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

Since [latex]V_{GS(off)}[/latex] = - 4 V, V P = 4 V. The minimum value of [latex]V_{DS}[/latex] or the JFET to be in its constant-current region is [latex]V_{DS} = V_{P} = 4 V[/latex] In the constant-current region with [latex]V_{GS}[/latex] = 0 V, [latex]I_{D} = I_{DSS} = 12 mA[/latex] The drop across the drain resistor is [latex]V_{R_{D} } = (12 mA) (560 \Omega ) = 6.7 V[/latex] Applying Kirchhoff’s law around the drain circuit gives [latex]V_{DD} = V_{DS} + V_{R_{D} } = 4 V + 6.7 V[/latex] = 10.7 V This is the minimum value of [latex]V_{DD}[/latex] to make [latex]V_{DS}= V_{P}[/latex] and to put the device in the constant-current region.

For the n -channel JFET in Figure 4–8 , VGS(off) = -4 V and IDSS = 12 mA. Determine the minimum value of VDD required to put the device in the constant-current region of operation.

Q. 4.1

Chapter 4 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.

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