Frequency Response of Amplifier

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Question 1

The Miller effect in the context of a Common Emitter amplifier explains

A	an increase in the low-frequency cutoff frequency
B	an increase in the high-frequency cutoff frequency
C	a decrease in the low-frequency cutoff frequency
D	a decrease in the high-frequency cutoff frequency

GATE EC 2017-SET-1 Analog Circuits

Question 1 Explanation:

Miller effect increases input capacitance and thereby decreases the higher cut-off frequent

Question 2

Which one of the following statements is correct about an ac-coupled common-emitter amplifier operating in the mid-band region?

A	The device parasitic capacitances behave like open circuits, whereas coupling and bypass capacitances behave like short circuits.
B	The device parasitic capacitances, coupling capacitances and bypass capacitances behave like open circuits.
C	The device parasitic capacitances, coupling capacitances and bypass capacitances behave like short circuits
D	The device parasitic capacitances behave like short circuits, whereas coupling and bypass capacitances behave like open circuits

GATE EC 2016-SET-2 Analog Circuits

Question 3

The ac schematic of an NMOS common-source stage is shown in the figure below, where part of the biasing circuits has been omitted for simplicity. For the n -channel MOSFET M, the transconductance

g_{m}=1 mA/V

and body effect and channel length modulation effect are to be neglected. The lower cutoff frequency in Hz of the circuit is approximately at

A	$8$
B	$32$
C	$50$
D	$200$

GATE EC 2013 Analog Circuits

Question 3 Explanation:

f_{L}=\frac{1}{2 \pi R C}=\frac{1}{2 \times 3.14 \times(10 k+10 k) \times 1 \mu F}=8 Hz

Question 4

A bipolar transistor is operating in the active region with a collector current of 1 mA. Assuming that the

\beta

of the transistor is 100 and the thermal voltage (

V_{T}

) is 25 mV, the transconductance (

g_{m}

) and the input resistance (

r_{\pi}

) of the transistor in the common emitter configuration, are

A	$g_{m}=25 mA/V \; and \; r_{\pi}=15.625 k\Omega$
B	$g_{m}=40 mA/V \; and \; r_{\pi}=4.0 k\Omega$
C	$g_{m}=25 mA/V \; and \; r_{\pi}=2.5 k\Omega$
D	$g_{m}=40 mA/V \; and \; r_{\pi}=2.5 k\Omega$

GATE EC 2004 Analog Circuits

Question 4 Explanation:

\begin{array}{l} g_{m}=\frac{I_{C}}{V_{T}}=\frac{1 \mathrm{mA}}{25 \mathrm{mV}}=0.04=40 \mathrm{mAV} \\ h_{\mathrm{fe}}=g_{m} \cdot r_{\pi}, h_{\mathrm{fe}}=\beta \\ r_{\pi}=\frac{\beta}{g_{m}}=\frac{100}{40 \times 10^{-3}}=2.5 \mathrm{k} \Omega \end{array}

Question 5

An npn BJT has

g_{m}=38m\, A/V

C_{\mu }=10^{-14}\,F

C_{\pi }=4\times 10^{-13}\,F

and DC current gain

\beta _{0}=90

. For this transistor

f_{T}

and

f_{\beta }

are

A	$f_{T}=1.64\times 10^{8}HZ$ and $f_{\beta }=1.47\times 10^{10}HZ$
B	$f_{T}=1.47\times 10^{10}HZ$ and $f_{\beta }=1.64\times 10^{8}HZ$
C	$f_{T}=1.33\times 10^{12}HZ$ and $f_{\beta }=1.47\times 10^{10}HZ$
D	$f_{T}=1.47\times 10^{10}HZ$ and $f_{\beta }=1.33\times 10^{12}HZ$

GATE EC 2001 Analog Circuits

Question 5 Explanation:

\begin{aligned} f_{T} &=\frac{g_{m}}{2 \pi\left(C_{\mu}+C_{\pi}\right)} \\ &=\frac{38 \times 10^{-3}}{2 \pi \times\left(10^{-14}+4 \times 10^{-13}\right)} \\ &=\frac{38 \times 10^{-3}}{2 \pi \times 10^{-13}(0.1+4)}=1.47 \times 10^{10} \mathrm{Hz} \\ &=\frac{f_{T}}{\beta_{0}}=\frac{1.47 \times 10^{10}}{90} \\ &=1.64 \times 10^{8} \mathrm{Hz} \quad&\left(\beta_{0}=h_{1}\right) \end{aligned}

There are 5 questions to complete.