Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 4 M

Year: May 2014

Fig1 DC analysis of circuit

1. Calculation for RB

$I_B= \frac{Ice}{βdc}$

$=\frac{2mA}{100}$

$=20µA$

$V_{BE}=V_B-V_E$

From the diagram, $V_E=0V$

Hence $V_B=0.7V$

$R_B=\frac{Vcc-VB}{I_B}$

$=\frac{10-0.7}{20µA}$

$R_B= 0.465MΩ$

$$\boxed{R_B= 0.465MΩ}$$

2. Calculation for $R_c$

• Applying KVL to collector-emitter loop

  $V_{cc}-I_{CE}R_c-V_{CE} =0$

  $10-2m*Rc-5=0$

  $2m*Rc =5$

  Hence, $R_c=\frac{5V}{2mA}=2.5kΩ$

  $$\boxed{R_c=2.5kΩ}$$