GATE ECE 2001 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2001

MCQ (Single Correct Answer)

-0.6

In the figure assume the Op-Amps to be ideal. The output V₀ of the circuit is GATE ECE 2001 Analog Circuits - Operational Amplifier Question 63 English

GATE ECE 2001 Analog Circuits - Operational Amplifier Question 63 English

10 $$\cos $$ (100 t)

10 $$\int\limits_0^t {\cos \,\,\left( {100\tau } \right)} \,\,d\tau $$ v

$${10^{ - 4}}\int\limits_0^t {\cos \,\,\left( {100\tau } \right)} \,\,d\tau $$

$${10^{ - 4}}{d \over {dt}}\cos \left( {100t} \right)$$

GATE ECE 2001

MCQ (Single Correct Answer)

-0.3

The current gain of a BJT is

$$g_mr_o$$

$$\frac{g_m}{r_o}$$

$$g_mr_\mathrm\pi$$

$$\frac{g_m}{r_\mathrm\pi}$$

GATE ECE 2001

MCQ (Single Correct Answer)

-0.6

The Oscillator circuit shown in the figure is GATE ECE 2001 Analog Circuits - Oscillators Question 5 English

Hartley Oscillator with $${f_{oscillation}}$$= 79.6 MHz

Colpitts Oscillator with $${f_{oscillation}}$$ =50.3 MHz

Hartley Oscillator with $${f_{oscillation}}$$= 159.2 MHz

Colpitts Oscillator with $${f_{oscillation}}$$=159.2 MHz

GATE ECE 2001

MCQ (Single Correct Answer)

-0.6

An npn BJT has g_m = 38 mA/V, $${C_\mu }\, = {10^{ - 14}}$$ F, $${C_\pi }\, = 4\, \times {10^{ - 13}}\,F$$ and DC current gain $$\beta \, = \,90$$. For this transistor f_T and $${f_\beta }$$ are

$$\eqalign{ & {f_T} = 1.64 \times {10^8}\,\,Hz\,and \cr & {f_\beta } = 1.47 \times {10^{10}}\,\,Hz \cr} $$

$$\eqalign{ & {f_T} = 1.47 \times {10^{10}}\,\,Hz\,and \cr & {f_\beta } = 1.64 \times {10^8}\,\,Hz \cr} $$

$$\eqalign{ & {f_T} = 1.33 \times {10^{12}}\,\,Hz\,and \cr & {f_\beta } = 1.47 \times {10^{10}}\,\,Hz \cr} $$

$$\eqalign{ & {f_T} = 1.47 \times {10^{10}}\,\,Hz\,and \cr & {f_\beta } = 1.33 \times {10^{12}}\,\,Hz \cr} $$

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