written 8.2 years ago by | • modified 8.2 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 16
written 8.2 years ago by | • modified 8.2 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 16
written 8.2 years ago by |
Step1: Draw the the Thevenin’s equivalent circuit
I = (VCC+VEE)(R1+R2)=(5+5)(12k+2k)=(10V)(14KΩ)= 0.714 mA
Vth=IR2−VEE= 0.714 mA x 2 kΩ - 5 V
= -3.5714 V
And RB = R1||R2 = 12k||2k = 1.714kΩ
Step2: Calculate IB
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.