EC.1) If the input to a comparator
is a sine wave, the output is a
(a) ramp
voltage
(b) sine
wave
(c)
rectangular wave
(d)
sawtooth wave
Answer is (c)
Explanation
Comparator compares the input value with a reference voltage and returns either high or low depending on the comparison operator (greater/lesser/greater than or equal to/less than equal to/equal to and so on). Thus the output waveform will have only two levels, high and low which further yields a rectangular wave.
Explanation
Comparator compares the input value with a reference voltage and returns either high or low depending on the comparison operator (greater/lesser/greater than or equal to/less than equal to/equal to and so on). Thus the output waveform will have only two levels, high and low which further yields a rectangular wave.
EC.2) Silicon is doped with boron to a concentration of 4x1017 atoms per cubic centimeters (atoms/cm3). Assume the
intrinsic carrier concentration of silicon to be 1.5x1010 per cubic
centimeters (atoms/cm3) and the value of kT/q to be 25mV at 300K. Compared to
undoped silicon , the Fermi level of doped silicon
(a) goes
sown by 0.13 eV
(b) goes up
by 0.13eV
(c) goes
down by 0.427 eV
(d) goes up
by 0.427 eV
Answer is (c)
Explanation:
${n_i} =1.5\times 10^{10}$
${V_T} = \frac{kT}{q}$
$ q \times {V_T} = kT$(q is charge of electron)
$kT = 25meV$
$ E_{F_P} $(Fermi Level of P-Type doped Si)
$E_{F_i}$(Fermi Level of intrinsic Si)
$E_{F_i} - E_{F_P} = kT\ln (\frac{N_A}{n_i})$
$E_{F_i} - E_{F_P} = 25\times10^{3}\ln(\frac{4\times10^{17}}{1.5\times10^{10}})$
$ E_{F_i} - E_{F_P} = 0.457eV \therefore E_{F_P}$ moves down by 0.457eV
EC.3) When
transistors are used in digital circuits they usually operate in the
(a) active
region
(b) breakdown
region
(c)
saturation and cutoff regions
(d) linear
region
Answer is (c)
Explanation
Digital circuits operate on the principle of binary logic i.e., ones and zeros, which can be represented by high and low voltage levels. Thus transistors in digital circuits operate in saturation and cut-off regions, representing ON and OFF states.
Explanation
Digital circuits operate on the principle of binary logic i.e., ones and zeros, which can be represented by high and low voltage levels. Thus transistors in digital circuits operate in saturation and cut-off regions, representing ON and OFF states.
EC.4) Semiconductor A has a higher band gap than
semiconductor B. If both
have same dimensions, the same number of electrons at a given temperature and
the same electron & hole mobilities, then
(a) A has
the same number of holes as B
(b) A has
larger number of holes than B
(c) A has
lesser number of holes than B
(d) all of
above
Answer is (b)
Explanation:
The empty energy levels refer to the Holes Filled in Valence Band
Thereby, an HIGHER BAND GAP implies Greater number of Holes.
EC.5) The effect of doping in intrinsic semiconductor is
to
(a) move
the Fermi level away from the center of forbidden band
(b) move the
Fermi level towards the center of forbidden band
(c) change
crystal structure of semiconductor
(d) keep the
Fermi level at the middle of forbidden band
Answer is (b)
EC.6) To operate
properly, a transistor's base-emitter junction must be forward biased with
reverse bias applied to which junction?
(a)
collector-emitter
(b)
base-collector
(c)
base-emitter
(d)
collector-base
Answer is (d)
EC.7) Most commonly used semiconductor material is
EC.7) Most commonly used semiconductor material is
(a) silicon
(b)
germanium
(c) mixture
of silicon and germanium
(d) none of
the above
Answer is (a)
EC.8) In which of these is reverse recovery time nearly zero?
EC.8) In which of these is reverse recovery time nearly zero?
(a) Zener
diode
(b) Tunnel
diode
(c)
Schottky diode
(d) PIN
diode
Answer is (c)
Reason
In Schottky diode, there is no accumulation of charges.
EC.9) For a transistor in cutoff state, Vce will be approximately equal to
Reason
In Schottky diode, there is no accumulation of charges.
EC.9) For a transistor in cutoff state, Vce will be approximately equal to
(a) Vcc
(b) Vb
(c) 0.2V
(d) 0.7V
Answer is (a)
Explanation
Applying KVL to the circuit, Vcc = IcRc-Vce
When the transistor is open, there is no current flow and thus Ic will be zero. Thus Vcc = Vce
EC.10)The electron & hole concentrations in a intrinsic semiconductor are Ni & Pi respectively, when doped with a p-type material, these change to N & P respectively , then:
Explanation
Applying KVL to the circuit, Vcc = IcRc-Vce
When the transistor is open, there is no current flow and thus Ic will be zero. Thus Vcc = Vce
EC.10)The electron & hole concentrations in a intrinsic semiconductor are Ni & Pi respectively, when doped with a p-type material, these change to N & P respectively , then:
a) N + P =
Ni +Pi
b) N + Ni
= P + Pi
c) N*Pi =
Ni*P
d) N*P =
Ni * Pi
Answer is (d)
EC.11) Current flow in a semiconductor depends on the phenomenon of
(a) drift
(b) diffusion
(c) recombination
(d) all of
above
Answer is (d)
Answer is (d)
EC.12) If a
transistor operates at the middle of the dc load line, a decrease in the
current gain will move the Q point:
(a) off the
load line
(b) nowhere
(c) up
(d) down
Answer is (d)
Explanation
Current gain is given by IC/IB. From Figure, it is clear that as the Q-point moves down the load line, IC decreases which inturn decreases the current gain.
Answer is (d)
Explanation
Current gain is given by IC/IB. From Figure, it is clear that as the Q-point moves down the load line, IC decreases which inturn decreases the current gain.
EC.13) Voltage gain of an emitter follower
is
(a) much less than one
(b) approximately equal to one
(c) greater than one
(d) zero
Answer is (b)
Explanation
Voltage gain is given by Vo/Vin
In emitter follower, Vo=Vin
Thus the voltage gain of the emitter follower is approximately equal to one.
Answer is (b)
Explanation
Voltage gain is given by Vo/Vin
In emitter follower, Vo=Vin
Thus the voltage gain of the emitter follower is approximately equal to one.
EC.14) Match List-
I with List - II & select the correct answer using CODE given below the
lists
List-I List-II
A. Drift current 1. Law of conservation of charge
B. Einstein's equation
C. Diffusion current
D. Continuity equation 4. Concentration gradient
CODE:
A B C D
(a) 2 3 4 1
(b)
2 1 4
3
(c)
4 1 2
3
(d)
1 3 4
2
Answer is (a)
Explanation
1. Drift Current is motion of charged particle due to Electric field
2. Einstein's Equation is related to Thermal Voltage which is given by kT/q
3. Diffusion current is due to concentration gradient
4. Continuity Equation is related to Law of Conservation of charge
Explanation
1. Drift Current is motion of charged particle due to Electric field
2. Einstein's Equation is related to Thermal Voltage which is given by kT/q
3. Diffusion current is due to concentration gradient
4. Continuity Equation is related to Law of Conservation of charge
EC.15) In an abrupt pn-junction the doping concentrations on the p-side and
n-side are Na=9x10^16/cm^3 and Nd=1x10^16/cm^3 respectively. The pn junction is
reverse biased and the total depletion width is 3 um. The depletion width on
p-side is
(a) 2.7 um
(b) 0.3 um
(c) 2.25
um
(d) 0.75
um
Answer is ()
EC.16) In an n channel JFET, the gate is
(a) n type
(b) p type
(c) either
n or p
(d)
partially n & partially p
Answer is (b)
EC.17) Characteristics of a Voltage Follower circuit
are:
(a) high
input impedance, low output impedance,unity gain
(b) high
input impedance, high output impedance, unity gain
(c) low
input impedance, high output impedance,
No unity
gain
(d) High
input impedance, low output impedance, no unity gain
Answer is (a)
EC.18) In p-n-p transistor the current Ie has two components
viz. Iep due to injection of holes from p-region to n-region and Ien due to
injection of electrons from n-region to p-region. Then
(a) Iep
and Ien are almost equal
(b) Iep
>> Ien
(c) Ien
>> Iep
(d) either
(a) or (c)
Answer is (b)
EC.19) Assertion (A): A p-n junction has high resistance in
reverse direction.
Reason
(R): When a reverse bias is applied to p-n junction, the width of depletion
layer increases.
(a) Both A
and R are true and R is correct explanation of A
(b) Both A
and R are true but R is not a correct explanation of A
(c) A is
true but R is false
(d) A is
false but R is true
Answer is (a)
EC.20) Impurity commonly used for realizing the base region of
a silicon n-p-n transistor is
(a) Gallium
(b) Indium
(c) Boron
(d)
Phosphorous
Answer is (c)
EC.21) For a stable operational amplifier-feedback
network, the ideal closed loop gain can be calculated if which of the following
conditions is true:
(a) A
negligible differential voltage applied at input terminals, produces a
significant output voltage
(b) The
current required at either of the op-amp terminals is negligible
(c) both
(a) and (b) are true
(d) only
(a) is true
Answer is ()
EC.22) At very high temperatures the extrinsic semiconductors
become intrinsic because
(a) drive
in diffusion of dopants and carriers
(b) band
to band transition dominants over impurity ionization
(c)
impurity ionization dominants over band to band transition
(d) band
to band transition is balanced by impurity ionization
Answer is (b)
EC.23)For the non-inverting integrating amplifier in the figure,the ideal closed-loop gain V_0(s)/V_i(s) equals
(a)1/(R1C1s+1) .( RCs+1)/RCs
(b)1/(R1C1s+1).RCs/(RCs + 1)
(c)1/(R1C1s).(RCs+1)/RCs
(d)(R1C1s+1).RCs/(RCs+1)
EC.24)Identify the op-amp circuit above and define the relationship between the input and output voltage ,
given the diode-current voltage :
i_D =I_S(e^[qv_D/kT] -1)
Here, I_S is a constant dependent on diode construction.
q is electron charge
k is Boltzmann's constant
T is absolute temperature.
Then,v_i at the input equals
(a)-RI_S(e^[qv_o/kT] -1)
(b)-RI_S(e^[-qv_o/kT +1)
(c)R/I_S(e^[qv_o/kT])
(d)insufficient data,cannot be determined
EC.25) When a voltage is applied to a semiconductor crystal then the free electrons will flow...
(a) towards positive terminal
(b) towards negative terminal
(c) either towards positive terminal or negative terminal
(d) towards positive terminal for 1 μs and towards negative terminal for next 1 μs
EC.26) An ideal opamp is an ideal
(a) Voltage controlled current source
(b) Voltage controlled voltage source
(c) Current controlled current source
(d) Current controlled voltage source
EC.27)The gain bandwidth product for a 741-opamp , given the closed loop gain, A_v=80dB and the cutoff frequency at that gain f_c=80Hz is
(a)800k
(b)80k
(c)8k
(d)none of the above
EC.28) The number of doped regions in PIN diode is
(a) 1
(b) 2
(c) 3
(d) 1 or 2
EC.23)For the non-inverting integrating amplifier in the figure,the ideal closed-loop gain V_0(s)/V_i(s) equals
(a)1/(R1C1s+1) .( RCs+1)/RCs
(b)1/(R1C1s+1).RCs/(RCs + 1)
(c)1/(R1C1s).(RCs+1)/RCs
(d)(R1C1s+1).RCs/(RCs+1)
EC.24)Identify the op-amp circuit above and define the relationship between the input and output voltage ,
given the diode-current voltage :
i_D =I_S(e^[qv_D/kT] -1)
Here, I_S is a constant dependent on diode construction.
q is electron charge
k is Boltzmann's constant
T is absolute temperature.
Then,v_i at the input equals
(a)-RI_S(e^[qv_o/kT] -1)
(b)-RI_S(e^[-qv_o/kT +1)
(c)R/I_S(e^[qv_o/kT])
(d)insufficient data,cannot be determined
EC.25) When a voltage is applied to a semiconductor crystal then the free electrons will flow...
(a) towards positive terminal
(b) towards negative terminal
(c) either towards positive terminal or negative terminal
(d) towards positive terminal for 1 μs and towards negative terminal for next 1 μs
EC.26) An ideal opamp is an ideal
(a) Voltage controlled current source
(b) Voltage controlled voltage source
(c) Current controlled current source
(d) Current controlled voltage source
EC.27)The gain bandwidth product for a 741-opamp , given the closed loop gain, A_v=80dB and the cutoff frequency at that gain f_c=80Hz is
(a)800k
(b)80k
(c)8k
(d)none of the above
EC.28) The number of doped regions in PIN diode is
(a) 1
(b) 2
(c) 3
(d) 1 or 2
EC.29) How much is the base-to-emitter voltage of a transistor in the "on" state?
(a) 0 V
(b) 0.7 V
(c) 0.7 mV
(d) Undefined
EC.30) Which of the following equipment can check the condition of a transistor?
(a) Current tracer
(b) Digital display meter (DDM)
(c) Ohmmeter (VOM)
(d) All of the above
EC.31) The Two Port Darlington impedance booster of figure uses identical transistors (hie = 1 KΩ, hfe = 100, hre = hoe = 0).
[13/07 4:54 pm] rashmi: Calculate the Z – 11 & Z-22 parameters of the network. Use relevant approximations.
a)10.3 MΩ , 2 KΩ
b)10.3 MΩ , 1 KΩ
c)20.3 MΩ , 2 KΩ
d) none of the above
d)none of above
EC.32)Match the following:
a)Hartley 1)Low frequency oscillator
b)Wein-bridge 2) High frequency oscillator
c)Crystal 3)stable frequency oscillator
4) Relaxation Frequency oscillator
5) Negative resistance oscillator
a) (a)-2 ,( b )- 4 , (c)- 3
b) (a)-2 ,( b )- 1 , (c)- 3
c) (a)-1 ,( b )- 4 , (c)- 5d)none of above
EC.33)Consider the following two statements :
Statement 1: Astable Multivibrator can be used for generating square wave.
Statement 2: Bistable Multivibrator can be used for storing binary information.
a. Only statement 1 is correct.
b. Only statement 2 is correct.
c. Both the statements 1 and 2 are correct.
d. Both the statements 1 and 2 are incorrect.
EC.34) For what kind of amplifications can the active region of the common-emitter configuration be used?
(a) Voltage
(b) Current
(c) Power
(d) All of the above
EC.35) What range of resistor values would you get when checking a transistor for forward- and reverse-biased conditions by an ohmmeter?
(a) 100 to a few k ohm, exceeding 100 k ohm
(b) Exceeding 100 k ohm, 100 ohm to a few k ohm
(c) Exceeding 100 k ohm, exceeding 100 k ohm
(d) 100 ohm to a few k ohm, 100 ohm to a few k ohm
EC.36) In which region are both the collector-base and base-emitter junctions forward-biased?
(a) Active
(b) Cutoff
(c) Saturation
(d) All of the above
EC.37) What does a reading of a large or small resistance in forward- and reverse-biased conditions indicate when checking a transistor using an ohmmeter?
(a) Faulty device
(b) Good device
(c) Bad ohmmeter
(d) None of the above
EC.38) In the JFET circuit shown, assume that R1/R2 = 1 MΩ and the total stray capacitance at the output to be 20 pF.Determine the upper cutoff frequency of the amplifier.
(a)3.18MHz
(b)2.46MHz
(c)3.45MHz
(d) None of the above
EC.39)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 parameter of the amplifier
a. Gain-bandwidth product
b. Slew rate
c. Upper 3 dB frequency
d.Lower 3 dB frequency
EC.40) A high – Q quartz crystal exhibits series resonance at the frequency ωsand parallel resonance at the frequency ωp. Then
(a)ωs is very close to, but less than ωp
(b)ωs << ωp
(c)ωs is very close to, but greater than ωp
( d.) ωs >> ωp
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