Basic Electronics Engineering Test Questions Set 10

ANSWER: MF5 filter

Explanation:
All the fundamental types of filters like low-pass, high-pass, band-pass, notch & all-pass filters can be easily designed with the help of MF% filter & hence, it is regarded as a universal filter. This MF5 is a second order filter which has the proficiency in realizing shape-desired response in addition to Butterworth, Chebyshev, Bessel & Elliptic. MF5 is also renowned as sampled data filter. Its one of the characteristic feature is that it can be cascaded so as to provide steep attenuation slopes.

ANSWER: They affect signals at frequencies greater than one-half the sampling frequency

Explanation:
If a signal at frequencies greater than one-half the sampling frequency is applied to an input of sampled data systems, then it will get reflected to a frequency less than one-half the sampling frequency. This effect ultimately give rise to a phenomenon known as 'Aliasing Effect'. It is feasible to minimize or eliminate this effect by limiting the input signal spectrum to less than one half the sampling frequency.

ANSWER: Between cut-off & saturation at high frequency

Explanation:
Efficiency of a regulator can be increased by using series-pass transistor as a switch. The configuration arrangement of series switching regulator is in such a way that the series-pass transistor is allocated between cut-off and saturation region. This is so because it leads to the generation of square wave output representing pulse width modulation. Low pass LC filters the modulated square wave output so as to produce average DC output voltage. Since the output voltage is proportional to pulse width & frequency, the efficiency can be seen improved upto approximately 95%. These regulators find their applications in many kinds of circuits like push-pull, feed-forward and so on.

ANSWER: Series Dissipative Regulator

Explanation:
LM317 is an adjustable & series dissipative regulator as they exhibit simulation of variable resistance between input and load. However, the mode of functionality is approved to be linear. These linear regulator are specifically designed to control or regulate constant output voltage by dissipation of excessive power in the form of heat when the input voltage and load current shows variations for certain extent. However, LM317 are considered as 'Series Dissipative Regulators'. The conversion efficiency in series dissipative regulator is inversely proportional to input/output differential value.
Thus, they are applicable to medium current applications where it becomes possible to manage dissipated power with heat sinks.

ANSWER: Both a & b

Explanation:
Monolithic components like Schottky-barrier diodes & thin-film resistors forms the basis for 8038 function generator. The main characteristic of Schottky diode is that it can perform the rectification for signal frequencies exceeding 300MHz. Besides this, thin film resistors are mainly built -up by using Silicon or alumina.
A very thin layer is deposited over the substrate of thin film resistor comprising two layers of conductivity and resistivity. However, though they are expensive but provides better electrical properties as compared to thick or any other kinds of resistors in the monolithic package.

ANSWER: 8038 Function Generator

Explanation:
8038 function generator is a 14-pin DIP which satisfies all the above mentioned specifications. LED temperature indicator, 555 Timers & voltage regulators consists of different configuration and specifications. It mainly comprises two comparators, two buffers, a flip-flop along with sine converter. These entire specifications are taken into consideration from an overall operational functions including 14 different pins.
Generation of triangular wave takes place by means of alternate charging the capacitor from one current source & then discharging it with another. This generated triangular wave is applied to both the comparators & then buffering becomes possible to sine converter. As a result, flip-flops and comparators play a significant role in converting triangular wave to square wave & also triangular wave gets converted to sine wave by means of sine converter.

ANSWER: A3, B1, C4, D2

Explanation:
Classification of Integrated Circuits is based on the number of components mounted on the same chip inclusive of gates too. Number of components mounted on chip of Small-Scale Integration (SSI) are less than 10. If the number of components configured on a chip are less than 100, then the chip belongs to the category of Medium Scale Integration (MSI).
Apart from these, if the number of components goes on increasing the size of chip to about greater than 100, then the chip is associated with Large Scale Integration (LSI) whereas if the number of components exceeds than 1000, then the chip gets specified to the category of Very Large Scale Integration (VLSI).

ANSWER: Positive Scalar

Explanation:
The necessity of using non-inverting amplifier arises whenever it is essential to obtain an output which should be equal to input with product of positive constant. Hence, it is also renowned by 'Positive Scalar' and this amplifier circuit utilizes negative feedback but along with the provision of desired output in terms of product of input as well as positive constant.
Negative Scaling is especially referred to 'Inverting Amplifier'. Remaining options mentioned above are meant for perplexities & has nothing to deal with non-inverting amplifier.

ANSWER: Unity Follower

Explanation:
Unity Follower circuit provides buffering or can be also used as an isolation amplifier which has a provision to obtain a gain at unity despite the requisite of any phase reversal. Emitter follower circuit is also semantic to unity follower only with an exception of gain which is in very much proximity of being absolutely equal to 1 (unity level).
Unity Follower is supposed to be used as buffer since it becomes quite feasible to transfer the input voltage in the form of output voltage along with the prevention of load resistance from loading down to input source simultaneously. This becomes possible because the value of input resistance tends to infinity & output resistance equals to zero.

ANSWER: Due to reduction in an input voltage at inverting terminal to zero

Explanation:
The notion of virtual ground originates from the condition when input voltage at the inverting terminal is insisted to reach a very minute value approximately to almost zero value. Thus, the point at which inverting terminal find its connection is significantly at the ground voltage and therefore, it is regarded at the virtual ground level. The reason behind getting reduction of input voltage to zero is the positive potential attained by point A and existence of the output voltage at the same time. Again, some part of output is fed back to input terminal out of phase. However, the sum of these two voltages become equal to zero.

ANSWER: Addition of input and feedback currents at point 'A'

Explanation:
Summing point is basically a point where all input as well as output or (may be feedback) are summed together. This point simply performs summation of all incoming currents coming towards it. In figure, point 'A' is a summing point. Since the configuration of Op-amp is exhibited to be of negative feedback in above drawn figure, so some part of output is connected at the input through point 'A' is also taken into consideration for the purpose of algebraic sum evaluation.
Generally, the concept of summing point and virtual ground are explained simultaneously for the semantic configuration requirement.

ANSWER: Adder

Explanation:
In any adder circuitry of Op-amp, the output voltage must be equivalent to the summation of all input voltages where each input voltage is a multiple of constant gain factor. This circuit is usually more similar to inverting scalar type of amplifier with an exception of possessing more than one input. Therefore, output voltage of adder or summer Op-amp is phase inverted.
Referring to subtractor, integrator and differentiator amplifiers, the output voltage of subtractor deals with the difference between two input signals, while integrator generates its output voltage in terms of integral of input voltage whereas the output voltage of differentiator is proportional to change in input in accordance to time.

ANSWER: PUT

Explanation:
Programmable Unijunction Transistor (PUT) is a switching device adopted by Op-amp integrator in VCO, which is generally connected in parallel to feedback capacitor so as to allow the termination of each ramp by adjusting reset conditions of the circuit. Thus, PUT comprises three terminals like an anode, a cathode and a gate where the biasing level of gate is supposed to be positive corresponding to the cathode.
An extravagant increase in an anode voltage as compared to gate voltage can be figured out by approximately 0.7 V. As a result, the PUT gets turn ON and starts operating like diode in forward biased stipulation. Thus, PUT gets turned OFF due to decrease in value of anode voltage below this level.

ANSWER: To block all frequencies that fall within its bandwidth

Explanation:
Notch filter mainly consists of a low pass filter, a high pass filter and a summing amplifier. The circuit configuration is in such a manner that cut-off frequency assigned by low-pass filter is lesser in value as compared to the cut-off frequency allocated by high-pass filter. Therefore, the gap between these both cut-off frequencies represents the bandwidth of the filter.
If the input frequency is lower than cut-off frequency of high-pass filter then output of high pass filter becomes zero. Hence, the total output of the filter will be equal to the output of low pass filter.
On the contrary, if the input frequency is higher than cut-off frequency of high pass filter, then output of low-pass filter tends to zero & hence, the net output under this condition will be definitely equal to the output of high pass filter.
As a result, it is perspicuous that notch filter permits to pass the frequencies below the cut-off frequency of low-pass filter and those above the cut-off frequency of high-pass filter, which can be termed as a crucial function or purpose of the notch filter.

ANSWER: All of the above

Explanation:
Designing of Op-amps is very crucial aspect since they need to be designed in such a way that external components should be connected to terminals so as to diverge or vary the external characteristics of Op-amps.
According to the technique of standard dual-in-line package, the ICs of Op-amp are sealed hermetically along with the two-sided packaging. Also, TO-5 case is well-known for 'Transistor Outline' and is concerned to a series of technical specifications. TO-5 type of packages belong to the header family metal case with the pitch of about 0.20'' via 90°.
Flat pack category of packaging mainly deals with PCB surface mount component packaging in the form of glass, metal or ceramics & hence provide greater defence to hermetic seals in order to prevent them from corrosion and moisture. Flat-pack type of packages are lighter and smaller as compared to the conventional round types of TO-5 cases.
All the three types mentioned above are highly confined for packaging dealt with Op-amps.

ANSWER: μ A 709

Explanation:
CA 741 CT & CA 741 T are the Op-amps developed or manufactured by Bharat Electronics Ltd (BEL), Bangalore. These are high-gain operational amplifiers which are absolutely applicable in summer, multivibrator, integrator, differentiator and so on.
μ A 709 is also a high-gain operational amplifier which is constructed on a single chip of silicon by an assistance of planar epitaxial mechanism by Semiconductors Limited, Pune. It's major areas of applications include DC servo systems, analog computers with high impedance values and low-level instrumentation prospects.

ANSWER: -38.3V

Explanation:
Given data:
V1 = -10V, R1 = 600k
V2 = 10V, R2 = 300K
V3 = 5V, R3 = Rf = 2M

The output voltage of an inverting adder or summer circuit is evaluated by,

V_o = -(K₁V₁ + K₂V₂ + K₃V₃)

where, 'K' represents the constant gain factor.
K₁ = R_f / R₁ = 2M/ 600K = 2000K/ 600K = 3.33
K₂ = Rf / R2 = 2M/300K = 2000K/ 300K = 6.66
K₃ = R_f / R₃ = 2M/ 2M = 1

Therefore, V_o = -(K₁V₁ + K₂V₂ + K₃V₃)
= -[3.33 x (-10) + 6.66 x (10) + 1 x 5]
= -[-33.3 + 66.6+ 5]
= -38.3 V

Hence, the final output value of voltage of an inverting amplifier is nothing but summation of all input voltages estimated to be in terms of negative voltage of about -38.3 V.

ANSWER: Phase shift & magnitude (gain)

Explanation:
It is the fundamental property of an amplifier that the value of gain (magnitude) goes on decreasing while phase-shift between input and output goes on increasing whenever there is rise in an operating frequency.
Change in the gain of an Op-amp due to variation in frequency is graphically estimated by 'Magnitude Plot', whereas change in phase with respect to variation in phase-shift is evaluated in terms of 'Phase Angle Plot'.
However, modification in an op-amp performance can be handled as per the requisite level by controlling the level of gain and phase-shift of the compensating networks.

ANSWER: Both a & b

Explanation:
Slew rate is the maximum rate of change of output voltage with respect to time. For an instance, suppose that if the slew rate appears to be 1V/microseconds, then the output rises or falls at the speed rating of 1V/microseconds. Slew rates are also referred as output voltage swing as a function of frequency or as a voltage follower large signal pulse response. Also, the slew rate has an inverse relation to temperature. As the temperature increases, the slew rate decreases.
Since slew rate belongs to the category of large signal phenomenon, current limiting and the saturation of intrinsic stages of op-amp leads to reasons of slew rate to greater extent. However, when a frequency & amplitude signal is applied, the resulting current is also at extravagant level so as to charge the network of capacitance compensation. It is obvious that charging & discharging of capacitor needs some finite amount of time and therefore, the internal capacitors are solely responsible in preventing the output voltage to respond instantaneously to the fast varying input.
Hence, capacitor charging rate affects the current limiting in slew rate, which indicates that the voltage across the capacitor is nothing but the output voltage.

ANSWER: Bandwidth

Explanation:
The significant AC parameters of Op-amp are transient response, bandwidth and slew rate but all of them have different results while becoming excessive. The transient response affects the settling time while slew rate give rise to distortion if exceeded. As a result, bandwidth reduces the output voltage when it is exceeded.
Since bandwidth is a small signal phenomenon, it represents the band of frequencies for which the gain remains constant & also highly dependable on compensating components including closed loop gain. But, it exhibits adverse consequences in the reduction of an output voltage when exceeded to greater extent.

ANSWER: In series with each emitter

Explanation:
External resistor of a differential amplifier is connected in series combination with each emitter. Due to this configuration, the dependability of voltage gain which varies in accordance to changes in emitter resistance can be ablated. Apart from this advantage, external resistor also possesses a beneficiary notion of increasing the linearity range of differential amplifiers.
However, the value of external resistor has an efficiently high value to swamp the consequences of emitter resistor of amplifier. As a result, this external resistor is also renowned as 'Swamping Resistor' because of its swamping strategy.

ANSWER: CA3086

Explanation:
CA3086 play a significant role in proper execution or operation of the circuit during the construction of mirror circuit. Here, the diode gets formed with an assistance of transistors placed adjacently. However, this type of transistor array ultimately leads to acquire the thermal stability and the requisite amount of collector current in a current mirror circuit.
Due to simplicity in fabrication and operation, current mirror circuit have wide range of applications in the aspects of differential and operational amplifiers.

ANSWER: 10.99 & 36.36 kHz

Explanation:
For non-inverting mode of an amplifier (voltage series feedback amplifier)

1) A_f = AB / 1 + AB
But, B = R1 / (R1 + Rf) = 2 k / (2 k + 20 k)
= 1 / 11
1 + AB = 1 + 400000 x (1 / 11)
= 36364.6
A_f = 400000 / 3636.4 = 10.99

2) f_F = (1 + AB) f_o = (36364.6) x 10Hz = 363646 = 36.36 KHz

ANSWER: R2, R3 = 33k Ω, C1 = C2 = 0.0047μ F, R1 = 27k Ω & Rf = 15.8k Ω

Explanation:
Since the value of high cut- off frequency is mentioned. Below enlisted are the steps to evaluate filter components of second order low pass filter for designing purpose.

Step 1 : Assume R2 = R3 = R & C1 = C2 = C.

Step 2 : Select the value of C = 1μ F
Suppose that C1 = C2 = C = 0.0047μ F

Step 3 : Determine R by using formula R = 1 / 2 π f_H C
R2 = R3 = 1 / (2 π x 10³ x 47 x 10^-10)
= 33.86k Ω
(consider the round figure value i.e. 33k Ω
As we know that Rf should be equal to 0.586 x R1, then R1 = 27k Ω)
Thus, Rf = 0.586 x 27 = 15.86k Ω
By taking into consideration the pot value of about 20k Ω, estimated resistances are
R2 = R3 = 33k Ω
C2 = C3 = 0.0047μ F
R1 = 27k Ω & Rf = 15.8k Ω

Step 4 : Since the values of resistors and capacitors are equal , the passband voltage gain must be equal to A_f = 1 + (R_f / R1) 1.586
However, the gain must satisfy the Butterworth response & hence, the value of R1 must be selected = 100k Ω in order to estimate the value of Rf.

ANSWER: -42000 π cos 2000 πt

Explanation:
Given data:
Voltage Peak Value = 7mV
Frequency = f = 1KHz
R = 1500K Ω
C = 2μF = 2 x 10^-6 F

An input voltage equation of a differentiator circuit can be expressed by,

V₁ = Asin2 πft
= 7 sin2π x 1000 t
= 7 sin2000π t mV

Scale Factor = -CR = - [2 x 10^-6 x 15 x 10⁵]
= - [30 x 10^-1 ]
= -3
Hence, the output voltage of differentiator circuit can be evaluated as:

Vo = -3 d / dt [7 sin 2000 π t]
= -3 x 7 x 2000 cos 2000 π t
= -42000 π cos 2000 π t mV.

ANSWER: 47.56 kHz & 20 Vpp

Explanation:
In accordance to given data, the ac input resistance of the amplifier can be given by,

R_if = (R2)¦(R3)¦ [R_i (1+AB)]

When [R_i (1+AB)] >> R2 or R3 then, the above equation becomes

R_if ≈ (R2)¦(R3)
≈ 200K ¦ 200K
= 100K Ω
The low frequency cut-off limit can be exaggerated using

f_L = 1 / [2 πC_i(R_iF + R₀)]

= 1 / [2 π (0.1 x 10 -6) (100K +40 )]
= 62.8 Hz
The gain of an amplifier is = 1 + (Rf / R1)
= 1 + (2M / 100K)
= 21

From the formula of Unity gain bandwidth,
UGB = Af x FH
Therefore, fH = UGB / Af = 1 M / 21 = 47.61 Khz

Bandwidth = FH – FL = 47.61 Khz – 0.0476 KHz = 47.56 KHz &
Ideal Maximum Output Voltage Swing = +Vcc = +20Vpp.

ANSWER: 0.3 mA & 4V

Explanation:
Given data :
From the circuit, the given values are:
V_in = 10V, R = 30K & V₁ = 2V

We know that load current depends on input voltage resistance R in case of voltage to current converter.

Thus, I_L = V_in / R
= 10V / 30K
= 0.3 mA

However, the output voltage (V_o) of voltage to current converter is given by,

V_o = 2 V₁ = 2 x 2 = 4V.

ANSWER: 30.47 kHz & 10 kHz

Explanation:
(a) By applying voltage -divider rule, initial voltage at terminal 5 is given by,

V_c = (20K) (15) / 11.5K = 26.09 V
Thus, the frequency of output waveform (nominal frequency) is approximated by,

f_o ≈ 2(+V - V_c) / R₁C₁ (+V)
= 2 (15 – 10.43) / (20K x 10^-9) x (15)
= 30.47 kHz

(b) Thus, the modulation in the output frequencies can be calculated by initially substituting V_c = 10V & then 11.5 V

f_o = 2 (15 – 10) / 20K x 10^-9 x 15
= 33.34 kHz

f_o = 2 (15 – 11.5) / 20K x 10^-9 x 15
= 23.34 kHz

Hence, the variation in an output frequency can be evaluated by the difference between these both frequencies.
i.e. 33.34 - 23.34 = 10 kHz.

ANSWER: Channel Separation

Explanation:
Channel separation is a measure of the amount of electrical coupling between the op-amps which are integrated on a semantic chip. This parameter is specifically highlighted in datasheet specifications of dual & quad Op-amps like μAF772 & μAF774 respectively.
Whenever there is an application of signal to input of only one op-amp, then some of the other signal appears on the output confined level of other op-amps due to physical vicinity of op-amps in dual and quad form of packages.
However, the amplitudes of all these signal are almost similar and therefore, can be evaluated with an assistance of channel separation. This conceptual characteristic of parameter specification & mechanism is supposed to be referred as 'amplifier to amplifier coupling'.

ANSWER: All of the above

Explanation:
No explanation is available for this question!

ANSWER: Emitter

Explanation:
No explanation is available for this question!

ANSWER: For maximum power dissipation

Explanation:
No explanation is available for this question!

ANSWER: A & B

Explanation:
No explanation is available for this question!

ANSWER: I_E = I_B + I_C

Explanation:
No explanation is available for this question!

ANSWER: Leakage Current

Explanation:
No explanation is available for this question!

ANSWER: emitter is positive w.r.t. base

Explanation:
No explanation is available for this question!

ANSWER: Alpha (α_dc)

Explanation:
No explanation is available for this question!

ANSWER: CE

Explanation:
No explanation is available for this question!

ANSWER: h_fe

Explanation:
No explanation is available for this question!

ANSWER: 6.01 mA

Explanation:
No explanation is available for this question!