present fixed depleted dopants).
 

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