Q41GATE 2005MCQ

For an npn transistor connected as shown in figure $V_{BE}$ =0.7 volts. Given that reverse saturation current of the junction at room temperature $300^{\circ}$ K is $10^{-13}$ A, the emitter current is

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📖 Explanation

When two terminals o a transistor are shorted it acts as diode

\begin{array}{l} I=I_{0}\left(\frac{V_{0}}{\eta^{\eta_{T}}}-1\right) \\ =10^{-13}\left(e^{\frac{0.7}{1 \times 26 \times 10^{-3}}} -1 \right)=49 \mathrm{mA} \end{array}

Q42GATE 2004MCQ

Assuming $V_{CEsat}$ = 0.2 V and $\beta$ = 50, the minimum base current ( $I_{B}$ ) required to drive the transistor in the figure to saturation is

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📖 Explanation

\begin{aligned} V_{C E} &=V_{C C}-I_{C} R_{C} \\ 0.2 &=3-I_{C} \times 1 \mathrm{k} \\ I_{C} &=2.8 \mathrm{mA} \\ \beta &=\frac{I_{C}}{I_{B}}, I_{B}=\frac{I_{C}}{\beta}=\frac{2.8 \mathrm{mA}}{50} \\ &=0.056 \mathrm{mA}=56 \mu \mathrm{A} \end{aligned}

Q43GATE 2004MCQ

Assume that the $\beta$ of transistor is extremely large and $V_{BE}$ = 0.7V, $I_{C}$ and $V_{CE}$ in the circuit shown in the figure are

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📖 Explanation

\begin{aligned} V_{Th}&=\frac{R_{2}}{R_{1}R_{2}}V_{C C}=\frac{1}{5} \times 5=1\\ R_{Th}&=R_{1} \| R_{2}=\frac{4}{5} k \Omega\\ I_{C}&\simeq \frac{1}{0.3 k \Omega}=1 \mathrm{mA}\left(\because \beta \text { is large } I_{B} \simeq 0\right)\\ V_{CE}&=V_{C C}-I_{C} R_{C}-I_{E} R_{E}\left(I_{E}=I_{C}\text { as } \right. \\ V_{CE}&=5-2.2 \times 1-0.3 \times 1 \\ V_{CE}&=2.5 \mathrm{V} \end{aligned}

Q44GATE 2003MCQ

Choose the correct match for input resistance of various amplifier configurations shown below :

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Q45GATE 2003MCQ

In the amplifier circuit shown in the figure, the values of $R_{1}$ and $R_{2}$ are such that the transistor is operating at $V_{CE}$ = 3 V and $I_{C}$ = 1.5 mA when its $\beta$ is 150. For a transistor with $\beta$ of 200, the operating point ( $V_{CE}$ , $I_{C}$ ) is

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📖 Explanation

\begin{aligned} V_{C E} &=V_{C C}-I_{C} R_{2} \\ 3 &=6-1.5 \mathrm{mA} \times R_{2} \\ 1.5 \mathrm{mA} \times \mathrm{R}_{2}&=3 \\ R_{2} &=2 \mathrm{k} \Omega \\ I_{B} &=\frac{I_{C}}{\beta}=\frac{1.5 \mathrm{mA}}{150}=0.01 \mathrm{mA} \\ \text { When } \beta&=200 \\ I_{C} &=\beta I_{B}\\ &\left(\text{as} R_{1} \text { is same } I_{B} \text{ remains same }\right)\\ &=0.01 \mathrm{mA} \times 200\\ I_{C} &=2 \mathrm{mA} \\ V_{C E} &=V_{C C}-I_{C} R_{2}=6-2 \mathrm{mA} \times 2 \mathrm{K} 2 \\ V_{C E} &=2 \mathrm{V} \end{aligned}

Q46GATE 2003MCQ

Generally, the gain of a transistor amplifier falls at high frequencies due to the

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Q47GATE 2002MCQ

If the transistor in Figure is in saturation, then

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Q48GATE 2002MCQ

Three identical RC-coupled transistor amplifiers are cascaded. If each of the amplifiers has a frequency response as shown in Figure, the overall frequency response is as given in

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📖 Explanation

\begin{array}{l} f_{L}=20 \mathrm{Hz} \\ f_{H}=1 \mathrm{kHz} \text { for single stage } \end{array}

For cascaded stage

\begin{array}{l} f_{L}^{*}=\frac{f_{L}}{\sqrt{2^{1 / n}-1}}=\frac{20}{\sqrt{2^{1 / 3}-1}}=39.2 \mathrm{H}_{2} \\ f_{H}^{*}=f_{H} \sqrt{2^{1 / n}-1}=0.5 \mathrm{k} \Omega \end{array}

Q49GATE 2001MCQ

The current gain of a BJT is

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📖 Explanation

$h_{f e}=g_{m} \cdot r_{\pi}$