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If IDSS= 10mA and Vp= -5V, when VDS=Vp, what is the value of ID and VGS?

Lets say I apply KVL on drain-source loop.

Then I get VDD-IDRD-VDS=0.

Substitute the value then I get ID=2.1x10-3. After that, I apply

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Solve the formula and get VGS=2.71V

Is the method I used correctly in solving this question?

  • 2
    \$\begingroup\$ But in your schematics, Vgs voltage is equal to Vgg thus Vgs = -1V \$\endgroup\$ – G36 Jun 21 '20 at 19:05
  • \$\begingroup\$ You have to make a design choice of the required Vds to determine the Id. 5V would be a reasonable value giving 0.5mA. \$\endgroup\$ – Kevin White Jun 21 '20 at 19:24
  • \$\begingroup\$ You're in the "just applying formulas" mode without looking at the circuit to see what actually is happening. As G36 comments, there's a Vgs applied. What does that Vgs mean for the JFET? \$\endgroup\$ – Bimpelrekkie Jun 21 '20 at 19:27
  • \$\begingroup\$ But isn't that there is a potential drop across the R1 resistor? Wouldn't this make Vgg not equal to Vgs? Or the resistor does not have effect on the circuit? \$\endgroup\$ – Nobu Nobu Jun 22 '20 at 14:02
  • \$\begingroup\$ the normal method is to use KVL in GS loop and then use the current formula. Now you have two-equation and two unknown and you may find ID and VGS. \$\endgroup\$ – arsalan ghasemian Jun 23 '20 at 4:10

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