Bipolar Junction Transistors (BJT): Important Example Solved Problems

Electron Devices

Chapter 4: Bipolar Junction Transistors

Important Example Solved Problems

Common Base Transistor Configuration

Ex. 1: Given I_E = 2.5 mA, α = 0.98 and I_CBO = 10 μA calculate I_B and Ic.

Solution:

I_C = αI_E + I_CBO

= 0.98 × 2.5 × 10^‒3 + 10×10^‒6 = 2.46 mA

I_B  = I_E‒I_C = 2.5×10³ − 2.46×10⁻³ = 40 μA

Common Emitter Transistor Configuration

Ex. 2: Calculate the collector and emitter current levels for a BJT with α_dc = 0.99 and I_B = 20 μA

Solution: :

B_dc = α_dc / (1 ‒ α_dc)

= 0.99 / (1‒0.99)

= 99

I_C = β_dc I_B = 99×20 μA = 1.98 mA

I_E = I_C+I_B = 1.98 mA + 20 μA = 2 mА

Ex. 3: Calculate the values of I_C and I_E for a BJT with α_dc = 0.97 and I_B = 50 μA. Determine β_dc for the device.

Solution: :

B_dc = α_dc / (1 ‒ α_dc)

= 0.97 / (1‒0.97)

= 32.33

I_C = β_dc I_B = 32.33 × 50 μA = 1.6165 mA

I_E = I_B + I_C = 50 μA + 1.6165 mA

= 1.6665 mA

I_CEO

From equation (4.4.4) we have,

I_C = (α_dc / 1‒α_dc )I_B + (I_CBO / 1‒α_dc )

I_C = β_dcI_B + (1+β_dc)I_CBO

            ………….(4.4.5)

Since

1+ β_dc = 1 + (α_dc / 1‒α_dc)

 = [ 1 ‒α_dc + α_dc ] / [ 1‒α_dc ]

= 1 / ( 1‒α_dc )

The terms (1 + β_dc) I_CBO in equation (4.4.5) is denoted as I_CEO and is the reverse saturation current for the CE configuration.

     (1+β_dc)I_CBO = I_CEO

 I_C = β_dcI_B + I_CEO

Ex. 4: The reverse leakage current of the transistor when connected in CB configuration is 0.2 μA and it is 18 μA when same transistor is connected in CE configuration. Calculate α_dc and β_dc of the transistor. (Assume I_B = 30 mA)

Solution:

Given: I_CBO = 0.2 μA, I_CEO = 18 μA

 I_CEO = (1 + β_dc)I_CBO

 (1 + β_dc) = I_CEO / I_CEO = 18 μA / 0.2 μA = 90

β_dc = 89

α_dc = β_dc / (1 + β_dc)

=  89 / (1+89) = 0.989

Ex. 5: Calculate the α_dc and β_ac for the given transistor for which I_C=5 mA, I_B =50 μA and I_CO = 1 μΑ.

Solution:

 I_C=5mA, I_B = 50 μA, I_CO =I_CBO = 1 μA

I_C = β_dcI_B +(1+β_dc)I_CBO

 5×10^‒3 = β_dc × 50×10⁻⁶ + (1+β_dc)×1×10^‒6

5×10^‒3 ‒ 1×10^‒6 = 51×10^‒6 β_dc

 β_dc = 4.999×10^‒3 / 51×10^‒6 = 98

 α_dc = β_dc / (1+ β_dc) = 98 / (1+98) = 0.9899

Ex. 6: A transistor has β = 150, find the collector and base current, if I_E = 10 mA.

Solution:

 I_B = I_E / (1+β) =  10mA / (1+150) = 66.225 μA

 I_C = βI_B = 150×66225 μA = 9.934 mA

Ex. 7: A transistor with I_B = 100 μA and I_C = 2 mA find

1) β of the transistor

2) α of the transistor

3) Emitter current I_E

4) If I_B changes by 25μA and I_C changes by 0.6 mA. Find the new value of β.

Solution: :

1) β = I_C/I_B

= 2×10^‒3 / 100×10^‒6

= 20

2) α = β / (1+β)

 = 20 / (1+20) = 0.9524

3) I_E = I_B + I_C = 100×10^‒6 + 2×10^‒3 = 2.1 mA

4) New value of I_B = 100 + 25 = 125 mA

New value of I_C = 2 + 0.6 = 2.6 mA

New value of β = 2.6×10^‒3 / 125×10^‒6

= 20.8