Unit-1

Small signal amplifiers

Q1) In a CB IE= 2mA, IC=1.5mA. Calculate IB?

A1)

IE =IB+IC

2= IB+1.5

IB=0.5mA

Q2) In a CB current amplification factor is 0.9. If emitter current is 1.2mA. Determine the value of base current?

A2)

α = 0.9

IE =1.2mA

α = IC/ IE

IC = α IE =0.9 x 1.2 = 1.08mA

IE =IB+IC

1.2= IB+1.08

IB= 0.12mA

Q3) In a CB connection IC=1.0mA and IB= 0.02mA. Find the value of current amplification factor?

A3) IE =IB+IC =1+0.02 = 1.02mA

α = IC/ IE

α = 1.0/1.02 = 0.98

Q4) In a CB connection the emitter current is 0.98mA. If the emitter circuit is open the collector current becomes 40A. Find total collector current. α =0.92

A4) ICBO=40A

IC = α IE+ICBO

    = (0.92 x 0.98x10-3) + 40x10-6

IC =0.94mA

Q5) In a common base connection, α = 0.95. The voltage drop across 3 kΩ resistance which is connected in the collector is 2.5 V. Find the base current.

A5) IC = 2.5/3000 = 0.83mA

α = IC/ IE

IE = IC/α =0.83/0.95=0.87mA

IE =IB+IC

0.87 =IB+0.83

IB=0.04mA

Q6) Find the value of β if (i) α = 0.9 (ii) α = 0.98 (iii) α = 0.99.

A6)

= α/1- α = 0.9/1-0.9 = 9

= α/1- α = 0.98/1-0.98 = 49

= α/1- α = 0.99/1-0.99 = 99

Q7) The collector leakage current in a transistor is 200 μA in CE arrangement. If now the transistor is connected in CB arrangement, what will be the leakage current? Given that β = 120.

A7) ICEO=200 μA

= 120

α = /1+= 120/121=0.99

ICEO=ICBO/1- α

ICBO= 1.6 μA

Q8) For a certain transistor, IB = 18 μA; IC = 2 mA and β = 60. Calculate ICBO.

A8)

IC = IB+ICEO

ICEO= IC - IB= 2x10-3-(60x18x10-6) = 0.92mA

α = /1+= 60/61=0.98

ICBO= (1- α) ICEO = (1-0.98)x 0.92=15.08 μA