I’ve been reading the book

Electronic Principles by Malvino

In the Voltage-Divider bias section of JFET, if gate current is negligible then how did Gate Voltage attain a negative value, given Vdd is a positive value.

And also, to calculate drain current Gate-Source potential difference was neglected compared to Gate voltage.

This a diagram for JFET VDB

How were the above conclusions drawn?

  • 1
    \$\begingroup\$ Would Electrical Engineering be a better home for this question? \$\endgroup\$
    – Qmechanic
    Commented Aug 19, 2018 at 14:02
  • 1
    \$\begingroup\$ Please provide a schematic. \$\endgroup\$
    – Oldfart
    Commented Aug 19, 2018 at 17:10
  • \$\begingroup\$ The author assumed that the Vg voltage is a much larger than |Vgs| voltage. And this is why he writes Id = (Vg - |Vgs|)/Rs ≈ Vg/Rs \$\endgroup\$
    – G36
    Commented Aug 19, 2018 at 17:46

1 Answer 1


For example, you have this circuit


simulate this circuit – Schematic created using CircuitLab

And the JFET parameter are:

\$I_{DSS} = 2\textrm{mA}\$ , \$V_P = -1V\$

And we want the drain current to be equal to \$1\textrm{mA}\$

The drain current describe this equation:

$$I_D = I_{DSS} \left(1 -\frac{V_{GS}}{V_P}\right)^2 $$

And we need to know \$V_{GS}\$ for \$I_D = 1\textrm{mA}\$

$$V_{GS} = V_P \left(1 -\sqrt{\frac{I_D}{I_{DSS}}}\right) = -1V \left(1 -\sqrt{\frac{1\textrm{mA}}{2\textrm{mA}}}\right) = -0.292V$$

This result tells us that we need to set the Gate voltage lower than the Source voltage by \$0.292V\$

So if I choose \$V_S\$ we can solve for source resistor

\$R_S = \frac{1V}{1\textrm{mA}} = 1\textrm{k}\Omega\$

And the voltage divider output voltage needs to be equal to:

\$V_G = V_S + Vgs = 1V + (-0.292V) = 0.708V\$

And now you can choose the voltage divider resistors values. For example

\$R_1 = 510\textrm{k}\Omega\$ and \$R_2 = 39\textrm{k}\Omega\$

Will met our requirements.

As you can see by choosing the proper \$V_G\$ voltage and \$R_S\$ we can set the drain current.

And If you would like to check it.

We need to solve this quadratic equation

$$I_D = I_{DSS}\left(1 - \frac{V_G - I_D\cdot R_S}{V_P}\right)^2 = 2\left(1 - \frac{10*\frac{39}{39+510} - I_D\cdot 1}{-1}\right)^2$$

And the solution is


\$I_D = 1.00242 \textrm{mA}\$

  • \$\begingroup\$ Pardon for asking such questions. How did you decide source voltage would be 1V? \$\endgroup\$ Commented Aug 20, 2018 at 1:32
  • \$\begingroup\$ I just pick this value randomly as a designer. I could pick Vs =2V but then Vg must be equal to 2V - 0.292V = 1.7V \$\endgroup\$
    – G36
    Commented Aug 20, 2018 at 13:07

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