written 8.3 years ago by | modified 3.2 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals
Marks: 5 Marks
Year: Dec 2013
written 8.3 years ago by | modified 3.2 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals
Marks: 5 Marks
Year: Dec 2013
written 8.3 years ago by |
Pinch off Voltage
At VGS=0, in response to a small applied voltage VDS, the n-type bar acts as a simple semiconductor resistor and the current ID increases linearly with VDS. As VDS increases, the voltage drop along the channel also increases. This increase in voltage drop increases the reverse bias on gate-source junction and causes the depletion regions to penetrate into the channel, reducing channel width. The effect of reduction in channel width provides more opposition to increase in drain current ID. Thus, rate of increases in ID with respect to VDS is now reduced. This is shown by the curved shape in the characteristics.
At some value of VDS, drain current ID cannot be increased further, due to reduction in channel width. Any further increase in VDS does not increase the drain current ID. ID approaches the constant saturation value. The voltage VDS at which the current ID reaches to its constant saturation level is called ‘Pinch-off Voltage’ VP.
The pinch-off voltage VP is given by,
|VP|=qND2εa2
Cut-off Voltage
As we know, for an n-channel JFET, the more negative VGS causes drain current to reduce and pinch-off voltage to reach at a lower drain current. When VGS is made sufficiently negative, ID is reduced to 0, as shown in the Fig. this is caused by the widening of the depletion region to a point where it completely closes the channel. The value of VGS at the cut off point is designated as VGS(OFF).