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.