Terminology:

- I
_{DS}= current from drain to source OR drain-source current - V
_{DS}= drain to source voltage - L = length of the channel

Now for the ideal case, in the saturation region, I_{DS} becomes independent of V_{DS} i.e. in the saturation region channel is pinched off at the drain end and a further increase in V_{DS} has no effect on the channel’s shape.

But in practice increase in V_{DS} does affect the channel. In the saturation region, when V_{DS} increases, the channel pinch-off point is moved slightly away from the drain, towards the source as the drain electron field “pushes” it back. The reverse bias depletion region widens and the effective channel length decreases by an amount of ∆L for an increase in V_{DS}.

Thus the channel no longer “touches” the drain and acquires an asymmetrical shape that is thinner at the drain end. This phenomenon is known as channel length modulation.

_{DS}) in the saturation region.

In large devices, this effect is negligible but for shorter devices ∆L/L becomes important. Also in the saturation region due to channel length modulation, I_{DS} increases with increase in V_{DS} and also increases with the decrease in channel length L.

The voltage-current curve is no longer flat in this region.

The drain current with channel length modulation is given by:

DERIVATION:

To account for the dependence of I_{D} on V_{DS} in the saturation region, replace L by L – ∆L. We know that in the saturation region, drain to source current (I_{DS} = I_{D}) is given by:

Assuming

Since ∆L increases with increase in V_{DS}

OR

where, = process technology parameter with unit *μm/V.*

therefore,

where,

= process technology parameter with unit V^{-1}

rahulsir i still don”t understand when it said that channel length modulation occurs when vds is higher than vgs, so what is the function of high voltage vds capability shown in the datasheet i.e. 600v it is about 10 times bigger than the vgs if the vgs is 15v?