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 $V_B$ and $V_E$

$I_E= (β+1) I_B$

$I_E=91×50.52µ$

$I_E=4.597mA$

$V_E=V_{EE}-I_ER_E$

$V_E=20-(4.597×2k)$

$V_E=9.194V$

$V_{BE}=V_B-V_E$

$0.7V=V_B-9.194V$

$V_B=9.894V$

3. Calculation for $V_C$ and $V_{CE}$

• From diagram we conclude that, $V_C=0V$

  $V_{CE}=V_C-V_E$

  $V_{CE}=-9.194V$

4. Calculation of $S_{ICO}$

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

Now differentiating w.r.t. $I_C$

$\frac{∆I_B}{∆I_C } =0$; since all term are constant no term depends upon $I_C$

$S_{ICO}=\frac{1+β}{1-β(\frac{∆I_B}{∆I_C })}$

$S_{ICO}=1+β=91$

$$\boxed{ I_B=50.52µA , I_C= 4.547mA , V_E=9.194V \\ V_C=0V , V_{CE}=-9.194V , V_B=9.894V \\ S_{ICO} =91 }\qquad$$