GATE Electronics and Communication Engineering - Analog Electronics
Exam Duration: 45 Mins Total Questions : 30
In the circuit shown below, D1 and D2 are ideal diodes. The currents l1 and l2 are
- (a)
Zero,4mA
- (b)
4 mA, zero
- (c)
zero, 8 mA
- (d)
8 mA, zero
The figure is shown below,
From above figure
I3>0 ,Vab < 5V and \({ V }_{ { D }_{ 1 } }\le 0\)
Hence, I1=0 and \({ I }_{ 2 }=\frac { 5-3 }{ 500 } =4mA\)
The circuit inside the box in figure shown below contains only resistor and diodes. The terminal voltage Va is connected to some point in the circuit inside the box.
The largest and smallest possible values of Va most nearly to, are respectively
- (a)
15V, 6V
- (b)
24V, zero
- (c)
24V, 6V
- (d)
15V, -9V
The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage.
Consider the circuit as shown below:
The output across capacitor C2 wi II be
- (a)
Vm
- (b)
zero
- (c)
2Vm
- (d)
None of these
During positive half cycle,
Charges across C1 is Vm as capacitor will charge through Vm
During negative half cycle,
No current flows through the circuit. Hence, voltage drop across capacitor C2 is zero.
A diode whose internal resistance is \(10\Omega \), is to power supply to \(1000\Omega \) load from 110 V ( rms ) source of supply. Calculate the DC and AC load currents.
For the above question, find the ripple factor.
- (a)
12.1
- (b)
0.121
- (c)
1.21
- (d)
1.01
\(Ripplefactor=\frac { rms\quad value\quad of\quad AC\quad component }{ DC\quad componnet\quad of\quad the\quad current } \\ \gamma =\sqrt { { \left( \frac { { I }_{ rms } }{ { I }_{ DC } } \right) }^{ 2 }-1 } =\sqrt { { \left( \frac { 77 }{ 49.02 } \right) }^{ 2 }-1 } =1.21\)
In the circuit shown beJowVB = -1 V
The value of β is
- (a)
103.4
- (b)
135.5
- (c)
134.5
- (d)
102.5
VB=-IBRB
\(\Rightarrow\ \ \ I_B={-V_B\over R_B}={1\over 500}=2.0\mu A\)
VE = -I - 0,7 = - 1.7 V
\(I_E={V_E-(-3)\over 4.8}\)
\(={-1.7+3\over 4.8}=0.271mA\)
\({I_E\over I_B}=(β+1)={0.271\over 2}\)
⇒ β=134.5
In the circuit shown below the parameters are gm =1m/V, r0 =50 kΩ The gain AV = Vo/Vs is
- (a)
-8.01
- (b)
8.01
- (c)
14.16
- (d)
-14.16
The small signal equivalent circuit is as shown below
\(V_{GS}={50\over 50+2}V_S\)
\(V_0=g_mV_{GS}(50||10)=-{50\over 52}V_s(8.33)\)
\(A_V={V_0\over V_s}=-8.01\)
A three stage cascaded amplifier of identical non-interacting FET common source stage has all over an voltage gain of 110001 and overall bandwidth of 25 x 106 rad/s.
Given, gm =5 mA/V, the shunt capacitance of each stage is given by
- (a)
1.6pF
- (b)
10pF
- (c)
20pF
- (d)
3.2pF
Gain/stage=\(\sqrt[3]{1000}=10\)
BW/stage = 2 x 25 X 106
\(C={g_m\over G(BW)}\)
\(={5\times10^{-3}\over 10\times50\times10^6}=10pF\)
Which class of amplifiers operates with least distortion?
- (a)
Class A
- (b)
Class B
- (c)
Class C
- (d)
Class D
In a class A amplifier, the operating point is in the centre of load line.The output signal waveform is exactly similar to input signal waveform, (as transistor operates over the linear portion of the load) hence, it have least distortion.
Which oscillator is characterized by a split capacitor in its tank circuits?
- (a)
RC phase shift oscillator
- (b)
Colpitts oscillator
- (c)
Wien bridge oscillator
- (d)
None of the above
Colpitts oscillator
The cascode amplifier is a multi-stage configuration of
- (a)
CC-CB
- (b)
CE-CB
- (c)
CB-CC
- (d)
CE-CC
In cascode amplifier has common emitter (CE) feeding a common base (CB).
An ideal op-amp is an ideal
- (a)
voltage controlled current source
- (b)
voltage controlled voltage source
- (c)
current controlled current source
- (d)
current controlled voltage source
The op-amp output voltage is controlled by input voltage.
The circuit given in figure is a
- (a)
low-pass filter
- (b)
high-pass fi Iter
- (c)
band-pass fi Iter
- (d)
band-reject fi Iter
This is second order low-pass active filter.
If the input to the ideal comparator shown in figure below is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of
- (a)
1/2
- (b)
1/3
- (c)
1/6
- (d)
1/12
The given input signal can be expressed as
Vo = Vi; Vi ≥ Vref
4 sin ωt = 2
\(sin\ \omega t={1\over 2}\)
\(\omega t={\pi\over 6},{5\pi\over 6}\)
Duty cycle \(={T_{ON}\over T}={{5\pi\over 6}-{5\pi\over 6}\over 2\pi}={4\pi\over 12\pi}={1\over 3}\)
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
- (a)
- (b)
- (c)
- (d)
Given, lower cut-off frequency fL = 20 Hz
Upper cut-off frequency fH = 1 kHz
When n identical amplifiers are cascaded, then overall
lower cut-off frequency,\(f_L={f_L\over \sqrt{2^{1/3}-1}}\)
For n= 3
\(f^*_L={20\over \sqrt{2^{1/3}-1}}\)
= 39.22 ≃ 40 Hz
overall upper cut-off frequency
\(f^*_H=f_H\times\sqrt{2^{1/n}-1}\)
For n=3
\(=1\times \sqrt{2^{1/3}-1}=0.5kHz\)
If the transistor parameters are β = 180 and early voltage Va =140 V and it is biased at ICQ = 2 mA, the values of hybird 1t parameters, gm' rπ and ro are, respectively
- (a)
14 A/V, 2.33 kΩ, 90 kΩ
- (b)
14 A/V, 90 kΩ, 2.33 kΩ
- (c)
77 A/V, 2.33 kΩ, 70 kΩ
- (d)
77.2 A/V, 70 kΩ, 2.33 kΩ
\(g_m={I_{CQ}\over V_t}={2m\over 0.0259}=77.2mA/V\)
\(r_\pi={β V_t\over I_{CQ}}={\beta\over g_m}={180\over 77.2m}=2.33k\Omega\)
\(r_0={V_A\over I_{CQ}}={140\over 2m}=70k\Omega\)
From a measurement of the rise time of the output pulse of an amplifier whose input is a small amplitude square wave, one can estimate the following parameters of the amplifier
- (a)
gain-bandwidth product
- (b)
slew rate
- (c)
upper 3 dB frequency
- (d)
lower 3 dB frequency
\(t_r={0.35\over f_H}\)
where, fH = upper 3 dB frequency.
How can we minimize the errors due to input bias current, input offset current?
- (a)
Use the resistance Rcomp=RF || R1
- (b)
Use of op-amp with small ratings of lios
- (c)
Keep the resistance value as small as possible
- (d)
All of the above
To minimize error,
Rcomp = RP II R1, small rating of lios and running R small.
An amplifier without feedback has a voltage gain of 50, input resistance of 1 kn and output resistance of 2.5 kn. The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is
- (a)
\(\frac { 1 }{ 11 } k\Omega \)
- (b)
\(\frac { 1 }{ 5 } k\Omega \)
- (c)
\(5k\Omega \)
- (d)
\(11k\Omega \)
The input resistance of current shunt feedback decreases by factor \((1+A\beta )\)
\({ R }_{ if }+\frac { { R }_{ t } }{ 1+A\beta } =\frac { 1\quad k\Omega }{ 1+50\times 0.2 } =\frac { 1 }{ 11 } k\Omega \)
The oscillator circuit shown in figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is
- (a)
\(\frac { 1 }{ \left( 2\pi \sqrt { 6 } RC \right) } \)
- (b)
\(\frac { 1 }{ \left( 2\pi RC \right) } \)
- (c)
\(\frac { 1 }{ \left( \sqrt { 6 } RC \right) } \)
- (d)
\(\frac { \sqrt { 6 } }{ \left( 2\pi RC \right) } \)
One RC pair gives phase shift 600 and hence total phase shift will be \(180°\)
This is a RC phase shift oscillator
\(f=\frac { 1 }{ 2\pi \sqrt { 6 } RC } \)
The input to full - wave rectifier shown below is Vi =120 sin 2\(\pi\)60t V. The diode cut - in voltage is 0.7 V. If the output voltage cannot drop below 100 V, the required value of the capacitor is
- (a)
61.2 \(\mu\)F
- (b)
41.2 \(\mu\)F
- (c)
20.6 \(\mu\)F
- (d)
30.6 \(\mu\)F
Full wave rectifier
Vs = Vi = 120 sin 2\(\pi\)60t V
Vmax = 120 - 0.7 = 119.3 V
Vrip = 119.3 - 100 = 19.3 V
C = \(\frac{V_{max}}{2fRV_{rip}}\)
= \(\frac{119.3}{2(60)2.5\times10^3\times14.4}\)
= 20.6 \(\mu\)F
For the op-amp circuit shown in the figure below, Vo is
- (a)
-2 V
- (b)
-1 V
- (c)
- 0.5 V
- (d)
0.5 V
\({ V }_{ A }=\frac { 1 }{ 1+1 } (1\quad V)=0.5\quad V\) ( from voltage divided rule )
Applying KCL at node B,
\(\frac { 1-{ V }_{ B } }{ 1k } =\frac { { V }_{ B }-{ V }_{ o } }{ 2k } \)
From virtual group concept,
VB=VA=0.5 V
\(\frac { 1-0.5 }{ 1k } =\frac { 0.5-{ V }_{ o } }{ 2k } \)
1=0.5-Vo
Vo=-0.5 V
A Zener diode regulator in figure is to be designed to meet the specifications IL =10 mA, V0=10 V and Vin varies from 30 V to 50 V. The Zener diode has Vz =10 V and IZK (knee current) =1mA. For satisfactory operation
- (a)
\(R\le 1800\Omega \)
- (b)
\(2000\Omega \le R\le 2200\Omega \)
- (c)
\(3700\Omega \le R\le 4000\Omega \)
- (d)
\(R>4000\Omega \)
\(\frac { { V }_{ in }-{ V }_{ 0 } }{ R } \ge { I }_{ Z }+{ I }_{ L }({ I }_{ 1 })\\ \Rightarrow { I }_{ Z }+{ I }_{ L }={ I }_{ 1 },\quad where\quad { V }_{ 1 }=30V\\ \frac { 30-10 }{ R } \ge (10+1)mA=\frac { 20 }{ R } \ge 11mA\\ R\le 1818,when\quad { V }_{ in }=50V\\ \quad \frac { 40 }{ R } \ge 11\times { 10 }^{ -3 }\\ R\le 3636,from\quad this\quad R\le 1818\)
The circuit using a BJT with \(\beta \) =50 and VBE =0.7 V shown in figure. The base current IB and collector voltage Vc are, respectively
- (a)
43\(\mu\)A and 11.4 V
- (b)
40\(\mu\)A and 16 V
- (c)
45\(\mu\)A and 11 V
- (d)
50\(\mu\)A and 10 V
Applying KVL in collector base circuit,
VCC - IBRB - VBE - I ERE = 0
IE = IC + IB
= \(\beta\)IB + IB = IB ( 1 + \(\beta\) )
VCC - IBRB - VBE - IB ( 1 +\(\beta\) ) RE = 0
\({ I }_{ B }=\frac { { V }_{ CC }-{ V }_{ BE } }{ (1+\beta ){ R }_{ E }+{ R }_{ B } } \)
= 40\(\mu\)A
IC = \(\beta\)IB = 2 mA
Collector voltage VC = VCC - ICRC
= 20 - 2 x 10-3 x 2 x 103
= 16 V
An n-p-n BJT has \({ g }_{ m }=38mA/V,{ C }_{ \mu }={ 10 }^{ -14 }F,{ C }_{ \pi }=4\times { 10 }^{ -13 }F,\) and DC current gain \({ \beta }_{ 0 }=90\). For this transistor fT and \({ f }_{ \beta }\) are
- (a)
\({ f }_{ T }=1.64\times { 10 }^{ 8 }Hz\quad and\quad { f }_{ \beta }=1.47\times { 10 }^{ 10 }Hz\)
- (b)
\({ f }_{ T }=1.47\times { 10 }^{ 10 }Hz\quad and\quad { f }_{ \beta }=1.64\times { 10 }^{ 8 }Hz\)
- (c)
\({ f }_{ T }=1.33\times { 10 }^{ 12 }Hz\quad and\quad { f }_{ \beta }=1.47\times { 10 }^{ 10 }Hz\)
- (d)
\({ f }_{ T }=1.47\times { 10 }^{ 10 }Hz\quad and\quad { f }_{ \beta }=1.33\times { 10 }^{ 12 }Hz\)
\({ F }_{ T }=\frac { { g }_{ m } }{ 2\pi ({ C }_{ \mu }+2) } \)
Putting the value FT=\(1.47\times { 10 }^{ 10 }Hz\)
From the relation
\({ f }_{ \beta }.{ h }_{ fe }<{ f }_{ T }\\ { f }_{ \beta }={ f }_{ T }/.{ h }_{ fe }\)
Putting the value
\(F_{ \beta }={ 1.64\times { 10 }^{ 8 } }Hz\)
An n-p-n transistor has a beta cut - off frequency f\(\beta\) of 1 MHz, and common - emitter short circuit low frequency current gain \(\beta_0\) of 200. If unity gain frequency f\(\gamma\) and the alpha cut-off frequency f\(\alpha\) respectively, are
- (a)
200 MHz, 201 MHz
- (b)
200 MHz, 199 MHz
- (c)
199 MHz, 200 MHz
- (d)
201 MHz, 200 MHz
ft = \(\beta_0\) , \(f_\beta\) = 200 X 1 MHz = 200 MHz
\(f_{\alpha}\) = \(\frac{f_{\beta}}{1-\alpha}\)
\(f_{\alpha}\) = \(\frac{f_\beta}{1-\frac{\beta}{1+\beta}}\)
\(f_{\alpha}\) = \(\frac{f_\beta}{1}\)(1+\(\beta\))
= 1 X 106 X 201
\(f_{\alpha}\) = 201 MHz
The transfer characteristic for the precision rectifier circuit shown below is (assume ideal op-amp and practical diodes)
- (a)
- (b)
- (c)
- (d)
This is an example of control precision. For Vi \(\ge\) - 5, diode D2 conducts and closes the negative feedback loop around the op-amp. A virtual ground therefore will appear at the inverting input terminal and the op-amp output will be damped at one diode drop below ground. This negative voltage will keep the diode D1 Off, and no current will flow in the feedback resistance R2 , that is, the rectifier output will be zero. As Vi goes negative, the voltage at the inverting input terminal will tend to go negative, causing the voltage at the op - amp output terminal to go to positive. This will cause D2 to be reverse biased and hence cut - off. The current through the feedback resistance R2 will be equal to the current through the input resistance R1. For R1 = R2, the output voltage
V0 = - Vi - 5 for Vi \(\ge\) - 5 V
Hence, transfer characteristics will be
The oscillator circuit shown in figure is
- (a)
Hartley oscillator with foscillation = 79.6 MHz
- (b)
Colpitts oscillator with foscillation = 79.6 MHz
- (c)
Hartley oscillator with foscillation = 159.2 MHz
- (d)
Colpitts oscillator with foscillation = 159.2 MHz
The tank circuit indicated in figure has capacitor divided circuit with inductor in parallel, hence it is a Colpitts oscillator.
\({ C }_{ eq }=\frac { { C }_{ 1 }{ C }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } } =1pF\\ f=\frac { 1 }{ 2\pi \sqrt { { LC }_{ eq } } } =79.6MHz\)
The phase-shift oscillator in figure below operates at f=80 kHz. The value of resistance RF is
- (a)
\(814\Omega\)
- (b)
\(236\Omega\)
- (c)
\(148\Omega\)
- (d)
\(438\Omega\)
Oscillation frequency,
\(f=\frac{1}{2\pi \sqrt{6}RC}\)
\(\therefore\ \ 80 k =\frac{1}{2\pi \sqrt{6}R(100)}\)
\(\Rightarrow\ \ \ R=\frac{1}{(80\ k)(2\pi \sqrt{6})(100)}=8.12 k\Omega\)
\(\Rightarrow \ \ R_F=(8.12k)(29)=236k\Omega\)
In an ideal differential amplifier shown in figure below, a large value of RB
- (a)
increases both the differential and common mode gains
- (b)
increases the common mode gain only
- (c)
decreases the differential mode gain only
- (d)
decreases the common mode gain only
In ideal differential amplifier, Common mode gain = \(-\frac{R_C}{2R_E}\)
Differential mode gain = - gmRC
(doesn't depend on RE )
Hence, by increasing RE the common mode gain will be decrease.
For the circuit shown below,assume that the Zener diode is ideal with a breakdown voltage of 6V. The wavwform observed across R is
- (a)
- (b)
- (c)
- (d)
When 0<Vi<6V;
Zener breakdown doesn't occur and
VR=0
6V<V1<12 V;
VR=Vin-6