Explain in your own words why the application of a positive voltage to the gate of an \(n\) -channel depletion-type MOSFET will result in a drain current exceeding \(I_{D S S}\).

Short Answer

Expert verified
The application of a positive voltage to the gate of an \(n\)-channel depletion-type MOSFET causes an increase in the number of charge carriers (electrons) in the gate. This increase enhances the conductivity of the gate, thus allowing more drain current to flow, exceeding \(I_{D S S}\).

Step by step solution

01

Understanding Depletion MOSFET Structure

The Depletion MOSFET has a physical channel already implanted during manufacturing. It's called a depletion-mode FET because we apply a voltage that depletes the number of carriers in the channel, reducing or turning off current (for negative voltage). However, applying positive voltage enhances the number of carriers (electrons in this case), therefore increasing current.
02

Relating Positive Voltage to Drain Current

In an \(n\)-channel depletion-type MOSFET, a positive voltage at the gate would attract more electrons (negative charges) towards the gate. This increases the population of carriers (electrons) inside the channel which enhances the conductivity. Increased conductivity allows more drain current to flow, thus the drain current exceeds \(I_{D S S}\).
03

Additional Information on \(I_{D S S}\)

\(I_{D S S}\) is the maximum drain current for zero gate-to-source voltage. As this voltage increases, it increases the drain current.

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