Module 3

Bipolar Junction Transistors and Electronic measuring instruments

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

Sol: 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?

Sol: 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?

Sol: 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

Sol: 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.

Sol: 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.

Sol: = 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.

Sol: 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.

Sol: 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

Q9) Compare CB CE and CC configuration?

Sol:

Q10) Draw and explain input and output characteristics of CB configuration?

Sol: Input Characteristic Curve

                                    Fig.6:  Input Characteristic Curve (Ref. 2)

• It is the relation between the input current IE to the input voltage VBE for various levels of output voltage VCB.
• It is also known as driving point characteristics.

Output Characteristic Curve

                                       Fig.7 : Output Characteristic Curve (Ref. 2)

• It is the relation between the output current IC to the output voltage VCB for various levels of input current IE.
• It is also known as collector set of characteristics.
• It has three basic regions:

1. Active Region

      Here, base-emitter junction is forward biased and collector-base junction is reverse biased.

      As input current IE increases above zero, output current IC increases to a magnitude equal to IE as determined by the basic transistor current relationship.

      So, the first approximation determined by the curve is

                                          IC ≈ IE

2.     Cut-off Region

      It is defined as the region where the collector current IC is equal to 0A.

      Here, the base-emitter junction and the collector-base junction both are in reverse bias.

3.     Saturation Region

      It is the region that lies towards the left of VCB = 0V.

      Here, the base-emitter junction and the collector-base junction both are in forward bias.