Consider a MOSFET tansister experiancing pinch-off near thedrain. by the equatio

Question

Consider a MOSFET tansister experiancing pinch-off near thedrain. by the equation I=Qv indicates that the charge density andcarrier velocity must change in opposite directions if the currentremains constant. How can this relationship be interpreted at thepinch off point, where the charge density approaches zero. thanks for the help and will rate lifesaver Consider a MOSFET tansister experiancing pinch-off near thedrain. by the equation I=Qv indicates that the charge density andcarrier velocity must change in opposite directions if the currentremains constant. How can this relationship be interpreted at thepinch off point, where the charge density approaches zero. thanks for the help and will rate lifesaver

Explanation / Answer

I am taking the example of a n-channel mosfet. Pinch-off occurs when the Vds reaches its saturationpoint. Since the drain end channel density hasbecome small, the current is much less dependent on Vds , butis still dependent on Vgs, since increased Vgs still increasesthe inversion layer density. But due to the high voltage difference between drain andsource, the horizontal electric field is very high. So theelectrons have high velocity and they reach the drain because ofthe momentum. So, in the equation I = Qv, Q is low near the drain,but v is very high because of the high electric field through whichthe electron is accelerated.Hence, increasing the Vds increases thefield which in turn increases the velocity but reduces the chargedensity near drain. So the current remain constant after pinch-offwith respect to Vds. But the current increses with increase inVgs.
I hope this helps you.

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