Basic Electronics Engineering Test Questions Set 6

ANSWER: Covalent Bond

Explanation:
When the junction is heavily doped with narrow depletion layers, breakdown occurs at a particular voltage which ultimately results in enhancing the strength of electric field (10⁸ V/m ). This electric field has the potential to break the covalent bonds by formation of electron – hole pairs. This entire mechanism is supposed to be called as 'Zener Breakdown'. However, large number of current carriers can be produced by a gradual increase in the reverse voltage. As a result, the junction experiences low resistance in the break-down region.

ANSWER: Reverse Biasing

Explanation:
Transition capacitance mainly occurs when P-N junction is in reverse-biased mode. This is so because P and N regions possesses low resistance on individual sides and therefore act like the plates of capacitor. However, the property by which level of capacitance variation can be maintained by application of biasing can be utilized for the construction of a device known as 'Varicap' or 'Varactor' or 'voltacap'. Basically, the capacitance is voltage dependent and can be expressed as,

C_T = K / (V_k + V_R ) ⁿ

ANSWER: 500 times C_T

Explanation:
Typical value of transition or (Space-charge) capacitance is 40 pF while typical value of diffusion capacitance is 0.02 μF which is absolutely 500 times the value of transition capacitance. Diffusion capacitance occurs when the junction is forward biased. Diffusion or (Storage capacitance) is directly proportional to forward current & hence this capacitance finds a greater impact on the applications of devices where the necessity of switching arises from forward to reverse biased operating modes.

ANSWER: 694 kHz to 1.83 MHz

Explanation:
As per the circuit configuration,

C_{T (min)} = 10 /2 = 5 pF & C_{T (max)} = 70 /2 = 35 pF

f_{r (max)} = 1 / 2π (√LC )

= 1 / 2π (√15 x 10^-3 x 5 x 10 ^-12)

= 1.83 MHz

f_{r (min)} = 1 / 2π (√LC )

= 1 / 2π (√15 x 10^-3 x 35 x 10^-12)

= 694 kHz

ANSWER: All of the above

Explanation:
It is highly not feasible for any diode to act like an ideal diode because no practical diode resemble to the behaviour of an ideal diode. Basically, ideal diode acts as a short-circuit in forward direction and acts as an open-circuit in reverse direction.

An ideal diode exhibits conductivity with zero resistance in forward direction and infinite resistance in reverse-biased condition without any voltage drop since a short has zero resistance. However, the configuration of an ideal diode closed in forward direction and open in reverse direction can be used as a better bistable switch.

ANSWER: 1 A

Explanation:
According to the circuit configuration, the diode D1 seems to be an open switch and connected in reversed biasing. Hence, diode D1 does not conduct and therefore, the current does not flow through it across the resistance of 20 ohm.
Another diode D2 connected in parallel configuration is forward-biased and therefore it acts as a closed switch. Hence, it conducts the current across the 10 ohm resistor. The current flowing through the circuit can be evaluated by;

I = V / R ….....( By Ohm's Law)
= V / ( R1 + R2)
= 15 / ( 5 + 10)
= 15 / 15 = 1 A

Thus, 1A is the only current flowing through the above drawn circuit configuration.

ANSWER: Surface- emitting LEDs

Explanation:
By taking into consideration the constructional arrangement of surface-emitting LEDs, P-type layer gets deposited over the N-type substrate by diffusion process.

P-layer is kept on the uppermost portion because of the generation of carrier recombination in the same layer itself.

The operational strategy of surface emitting LED can be confined with an assistance of below drawn schematic view.





However, surface LEDs emit light in accordance to the requisite forward current. The quantity of transmitted output power has a direct relationship with the forward current.

This conceptual notion can be exemplified in the graph that exhibit the relationship between transmitted output power and the forward current.

It is obvious from the above graph that maximum the forward current, greater is the light output.

ANSWER: Both a & b

Explanation:
Seven segment display exists in two types namely common anode and common cathode.

Generally, all the cathodes of the diodes are connected together to ground in common cathode type.

The reason behind this connectivity is that it becomes feasible to light any segment by applying forward biasing to that specific LED.

Therefore, only +5 V supply is sufficient to anode so as to light the segments at this level.

On the contrary, all the anodes in common anode mode or type are together connected to +5V and hence, ground is supposed to light the segments individually.

Both the common modes of anode and cathode types are usually taken into account while generating the seven segment display.

ANSWER: Green

Explanation:
Multicolored LEDs are the form of blinking LEDs that mainly comprise two or three color LEDs. Tri-color LEDs resemble to an ordinary LED but it undergoes the emission of light in specifically red, green and yellow light depending on the type of operating stipulation. Tri-color LEDs consists of two leads which functionally act as anode and cathode respectively.



Whenever the flow of DC current is unidirectional, then LED emits red light but as soon as the direction of DC current flow changes in opposite direction, emission of green light is lucidly visible. Apart from these conditions, LED exhibit yellow color in correspondence to AC current.

ANSWER: Energized regions absorb the incident light by generating localized black display

Explanation:
Basically, LCD does not generate its own illumination but it is dependent on external source of illumination for better visual effect.

Among the two categories of LCD display, field-effect display results in an absorption of incident light after getting energized. Besides this, it leads to the formation of localized black display.

Field-effect display is only concerned with an absorption of incident light irrespective of the reflection, refraction and scattering mechanisms.

Conversely, dynamic scattering display itself clarifies the notion of scattering the light in all directions by making the molecules of display to be extravagantly disruptive.

As a result, activated areas ultimately are taken on frosted glass appearance by creating silver display.

ANSWER: Rs = 3 KΩ & I_d = 0.15mA .0.19 mA

Explanation:
Applying Ohm's law,

I = V / R
I = V / ( R+ r) since 'r' is a cell resistance
I = 40 / (R + r)
Therefore, we can write,

R = (40 / I) – r
When a photoconductive cell is illuminated,
R = ( 40/ 8x 10 ^-3 ) - 2x 10³
= 3 kΩ

Also, the dark current can be given by,
I_d = V / (Rs + R_D )
= 40 / (3 + 200) x 10³
= 0.19 x 10 ^-3 A
= 0.19 m A

ANSWER: Photo-conductive Cell

Explanation:
Photo-conductive Cell is a semiconductor device whose resistance is inversely proportional to intensity of light that falls on it.

Hence, it is also known as 'Photo-resistive Cell' since it works according to the phenomenon of photo-resistivity.

Conceptually, the resistivity of any semiconductor device is dependent on the number of free charge carriers present in it.

In the absence of illumination or light, the number of charge carriers are extremely less and resistivity is tremendously high.

On the contrary, the photons strikes the semiconductor in presence of light and then eventually transfer energy to it.

If the energy of photon exceeds the energy band gap of semiconductor, then liberation of free mobile charge carriers takes place which is also another reason in reducing the value of resistivity.

ANSWER: 528

Explanation:
It is essential to connect the solar cells in series to obtain requisite amount of voltage.
At the same time, it is also crucial to connect such groups in parallel combination for the provision of essential current.

It is almost a prior condition that the charging voltage must be greater than battery voltage.

Number of solar cells connected in series combination
= Solar bank Voltage /Continuous Current

Assuming that the solar voltage to be 16.5 V, we can write

No. of solar cells in series connection = 16 /2 = 8

Charge of battery within 24 hour duration = 24 x 2 = 48 Ah

It is mandatory for solar cells to supply the charge of 48 Ah over the similar duration.

Conceptually, solar cells has an ability to supply the current only under the influence of illumination.

This implies that current can be delivered only for 12 hours in every 24 hours.
Thus, the required charging current from solar cells = 48 Ah / 12 h
= 4 A
Consequently, total no. of groups in parallel combination equals to the ratio of output current to the cell current.

Total No. of groups = 4 / (60 x 10^-3)

= 66

As a result, the total number of solar cells necessary for an earth satellite

= No. of solar cells in series combination x total no. of groups of solar cells in parallel combination

= 8 x 66

= 528

ANSWER: Clipping Circuits

Explanation:
Clipping circuits are used to clip off the input waveform at different levels by simply varying the battery voltage as well as by interchanging position of several elements in the circuit.

These clipper circuits are distinctively used in devices where it is extremely essential to remove signal voltages above or below the specified voltage level. Hence, the clippers are also renowned as voltage (or current limiters ), amplitude selectors or slicers. It is almost referred as a slicer because the output comprises some portion or slice of input in the midst of two reference levels.

ANSWER: Rectangular

Explanation:
Basically, the sampling gate is a transmission circuit where the output is an exact reproductivity of an input waveform within a specified time interval and can also be zero otherwise.

The time interval for the transmission purpose is adopted with the support of an external signal called as control signal or gating signal. These control signals are usually defined in terms of rectangular shape.

However, sampling gates are most oftenly renowned as 'Transmission Gates' or 'Time Selection Circuits'.

ANSWER: Step Recovery Diode

Explanation:
Step Recovery Diode belongs to the category of Voltage Variable Capacitor (VVC) diode whose doping density reduces in the vicinity of junction.
This gives rise to the generation of strong electric field on both sides of the junction.

As usual, this diode also conducts in forward direction similar to other diodes.Thus, the reverse current flows during the reverse half-cycle because of draining of the stored charge and consequently the current suddenly falls to zero.

This kind of operational strategy resemble like a diode that has snapped open suddenly during an inceptional portion of reverse cycle.

As a result, step recovery diode is also known as 'Snap diode'. Also, the property of diode to exhibit sudden recovery from reverse ON current to reverse OFF current nominates it as a 'Step Recovery Diode'.

ANSWER: Avalanche Breakdown

Explanation:
Breakdown devices are specifically renowned by thyristor which act as a semiconductor switch whose bistable action is dependent on P-N-P-N regenerative feedback mechanism.

UJT, LASCR, Diac, Triac, SCR & SCS are some of the widely applicable breakdown devices whose working strategy is confined by the consequences of an avalanche breakdown despite of Zener and PN junction breakdown phenomena which occurs specifically in cases of Zener & PN junction diodes respectively.

These devices performs the function of latching since they consist of two or more junction & can be easily switched ON or OFF at instantaneously rapid speed.

ANSWER: large ac power with a small signal

Explanation:
Unijunction transistor comprises three terminals with a single PN junction diode & is different from conventional transistors that it does not undergo amplification. It possesses a negative resistance characteristic that plays a significant role to function it as an oscillator by controlling large ac power with a small signal.

ANSWER: Intrinsic Stand-off Ratio

Explanation:
Intrinsic stand-off ratio is said to have a value which is always supposed to be less than 1.

This is so because it specifies the voltage division factor which is usually a ratio of base resistances of first transistor to the additional sum of base resistances concerned to another transistor.

Stand-off ratio = Rb1 /(Rb1+ Rb2)

However, it can also be represented in terms of voltage values as,

V_A = stand-off ratio x V_BB

This eventually results in the voltage drop at the base resistance of another (second) transistor.

ANSWER: V_E decreases & I_E increases

Explanation:
When emitter voltage becomes exactly equal to peak voltage, the current (I_p ) flows through the resistance to ground causing the UJT to get turned ON.

It is obvious that as V_E & I_E rises beyond current, the values of base resistance and intrinsic stand-off ratio decreases.

Thus, the process becomes more regenerative by further reduction in emitter voltage since emitter current increases.

With an increase in emitter current , the emitter voltage decreases. Due to this, UJT possesses negative resistance .

Furthermore ,the consequences shows variations when UJT is in saturation mode beyond the valley point and therefore, emitter voltage increases.