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Right CHOICE is (a) Beta increases

Explanation: Input impedance of the BASE is GIVEN as= beta * rE.

Therefore, beta(GAIN ratio) is directly proportional input impedance of the base.

Correct option is (d) Reduce Distortion

To ELABORATE: Since the negative feedback to any amplifier reduces its OVERALL gain; hence any NOISE and distortion in the amplifier is ALSO reduced.

The correct CHOICE is (a) True

Best explanation: One of the MAIN functions of the coupling device in a differential amplifier with active load is to block the DIRECT current from passing through to the input of NEXT stage from the output of the first stage. Another function of the coupling device is to transfer the current from the output of first stage to the input of the SECOND stage.

Correct choice is (b) 4

For explanation I would SAY: There are four types of COUPLING MECHANISMS used. They are Resistance – CAPACITANCE coupling, thermal coupling, impedance coupling and TRANSFORMER coupling. Resistance – capacitance coupling is the most vastly used mechanism.

The correct answer is (b) only if the BASE current is MUCH smaller than the current through R2 (the lower bias RESISTOR)

Explanation: The input resistance can be NEGLECTED if the values of R1 and R2 are very large. Therefore the base current becomes very small.

Right choice is (C) VO = Ad × Vd + AC × VC

The best EXPLANATION: The differential GAIN of a differential amplifier with active load is defined as the gain OBTAINED at the output signal with respect to the difference in the input signals applied. To calculate the output voltage of a differential amplifier with active load we use Ad × Vd + AC × VC. Where, Ad is the differential gain and AC and VC represent the COMMON mode gain.

Right answer is (C) VC = 8.75 V

Easiest EXPLANATION: VCE=Vcc-IC Rc(VCE = VC , β= IC/IB => IC= β*IB)

Vc = Vcc – β* IB * Rc

=20-100x75x10^-6 x1.5x 10^-3

=8.75V.

The correct answer is (a) voltage-divider bias.

The best I can explain: Voltage divider biasing is commonly used because of the main reason that the transistor under this biasing always remains in the ACTIVE region.

In voltage divider biasing, the voltages at the transistor’s base, emitter and COLLECTOR all depend upon the external circuit i.e. the biasing resistors R1 and R2 whose valued are fixed thus variation with beta is not present here.

 HOWEVER, voltage divider biasing has the ADVANTAGE that its stability FACTOR is greater than that of collector feedback biasing, that’s why it is used.

Correct option is (d) halfway between cut off and saturation

The best I can explain: If Q-point is NEAR to saturation then positive clipping of input SIGNAL, and to cutoff then negative clipping of input signal, if IC is maximum then Q-point is in saturation region. Linear operation means OUTPUT varies according to input without any DISTORTION.

The correct choice is (b) REDUCE distortion

To elaborate: In a negative feedback amplifier, when emitter feedback resistance is INCREASED, the magnitude of the gain of the amplifier reduces;but remains stable. Since the gain reduces the distortion and the noise in the SIGNAL is also reduced; as the overall signal we GET at OUTPUT is the input signal multiplied by the gain of the amplifier.Therefore, feedback in an amplifier helps in reducing distortion.

Right answer is (c) Permanent/semi-permanent

For explanation: The EPOXY bond between the heat sink and component is semi-permanent/permanent. This is done to make the re-work very difficult and at TIMES impossible. The most TYPICAL DAMAGE CAUSED by rework is the separation of the component die heat spreader from its package.

Right OPTION is (d) both positive and NEGATIVE supply voltage

To EXPLAIN I would say: If a dual power supply is there, then it is the most useful, and PROVIDES zero bias voltage at the EMITTER or collector for load. The negative supply is used to forward-bias the emitter junction through emitter resistor. The positive supply is used to reverse-bias the collector junction. Also has good stability.

The correct OPTION is (a) Field controlled diode

To elaborate: The static induction THYRISTOR (SITHs) is a thyristor with a BURIED gate structure in which the gate electrodes are PLACED in n-base region.They are normally on-state, gate electrodes must be NEGATIVELY biased to hold-off state. It is a self controlled GTO like device. Hence it is sometimes called the field controlled diode.

Right choice is (a) Push-pull amplifier

Easiest explanation: In push-pull mode; one transistor CONDUCTS one HALF cycle and the other transistor conducts the other half cycle. The OUTPUT from both TRANSISTORS are then combined together to get a SCALED replica of the input.

Correct OPTION is (b) Infinity

To explain: Ideally the VALUE of input resistance must be infinite while the value of OUTPUT resistance must be zero in ORDER to YIELD maximum amplified voltage gain.

But practically, it can be implied that the voltage must have large value of input resistance and small value of output resistance.

The CORRECT option is (c) larger

The explanation: A SWAMPED amplifier reduces variations in voltage gain by increasing the ac resistance of the emitter circuit to get a stable voltage gain. It is ALSO referred as gain-stabilized amplifier. That’s why the INPUT impedance is larger.

The CORRECT option is (c) Negative feedback

To EXPLAIN I would say: A swamped amplifier reduces variations in voltage gain by INCREASING the ac resistance of the emitter CIRCUIT to get a stable voltage gain. As it involves reduction in gain, it is a negative feedback.

Right answer is (b) Low EFFICIENCY

To elaborate: RC coupled amplifiers have low voltage and power gain. It is because the low resistance presented by the INPUT of each STAGE to the PRECEDING stage decreases the effective load resistance and hence the gain. Thus its efficiency is reduced.

Correct answer is (a) 20W

Best explanation: MAXIMUM AC power is GIVEN as;

P0 =(VCEQ . ICEQ)/2

So the MINIMUM power rating USED should be,

10W * 2 = 20W.

Right answer is (B) DC switch

To explain I would say: THYRISTOR is a unidirectional DEVICE, that is it will only conduct current in ONE direction only, but unlike a DIODE, a thyristor can be operate as either ran open circuit switch or as a rectifying diode depending on how the thyristor gate is triggered.In other words, the thyristor can operate only in switching mode.

Correct answer is (a) Small

Explanation: A POWER amplifier has relatively small gain β, As COMPARED to voltage amplifier. Both current and voltage amplifiers have greater gain as compared to power amplifiers. Power AMPLIFIES HANDLE large signals as compared to voltage amplifiers. Power amplifiers have the following stages- A, B, C, D, AB.

The correct ANSWER is (b) Class A

The BEST I can explain: The efficiency of this TYPE of circuit is very low (less than 30%) and delivers small power OUTPUTS for a large drain on the DC power supply. A Class A amplifier stage passes the same load current even when no input signal is applied so large heat sinks are needed for the output transistors.

Correct choice is (d) Unimportant

To EXPLAIN: To make the voltage gain stable, a resistance rE is inserted in series with the emitter and THEREFORE emitter is no longer AC grounded.

Because of this the ac emitter current flows through rE and PRODUCES an ac voltage at the emitter. If rE is much greater than r’e (emitter diode resistance) almost the entire ac input signal appears at the emitter, and the emitter is bootstrapped to the base for ac as well as for dc.

Thus the EFFECTS of the emitter diode become unimportant in swamping amplifier.

Right OPTION is (b) Emitter feedback resistance

To explain I would SAY: In a negative feedback AMPLIFIER, when emitter feedback resistance is increased, the magnitude of the gain of the amplifier reduces;but remains STABLE. Since the gain reduces the distortion and the noise in the signal is also reduced; as the overall signal we get at OUTPUT is the input signal multiplied by the gain of the amplifier.

Correct choice is (d) Poor

For explanation I WOULD SAY: The transformer coupling has a poor frequency response. The gain VARIES considerably with frequency. The gain is constant only over small range of frequencies. Thus transformer coupling introduces frequency distortion; DUE to which its frequency response is poor.

The correct ANSWER is (a) Turn On

To elaborate: Pulse repetition rate (PRR) = 1/(2.5 – 10^3) = 0.4 ms = 400 µs.

Mark/space RATIO = 1/10.

Pulse width = 400/11 = 36.4 µs.

The SCR will ‘turn on’ as the pulse width is greater than SCR turn on time.

Right answer is (b) DRAIN, source, gate

Easy explanation: A power MOSFET has three terminals called Drain, source and gate in PLACE corresponding to the three terminals Collector, EMITTER and base for BJT.

The correct choice is (c) the JFET has a pn JUNCTION

The best explanation: A MOSFET has no pn junction, instead at the gate of the constituent there is an insulated SILICON dioxide layer that insulates from the channel.A JFET is a DEPLETION MODE transistor. A JFET has diode junction between the gate and the channel.

The correct choice is (b) 78.5%

The best I can explain: The conduction angle is 180° for a Class B amplifier. Since the active device is switched off for HALF the INPUT cycle, the active device dissipates less POWER and hence the efficiency is ENHANCED. Theoretical maximum efficiency of Class B power amplifier is 78.5%.

Right choice is (d) 2

The EXPLANATION: The total GAIN of a differential AMPLIFIER with active load is the ratio of the output of second stage amplifier to the input of first stage amplifier. Where, Voutput = 100V and Vinput = 50V.

AV = Voutput / Vinput = 100V / 50V = 2

Correct option is (a) Pin, straight pin, FLARED pin

For explanation: In general pin fins are CLASSIFIED as straight pin and flared pin. For the straight FIN it was 44 °C or 6 °C better than the pin fin.Pin fin heat sink performance is significantly better than straight fins when used in their intended application where the fluid flows axially ALONG the pins RATHER than only tangentially across the pins.

The correct choice is (d) 100A

Explanation: GENERALLY, anode CURRENT required of GTO is four TIMES of turn off current. So the amount of current required to turn off GTO is 400/4 = 100A.

Right choice is (c) Push-pull

To ELABORATE: When CLASS B amplifier is to be used with a pair of active devices (transistors); it is ARRANGED in a push-pull mode where ONE TRANSISTOR conducts one half cycle and the other transistor conducts the other half cycle.

Right CHOICE is (a) decrease

For explanation: The emitter bypass capacitor is connected in parallel to the emitter resistance. The use of this connection is to provide an ac ground at the emitter terminal of the transistor. This has the effect of increasing the circuit voltage gain.

The bypass capacitor is normally a high-value COMPONENT that provides LITTLE REACTANCE at the lowest circuit operating FREQUENCY. Therefore, in the absence of this bypass capacitor, the circuit voltage gain will reduce.

Right choice is (C) drain and source

Easy explanation: The field effect transistor is a three terminal device that is constructed with no PN-junctions WITHIN the main current carrying PATH between the Drain and the Source terminals. The current path between these two terminals is CALLED the “channel” which may be made of WHICHEVER a P-type or an N-type semiconductor material.

The CORRECT choice is (a) current to voltage convertor

The explanation is: A transconductance amplifier converts an input of voltage to an output of current. It is also CALLED a current to voltage CONVERTER or I to V converter.There is usually an additional input for a current to CONTROL the amplifier’s transconductance.

The correct CHOICE is (C) has an ac voltage

Best explanation: The resistance of the emitter diode r’e equals 25mV / IE and depends on the temperature. Any change in r’e will change the voltage gain in CE amplifier. In some applications, a change in voltage is acceptable. But in many applications a stable voltage gain is required.

To MAKE it stable, a resistance rE is inserted in series with the emitter and therefore emitter is no LONGER ac grounded.

Because of this the ac emitter current flows through rE and PRODUCES an ac voltage at the emitter. If rE is much greater than r’e almost the entire ac input signal appears at the emitter, and the emitter is bootstrapped to the base for ac as well as for dc.

The correct option is (b) Low INPUT RESISTANCE

The EXPLANATION is: The problem with a single operational differential amplifier with ACTIVE load is its low input resistance. This low input resistance results in a loss of a major component of the SIGNAL. Operational Amplifier is internally a Differential amplifier with active load with features like High Input Impedance, Low Output Impedance.

Right choice is (a) base bias

For explanation: In base bias IB, RB and IC are FIXED therefore it is also called fixed bias transistor. When the TEMPERATURE is vary then the Q point is also varied. Therefore base bias has no stabilization.

Correct choice is (c) Voltage divider bias

The explanation is: When the TRANSISTOR starts operating, temperature at the junction increases. Hence IC increases. As a result of which, Ie increases. DUE to this increase in Ie, voltage drop across Re increases. This reduces the forward voltage across the emitter. Thus Ib reduces.

We know that, for any value of ICO,IC = β * Ib +(1+ β) ICO

whereICO =reverse saturation current(increases with temperature)

β = gain

Therefore, βand ICO increases and at the same TIME there is decrease in Ib. Hence above equation confirms that IC can be maintained within limits. Thus the circuit is more thermally stable and the operating POINT is more stable.

Right option is (a) POSITIVE gate signal

The explanation is: A Gate TURN Off (GTO) like and SCR is a device four layer, three JUNCTION semiconductor device with three external terminal (anode, cathode, and gate). GTO can be turned ON and OFF by positive pulse or signal RESPECTIVELY, to the gate terminal.

Right option is (d) Environment Monitoring Systems

The explanation: Heat dissipation is an unavoidable by-product of electronic devices and circuits. LED performance and lifetime are STRONG functions of their temperature. Effective cooling is therefore essential. A case STUDY of a LED BASED down lighter shows an example of the calculations done in order to calculate the required heat sink necessary for the effective cooling of lighting system. Temporary heat sinks are sometimes used while soldering circuit boards, preventing excessive heat from damaging sensitive nearby electronics.

The correct CHOICE is (b) three layer, two junction device

To EXPLAIN I would say: A POWER BJT has three LAYERS, p-n-p or n-p-n forming two junctions.

p-n-p: two positive (p) layers and one negative (n) layers in between them.

n-p-n:two negative(n) layers and one positive(p) layers.

Correct choice is (c) because the COLLECTOR terminal is the critical terminal for heat DISSIPATION

To explain I would say: TRANSISTOR case is used for power TRANSISTORS as the power dissipated at their collector junction is large. If heat dissipation is not done, this will cause large increases in junction TEMPERATURE.

The CORRECT CHOICE is (b) CATHODE

Explanation: Electron-flow from the source terminal towards the drain terminal is influenced by an applied VOLTAGE. In vacuum tube, the source terminal of JFET corresponds to the cathode of vacuum tube.

Right choice is (d) More than one

Easy explanation: A multistage amplifier circuit affects the high input impedance of a common source stage combined with the input to OUTPUT ISOLATION of a common gate stage. A radio receiver has more than one stage of amplification because it is required to restore the CHARACTERISTICS of a radio signal over VARIOUS channels.

The correct option is (b) POWER MOSFET

For EXPLANATION: Power MOSFET has a low turn off TIME. So it can be operated in a frequency range of 1 to 10 MHz.