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  • A N-P-N transistor in CE configuration in which collector supply is 8V and the voltage drop across the load resistance of <img alt="\mathrm{800\Omega }" src="https://entrancecorner.oncodecogs.

A N-P-N transistor in CE configuration in which collector supply is 8V and the voltage drop across the load resistance of \mathrm{800\Omega } connected in the collector circuit is 0.8V. If current amplification factor (\mathrm{\alpha }-is 25/26, determine collector-emitter voltage and base current. If the input resistance of the transistor is 200 The power gain is:

Option: 1

1000


Option: 2

1500


Option: 3

2000


Option: 4

2500


Answers (1)

best_answer

With reference to the circuit arrangement shown in the figure,

\mathrm{\mathrm{I}_{\mathrm{c}}=\frac{\mathrm{V}_0}{\mathrm{R}_2}=\frac{0.8}{800}=10^{-3} \mathrm{~A}=1 \mathrm{~mA}}

The output equation for the circuit is given by

\mathrm{\begin{aligned} & \Rightarrow V_{C E}=V_{C C}-I_{C} R_L \\ & \Rightarrow V_{C E}=(8-0.8) \mathrm{V} \\ & \Rightarrow V_{C E}=7.2 \mathrm{~V} \end{aligned}}

The d.c. current gain of the transistor is

\mathrm{\begin{aligned} & \beta=\frac{\alpha}{1-\alpha}=\frac{25 / 26}{1-25 / 26}=\frac{(25 / 26)}{\left(\frac{1}{26}\right)}=25 \\ & \text { Now, } \beta=\frac{I_C}{I_B} \Rightarrow \quad I_B=\frac{I_C}{\beta}=\frac{10^{-3}}{25} \end{aligned}}

The power gain is given by

\mathrm{\begin{aligned} A_p & ==\frac{I_i^2 R_L}{I_B^2 R_i}=\frac{\left(10^{-3}\right)^2 \times 800}{\left(4 \times 10^{-5}\right)^2 \times 200}=\frac{10^{-6} \times 4}{16 \times 10^{-10}} \\ & =\frac{1}{4} \times 10^4=25 \times 10^2=2500 \end{aligned}}

Posted by

SANGALDEEP SINGH

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