shown in fig. for $β=100.$ -

Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 10 M

Year: May 2014

1. Determining $V_{th}$ and $R_{th}$

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

• From the Fig1,

  $R_{th} =R_1||R_2$

  $R_{th}=35k||20k$

  $R_{th}=12.72kΩ$

  $I=\frac{2-(-5)}{R_1+R_2 }=\frac{7}{55k}=0.127mA$

  $V_{th}-(-5V) =I R_2$

  $V_{th} = (0.127m×20k)-5V$

  $V_{th}= -2.46V$

• The reduced circuit is as shown in fig2.

Fig2 Thevenin’s equivalent circuit

2. Calculation for ICQ

• Applying KVL to base-emitter loop,

  $-V_{th}-I_BR_{th}-V_{BE}-I_ER_E-V_{EE}=0$

  $V_{EE}- V_{BE}-V_{th} = I_BR_{th}- I_ER_E$

  Put $I_E= (β +1)I_B$

  $V_{EE}- V_{BE}-V_{th} = I_BR_{th}-(β +1)I_BR_E$

  $I_B=\frac{V_{EE}-V_{BE}-V_{th}}{R_{th}+(β +1)R_E}$

  $I_B=\frac{10-0.7-2.46}{12.72k+(100 +1)0.5}$

  $I_B=0.108mA$

  $I_{CQ}= βI_B$

  $I_{CQ}= 100×0.108mA$

  $I_{CQ}=10.8mA$

  $I_E=10.908mA$

3. Calculation for $V_{CEQ}$

• Applying KVL to collector-emitter loop

  $10-I_CR_C-V_{CE}-I_ER_E -V_{EE}=0$

  $10+10-(10.8m×0.8k)-(10.908m×0.5k)=V_{CEQ}$

  $V_{CEQ}=14.094V$

$$\boxed {I_{CQ}=10.8mA , V_{CEQ}=14.094V}$$