power electronics - How could the Gate-Source potential difference be neglected?

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.

How were the above conclusions drawn?

Answer

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

http://www.wolframalpha.com/input/?i=x+%3D+2+(1+-+(10*39%2F(39%2B510)-x*1)%2F-1.0)%5E2

$I_D = 1.00242 \textrm{mA}$