BDL MT Electronics Mock Test - 2 - Electronics and Communication Engineering (ECE) MCQ

High electron mobility transistor:

• The High Electron Mobility Transistor is a form of Field Effect Transistor (FET) that uses a junction between two materials with different bandgaps (i.e. a heterojunction) as the channel.
• The most common material used is aluminum gallium arsenide (AlGaAS) and gallium arsenide (GaAs)
• Gallium Arsenide (GaAs) is generally used because it provides high-level basic electron mobility.
• HEMT is used in microwave circuits as it provides low noise and high gain.
• HEMT has a high gain at microwave frequencies because the charge carriers are almost exclusively the majority carriers. Minority carriers are not significantly involved.

Accelerometer chip:
These are devices that convert acceleration into electrical signals.
MEM's devices:
Micro-electromechanical devices involve the conversion of mechanical energy into electrical energy and vice-versa.
Optoelectronic devices:

• Optoelectronics deals with the science and technology of electronic devices which involves interaction with light.
• Optoelectronic devices include LED which involves the emission of light when current is passed through a forward bias LED.

FET:
1. As the input circuit of FET is reverse biased, FET exhibits a much higher input impedance ( in the order of 100 MΩ) and lower output impedance and there will be a high degree of isolation between input and output.
2. The FET (field – effect transistor) is a majority carrier (current flows due to majority carrier i.e. electron for N-type FET and hole for p-type FET) device. It is also called a unipolar transistor
3. They are also considered low–power consumption configuration with a good frequency range and minimal size and weight. FET is a voltage control device.

Explanation
FETs are majority-charge-carrier devices, in which the current is carried predominantly by majority carriers. So statement 1 is correct and statement 4 is incorrect.
In FET gate source is reverse biased so its input impedance is high. Due to high input impedance in FET gate current is zero. So statement 2 is correct.
Now, For FETs

Cg_s = gate to source capacitance
Cg_d = gate to drain capacitance
FET has low transconductance (g_m) in comparison to BJT that’s why FET is not suitable for high frequency. So statement 3) is incorrect.
In summary → FET → low value of g_m → f_T ↓
At high-frequency f > f_T then FET will not provide amplification.

IMPORTANT POINTS ABOUT FET:

1. The field-effect transistor (FET) is a type of transistor that uses an electric field to control the flow of current.
2. FETs are also known as unipolar transistors since they involve single-carrier-type operations.
3. FET has better thermal stability than a BJT.
4. FET is less noisy.
5. No minority carrier in FET.
6. FET offers a large Bandwidth but small gain.
7. FET is a voltage-controlled device.

The current flowing through 5Ω resistance is

I = 50/5 = 10 A 

Applying KVL in loop (1) we get

∴ KVL is violated
∴ It is not possible to find the power

The bode magnitude plot is shown

At resonant frequency, (ωp), magnitude is maximum and hence slope is 0.

Concept:
Another number system to Decimal
In each and every representation of numbers with different bases, the maximum value in a number system with the base ‘r’ is r – 1. Since numbers vary from 0 to r – 1.
To convert any number which is in the different base to decimal number system we use binary-weighted representation.
Eg: let the number be ( abc⋯ ⋯ yz)r
Now to convert the above number into the decimal number system
a × rn-1 + b × rn-2 + ⋯ ⋯ + y × r1 + z × r0
If we convert all numbers into decimal then we can perform normal addition and subtraction etc.

Application:
Given:
(110)_x = (132)₄,
The decimal equivalent of this number will be:
1 x x² + 1 x x¹ + 0 x x⁰ = 1 x 4² + 3 x 4 + 2 x 4⁰
x² + x + 0 = 16 + 12 + 2
x² + x - 30 = 0
On solving this quadratic equation we'll get:
x = 5, -6
Base can't be a negative so;
x = 5

Combinational Logic circuits are circuits for which the present output depends only on the present input, i.e. there is no memory element to store the past output.

A combinational circuit can have ‘n’ number of inputs and ‘m’ number of outputs as shown:

Combinational circuits are:

• Multiplexer/Demultiplexer
• Encoder/Decoder
• Adders
• Subtractors
• Code Converters

Important Point:

In a sequential circuit, the output depends on both the present and the past values. The circuit diagram is as shown:

Examples of sequential circuits:

• Shift Registers
• Flip flops
• Counters

Concept:
If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.


Similarly, when we pass the input signal into an n-bit flip flop counter, the output frequency (fout) will be:

Application:
Given Input frequency f = 100 Hz

fout = 12.5 Hz

Resolution is the smallest change in the analog output for a change in digital input, also referred to as step size
Resolution =
Given:

⇒ 2^N - 1 = 1000/4 = 250
⇒ 2^N = 251
N ≈ 8

JEET is considered as a voltage-controlled device because the drain current is controlled by the gate voltage.
Explanation:
JFET:

• It is affiliated as a junction field-effect transistor
• The types of JFET are n-channel FET and p-channel FET
• A p-type material is added to the n-type substrate in n-channel FET, whereas an n-type material is added to the p-type substrate in p-channel FET.

• JFET is a three-terminal device and since gate voltage controls the drain current, JFET is called a voltage-controlled device.
• JFET’s have only a depletion mode of operation.
• When there is no potential applied between gate and source terminals and a potential VDD is applied between drain and source, then a current I0 flows from drain to source terminals at its maximum as the channel width is more.
• When the voltage applied between gate and source terminal VGG is reverse biased, this decreases the depletion width, thus the layers grow and the cross-section of the channel decreases and hence the drain current ID also decreases.
• Thus drain current in JFET is controlled by changing the channel width.
• Gate source p-n junction is always reverse biased because if it is forward then all the channel current will flow to the Gate and not to the source, ultimately damaging JFET.

Tunnel Diode:

• Symbol:

• It is a highly doped PN Junction diode, used for low voltage high-frequency switching applications.
• It works on the tunneling principle.
• It has a high charge carrier velocity.
• When compared to a normal p-n junction diode, the tunnel diode has a narrow depletion width.
• In normal forward-biased operation, it exhibits the “Negative resistance region” as shown:

• The negative differential resistance in their operation, allows them to be used as oscillators, amplifiers, and switching circuits.
• Their low capacitance allows them to function at microwave frequencies.
• Microwave frequencies range between 109 Hz (1 GHz) to 1000 GHz
• Tunnel diodes are not good rectifiers, as they have relatively high leakage current when reverse biased.

The feedback amplification factor is given by A _f = A/1 + Aβ
where A is open-loop gain and βA is loop gain.
As feedback increases the gain decreases thereby bandwidth increases.
The negative feedback in amplifiers causes

1. reduced the voltage gain and increases the stability in gain
2. increases the bandwidth by the factor (1 + Aβ) to maintain constant gain-bandwidth product
3. Reduces the distortion and noise in the amplifier
4. but the signal to noise ratio is not affected

Explanation:
1. A Darlington pair configuration has high input impedance and low output impedance.
2. Thus it is used as a buffer circuit for impedance matching with the circuits that require high input impedance at the input.
3. In addition to providing high input impedance, the Darlington pair also provides high current gain
Hence a Darlington emitter follower circuit is sometimes used in the output stage of the TTL gate to increase speed.
Note:
Darlington Pair: CC – CC

• A Darlington pair is a two-transistor circuit with the emitter of one transistor is connected to the base of other transistors, while both collector terminals are connected to the common terminal
• It has a high current gain β; (equal to the product of the current gain of individual transistors) and is useful in applications where current amplification or switching is required.
• It also has high input impedance and low output impedance.
• The current gain of a Darlington pair in common emitter configuration is approximate β2

The symbolic diagram of the Darlington pair is shown below

The frequency spectrum of an Amplitude Modulated wave is given by:

Bandwidth in AM = 2f_m = 6 kHz
f_m = 3 kHz
Upper sideband = f_c + f_m = 600 kHz
∴ The carrier frequency is:
f_c = 600 – f_m = 600 - 3 kHz
f_c = 597 kHz

Derivation:
Given, In FSK modulation, let f1, f2 be frequencies for bit ‘0’
s (t) = A sin (2πf1t)
For bit ‘1’
s (t) = A sin (2πf2t)
Let Tb be bit duration
For both waveforms to be orthogonal,

It is possible if both 2π(f1 - f2)Tb and 2π(f1 + f2)Tb are integral multiplies of π(i.e. = nπ), i.e.

Application:
Given, f1 = 10 kHz and f2 = 25 kHz
15 kHz = n/t_b and 35 kHz = m/t_b
It is possible for 1/t_b = 5kHZ , with minimum m and n value, i.e.
i.e. Tb = 200 μs. (minimum value)

HL contains 0107 H and continents of 0107H is 20 H.
Thus after execution of SUB the data of A is 20 H – 20 H = 0.

ALE: (Address Latch Enable):

• It is an active high output signal.
• It indicates that the bits on AD7 - AD0 are address bits
• This signal is used to latch the lower order address from the multiplexed bus and generate a separate set of eight address lines A7 – A0
• It is provided by the microprocessor to latch the address into the 8282 / 8283 address latch.
• It is a high pulse active during T1 of any machine cycle. The T1 state is sufficient enough to latch the address on the multiplexed AD0 - AD15 lines.
• Whenever the processor sends a valid address on the multiplexed lines it also makes the ALE High.
• ALE is never floated.
• ALE is used to demultiplexed/multiplexed data buses and address buses.

Note:
The 8085 microprocessor signals can be classified in various groups, namely:

1. address bus,
2. data bus
3. control bus
4.  status signals,
5. externally initiated signals
6. acknowledgement, power and frequency,
7. serial I/O signals.

Concept:

Poisson distribution is applied when the number of trials is very large and the probability of success is small.

Calculation:

Let's say we have an event E such that the success of the event is “Ringing a call at a time t₀” and failure is “Not ringing a call at time t₀”.

Since the total number of success or failure of the event is unknown. Therefore, we can say that the event is a random variable.

1 hour = 3600 sec = 3600000 msec = 3.6 × 10⁹ μsec

The event can occur a large number of times for any given time interval and the probability of success is very less.

Explanation:
The following set of equations can be written in the matrix form
AX = 0

Where,

Performing row operations,
R₂ → R₂ -(3/2)R₁
R₃ → R₃ -(R₁/2)

R₃ → R₃ -(R₂/5)

∴ Rank (A) = 2, which is less than the number of variables.
∴ The solution is consistent and there are infinite numbers of non-trivial solutions.

Concept:
If real valued function f (x)
(i) is continuous in [a, b]
(ii) is differentiable on (a, b)
(iii) f'(c) =  [f (b) - f(a)]/(b - a)
Calculation:
(i) is continuous in [a, b]
(ii) is differentiable on (a, b)
f (x) = x (x - 4)
f (a) = f (1) = 1 (1 - 4) = 1 (-3) = -3
f (b) = f (4) = 4 (4 - 4) = 4 0 = 0
f'(c) = [f (b) - f (a)]/(b - a) = [0- (-3)]/(4 - 1) = 3/3 = 1
As per the mean value theorem statement, there is a point c ∈ (1, 4) such that f'(c) = [f (b) – f (a)]/ (b – a), i.e. f'(c) = 1.
2c - 4 = 1

2c = 1 + 4

c = 5/2

Given:

The length and breadth of a room is 16 m and 12 m respectively

The area of four walls = 784 m²

Formula used:

Area of four walls = 2(L + B) × H

Where,

L = Length of the room

B = Breadth of the room

And, H = Height of the room

Calculation:

According to the formula, we have

{2(16 + 12) × H} = 784 m2

⇒ 2 × 28 × H = 784

⇒ H = (784/56)

⇒ 14 m

∴ The height of the room is 14 m

Other Related Points

In a room, the area of two walls consists of length and height, and the other two wall's area consists of breadth and height.

Shortcut TrickP ∝ I
P ∝ T
Now,
P ∝ (I × T)

⇒ Profit = 9 ∶ 6 ∶ 4
Total profit = Rs. 5700
⇒ 19 units ≡ Rs. 5700
⇒ 1 unit ≡ Rs. 300
For A,
⇒ 9 units ≡ Rs. 2700
∴ The profit share of A is Rs. 2700.Alternate Method
Given:
Investment of A = Rs. 15000, Time = 12 monthsInvestment of B = Rs. 12000, Time = 10 months
Investment of C = Rs. 10000, Time = 8 months
Concept used:
1) P ∝ I
2) P ∝ T
Where, P = Profit, I = Investment, T = Time
Calculation:
Investment of A × Time ∶ Investment of B × Time ∶ Investment of C × Time
⇒ (15000 × 12) ∶ (12000 × 10) ∶ (10000 × 8)
⇒ (15 × 12) ∶ (12 × 10) ∶ (10 × 8)
⇒ 9 ∶ 6 ∶ 4
The ratio of the profit is directly proportional to the multiplication of investment to time.
⇒ P_A ∶ P_B ∶ P_c = 9 ∶ 6 ∶ 4
The total profit after 1 year = Rs. 5700
⇒ 9k + 6k + 4k ≡ Rs. 5700
⇒ 19k ≡ Rs. 5700
⇒ k ≡ Rs. 300
For 'A'
⇒ 9k ≡ Rs. 2700
∴ The profit share of A is Rs. 2700.
Important Point
The ratio of the profit is directly proportional to the multiplication of investment to time.

Calculation:

In 20 litres of mixture

Quantity of milk

⇒ (3/4) × 20 = 15 litres

Quantity of water

⇒ (1/4) × 20 = 5 litres

Let the amount of milk to be added is x litres

⇒

⇒ 15 + x = 20

⇒ x = 5 litres.

∴ The answer is 5 litres.

Given:

Selling price is doubled then profit is tripled

Formula used:

Profit = SP - CP

Profit % = (Profit/Cost Price) × 100

Calculation:

Let the selling price be x and cost price be y

Initially Profit (P) = x - y

If the selling price is doubled profit will be tripled

2x - y = 3P

⇒ 2x - y = 3(x - y)

⇒ 2x - y = 3x - 3y

⇒ x = 2y

Profit % = (Profit/Cost Price) × 100

Profit % = [(x - y)/y] × 100

Profit % = [(2y - y)/y)] × 100

Profit % = 100%

∴ Profit percentage will be 100%

The correct answer is 'Carefree'.
Explanation

• ​The word Anxious(adjective) means- "worried and nervous". 
  • E.g: I saw my sister's anxious face at the window.​
• ​The antonyms of the given word Anxious are- 'Carefree', 'confident', 'sure', 'self-assured', 'controlled, etc.
• From the antonyms of the given word, we can say that the word 'Carefree' is the opposite in meaning.
• The word Carefree(adjective) means- "having no problems or not being worried about anything". 
  • E.g: I remember my carefree student days.​
• ​Hence, the correct option is Option 1).
• Let's have a look at the meaning of other given options:
  • Active(adjective)- "busy with a particular activity". 
  • Taste(noun)- "the flavor of something, or the ability of a person or animal to recognize different flavors". 
  • Rise(verb)- "to move upwards". 

Other Related Points

• The synonyms of the given word Anxious are- 'Agitating', 'distressful', 'tense', 'nervous, etc.
• Example of 'Anxious' in a sentence:
  • The drought has made farmers anxious about the harvest.

Here the correct answer is too.
Explanation

• 'Too' is an adverb meaning to a higher degree than is desirable, permissible, or possible; excessively.
• Too - besides, also.
• For Examples:-
  • Sell the house and furniture too.
  • He was driving too fast.
• In the 4^th line of the passage, the Delhi police also deployed three companies of paramilitary force at the border of UP and Haryana.
• Therefore, 'too' will be used.

Here the correct answer is Across.
Explanation

• 'Across' will be used after roads because across is a preposition meaning from one side to other of something with clear boundaries.
• For Examples:-
  • Across a road.
  • Across a bridge.
• 'Through' will not be used because it is used to indicate the movement from one end to the other.
  • For Example - They ran swiftly through the woods.
• In the above given passage in the 3^rd line, the movement is across a state which has clear boundaries.
• Therefore, 'across' will be used.

Here the correct answer is Breaking through.
Explanation

• 'Break through' is a phrasal verb meaning make or force a way through a barrier.
  • For Example - The crowd tried to break through the gates.
• Here, in the 1^st line the farmers along with the leaders of Bhartiya Kisan Union made a way through the two barriers on NH-44 on Wednesday.

Thus, the correct phrasal verb for the blank is 'Breaking through'.

The correct answer is- 'than'.

Explanation

• No sooner than: is used to show that one thing happens immediately after another thing.
• It is often used with the past perfect.
• It is followed by 'than'.
  • Example: We had no sooner started cooking than there was a power cut.
• When 'no sooner' is used in the front position, we invert the order of the auxiliary verb and subject.
  • Example: No sooner had they stepped out of the house than it started to rain.
• From the above explanation, it is clear that 'than' should come in place of 'then' in the given sentence.
• Hence, option 3 is the correct answer.

The correct sentence is: No sooner had I eaten the pizza than I started feeling sick.

Five friends: Satish, Kishore, Mohan, Anil and Rajesh.

1) Satish is shorter than Kishore but taller than Rajesh.

Kishore > Satish > Rajesh

2) Anil is a little shorter than Kishore but a little taller than Satish.

Kishore > Anil > Satish

3) Mohan is the tallest.

Combining all these statements;

Mohan > Kishore > Anil > Satish > Rajesh

So, Satish is taller than Rajesh but shorter than Anil.

Hence, "Satish" is the correct answer.

Arranging the given words according to the English dictionary order :
4. Navigate
1. Navigation
2. Necklace
3. Necklet

The reverse order will be :
3. Necklet
2. Necklace
1. Navigation
4. Navigate
Navigation is the second word from the bottom end.
Hence, "Navigation" is the correct answer.