shown in Fig. $β=90$ -

Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 10 M

Year: Nov 2013

Fig1 DC equivalent circuit

1. Calculation $I_B$ and $I_C$

• Applying KVL to emitter-base loop

  $-200I_B-V_{BE} -2kI_E + V_{EE}=0 ………………………(1)$

  $I_E=I_C+I_B$

  $I_E= βI_B+I_B$

  $I_E= (β+1) I_B$

  Put value of $I_E$ in equation (1)

  $-R_BI_B-V_{BE}-R_E(β+1)I_B + V_{EE}=0$

  $I_B=\frac{V _{EE}-{V_{BE}}}{RE(β+1)+RB}$

  $I_B=\frac{V _{EE}-{V_{BE}}}{2k(β+1)+200k}$

  $I_B=\frac{20-0.7}{2k×91+200k}$

  $I_B=\frac{19.3}{382k}$

  $I_B=50.52µA$

  $I_C=βI_B$

  $I_C=90×50.52µ$

  $I_C= 4.547mA$

2. Calculation for $V_B$ …