"Channel Length Modulation" - Decreases the effective channel length in a MOS. Still current flows.. Why??

Question

ie. there is a region under the gate where there is no channel. but still the charges flow from drain to source.. Why??

Answer 1

That region you speak of still has finite resistivity, and therefore can conduct. Since the region is highly resistance, the current is small for a fixed voltage applied. Resistance, is about rho*l/A where l is the length, A is the area, and rho is ther resistivity.

Also, there are capacitances in the FET, but they conduct only currents when the voltage is changing. The channel-length modulation refers to what is normally a steady-state (DC) characteristic.

Answer 2

The first answerer is correct; I would only add that in a junction FET, when pinch off is achieved, there will not be any significant current flow. It may be that without additional passivating layers (requiring additional process steps and adding greatly to the cost of the device), so leakage is almost inevitable. Please note that if the channel length alone is modulated, while the channel width remains unchanged, that may not completely cut off the current.

Probably there is an unfairly excluded region in the bulk or substrate material where the first doping was implanted, a region that looks like a sharp line on the macroscopic scale, but on the level of the device maybe fuzzy. If some doping impurities exist in this region, if there is not perfect contiguity of the differently doped regions or doped regions and the adjoining passivation layers, it may be possible for leakage currents to flow, analogous to ground currents in a larger circuit.

Answer 3

There are capacitances everywhere in a s MOS. A MOS has a gate, drain, source and bulk (usually ignored). So there are (4 choose 2) combinations of capacitances.

Charge flows from drain to source through that particular capacitance.