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  • Consider the circuit arrangement shown in Fig 14.16 (a) for studying input and output characteristics of n-p-n transistor in CE configuration.Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point

Consider the circuit arrangement shown in Fig 14.16 (a) for studying input and output characteristics of the n-p-n transistor in CE configuration.


Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in Fig. 14.16 (b).

Given that the input impedance of the transistor is very small and VCC = VBB = 16 V, also find the voltage gain and power gain of the circuit making appropriate assumptions.

Answers (1)

Explanation:-

According to the problem at point Q, from the graph V_{BE}=0.7V,V_{CC}=V_{BB}=16V and V_{CE}=8V

The circuit consists of an NPN transistor in the Common Emitter (CE) configuration. Vb? is the base bias voltage, and Vc is the collector supply voltage, both given as 16 V. The transistor's base-emitter voltage Vbe=0.7 V

I_{C}=4mA=4\times 10^{-3}A\\ I_{B}=30\mu A=30\times 10^{-6}A\\ Since\: V_{CC}=I_{C}R_{C}+V_{CE}\\ R_{C}=\frac{V_{CC}-V_{CE}}{I_{C}}=\frac{16-8}{4 \times 10^{-3}}=\frac{8 \times 1000}{4}=2k\Omega\\ Similarly, V_{BB}=I_{B}R_{B}+V_{BE}\\ R_{B}=\frac{V_{BB}-V_{BE}}{I_{B}}=\frac{16-0.7}{30 \times 10^{-6}}=510 \times 10^{3}\Omega=510k\Omega\\ Current \: gain=\beta =\frac{I_{C}}{I_{B}}=\frac{4 \times 10^{-3}}{30 \times 10^{-6}}=133.3\\ Voltage \: gain=\beta =\frac{R_{C}}{R_{B}}=\frac{133 \times 2 \times 10^{3}}{510 \times 10^{3}}=0.52\\ Power \: gain=\beta \times voltage\: gain=133 \times 0.52=69

 

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