Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: May 16

Step1: Draw the the Thevenin’s equivalent circuit

I = $\frac{(V_CC+ V_EE)}{(R_1 + R_2 )} = \frac{(5+ 5)}{(12k +2k)} = \frac{(10 V)}{(14 KΩ)} $= 0.714 mA

$V_th = IR_2 - V_EE $= 0.714 mA x 2 kΩ - 5 V

= -3.5714 V

And $R_B$ = $R_1 ||R_2$ = 12k||2k = 1.714kΩ

Step2: Calculate $I_B$

By redrawing network as shown in fig and apply KVL,

Substituting $I_E = (1+β) I_B$ gives,

$V_EE-V_th - V_BE- (1+β) I_B R_E - I_B R_th = 0$

$I_B = \frac{(V_EE -V_th - V_BE)}{( R_th+ (1+β) R_E )}$

= $\frac{(5-3.571-0.7)}{(1.714 kΩ+101x0.5kΩ)} = \frac{(0.729 V)}{(52.214 kΩ)}$ = 13.96 µA

Step3: Calculate $I_CQ$ and $V_CEQ$

$I_(C Q)= βI_B $= 100 x 13.96µA = 1.396 mA

$I_E =(1+β) I_B$ = 101 x 13.96 µA = 1.409 mA

$V_CEQ= V_CC-V_EE - I_CQ R_C -I_E R_E = 10 - 1.396 x 10^(-3) x 5 x 10^3 - 1.409 x10^(-3) x 0.5 x 10^3$

= 10 – 7.6845

= 2.315 V

$V_CEQ$ = 2.315 V

Step4: Draw DC load line

For Load Line, consider $V_CEQ$ = 0

$V_CEQ= V_CC-V_EE - I_CQ R_C -I_E R_E$

0 = $V_CC-V_EE - I_CQ R_C -I_E R_E$

= 10 - $ I_(C ) x 5 x 10^3 -1.409 x10^(-3) x 0.5 x 10^3$

$I_(C )= \frac{(10- 1.409 x 10^(-3) x 0.5 x 10^3)}{(5 x 10^3 )} = \frac{(10-0.7045)}{(5 x 10^3 )} = \frac{9.2955}{(5 x 10^3 )}$

= 1.85 mA

Hence, $I_(C)$=1.85 mA

And $V_CE= V_(CC )$= 10 V

The load line and the calculated Q point are shown in fig.