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## Q. 4.1

For the n-channel JFET in Figure 4–8, $V_{GS(off)}$ = 4 V and $I_{DSS}$ = 12 mA. Determine the minimum value of $V_{DD}$ required to put the device in the constant- current region of operation. ## Verified Solution

Since $V_{GS(off)}$ = 4 V, VP = 4 V. The minimum value of $V_{DS}$ or the JFET to be in its constant-current region is

$V_{DS} = V_{P} = 4 V$

In the constant-current region with $V_{GS}$ = 0 V,

$I_{D} = I_{DSS} = 12 mA$

The drop across the drain resistor is

$V_{R_{D} } = (12 mA) (560 \Omega ) = 6.7 V$

Applying Kirchhoff’s law around the drain circuit gives

$V_{DD} = V_{DS} + V_{R_{D} } = 4 V + 6.7 V$ = 10.7 V

This is the minimum value of $V_{DD}$ to make $V_{DS}= V_{P}$ and to put the device in the constant-current region.