Question

An n-p-n transistor in a common - emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The positive terminal of a 8 V battery is connected to the collector through a load resistance R_L and to the base through a resistance R_B . The collector - emitter voltage V_(CE) = 4 V , the base - emitter voltage V_(BE) = 0.6 V and the current amplification factor beta = 100 . Calculate the values of R_L and R_B .

Answer 1

The values of Rb and Rl is :

• Given : Collector Current, Ic = 4 mA

Collector-emitter voltage Vce = 4 V , the base - emitter voltage Vbe = 0.6 V, the current amplification factor beta = 100 and collector voltage Vcc = 8V

• Applying Kirchhoff's loop law at collector side,

Vce = Vcc - Ic×RL

• RL = Vcc - Vce / Ic = 8-4 / 4×10^-3

= 10^3 ohm = 1 Kohm

•Also given that Beta = Ic / Ib

Ib = Ic / beta = 4×10^-3 / 100 = 4×10^-5 A

• Again applying Kirchoff's loop law at base side,

Vbe = Vcc - Ib×Rb

Rb = Vcc - Vbe / Ib = 8-0.6 / 4×10^-5

= 1.85×10^-5 ohm

• Hence, RL = 10^3 ohm and Rb = 1.85×10^-5 ohm