← MES Jr. Grade
Past Questions
Which of the following devices has the ‘negative resistance’ characteristic?
A)
Reflex klystron
B)
Gunn diode
C)
P-N-P transistor
D)
Magnetron
The correct answer is the Gunn diode.
1. **Gunn Diode**: This device exhibits a phenomenon known as the 'Transferred Electron Effect'. Unlike standard diodes, it does not have a P-N junction. When the applied electric field exceeds a certain threshold, electrons are transferred from a high-mobility lower valley to a low-mobility upper valley in the conduction band of the semiconductor (typically GaAs). This causes the current to decrease as the voltage increases, resulting in a 'Differential Negative Resistance' (DNR) characteristic, which allows the device to act as a microwave oscillator.
2. **Reflex Klystron**: While it is used as an oscillator, it achieves oscillation through velocity modulation and bunching of electrons in a cavity, not through an inherent negative resistance characteristic of the material itself.
3. **P-N-P Transistor**: This is a standard bipolar junction transistor used for amplification and switching. It exhibits positive resistance in its active regions.
4. **Magnetron**: This is a high-power vacuum tube that uses crossed electric and magnetic fields to create electron 'spokes'. While it acts as an oscillator, the mechanism is based on interaction between electron beams and cavity resonators, not the negative resistance characteristic associated with semiconductor materials like the Gunn diode.
Write an assembly language program for carrying out multiplication of two 8 bit numbers.
Define critical angle and numerical aperture associated with optical fiber communication. For a silica optical fiber, the refractive index of core layer and cladding layer is 1.56 and 1.35 respectively. Calculate critical angle and numerical aperture.
An 8085 microprocessor based system uses a 4K × 8-bit RAM whose starting address is AA00H. The address of the last byte in this RAM is :
A)
AFFFH
B)
B9FFH
C)
BFFFH
D)
A9FFH
Step-by-step calculation:
1. **Determine the size of the RAM**: The RAM size is given as 4K × 8-bit. In computing, 1K = 1024 bytes. Therefore, 4K = 4 × 1024 = 4096 bytes.
2. **Calculate the offset**: Since the first byte is at address 0, the last byte of a memory block of size N is at offset (N - 1).
Offset = 4096 - 1 = 4095 (decimal).
3. **Convert the offset to Hexadecimal**:
4095 divided by 16 = 255 with remainder 15 (F)
255 divided by 16 = 15 with remainder 15 (F)
15 divided by 16 = 0 with remainder 15 (F)
So, 4095 in decimal is FFFH in hexadecimal.
4. **Calculate the ending address**:
Ending Address = Starting Address + Offset
Ending Address = AA00H + 0FFFH
Performing the hexadecimal addition:
- 0 + F = F
- 0 + F = F
- A + F = 10 + 15 = 25. In hex, 25 = 1 × 16 + 9. So, we write 9 and carry 1.
- A + 0 + 1 (carry) = 11. In hex, 11 = B.
Result = B9FFH.
Comparing with options:
- a. AFFFH: Incorrect (This would be for a 2K RAM starting at AA00H).
- b. B9FFH: Correct.
- c. BFFFH: Incorrect (This would be for a 6K RAM starting at AA00H).
- d. A9FFH: Incorrect (This address is lower than the starting address).
In a waveguide, attenuation near the cut-off frequency is :
A)
Low
B)
High
C)
Very high
D)
Zero
In a waveguide, the attenuation constant (α) consists of two primary components: conductor loss (due to the finite conductivity of the walls) and dielectric loss. The attenuation due to conductor loss is inversely proportional to the group velocity of the wave. As the operating frequency (f) approaches the cut-off frequency (fc), the group velocity (vg) tends toward zero (vg = c * sqrt(1 - (fc/f)^2)). Since the wave slows down significantly near the cut-off frequency, the electromagnetic field interacts more intensely and for a longer duration with the lossy walls of the waveguide, leading to a sharp increase in attenuation. Therefore, attenuation is very high (or simply 'high' in standard terminology) near the cut-off frequency and decreases as the frequency increases further away from fc. While 'Very high' (option c) might seem tempting, in standard textbook terminology for these exams, 'High' is the conventional descriptor for the behavior at the cut-off boundary.
The main advantage of DSB and SSB is :
A)
More carrier power
B)
Double the side band power
C)
Higher power can be put into sidebands
D)
Less carrier power
In standard Amplitude Modulation (AM), a significant portion of the total transmitted power is wasted in the carrier wave, which carries no information. The carrier typically consumes at least 66.7% of the total power (for 100% modulation). Double Sideband Suppressed Carrier (DSB-SC) and Single Sideband (SSB) modulation techniques are designed to suppress or eliminate the carrier. By removing the carrier, the power that would have been spent on the carrier can be redistributed into the sidebands (which contain the actual information). This results in a much higher power efficiency, allowing the transmitter to put more power into the information-bearing components of the signal, thereby increasing the range or improving the signal-to-noise ratio at the receiver.
In a commercial TV broadcast station, the sound portion of TV signal is frequency modulated with a maximum frequency deviation of 50 kHz and a maximum modulating-signal frequency of 15 kHz. The deviation ratio is :
A)
3.3
B)
0.3
C)
750
D)
100
To find the deviation ratio of a frequency modulated (FM) signal, we use the standard formula for Deviation Ratio (DR).
1. **Given Data:**
- Maximum frequency deviation ($\Delta f$) = 50 kHz
- Maximum modulating-signal frequency ($f_m$) = 15 kHz
2. **Formula:**
The deviation ratio is defined as the ratio of the maximum frequency deviation to the maximum modulating frequency:
$$\text{Deviation Ratio} = \frac{\Delta f}{f_m}$$
3. **Calculation:**
$$\text{Deviation Ratio} = \frac{50\text{ kHz}}{15\text{ kHz}}$$
$$\text{Deviation Ratio} = \frac{50}{15} = \frac{10}{3} \approx 3.333...$$
4. **Conclusion:**
Rounding to one decimal place as per the options provided, we get 3.3.
- Option (b) 0.3 is incorrect as it is the reciprocal of the correct ratio.
- Option (c) 750 is incorrect as it is the product of the two values (50 * 15).
- Option (d) 100 is incorrect and unrelated to the given parameters.
The number of out pins of a 8085 microprocessor are :
A)
40
B)
27
C)
21
D)
19
The Intel 8085 is a 40-pin microprocessor. To determine the number of 'out pins' (output pins), we must categorize the pins based on their directionality:
1. **Input Pins (19 pins):** These include the Address/Data bus pins when acting as inputs (AD0-AD7), the Interrupts (TRAP, RST 7.5, 6.5, 5.5), INTR, READY, RESET IN, and the clock input (X1, X2).
2. **Output Pins (21 pins):** These include the higher-order address/data bus (A8-A15), the ALE (Address Latch Enable) signal, the Read/Write control signals (RD, WR), the IO/M signal, the status signals (S0, S1), the clock output (CLK OUT), and the reset output (RESET OUT).
Specifically, the breakdown of the 40 pins is generally viewed as:
- 19 Input pins
- 21 Output pins
Therefore, the number of output pins is 21.
To prevent oscillations in the TWT which one of the following is resorted to :
A)
Bunching defocusing mechanism is used
B)
Attenuator is used
C)
External magnetic field is provided
D)
Helix is used
A Traveling Wave Tube (TWT) is a wideband amplifier. However, because it has high gain over a broad frequency range, it is susceptible to parasitic oscillations if any part of the output signal is reflected back toward the input (feedback). To prevent these oscillations and ensure stability, an attenuator (specifically an isolator or a fixed attenuator) is used at the input or output stages to absorb reflected waves and break the feedback loop.
Analysis of other options:
- Option (a): Bunching is the fundamental mechanism of amplification in a TWT, not a method to prevent oscillations.
- Option (c): The external magnetic field is used to focus the electron beam and keep it traveling in a straight line along the axis of the slow-wave structure; it does not prevent oscillations.
- Option (d): The helix is the slow-wave structure used to synchronize the velocity of the RF wave with the electron beam; it is the cause of the gain, not a mechanism to prevent oscillations.
A narrowband FM does not have the following feature :
A)
it does not show any amplitude variations
B)
it has two sidebands
C)
both sidebands are equal in amplitude
D)
both sidebands have identical phase difference with respect to carrier
To solve this, we must analyze the characteristics of Narrowband Frequency Modulation (NBFM). NBFM occurs when the modulation index β is very small (typically β << 1).
1. Mathematical Representation: The expression for an NBFM signal is approximately: s(t) = Ac cos(ωc t) - (β Ac sin(ωc t)) sin(ωm t). Using trigonometric identities, this can be rewritten as: s(t) = Ac cos(ωc t) + (β Ac / 2) [cos(ωc + ωm)t - cos(ωc - ωm)t].
2. Analysis of Options:
- Option (a): In FM (both Wideband and Narrowband), the amplitude of the carrier remains constant. Therefore, it does not show amplitude variations. This is a feature of NBFM.
- Option (b): As seen in the expanded equation, NBFM consists of the carrier and two sidebands (at ωc + ωm and ωc - ωm). This is a feature of NBFM.
- Option (c): The amplitudes of the two sidebands are both equal to (β Ac / 2). This is a feature of NBFM.
- Option (d): Looking at the expanded equation, the upper sideband has a positive sign (cos) and the lower sideband has a negative sign (-cos). A negative sign in trigonometry represents a phase shift of 180 degrees (π radians). Therefore, the two sidebands have opposite phases (they are in anti-phase), not identical phase differences.
Since the question asks which feature NBFM does NOT have, option (d) is the correct answer.
Decimal number 9 in Gray code is :
A)
1100
B)
1101
C)
110
D)
1111
To convert a decimal number to Gray code, we first convert the decimal number to its binary equivalent and then convert that binary number to Gray code.
Step 1: Convert Decimal 9 to Binary
9 divided by 2 = 4 remainder 1
4 divided by 2 = 2 remainder 0
2 divided by 2 = 1 remainder 0
1 divided by 2 = 0 remainder 1
Reading the remainders from bottom to top, the binary representation of 9 is 1001₂.
Step 2: Convert Binary (1001) to Gray Code
- The Most Significant Bit (MSB) of the Gray code is the same as the MSB of the binary number: MSB = 1.
- The next bit is found by XORing the first and second binary bits: 1 ⊕ 0 = 1.
- The next bit is found by XORing the second and third binary bits: 0 ⊕ 0 = 0.
- The last bit is found by XORing the third and fourth binary bits: 0 ⊕ 1 = 1.
Combining these results, the Gray code for decimal 9 is 1101.
Verification of options:
- a. 1100: This is the Gray code for decimal 8 (Binary 1000).
- b. 1101: Correct.
- c. 110: This is a 3-bit number and cannot represent decimal 9.
- d. 1111: This is the Gray code for decimal 10 (Binary 1010).
What is the need of low pass filter and sample-hold circuit in PCM generator? Explain slop over load error and granular noise associated with Delta Modulation (DM).
In which of the following mode of propagation the waves are guided along the surface of the earth?
A)
Ground wave
B)
Sky wave
C)
LOS
D)
Space wave
The question asks for the mode of propagation where waves are guided along the surface of the earth.
1. **Ground Wave Propagation (Surface Wave):** In this mode, the electromagnetic waves travel along the curvature of the earth's surface. Due to the earth's conductivity, the waves are 'guided' or 'clung' to the surface, allowing them to travel beyond the visual horizon. This is typically used for Low Frequency (LF) and Medium Frequency (MF) communications (e.g., AM radio).
2. **Sky Wave Propagation:** These waves are radiated upwards and reflected (refracted) back to earth by the ionosphere. They do not follow the surface of the earth.
3. **LOS (Line-of-Sight):** This refers to propagation where the transmitter and receiver must be visible to each other without obstructions. It is a straight-line path.
4. **Space Wave Propagation:** This is a general term that includes LOS and waves that travel through the atmosphere/space. While it includes the direct path, it does not describe the specific phenomenon of being 'guided along the surface' like ground waves do.
Therefore, Ground Wave propagation is the correct answer.
The maximum improvement in power efficiency and reduction in bandwidth of an AM signal can be achieved by :
A)
having modulation index closer to 100%
B)
having modulation index closer to 10%
C)
removing the carrier signal
D)
removing the carrier signal and its one of the Sidebands
To determine the maximum improvement in power efficiency and reduction in bandwidth, we must analyze the components of a standard Amplitude Modulation (AM) signal.
1. **Power Efficiency**: In standard AM, the total power is the sum of the carrier power and the sideband power ($P_t = P_c + P_{usb} + P_{lsb}$). The carrier contains no information but consumes the majority of the power. Removing the carrier signal (converting AM to DSB-SC) significantly increases power efficiency because 100% of the transmitted power is then dedicated to the information-bearing sidebands.
2. **Bandwidth**: A standard AM signal (and DSB-SC) has a bandwidth of $2f_m$ (where $f_m$ is the highest modulating frequency) because it contains two redundant sidebands (Upper Sideband and Lower Sideband).
3. **Combining both**:
- Removing the carrier (DSB-SC) improves power efficiency but keeps the bandwidth at $2f_m$.
- Removing the carrier AND one of the sidebands results in **Single Sideband Suppressed Carrier (SSB-SC)** modulation.
- In SSB-SC, power efficiency is maximized because only one sideband is transmitted, and bandwidth is reduced by half (from $2f_m$ to $f_m$).
Comparing the options:
- Option (a) and (b) only change the modulation index, which affects efficiency slightly but does not reduce bandwidth.
- Option (c) (DSB-SC) improves power efficiency but does not reduce bandwidth.
- Option (d) (SSB-SC) provides both the maximum power efficiency and the minimum bandwidth requirement.
Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance :
A)
In series with a current source
B)
In parallel with a current source
C)
In series with a voltage source
D)
In parallel with a current source
Norton's Theorem is a fundamental network theorem used to simplify complex linear circuits. It states that any linear circuit containing several voltage sources and resistances can be replaced by an equivalent circuit consisting of a single current source (Norton current, In) in parallel with a single impedance (Norton impedance, Zn).
Step-by-step breakdown:
1. Norton's Theorem focuses on the current delivered to a load.
2. The Norton equivalent circuit consists of a constant current source (In) connected in parallel with an equivalent resistance or impedance (Zn).
3. This is the dual of Thévenin's Theorem, which replaces a network with a voltage source in series with an impedance.
Analysis of options:
- Option (a) is incorrect because a current source in series with an impedance is not the standard Norton equivalent (and would be impractical as the impedance wouldn't affect the current flow from the source).
- Option (b) is correct as it describes the standard Norton equivalent circuit.
- Option (c) describes a Thévenin equivalent circuit, not a Norton equivalent.
- Option (d) is identical to option (b) in the provided text, but the conceptual answer remains 'In parallel with a current source'.
When the diode is forward biased, it is equivalent to :
A)
An off switch
B)
An On switch
C)
A high resistance
D)
A low resistance
A diode is a semiconductor device that allows current to flow in only one direction.
1. **Forward Bias Condition**: When the positive terminal of a voltage source is connected to the P-type material (anode) and the negative terminal to the N-type material (cathode), the diode is said to be forward biased.
2. **Mechanism**: In forward bias, the external voltage overcomes the internal barrier potential (approximately 0.7V for Silicon). This narrows the depletion region, allowing majority charge carriers to flow across the junction.
3. **Equivalent Circuit**: In an ideal model, a forward-biased diode acts as a closed switch (an 'On' switch) with zero resistance, allowing current to flow freely. In a practical model, it acts as a small voltage source in series with a very low resistance.
**Analysis of other options:**
- **Option a (An off switch):** This describes a diode in reverse bias, where it blocks current flow.
- **Option c (A high resistance):** This is characteristic of a reverse-biased diode.
- **Option d (A low resistance):** While a forward-biased diode does have low resistance, the term 'On switch' is the standard conceptual equivalent used in circuit analysis to describe the state of conduction.
High resistivity materials are used in :
A)
LED light
B)
Heating instruments
C)
Tube light starter
D)
Cool down instruments
The principle behind heating instruments (like electric heaters, irons, and toasters) is Joule Heating. According to Joule's Law, the heat produced (H) in a conductor is given by the formula H = I²Rt, where I is the current, R is the resistance, and t is the time. To generate a significant amount of heat from an electrical current, a material with high electrical resistivity is required. Materials such as Nichrome (an alloy of nickel and chromium) are used because they have high resistivity and a high melting point, allowing them to convert electrical energy into thermal energy efficiently without melting.
Analysis of other options:
- a. LED light: These are semiconductor devices (p-n junctions) that rely on electroluminescence, not high resistivity for heat generation.
- c. Tube light starter: A starter typically uses a bimetallic strip and a neon bulb to create a momentary short circuit; while it involves specific material properties, the primary function of the device is switching, not utilizing high resistivity for its core operation.
- d. Cool down instruments: Cooling systems (like refrigerators or ACs) rely on thermodynamic cycles and refrigerants, not the electrical resistivity of materials to function.
A Zener diode works on the principle of :
A)
tunnelling of charge carriers across the junction
B)
thermionic emission
C)
diffusion of charge carriers across the junction
D)
hopping of charge carriers across the junction
A Zener diode is specifically designed to operate in the reverse breakdown region. The 'Zener effect' occurs in heavily doped p-n junctions where the depletion region is very narrow. Due to the high doping concentration, a very strong electric field is created across this narrow junction when reverse-biased. This strong field allows electrons to 'tunnel' directly from the valence band of the p-side to the conduction band of the n-side (or vice versa), a quantum mechanical phenomenon known as Zener Tunneling.
Why other options are incorrect:
- b. Thermionic emission: This refers to the emission of electrons from a heated filament or cathode (used in vacuum tubes), not the breakdown mechanism of a diode.
- c. Diffusion: Diffusion is the primary mechanism for current flow in a forward-biased diode, where carriers move from regions of high concentration to low concentration.
- d. Hopping: Hopping conduction typically occurs in disordered materials or semiconductors with high impurity levels at very low temperatures, but it is not the operating principle of a Zener diode.
The resistivity of a metal is a function of temperature because :
A)
The electron density varies with temperature
B)
The electron gas density varies with temperature
C)
Collision of electrons increases as temperature increases
D)
The lattice vibration increases with temperature
In metals, electrical resistivity is primarily caused by the scattering of conduction electrons as they move through the crystal lattice. According to the Drude model and quantum mechanical theories of conductivity, the resistivity (ρ) is proportional to the frequency of collisions between electrons and the lattice ions. As the temperature of a metal increases, the thermal energy causes the ions in the lattice to vibrate with greater amplitude around their equilibrium positions (these vibrations are quantized as phonons). These increased lattice vibrations increase the effective 'cross-section' for collisions, making it more likely that a conduction electron will be scattered. Therefore, the mean free path of the electrons decreases, leading to an increase in resistivity.
Why other options are incorrect:
- Options (a) and (b): In metals, the concentration of conduction electrons (electron density) is essentially constant and does not change significantly with temperature. This is a key difference between metals and semiconductors (where carrier density increases with temperature).
- Option (c): While it is true that collisions increase, 'collision of electrons' usually refers to electron-electron scattering. While electron-electron scattering occurs, the dominant mechanism for the temperature dependence of resistivity in metals is electron-phonon scattering (electron-lattice interaction), which is directly caused by the lattice vibrations described in option (d). Option (d) provides the fundamental physical cause.
The covalent bond is formed by :
A)
Transfer of electrons between atoms
B)
Sharing of electrons between atoms
C)
Sharing of variable number of electrons by a variable number of atoms
D)
Sharing of nucleus between atoms
A covalent bond is a type of chemical bond that occurs when two atoms share one or more pairs of valence electrons to achieve a stable electron configuration (usually an octet). In the context of semiconductor physics, elements like Silicon (Si) and Germanium (Ge) have four valence electrons and form covalent bonds with four neighboring atoms to create a stable crystal lattice structure.
Analysis of options:
- Option (a) describes an 'Ionic Bond', where electrons are transferred from a metal to a non-metal.
- Option (b) is the correct definition of a 'Covalent Bond'.
- Option (c) is incorrect because while the number of shared electrons can vary (single, double, or triple bonds), the fundamental mechanism is the sharing of electrons between specific atoms to reach stability, not a 'variable number of atoms' in a random sense.
- Option (d) is physically impossible; nuclei are held together by strong nuclear forces in the nucleus, not by chemical bonding between atoms.
The rms value of the voltage defined by v(t) = 5 + 5Sin(314t + pi/6) is :
A)
5 V
B)
2.5 V
C)
6.12 V
D)
10 V
The given voltage expression is v(t) = 5 + 5sin(314t + π/6). This is a composite signal consisting of a DC component and an AC sinusoidal component.
1. Identify the components:
- DC component (Vdc) = 5 V
- AC peak voltage (Vm) = 5 V
2. Calculate the RMS value of the AC component (Vrms_ac):
For a pure sine wave, Vrms_ac = Vm / √2
Vrms_ac = 5 / √2 ≈ 3.535 V
3. Calculate the total RMS value (Vrms_total):
The total RMS value of a signal containing both DC and AC components is given by the square root of the sum of the squares of the DC value and the RMS value of the AC components:
Vrms_total = √(Vdc² + Vrms_ac²)
Vrms_total = √(5² + (5/√2)²)
Vrms_total = √(25 + 25/2)
Vrms_total = √(25 + 12.5)
Vrms_total = √37.5
4. Final Calculation:
√37.5 ≈ 6.1237 V
Comparing this result to the options, 6.12 V is the correct value. Option (a) only considers the DC part, option (b) is incorrect, and option (d) is the peak-to-peak value of the AC part added to DC, which is not the RMS value.
An avalanche photo diode is operated under :
A)
Small forward bias
B)
Small reverse bias
C)
Independent of bias
D)
Large reverse bias
An Avalanche Photo Diode (APD) is a highly sensitive photodetector designed to provide internal gain. To achieve this gain, the diode must be operated under a large reverse bias voltage. When a high reverse bias is applied, a strong electric field is created within the depletion region. When a photon is absorbed and creates an electron-hole pair, these carriers are accelerated by the intense electric field to very high kinetic energies. These high-energy carriers collide with the crystal lattice, knocking more electrons out of the valence band and creating additional electron-hole pairs. This process is known as 'impact ionization,' leading to an 'avalanche' effect that multiplies the initial photocurrent.
Why other options are incorrect:
- (a) Small forward bias: Forward biasing allows current to flow easily and would saturate the diode, preventing the detection of light-induced carriers.
- (b) Small reverse bias: While a standard PIN photodiode operates under reverse bias, a 'small' reverse bias is insufficient to trigger the impact ionization required for the avalanche effect.
- (c) Independent of bias: The operation of an APD is strictly dependent on the bias voltage to create the necessary electric field for carrier multiplication.
What is photo diode? With proper characteristics curves and relevant diagrams explain the operation of photo diode. Explain the concept of photo conductivity.
Zener diode is ___________ device.
A)
Non-linear
B)
Amplifier
C)
Linear
D)
Transistor
A Zener diode is a semiconductor device. In electronics, a linear device is one where the output is directly proportional to the input (following Ohm's Law, V = IR, where resistance is constant). A non-linear device is one where the relationship between voltage and current is not a straight line.
1. **V-I Characteristics**: The current-voltage (V-I) characteristic curve of a Zener diode is highly non-linear. It remains almost zero until the threshold voltage is reached, and then it increases exponentially in forward bias. In reverse bias, it remains nearly zero until it reaches the Zener breakdown voltage (Vz), at which point the current increases sharply while the voltage remains almost constant.
2. **Why other options are incorrect**:
- **Amplifier (b)**: A Zener diode is a passive device used for voltage regulation; it cannot amplify a signal (which requires an active device like a transistor or Op-Amp).
- **Linear (c)**: As explained, its V-I curve is not a straight line passing through the origin.
- **Transistor (d)**: A Zener diode is a two-terminal diode, whereas a transistor is a three-terminal device (e.g., BJT or MOSFET) used for switching or amplification.
Mobility is defined as :
A)
Diffusion velocity per unit electric field
B)
Drift velocity per unit electric field
C)
Displacement per unit electric field
D)
Distance per unit electric field
In semiconductor physics, mobility (μ) is a measure of how quickly a charge carrier (electron or hole) can move through a metal or semiconductor when acted upon by an electric field. Mathematically, it is defined as the ratio of the drift velocity (vd) to the applied electric field (E). The formula is expressed as: μ = vd / E.
- Option (a) is incorrect because diffusion velocity is driven by a concentration gradient, not by an external electric field.
- Option (c) and (d) are incorrect because mobility describes the rate of flow (velocity), not the total distance or displacement traveled.