GATE ECE 2003 | Frequency Response Question 6 | Analog Circuits

GATE ECE 2003

MCQ (Single Correct Answer)

-0.6

An ideal sawtooth voltage waveform of a frequency 500 Hz and Amplitude 3 V is generated by charging a capacitor of 2 $$\mu F$$ in every cycle the charging requires

Constant voltage source of 3 V 1 ms

Constant voltage source of 3 V 2 ms

Constant current source of 3mA for 1 ms

Constant current source of 3mA for 2 ms

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

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

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

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

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

GATE ECE 1999

MCQ (Single Correct Answer)

-0.6

An amplifier is assumed to have a single pole high frequency transfer function. The rise time of its output response to a step function input is 35 nsec. The upper 3 dB frequency (in MHz) for the amplifier to a sinusoidal input is approximately at

4.55

28.6

GATE ECE 1999

MCQ (Single Correct Answer)

-0.6

An NPN transistor (with $${C_\pi }\,\, = \,\,0.3\,\,$$ pf) has a unity gain cutoff frequency $${f_T}$$ of 400 MHz at a DC-bias current I_c = 1mA. The value of its $${C_\mu }$$ (in pf) is approximately $$\left( {{V_T}\, = \,26\,mV} \right)$$.