Given the measured value of VD in Fig. 7.77, determine:a. ID.b. VDS.c. VGG.FIG. 7.77

$$ V_{D S}=V_{D}-V_{S} \ldots \ldots \text { (2) } $$

Here,

Source voltage \(\left(V_{s}\right)\) is \(0 \mathrm{~V}\).

Substitute \(0 \mathrm{~V}\) for \(\left(V_{s}\right)\) and \(6 \mathrm{~V}\) for \(\left(V_{D}\right)\) in equation (2).

$$ \begin{aligned} V_{D S} &=6-0 \\ &=6 \mathrm{~V} \end{aligned} $$

(c)

Write the equation of drain current.

$$ \begin{aligned} &I_{D}=I_{D S S}\left(1-\frac{V_{G S}}{V_{P}}\right)^{2} \\ &V_{G S}=V_{P}\left(1-\sqrt{\frac{I_{D}}{I_{D S s}}}\right) \ldots \cdots(3) \end{aligned} $$

Substitute \(8 \mathrm{~mA}\) for \(\left(I_{D S S}\right), 2.727 \mathrm{~mA}\) for \(I_{D}\) and \(-4 \mathrm{~V}\) for \(\left(V_{P}\right)\) in equation (3).

$$ \begin{aligned} V_{G S} &=(-4)\left(1-\sqrt{\frac{2.727 \times 10^{-3}}{8 \times 10^{-3}}}\right) \\ &=-1.66 \mathrm{~V} \end{aligned} $$

In this circuit configuration, the value of the voltage \(V_{G G}\) is equal to the voltage \(V_{G S}\).

$$ \begin{aligned} V_{G G} &=-V_{G S} \\ &=-(-1.66 \mathrm{~V}) \\ &=1.66 \mathrm{~V} \end{aligned} $$