Problem

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

Given the measured value of VD in Fig. 7.77, determine:

a. ID.

b. VDS.

c. VGG.

FIG. 7.77

#### Step-by-Step Solution

Solution 1

(a)

Calculate drain current $$I_{D}$$.

$$I_{D}=\frac{V_{D D}-V_{D}}{R_{D}} \ldots \ldots(1)$$

Substitute $$12 \mathrm{~V}$$ for $$\left(V_{D D}\right), 6 \mathrm{~V}$$ for $$\left(V_{D}\right)$$ and $$2.2 \mathrm{k} \Omega$$ for $$\left(R_{D}\right)$$ in equation (1).

\begin{aligned} I_{D} &=\frac{12-6}{2.2 \times 10^{3}} \\ &=2.727 \times 10^{-3} \\ &=2.727 \mathrm{~mA} \end{aligned}

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

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}