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Semiconductor Electronic: Material, Devices And Simple Circuits — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSemiconductor Electronic: Material, Devices And Simple Circuits5 Marks
Question

Consider the circuit arrangement shown in Fig (a) for studying input and output characteristics of npn transistor in CE configuration.

Select the values of $R_B$ and $R_C$ for a transistor whose $V_{BE} = 0.7V,$ so that the transistor is operating at point Q as shown in the characteristics shown in Fig. (b).
Given that the input impedance of the transistor is very small and $V_{CC} = V_{BB} = 16V,$ also find the voltage gain and power gain of circuit making appropriate assumptions.
✓

Answer

According to the problem, at point Q, from graph
$V_{BE} = 0.7V, V_{CC} = V_{BB} = 16V$ and $V_{CE} = 8V. IC = 4mA = 4 \times 10^{-3}A$
$\text{I}_\text{B}=30\mu\text{A}=30\times10^{-6}\text{A}$
Since, $V_{CC} = I_CR_C + V_{CE} $
$\text{R}_\text{C}=\frac{\text{V}_\text{CC}-\text{V}_\text{CE}}{\text{I}_\text{C}}=\frac{16-8}{4\times10^{-3}}=\frac{8\times1000}{4}=2\text{k}\Omega$
Similarly, $V_{BB} - I_BR_B + V_{BE}$
$\text{R}_\text{B}=\frac{\text{V}_\text{BB}-\text{V}_\text{BE}}{\text{I}_\text{B}}=\frac{16-0.7}{30\times10^{-6}}$
$=510\times10^3\Omega=510\text{k}\Omega$ Current gain,
$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}=\frac{4\times10^{-3}}{30\times10^{-6}}=133.3$
Voltage gain $=\beta\frac{\text{R}_\text{C}}{\text{R}_\text{B}}=\frac{133\times2\times10^3}{510\times10^3}=0.52$
$\text{Power gain}=\beta\times\text{Voltage gain}=133\times0.52=69$

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