Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals

Marks: 5 Marks

Year: Dec 2013

Pinch off Voltage

At $V_{GS}=0$, in response to a small applied voltage $V_{DS}$, the n-type bar acts as a simple semiconductor resistor and the current $I_D$ increases linearly with $V_{DS}$. As $V_{DS}$ 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 $I_D$. Thus, rate of increases in $I_D$ with respect to $V_{DS}$ is now reduced. This is shown by the curved shape in the characteristics.

At some value of $V_{DS}$, drain current $I_D$ cannot be increased further, due to reduction in channel width. Any further increase in $V_{DS}$ does not increase the drain current $I_D$. $I_D$ approaches the constant saturation value. The voltage $V_{DS}$ at which the current $I_D$ reaches to its constant saturation level is called ‘Pinch-off Voltage’ $V_P$.

The pinch-off voltage $V_P$ is given by,

$$|V_P|=\dfrac{qN_D}{2 \varepsilon}a^2$$

Cut-off Voltage

As we know, for an n-channel JFET, the more negative $V_{GS}$ causes drain current to reduce and pinch-off voltage to reach at a lower drain current. When $V_{GS}$ is made sufficiently negative, $I_D$ 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 $V_{GS}$ at the cut off point is designated as $V_{GS(OFF)}$.