Electrical Engineering - Analog Electronics
Exam Duration: 45 Mins Total Questions : 30
The diodes in the circuit shown below have linear parameters of VD = 0.6 V and rf = 0.
If V2 = 0, then output voltage Vo is
- (a)
6.43 V
- (b)
9.43 V
- (c)
7.69 V
- (d)
8.93 V
The circuit shown in figure with output voltage across AB is that of
- (a)
voltage quadrupler
- (b)
voltage tripler
- (c)
voltage doubler
- (d)
None of these
The diodes in the figure are ideal.
The transfer characteristic for -20 V \(\le \) V1 \(\le \) 20 V is
- (a)
- (b)
- (c)
- (d)
None of the above
The hybrid -\(\pi \) parameter values of gm,r\(\pi \) and r0 are
- (a)
24 mA/V, \(\infty \) , 5k\(\Omega \)
- (b)
24 mA/V, 5k\(\Omega \),\(\infty \)
- (c)
48 mA/V, 10 k \(\Omega \),18.4 k\(\Omega \)
- (d)
48 mA/V, 18.4 k \(\Omega \) ,10\(\Omega \)
The nominal quiescent collector current of a transistor is 1.2 mA. If the range of \(\beta \) for this transistor is 80 < \(\beta \) < 120 and if the quiescent collector current changes by \(\pm \) 10%, the range in value for r\(\pi \) is
- (a)
1.73 k\(\Omega \) < r\(\pi \) < 2.59 k\(\Omega \)
- (b)
1.93 k\(\Omega \) < r\(\pi \) < 2.59 k\(\Omega \)
- (c)
1.73 k\(\Omega \) < r\(\pi \) < 2.59 k\(\Omega \)
- (d)
1.56 k\(\Omega \) < r\(\pi \) < 2.88 k\(\Omega \)
Consider the circuit shown in figure below. What is the value of voltage V? Assume, Is = 6 \(\times \) 10-16 A, VT = 26 \(\times \) 10-13 V, \(\beta \) >> 1
- (a)
775 mV
- (b)
800 mV
- (c)
695 mV
- (d)
215 mV
In the circuit shown below, voltage VE = 4 V. The values of \(\alpha \) and \(\beta \) are respectively
- (a)
0.943, 17.54
- (b)
0.914, 17.54
- (c)
0.914, 11.63
- (d)
0.914, 10.63
Given, for an FET, gm=95 mA/V, total capacitance is 500 pF.For a voltage gain of -30, the bandwidth will be
- (a)
19MHz
- (b)
6.3MHz
- (c)
100kHz
- (d)
3MHz
The circuit shown below is
- (a)
clamper
- (b)
clipper
- (c)
log amplifier
- (d)
anti-log amplifier
The cut-in voltage for each diode in figure is Vg = 0.6 V. Each diode current is 1 mA. The value of R1,R2 and R3 will be respectively
- (a)
7 k\(\Omega \), 4 k\(\Omega \) and 1.47 k\(\Omega \)
- (b)
6 k\(\Omega \), 3 k\(\Omega \) and 3.43 k\(\Omega \)
- (c)
5 k\(\Omega \), 6 k\(\Omega \) and 4.93 k\(\Omega \)
- (d)
6 k\(\Omega \), 8 k\(\Omega \) and 6.43 k\(\Omega \)
The diodes in the circuit in figure have linear parameters of V\(\gamma \) = 0.6 V and rf = 0.
If V1 = V2 = 0, then output voltage V0 is
- (a)
0.964 V
- (b)
1.07 V
- (c)
10 V
- (d)
1.29 V
A voltage signal 10 sin \(\omega t\) is applied to the circuit with ideal diodes, as shown in figure. The maximum and minimum values of the output waveform Vout of the circuit are respectively
- (a)
+ 10 V and - 10 V
- (b)
+ 4 V and - 4 V
- (c)
+ 7 V and - 4 V
- (d)
+ 4 V and - 7 V
Two perfectly matched silicon transistors are connected as shown in the figure. Assuming the \(\beta\) of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current is
- (a)
zero
- (b)
3.6 mA
- (c)
4.3 mA
- (d)
5.7 mA
For the circuit the base voltage \(V_{ B }\)is
- (a)
10.2 V
- (b)
8 V
- (c)
9.6 V
- (d)
None of these
Transistor transconductance \(g_{ m }\) is
- (a)
directly proportional to current and inversally proportional to temperature
- (b)
directly proportional to current and directly proportional to temperature
- (c)
inversally proportional to current and directly proportional to temperature
- (d)
inversally proportional to both
A n-p-n transistor has a beta cut-off frequency \(f_{ \beta }\) of 1 MHz and emitter short-circuit low frequency current gain \(\beta _{ 0 }\) of 200.The unity gain frequency \(f_{ T }\) and the alpha cut-off frequency \(f_{ \alpha }\) and the alpha cut-off frequency \(f_{ \alpha }\) respectively are
- (a)
200 MHz,201 MHz
- (b)
201 MHz,200 MHz
- (c)
200 MHz,199 MHz,199 MHz
- (d)
199 MHz,200 MHz
Negative feedback in amplifiers
- (a)
improves the signal to noise ratio at the input
- (b)
improves the signal to noise ratio at the output
- (c)
does not affect the signal to noise ratio at the output
- (d)
reduces distortion
A small signal source Vi (t) = A cos 20t + B sin 106 t is applied to a transistor amplifier as shown below. The transistor has \(\beta\) = 150 and hie = 3 k\(\Omega\). Which expression best approximates Vo(t)?
- (a)
Vo(t) = -1500 (A cos 20t + B sin 106 t)
- (b)
Vo(t) = - 150 (A cos 20t + B sin 106 t)
- (c)
Vo(t) = - 1500 B sin 106 t
- (d)
Vo(t) = - 150 B sin 106 t
Negative current feedback
- (a)
decreases input impedence
- (b)
increases output impedence
- (c)
increases bandwidth
- (d)
All of the above
RC network shown in the given figure can provide a maximum theoratical phase shift of
- (a)
\(360^{ 0 }\)
- (b)
\(180^{ 0 }\)
- (c)
\(270^{ 0 }\)
- (d)
\(90^{ 0 }\)
For the transistor in figure \(\beta =50\) and \(V_{ CEQ }=10\quad V\) The value of \(V_{ 1 }\) is
- (a)
2.1 V
- (b)
6.43 V
- (c)
4.8 V
- (d)
3.73 V
The circuit shown in figure uses an ideal op-amp worki8ng with +5 V and -5 V power supplies the output voltage \(V_{ 0 }\) is equal to
- (a)
+5 V
- (b)
-5 V
- (c)
+1 ?V
- (d)
-1 V
Assume that the threshold voltage of the n-channel MOSFET shown in figure is + 0.75 V. The output characteristics of the MOSFET are also shown.
The voltage gain of the amplifier is
- (a)
+ 5
- (b)
- 7.5
- (c)
+ 10
- (d)
- 10
Consider the common emitter amplifer shown below with the following circuit parameters:
\(\beta =100,\quad g_{ m }=0.3861\quad A/V\)
\({ r }_{ o }=\infty ,\quad \pi =259k\Omega ,\quad R_{ S }=250\Omega \)
\({ R }_{ B }=93k\Omega ,\quad R_{ C }=250\Omega \)
\(R_{ L }=1k\Omega ,\quad C_{ 1 }=\infty \quad and\quad C_{ 2 }=4.7mF\)
The lower cur-off frequency due to C2 is
- (a)
33.9 Hz
- (b)
27.1 Hz
- (c)
13.6 Hz
- (d)
16.9 Hz
A regulated power supply shown in figure below, has an Unregulated (UR)input of 15V and generates a regulated output Vout Use th component values shown in the figure.
In the figure above, the ground has been shown by the symbol \(\triangledown \)
If the unregulated voltage increases by 20%, the power dissipation across the transistor Q1
- (a)
increases by 20%
- (b)
increases by 50%
- (c)
remains unchange
- (d)
decreases by 20%
Consider the circuit shown below. Assume that diodes are ideal
If V1=-5V and V2=5V, then V0 is
- (a)
9.474V
- (b)
8.943V
- (c)
4.5V
- (d)
9V
The diode in the circuit shown below has the non-linear terminal characteristics as shown in figure. Let the voltage be Vs=cos\(\omega\)t V
The voltage VD is
- (a)
0.25(3+cos\(\omega\)t)V
- (b)
0.25(1+3cos\(\omega\)t)V
- (c)
0.5(3+1cos\(\omega\)t)V
- (d)
0.5(2+3cos\(\omega\)t)V
For the transistor in circuit shown below, \(\beta\)=200
If VB = 2 V, the value of Vc is
- (a)
-7V
- (b)
1.5V
- (c)
2.6V
- (d)
None of these
The transistor in circuit shown below has \(\beta\)=200
If VBB=0, the value of voltage V0 is
- (a)
2.46V
- (b)
1.83V
- (c)
3.33V
- (d)
4.04V
In the given Darlington pair circuit of figure P 120, transistor Q1 and Q2 parameters are \(\beta _{ 1 },{ r }_{ \pi 1 }\quad and\quad \beta _{ 2 },{ r }_{ \pi 2 }\) respectively.
Input impedance Rin is
- (a)
\({ r }_{ \pi 1 }+{ r }_{ \pi 2 }\)
- (b)
\({ r }_{ \pi 1 }+(1+\beta _{ 1 }){ r }_{ \pi 2 }\)
- (c)
\({ \beta }_{ 1 }{ r }_{ \pi 1 }+{ \beta }_{ 2 }{ r }_{ \pi 2 }\)
- (d)
\((1+{ \beta }_{ 1 }){ r }_{ \pi 1 }+{ r }_{ \pi 2 }\)