Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Given:

$I_{DSS} = 7mA$

Vp = -2.5V

We know that,

$I_{DS}=I_{DSS}(1-\frac{V_{GS}}{V_{p}})^2$

$I_{DS}=7m(1+\frac{VGS}{2.5})^2$-----------(1)

Put different values of VGS and obtain IDS in equation (1),

Figure 1 shows the Graph of $V_{GS}V_s I_{DS}$

Apply KVL from R2 to ground through Gate & Source

$V_{R2} – V_{GS} – (I_{DS}×R{S})=0 …..(2)$

Put $I_{DS}=0$ in equation (2)

$V_R=V_{GS}=\frac {V_{DD}\times{R2}}{R1+R2}$

=$\frac{20\times1M\Omega}{1M\Omega+2M\Omega}$

=6.66mAwill be the point on Y-axis

From Figure 1, we obtain Q-point,

$I_{DSQ}=6.75mA$

$V_{GSQ}= - 0.05V$