Basic Electronics Engineering Test Questions Set 7

ANSWER: Spike waveforms

Explanation:
Relaxation oscillator basically comprises UJT and a capacitor that gets charged through resistance by switching ON V_BB .

UJT fires when capacitor voltage reaches to the value of peak voltage in time 't_s' & immediately discharges till the voltage value decreases to lowermost value (Vv),

As a result, the device gets turned off and the capacitor begins to charge again and hence it leads to the repetition of cycle ceaselessly by generating sawtooth waveform across the capacitor.

In addition to this, the involvement of external resistances connected serially in the circuit results in the generation of spike waveforms despite of sawtooth, square or triangular ones.

ANSWER: 30 & 2.7 seconds respectively

Explanation:
According to the given data,

Time constant has a least value of about 50000 x 40 x10^-6 = 2 seconds

Considering the utilization of an entire potentiometer, the maximum value of this time constant is given by,
= (40 + 500) x 10³ x 40 x 10^-6
= 21.6 seconds

Maximum number of blade strokes per minute = 60 / 2 = 30 seconds

Minimum number of blade strokes per minute = 60 / 21.6 = 2.7 seconds

ANSWER: Anode voltage

Explanation:
Programmable UJT (PUT) is also PNPN device comprising four layers along with terminal connected to N-region adjacently to anode & also exhibit similar V-I characteristics as that of UJT.

This PN junction undergoes in forward biased mode whenever the value of anode voltage goes beyond the gate voltage by 0.7 V, by ultimately turning ON the PUT.

On the contrary, the PUT gets turned OFF when the value of anode voltage decreases below this level.

The condition at which the anode voltage crosses this level is an eventual stage of turning ON the PUT & also the level is termed as programmable level.

Besides this, gate biasing can be adjusted to any desired level with an assistance of externally connected voltage-divider circuit and hence it is completely independent of programmable level.

ANSWER: 53.33 mA

Explanation:
According to given data:

Gain of PNP transistor = a₁ = 0.3

Gain of NPN transistor = a₂ = 0.4

Gate current = I_G = 40 mA = 0.04 A

According to the formula of anode current,

Anode current = I_A = a₂ I_G / [1- (a₁ + a₂ )]
= (0.4 x 0.04 ) / [1 - (0.3 + 0.4)]
= 0.016 / 0.3
= 53.33 mA

ANSWER: Triac

Explanation:
Triac is a bi-directional device comprising 5 layers and two SCRs connected in parallel combination upside down with respect to each other. This implies that the anode of one SCR is connected to the cathode of the other and the associated gates are directly connected together. Hence, it is possible to trigger the triac into conduction by both positive and negative voltages at its anodes as well as with both positive and negative triggering pulses at its gate.

Control of ac power supplied to load by switching ON & OFF modes during positive and negative half cycles is the major application of triac but is limited to the use of ac supply frequencies upto 400 Hz since it consumes more duration to recover to OFF state.

ANSWER: 1.9 V

Explanation:
In accordance to the given data, it is mandatory to estimate the value of input voltage in such a manner that it should overcome the barrier voltage of 0.7 V.

Secondly, it should allow the current of 6 mA to flow through the resistor of about 200 Ω.

Hence, the value of input voltage can be estimated as,

Input Voltage = Gate to Cathode Voltage + (Gate Current x Resistance)

Thus, V_in = V_GC + I_G R
= 0.7 +( 6 x 10_-3 x 200 )
= 0.7 + (1200 x 10_-3 )
= 1.9 V

ANSWER: Rectifier

Explanation:
Rectification is the process of converting AC to DC & basically diode act as a rectifier. Rectifier is a circuit which comprises one or more diodes so as to convert ac voltage into pulsating dc voltage.
On the contrary, inverter plays a crucial role in converting dc to ac. Chopper performs the functioning of DC to DC conversion while clamper circuits introduces a DC level into an ac signal & usually referred as DC restorers.

ANSWER: 3.6% & 42.06%

Explanation:
Considering the half-wave rectifier circuit,

Voltage Regulation (V_R) = R₀ / R_L
But, R_o = emf resistance + forward resistance
= 0.4 + 0.15
= 0.55

V_R = R₀ / R_L
= 0.55 / 15 = 0.036 or 3.6 %

Efficiency (η) = P_dc / P_in = 40.6 / (1 + Ro / R_L )
= 40.6 (1 + 0.55 / 15)
= 42.06 %

ANSWER: Transformer Utilization Factor (TUF)

Explanation:
While designing any regulated or unregulated type of power supply, it is extravagantly essential to estimate the ratings of transformer in terms of Transformer Utilization Factor (TUF).

TUF can be evaluated on the basis of the quantity of power delivered to load as well as the type of rectifier circuit employed by it. Simply, it can be defined as :

TUF = dc power delivered to load / ac rating of transformer secondary

= P_dc / P_{ac . Rated}
It is slightly different from conversion efficiency. Conversion efficiency measures the ratio of output dc power to total amount of input power.

η = P_dc / P_in

Apart from these, peak inverse voltage is the maximum voltage that occurs across the rectifying diode in reverse direction; while ripple factor can be defined in terms of rms value of ac voltage to that of the dc load voltage.

ANSWER: All of the above

Explanation:
Full-wave bridge rectifier circuit consists of four diodes without the necessity of centre- tapped transformer winding. In addition to this, small transformers are sufficient to utilize for controlling the same output level despite of using large transformers as in case of full-wave rectifier circuits with two diodes. Less PIV rating per diode is another important characteristic feature that serves to be quite beneficiary.

However, it becomes more flexible and easier to perform the rectification level by preferring full wave bridge rectifiers especially for high voltage applications.

ANSWER: Reduction in magnitude of ripple voltage

Explanation:
Basically, a capacitor opposes the change in voltage. The main function of filter circuit is to convert a pulsating output from a rectifier into a steady dc level since it smoothens out the pulstations at the output. A bigger capacitor is more likely to minimize the content or magnitude of ripple voltage.

With an increase in capacitor size, the peak current in the diode and dc voltage towards its limiting value also increases whereas magnitude of ripple voltage along with the time of flow current pulse in the diode goes on decreasing.

ANSWER: Thyristor

Explanation:
Controlled rectification process allow to vary the output of rectifier by controlling the point in ac cycle where the circuit gets turned ON during specific firing conditions. Thyristor like SCR is specifically used to control ( or trigger) the level of conduction angle in the cycle under the proper firing conditions. Only under the application of control signal to turn on the rectifier at a particular point, the load current flows through an ac cycle.

However, an half wave rectifier remains in conduction phase for the remaining positive half-cycle.

Due to this, determination of firing point is possible by an angle of delay while the firing signal is applied by the control circuit. Increase in delay angle ultimately results in delay of conduction phase in the cycle along with the decrement in load current.

ANSWER: 3.240 W & 1.097 W respectively

Explanation:
> Given data:
V_m = 300 V , R_L = 2 kΩ

For θ = 450 , the power delivered can be given by,
P = V_dc x I_dc

But, for a half-wave controlled rectifier,

V_dc = V_m / 2Л (1 + cos θ )

= V_m / 2Л (1+ cos 45)

= V_m / 2Л ( 1+ 0.707)

V_dc = 0.27 V_m

= 0.27 x 300

V_dc = 81 V

I_dc = V_dc / R_L = 81 / 2000 = 40 mA

Therefore, P = V_dc x I_dc = 81 x 40 = 3240 mW = 3.240 W

Similarly, the power delivered at an angle θ = 900 can be estimated as,

V_dc = V_m / 2Л (1 + cos θ )

= V_m / 2Л (1 + cos 900)

= V_m / 2Л (1 + 0)

= V_m / 2Л

= 0.159 V_m

= 0.159 x 300

V_dc = 47.7 V

I_dc = V_dc / R_L = 47.7 / 2000 = 23 mA

Hence, P = V_dc x I_dc = 47.7 x 23 = 1097.1mW = 1.097 W

ANSWER: 285 V

Explanation:
According to given data,

T_ON = 20 ms = 20 x 10 ^-3 s

T_OFF = 15 ms = 15 x 10 ^-3 s

T = T_ON + T_OFF

T = (20 x 10 ^-3 + 15 x 10 ^-3 ) = 35 x 10^-3 s

Therefore, duty cycle of the chopper is given by,

T = T_ON / T = 20 x 10 ^-3 / 35 x 10^-3 = 0.57

Thus, average dc output voltage of chopper can be evaluated by,

V_L = V_dc x duty cycle = 500 x 0.57 = 285 V

ANSWER: Bleeder Resistor

Explanation:
Bleeder resistor is most often configured across the output of filter so as to improve or enhance the voltage regulation. This resistor results in initial voltage drop by acting as a pre-load on the supply. Therefore, small amount of additional voltage drop gets generated due to connection of real load in the circuit configuration.

Thus, it becomes easier to figure the difference between no-load and full-load voltage and ultimately reduce it by improving the regulation level. Bleeder resistor renders a path for filter capacitor for discharging purpose in the absence of power supply ( or when power supply is switched off) and hence it is generally referred as a 'Bleeder Resistor'.

One major advantage of the bleeder resistor is that it invigorates the capability to improve the filtering action by controlling minimum current through the choke.

ANSWER: C-L-C Filter

Explanation:
C-L-C filter (or Pi- filter) comprises one inductor and two capacitors connected in Л configuration. Hence, it is also known as capacitor input Л filter.

The function of filtering is more better as compared to L-C filter circuit. Generally, the connection of capacitor C1 is configured directly across the supply.

Due to this, the necessity of high pulse current arises because of the generation of load current to greater extent.

The presence of these high peak current pulses may prove to be pernicious to damage the rectifier diode.

However, it is obligatory for C-L-C filters to be conditionally used with low-current equipment.

ANSWER: Peak Current

Explanation:
The output voltage is always greater than the input voltage for an IC connected to external components under the category of step-up switching regulator configuration. Pulse – width and frequency of an oscillator used in regulatory circuit are controlled by timing capacitor (C_T) .

Variations in duty-cycle can be controlled only on the basis of desired peak current and completely independent of peak voltage. This is so because the necessary voltage required by an oscillator internally is generally obtained across the current sensing resistor. Eventually, the regulating action is in addition and accordance to the output produced by the variations exhibited in duty-cycle of an oscillator.

ANSWER: Current Limiter

Explanation:
The current sensing resistor sets base-to-emitter voltage of transistor in current-limiting circuit, by evaluating the difference between the voltage across the external resistor and voltage across limiting resistor due to opposite polarities. If the value of current through the external pass transistor increases beyond maximum value due to shorted output, then the value of voltage across the limiting resistor reaches to 0.7V by turning ON the current limiting transistor.

Hence, it becomes feasible for the current limiting transistor to conduct the current away from the external pass transistor as well as regulator. This ultimately results in an occurrence of thermal overload to some extent by shutting down the regulator. Factually, overall internal protection of IC voltage regulator from thermal overload is a major stage taken into consideration from designing point of view.

ANSWER: 31.49 V & 1.25 V

Explanation:
According to the equation of output voltage for IC voltage regulator,

V_out = V_ref [1+ (R2/ R1)] + I_ADJ R₂

Considering R2 at its minimum value = 0 ohm, the output voltage is evaluated by,

V_{out (min)} = 1.25 [1 + (0 / 210 )] + (98.5 x10^-6) x 0
= 1.25 V

For maximum value of R2 = 5 k ohm, the output voltage is,

V_{out (max)} = 1.25 [1 + (5 x 10³ / 210)] + (98.5 x 10^-6 ) x 5 x 10³

= 1.25 [1+ (5000 / 210)] + (98.5 x 10 ^{- 6}) x 5000
= 31 + 0.4925
= 31. 49 V

ANSWER: Step-Down Regulator

Explanation:
Non-inverting Regulator

The output voltage in step-down regulator mainly depends on input voltage and pulse-width. Pulse width refers to ON time of transistor. With an increase in ON-time relative to OFF-time, charging capacity of capacitor increases by simultaneously increasing the output voltage & vice-versa. Hence, a wide range of variation can be observed in output voltage by adjusting the duty-cycle ( the ratio of T_ON / T ) of the transistor by,

V_OUT = V_in (T_ON / T)

where, the 'T' represents the period of ON-OFF cycle of transistor & has a reciprocal relative significance with the frequency in the form of T = 1 / f

Also, Total T = T_ON + T_OFF & hence the ratio of T_ON / T is precisely known as 'Duty Cycle'.