![]() ![]() |
When no gate voltage is applied and
is small, a small current flows in the channel. The magnitude of the current is
given by
. Therefore, the current varies linearly with the drain voltage. Of
course for any given drain voltage, the voltage along the channel increases from zero at the
source to
at the drain. Thus, the Schottky barrier becomes increasingly reverse
biased as we proceed from the source to the drain. As
is increased,
increases, and the average cross-sectional area for the current flow is reduced. The channel
resistance
also increases. As a result, the current increases at a slower rate.
As the drain voltage further increases, eventually the depletion region touches the
semi-insulating substrate. That happens when at the drain. We can obtain the
corresponding value of the drain voltage, called the saturation voltage
,
from
When a gate voltage is
applied to reverse bias the gate contact, the depletion-layer width increases. For small
, the channel again acts as a resistor but its resistance is higher because the
cross-section area available for current flow is decreased. When
is increased
to a certain value, the depletion region again touches the semi-insulating substrate. The value
of this
is given by