Calculate β for two transistors for which α=0.99 and 0.98. For collector currents of 10 mA, find the base current of each transistor.
The Correct Answer and Explanation is:
To calculate β\betaβ (current gain) for two transistors given their α\alphaα values, we use the following relationship between α\alphaα and β\betaβ:β=α1−α\beta = \frac{\alpha}{1 – \alpha}β=1−αα
For the first transistor with α=0.99\alpha = 0.99α=0.99:
β=0.991−0.99=0.990.01=99\beta = \frac{0.99}{1 – 0.99} = \frac{0.99}{0.01} = 99β=1−0.990.99=0.010.99=99
For the second transistor with α=0.98\alpha = 0.98α=0.98:
β=0.981−0.98=0.980.02=49\beta = \frac{0.98}{1 – 0.98} = \frac{0.98}{0.02} = 49β=1−0.980.98=0.020.98=49
So, the current gains for the two transistors are:
- For α=0.99\alpha = 0.99α=0.99, β=99\beta = 99β=99
- For α=0.98\alpha = 0.98α=0.98, β=49\beta = 49β=49
Now, let’s calculate the base current for each transistor, given that the collector current IC=10 mAI_C = 10 \, \text{mA}IC=10mA.
Using the relationship between the collector current ICI_CIC, the base current IBI_BIB, and β\betaβ:IC=β⋅IBI_C = \beta \cdot I_BIC=β⋅IB
Solving for IBI_BIB:IB=ICβI_B = \frac{I_C}{\beta}IB=βIC
For the first transistor with β=99\beta = 99β=99:
IB=10 mA99≈0.101 mAI_B = \frac{10 \, \text{mA}}{99} \approx 0.101 \, \text{mA}IB=9910mA≈0.101mA
For the second transistor with β=49\beta = 49β=49:
IB=10 mA49≈0.204 mAI_B = \frac{10 \, \text{mA}}{49} \approx 0.204 \, \text{mA}IB=4910mA≈0.204mA
Summary:
- For the transistor with α=0.99\alpha = 0.99α=0.99, β=99\beta = 99β=99, and the base current IB≈0.101 mAI_B \approx 0.101 \, \text{mA}IB≈0.101mA.
- For the transistor with α=0.98\alpha = 0.98α=0.98, β=49\beta = 49β=49, and the base current IB≈0.204 mAI_B \approx 0.204 \, \text{mA}IB≈0.204mA.
This shows how variations in α\alphaα affect both the current gain β\betaβ and the base current.
