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Correct ANSWER is (b) Full rectified

Easiest EXPLANATION: The RIGHT diode CONDUCTS during the positive half of the input cycle and the left diode conducts during the negative half of the input cycle. Hence, the output will be a fully rectified WAVE.

Right CHOICE is (c) 5 V

To EXPLAIN: As is CLEAR from the circuit that v0 = 0.5 vi and hence peak value of the OUTPUT wave is equal to half of the peak value of the INPUT wave and hence Vp = 10/2 = 5 V.

The correct answer is (d) Differential Model

The best I can EXPLAIN: Differential model doesn’t DEFINE any diode. It DEFINES an OPERATIONAL AMPLIFIER.

The correct option is (a) 7 ohms

Best explanation: RAV is the average AC resistance, foundbetween two MARKED points in the I-V graph. Here rav=0.1/14mA=7Ω.

The correct option is (b) R1 > r2

The BEST I can explain: r1 > r2 as AC resistance is INVERSELY proportional to the diode current.

Right answer is (C) Simplified MODEL

Best explanation: In simplified model, the value of RD is neglected and hence, we get a high value of current for voltage greater than or EQUAL to VT.

Right answer is (d) -10 V

Explanation: -10 V. The OPERATING CHARACTERISTIC of a diode operating in the reverse bias region SUGGESTS that BEYOND a certain threshold CURRENT, the voltage across the diode is nearly constant = reverse breakdown voltage.

The correct answer is (C) 4.3 mA

Best explanation: 4.3 mA. The silicon DIODE in the circuit is forward biased. Considering a voltage DROP of 0.7 V across the diode, the voltage drop across the resistor is (5-0.7) V = 4.3 V. HENCE, Current in the circuit = Current through the resistor = (4.3/1000) A = 4.3 mA.

Right answer is (b) 8 μA

To ELABORATE: The reverse saturation current for a silicon diode DOUBLES its value for every 10 K RISE in temperature. Hence Is at 313 K=2 x 2 x 2 = 8 uA.

The correct OPTION is (a) True

For EXPLANATION I would say: During reverse bias operation of a diode, the width of the depletion region INCREASES leading to extremely high values of resistance and HENCE negligible values of CURRENT.

The correct answer is (a) V1 > V2

To explain I WOULD say: TEMPERATURE in India is greater than that in Antarctica. Hence, the REVERSE saturation CURRENT measured in India will be higher as compared to that measured in Antarctica. Now, as Is and VT are inversely related, V1 >V2.

Correct answer is (a) 10 mA

Explanation: As both the DIODES are reverse biased. Voltage DROP across the resistor = 10 V. Hence, current = 10 V/1 k = 10 mA.

Right choice is (c) 8.6 mA

Easiest EXPLANATION: Here, CURRENT flowing through R2 = (10-0.7)/1k = 9.3 mA. Also the current through the PARALLEL resistor=0.7 mA. HENCE the current through the forward biased diode = (9.3-0.7) = 8.6 mA.

The correct ANSWER is (a) 0 mA

To elaborate: As the DIODE D2 is reverse BIASED, the current flowing through it = 0 mA.

The correct ANSWER is (a) 9.3 mA

The explanation: The voltage across the diodes is 0.7 V as they are forward BIASED. Hence, the current through the series resistor = (10-0.7)/1K = 9.3 mA.

The CORRECT ANSWER is (b) 0.3 mA

For explanation: The source VOLTAGE initially increases from 0-10 V. As the source voltage increases, the voltage across the DIODES also increases. When the voltage across the diodes reaches a value of 0.3 V, the Germanium diode starts conducting, whereas the silicon diode is still in the off state. Hence, the voltage across R = 0.3 V. Hence, current = 0.3 mA.

Correct choice is (B) 9.7 mA

Best explanation: Here, the voltage ACROSS R=0.3 V, hence, current through R2 = (10-0.3) V/ 1 K = 9.7 mA.

The correct option is (d) 0 mA

Best EXPLANATION: The source voltage initially increases from 0-10 V. As the source voltage increases, the voltage across the diodes also increases. When the voltage across the diodes reaches a value of 0.3 V, the Germanium diode starts conducting, whereas the silicon diode is still in the off STATE. Hence, the voltage across silicon diode = 0.3 V and hence, the current ACCORDING to the approximate equivalent circuit model is equal to ZERO.

Correct option is (a) 9.4 mA

The BEST I can EXPLAIN: Current through R2 = 9.7 mA. Also, current through R = 0.3 mA, current through the silicon diode = 0 mA. Hence, from the KCL, we get the current through the germanium diode to be 9.4 mA.

The correct choice is (c) 0.4 mA

To explain I WOULD say: When the germanium diode is forward biased, current through R = 0.3 mA. When the Germanium diode is reverse biased, current through R = 0.7 mA. Hence, CHANGE in current = 0.4 mA.

The correct option is (c) ID2 > ID3; ID2 = ID1

Easiest explanation: Voltage across the parallel diodes = 0.3 V. HENCE, current through D3 = 0. Hence ID2 > ID3. Also, as both the parallel silicon diodes are in parallel and in off-state, the diodes 1 and 2 are in series and hence, current through them is same.

The CORRECT choice is (c) ID1 = ID4

Explanation: In the given circuit, the SILICON diodes are in off-state as the voltage ACROSS them is 0.3 V. HENCE, they act as open circuit and the diodes 1 and 4 are in series and hence the CURRENT through diodes 1 and 4 is same.

Right OPTION is (d) CRO

To explain I would say: All the methods may be USED to TEST a diode for its PROPER functioning except CRO.

The correct ANSWER is (c) Open Loop Indication

The EXPLANATION is: The diode CHECKING function on the meter when used causes a current of 2 mA to flow through the diode and HENCE under reverse bias, the result OBTAINED is the open loop indication.

The correct answer is (b) Diode is faulty

To ELABORATE: A diode allows electrical conduction in just one DIRECTION. Hence, if an instrument shows an open loop INDICATION in both the directions, then the diode is faulty.

Right option is (c) Diode is short-circuited

For explanation I would say: A proper diode GIVES a LOW RESISTANCE READING along one direction and high resistance reading along the other. HENCE, if diode gives low resistance readings in both directions, then the diode must be short-circuited. It’s noteworthy that a faulty diode gives a high resistance reading along both paths.

The correct choice is (a) Diode is faulty

The best EXPLANATION: A proper diode gives a low resistance READING along one direction and high resistance reading along the other. HENCE, if diode gives high resistance READINGS in both directions, then the diode MUST be faulty.

The CORRECT choice is (B) By changing the DOPING concentration of the diode

The explanation is: An increase in doping will lead to an increase in the concentration of IMPURITIES, which would further lead to a change in VZ and hence change in the Zener region.

Correct answer is (c) Ideal DIODE

To EXPLAIN I would say: The EQUIVALENT circuit diagram of a Zener diode doesn’t involve an ideal diode as it is used under reverse BIAS and an ideal diode doesn’t conduct under reverse bias.

Correct choice is (a) Silicon

The BEST EXPLANATION: Because of its high heat and current handling CAPACITY, Silicon is GENERALLY used for the MANUFACTURE of Zener diodes.

The correct choice is (b) GaAsP

The explanation is: In SILICON and GERMANIUM, the major CHUNK of energy is given off as heat the hence the emitted light isn’t significant. In Gallium ARSENIDE Phosphide (GaAsP), the number of photons are ENOUGH to create a visual source of light.

The CORRECT option is (b) 10^-9

To elaborate: The response time of a LED is typically in the ORDER of nanoseconds.

The correct choice is (c) 2.7 V

Best EXPLANATION: The SAFE and sufficient operating VOLTAGE range for LED is between 1.7 V to 3.3 V. Hence, 2.7 V is the best option.

The correct ANSWER is (d) No need for a heat sink in LONG RUN

To explain: The response time of a LED is typically in the order of nanoseconds. The operating voltage range for LED is between 1.7 V to 3.3 V which makes it compatible with solid-state circuits. Also, the semiconductor construction makes them RUGGED. However, in the long run, an adequate heat sink is required for longevity.

The correct CHOICE is (b) Electrically controls electron flow

Easy EXPLANATION: An active DEVICE is any type of circuit COMPONENT with the ability to electrically CONTROL electron flow (electricity controlling electricity). For a circuit to be properly called electronic, it must contain at least one active device.

Correct OPTION is (b) False

The best explanation: A LIGHT Emitting Diode WORKS just like any other diode and HENCE it does not conduct under REVERSE bias operation and hence the LED will not glow.

The correct answer is (C) Transformers, Inductors and Diodes

Easiest explanation: A passive device is any TYPE of circuit COMPONENT which cannot control the FLOW of electrons by MEANS of any electric control. All these devices do not have the ability to electrically control the flow of electrons.

The correct choice is (a) In the base region movement of charge carrier is because of the ELECTRONS which are minority charge carrier in the base region

The explanation is: In the base region movement of charge carrier is because of the electrons which is minority charge carrier in the base region. So, a BJT can said to be minority current CONTROLLED DEVICE. Base current flows between base and emitter in the BJT to induce a larger current flow between emitter and COLLECTOR.

Right choice is (a) Amplifiers

To explain I would SAY: Active devices may be EMPLOYED to a govern large amount of power by APPLICATION of small amount of power. This behaviour is known as amplification. Therefore, active devices can be USED as amplifiers.

Correct option is (d) They AMPLIFY the SIGNAL by taking an input from an external source

To ELABORATE: The Law of Conservation of Energy is not violated because the additional power is SUPPLIED by an external source, usually a DC battery or equivalent. The amplifier neither creates nor destroys energy, but merely reshapes it into the waveform desired.

The correct choice is (a) its NEGATIVE resistance region is used

Explanation: It is an active device since its impedance is positive and the V-I characteristics lie in the FIRST & SECOND quadrants, tunnel DIODES can be used as an active device even though it is a diode which falls under the category of passive DEVICES.

The correct choice is (b) False

The best I can EXPLAIN: Attenuators weaken or attenuate the high-level output of a SIGNAL generator and thus cannot control the SIGNALS electronically. This makes them passive devices and they can only be PRESENT in a circuit with an active device.

Right option is (C) Its unit less

Explanation: Gain is the RATIO of same type of values i.e. either volt/volt or current(amp)/current(amp) or watts/watts thus this MAKES it unit less. If it is expressed in DECIBELS i.e. on a logarithmic scale, then they have the unit DB.