Frequency Response of Amplifier

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.
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