Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 8 M

Year: Nov 2013

• Calculation of $R_{th}$ and $V_G$

Fig1 (a) determining $R_{th}$ (b) determining $V_G$

• Calculation for $R_{th}$

  $R_{th}=R_1||R_2$

  $=910k||110k$

  $=98.13kΩ$

• Calculation for $V_G$

  $V_G =\frac{V _{DD} ×R_2}{R _1+R _2}$

  $=\frac{20×110k}{ 910k+110k}$

  $=2.16V$

• Simplified circuit

Fig2 thevenin’s equivalent circuit

• Applying KVL to gate-source loop

  $I_GR_{th}+V_G-V_{GS}-I_DR_S=0$

  Put $I_G=0$

  $V_{GS}=V_G- I_DR_S ……………(1)$

  Put $I_D$=0 in eq(1), then $V_{GS}=V_G=2.16V$

  Put $V_{GS}=0$ in eq(1), then $I_D=\frac{V_{GS}}{R_S} =\frac{2.16}{0.51k}=4.235mA$

  $I_D=I_{DSS}(1-\frac{V _{GS}}{V_p})^2$

  $I_D=10m (1+\frac{V _{GS}}{3.5})^2$

  • Plot above curve and line joining $I_D$ and $V_{GS}$ on graph
  • Point of intersection of two curves is Q-point

  $I_{DQ}=5.6mA, V_{GSQ}=-0.7V$

• Applying KVL to drain-source loop

  $V_{DD} -V_{DS}-I_D (R_S+R_D) =0 ……………(2)$

  Put $I_D=0$ in eq(2), then $V_{DS}=V_{DD}=20V$

  Put $V_{DS} =0$ in eq(2), then $I_D=\frac{V_{DS}}{R_S+R_D}=\frac{20}{0.51k+2.2k}=7.38mA$

  • Plot straight line passing through this two points
  • Mark IDQ point on the graph and corresponding point on x-axis is $V_{DSQ}$

  $V_{DSQ}=4.4V$