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Determine Q Point and draw d.c load line for the amplifier shown

enter image description here

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

Marks: 10M

Years: May 16

1 Answer
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Step1: Draw the the Thevenin’s equivalent circuit

I = (VCC+VEE)(R1+R2)=(5+5)(12k+2k)=(10V)(14K)= 0.714 mA

Vth=IR2VEE= 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.

enter image description here

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