Comparator MCQ Quiz - Objective Question with Answer for Comparator - Download Free PDF
Last updated on May 30, 2025
Latest Comparator MCQ Objective Questions
Comparator Question 1:
If the open-loop gain of an op-amp is 105 and the input voltage is 100 mV, the value of output voltage is ____________ assuming a supply of ±15 V.
Answer (Detailed Solution Below)
Comparator Question 1 Detailed Solution
Explanation:
The open-loop gain of an operational amplifier (op-amp) refers to the amplification factor it provides without any external feedback applied to the circuit. This gain is typically very high for op-amps, which allows them to amplify very small input signals to a much larger output signal. In the given problem, the open-loop gain (AOL) of the op-amp is 105 (100,000), and the input voltage (Vin) is 100 millivolts (0.1 volts).
The formula to calculate the output voltage (Vout) in an open-loop configuration is given by:
Vout = AOL × Vin
Substituting the given values:
Vout = 105 × 0.1 V
Vout = 100,000 × 0.1 V
Vout = 10,000 V
However, it is essential to consider the power supply voltage of the op-amp. In this case, the op-amp has a power supply of ±15 volts. This means that the maximum output voltage the op-amp can provide is limited to +15 volts and the minimum output voltage is limited to -15 volts, regardless of the calculated value.
Since the calculated output voltage (10,000 V) far exceeds the maximum supply voltage of the op-amp, the actual output voltage will be limited to the maximum supply voltage, which is +15 volts.
Therefore, the correct value of the output voltage is 15 V.
Important Information:
Analyzing the other options:
- Option 1 (104 V): This is the result of the calculation without considering the power supply limitation. This value is impractical because it exceeds the op-amp's supply voltage.
- Option 3 (0 V): This value would be correct if the input voltage was 0 V or if the op-amp was not powered. However, neither condition applies here.
- Option 4 (100 V): This value does not consider the power supply limitation and is also much higher than the maximum output voltage the op-amp can provide.
Comparator Question 2:
What type of feedback is used in the OP-AMP for it to operate as a comparator?
Answer (Detailed Solution Below)
Comparator Question 2 Detailed Solution
Explanation:
Operational Amplifier (OP-AMP) as a Comparator
Definition: An operational amplifier (OP-AMP) can be configured to function as a comparator, which is a device that compares two voltages or currents and outputs a digital signal indicating which is larger. The OP-AMP operates in an open-loop configuration when used as a comparator, meaning there is no feedback loop connecting the output to the input.
Working Principle: When an OP-AMP is used as a comparator, it compares the voltage at its inverting input (-) with the voltage at its non-inverting input (+). If the voltage at the non-inverting input is higher than the voltage at the inverting input, the output of the comparator goes to its positive saturation level (close to the supply voltage). Conversely, if the voltage at the inverting input is higher, the output goes to its negative saturation level (close to the ground or negative supply voltage). This creates a digital signal representing the result of the comparison.
Advantages:
- High speed operation due to the lack of feedback, allowing the output to switch rapidly between saturation levels.
- Simplicity in design, requiring minimal external components.
- Versatility in applications such as zero-crossing detectors, level shifters, and waveform generators.
Disadvantages:
- High sensitivity to noise, which can cause false triggering.
- Indeterminate output state when the input voltages are very close or equal.
Applications: OP-AMP comparators are widely used in various applications including analog-to-digital converters (ADCs), pulse-width modulation (PWM) signals, function generators, and various monitoring and control systems.
Correct Option Analysis:
The correct option is:
Option 3: No feedback
This option correctly describes the configuration of an OP-AMP when used as a comparator. In this mode, the OP-AMP operates in an open-loop configuration, meaning there is no feedback loop from the output to either of the inputs. This allows the OP-AMP to compare the input voltages and provide a digital output based on the comparison.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: Regenerative feedback
Regenerative feedback is a positive feedback mechanism used to create oscillations or to reinforce a signal. It is not used in OP-AMPs configured as comparators because it would cause the output to latch in one state or oscillate, which is not desirable for comparison purposes.
Option 2: Negative feedback
Negative feedback is commonly used in OP-AMP configurations such as amplifiers, integrators, and filters to stabilize the gain and improve linearity. However, in a comparator configuration, negative feedback is not used. The comparator relies on the open-loop gain of the OP-AMP to switch the output between saturation levels quickly.
Option 4: Positive feedback
Positive feedback can be used in certain comparator applications to create hysteresis, which helps to eliminate noise and provide a more stable output. However, the basic comparator operation does not inherently require positive feedback. The fundamental configuration involves no feedback loop.
Conclusion:
Understanding the operation of an OP-AMP as a comparator is crucial for correctly configuring and applying it in various electronic circuits. The correct answer to the question is that no feedback is used in the OP-AMP for it to operate as a comparator. This open-loop configuration allows the OP-AMP to effectively compare input voltages and produce a digital output indicating the result of the comparison. While other feedback mechanisms have their respective uses in different OP-AMP configurations, they are not applicable in the basic comparator setup.
Comparator Question 3:
If an op-amp comparator has a gain of 100,000, an input difference of 0.2 mV above reference. and a supply of ±12 V. the output will be:
Answer (Detailed Solution Below)
Comparator Question 3 Detailed Solution
Concept:
For an open loop Opamp generally output should be ± VSAT
Now,If Vd = V+ - V- > 0 then VO = + VSAT
Vd = V+ - V- < 0 then VO = - VSAT
Calculation:
Given,
Vd = 0.2 mV
VO = +12 V
Comparator Question 4:
Introduction of hysteresis in a comparator makes it
Answer (Detailed Solution Below)
Comparator Question 4 Detailed Solution
Hysteresis comparator:
- A hysteresis comparator is operated by applying a positive feedback to the comparator.
- The potential difference between the High and Low output voltages and the feedback resistor are adjusted to change the voltage that is taken as a comparison reference to the input voltage for the +IN terminal. The width of variation in the reference voltage is the hysteresis width. The signal is input to the ‐IN terminal ,the output is inverted.
- Hysteresis is important for producing stable switching behavior in a comparator circuit.
- This hysteresis is added by including a positive feedback loop between the output and one of the inputs, which then defines the threshold for switching as the input signal rises and falls.
- Noise on the input signal in a comparator circuit can produce multiple transitions as the input signal rises.
- Intentionally adding hysteresis to a comparator circuit is useful for suppressing this unintended switching due to noise.
- An analogous application is to eliminate contact bounce in mechanical switches, which will also produce unintended switching.
Note: A comparator cannot be operated as a hysteresis comparator when a negative feedback is applied.
Comparator Question 5:
If the input to a comparator is a sine wave, the output is a
Answer (Detailed Solution Below)
Comparator Question 5 Detailed Solution
The comparator of the op-amp does not contain any feedback and it will generate a rectangular wave at the output with saturation voltage levels when the input is a sin wave.
When V+ > V- the output is + Vsat
when V+ < V- the output is - Vsat
Thus a rectangular wave is produced at the output by driving the op-amp into saturation.
The output waveform for an applied sin wave is as shown:
Top Comparator MCQ Objective Questions
Clamp diodes used in comparators ______
Answer (Detailed Solution Below)
Comparator Question 6 Detailed Solution
Download Solution PDFA comparator is a circuit, which compares a signal voltage on one input of an op-amp with a known reference voltage on the other input.
The above circuit is called a non-inverting comparator circuit as the sinusoidal input signal Vin is applied to the non-inverting terminal. The fixed reference voltage Vref is connected to the inverting terminal of the op-amp.
Clamp diodes:
- The circuit diagram shows the diodes D1 and D2. These two diodes are used to protect the op-amp from damage due to an increase in input voltage.
- These diodes are called clamp diodes as they clamp the differential input voltages to either 0.7V or - 0.7V.
- Most op-amps do not need clamp diodes as most of them already have built-in protection.
- Resistance R1 is connected in series with the input voltage Vin and R is connected between the inverting input and reference voltage Vref.
- R1 limits the current through the clamp diodes and R reduces the offset problem.
Note:
Clamper:
- A clamper is an electronic circuit that fixes either the positive or the negative peak excursions of a signal to a defined value by shifting its DC value.
- The clamper does not restrict the peak-to-peak excursion of the signal, it moves the whole signal up or down so as to place the peaks at the reference level.
- A diode clamp (a simple, common type) consists of a diode, which conducts electric current in only one direction and prevents the signal exceeding the reference value; and a capacitor, which provides a DC offset from the stored charge.
- The capacitor forms a time constant with the resistor load, which determines the range of frequencies over which the clamper will be effective.
If the input to a comparator is a sine wave, the output is a
Answer (Detailed Solution Below)
Comparator Question 7 Detailed Solution
Download Solution PDFThe comparator of the op-amp does not contain any feedback and it will generate a rectangular wave at the output with saturation voltage levels when the input is a sin wave.
When V+ > V- the output is + Vsat
when V+ < V- the output is - Vsat
Thus a rectangular wave is produced at the output by driving the op-amp into saturation.
The output waveform for an applied sin wave is as shown:
If an op-amp comparator has a gain of 100,000, an input difference of 0.2 mV above reference. and a supply of ±12 V. the output will be:
Answer (Detailed Solution Below)
Comparator Question 8 Detailed Solution
Download Solution PDFConcept:
For an open loop Opamp generally output should be ± VSAT
Now,If Vd = V+ - V- > 0 then VO = + VSAT
Vd = V+ - V- < 0 then VO = - VSAT
Calculation:
Given,
Vd = 0.2 mV
VO = +12 V
A comparator circuit is used to
Answer (Detailed Solution Below)
Comparator Question 9 Detailed Solution
Download Solution PDFComparator:
A comparator circuit is one which may be used to mark the instant when an arbitrary waveform attains some reference level.
Applications of comparators:
- Measurement of time delays
- Timing markers generated from sine wave
- Phase meter
- Square waves from sine waves
If the open-loop gain of an op-amp is 105 and the input voltage is 100 mV, the value of output voltage is ____________ assuming a supply of ±15 V.
Answer (Detailed Solution Below)
Comparator Question 10 Detailed Solution
Download Solution PDFExplanation:
The open-loop gain of an operational amplifier (op-amp) refers to the amplification factor it provides without any external feedback applied to the circuit. This gain is typically very high for op-amps, which allows them to amplify very small input signals to a much larger output signal. In the given problem, the open-loop gain (AOL) of the op-amp is 105 (100,000), and the input voltage (Vin) is 100 millivolts (0.1 volts).
The formula to calculate the output voltage (Vout) in an open-loop configuration is given by:
Vout = AOL × Vin
Substituting the given values:
Vout = 105 × 0.1 V
Vout = 100,000 × 0.1 V
Vout = 10,000 V
However, it is essential to consider the power supply voltage of the op-amp. In this case, the op-amp has a power supply of ±15 volts. This means that the maximum output voltage the op-amp can provide is limited to +15 volts and the minimum output voltage is limited to -15 volts, regardless of the calculated value.
Since the calculated output voltage (10,000 V) far exceeds the maximum supply voltage of the op-amp, the actual output voltage will be limited to the maximum supply voltage, which is +15 volts.
Therefore, the correct value of the output voltage is 15 V.
Important Information:
Analyzing the other options:
- Option 1 (104 V): This is the result of the calculation without considering the power supply limitation. This value is impractical because it exceeds the op-amp's supply voltage.
- Option 3 (0 V): This value would be correct if the input voltage was 0 V or if the op-amp was not powered. However, neither condition applies here.
- Option 4 (100 V): This value does not consider the power supply limitation and is also much higher than the maximum output voltage the op-amp can provide.
The comparators (output = ‘1’, when input ≥ 0 and output = ‘0’, when input < 0), exclusive-OR gate and the unity gain low-pass filter given in the circuit are ideal. The logic output voltages of the exclusive-OR gate are 0 V and 5 V. The cutoff frequency of the low-pass filter is 0.1 Hz. For V1 = 1 sin (3000t + 36o) V and V2 = 1 sin (3000t) V, the value of VO in volt is ______.
Answer (Detailed Solution Below) 1
Comparator Question 11 Detailed Solution
Download Solution PDFVo = average value of waveform of Vx
\({V_0} = \frac{1}{{2\pi }}\left[ {\mathop \smallint \limits_0^{\frac{\pi }{5}} Adt + \mathop \smallint \limits_\pi ^{\frac{{6\pi }}{5}} Adt} \right]\)
We have, The logic output voltages of the exclusive-OR gate are 0 V and 5 V
Hence, A = 5 volts
\({V_0} = \frac{1}{{2\pi }}\left[ {\mathop \smallint \limits_0^{\frac{\pi }{5}} 5dt + \mathop \smallint \limits_\pi ^{\frac{{6\pi }}{5}} 5dt} \right]\)
\(= \frac{1}{{2\pi }}\left[ {5\left( {\frac{\pi }{5} - 0} \right) + 5\left( {\frac{{6\pi }}{5} - \frac{\pi }{5}} \right)} \right]\)
\( = \frac{1}{{2\pi }}\left[ {2\pi } \right] = 1\;V\)
Comparator Question 12:
The input to op amp open loop comparator is a sinusoidal signal, the output of the comparator is:
Answer (Detailed Solution Below)
Comparator Question 12 Detailed Solution
- An analog comparator has two inputs. One is a constant reference voltage VR and the other is time-varying signal Vi and one output Vo.
- When the non-inverting voltage is larger than the inverting voltage the comparator produces a high output voltage (+Vsat).
- When the non-inverting voltage is less than the inverting voltage the comparator produces a low output voltage (-Vsat).
Comparator Question 13:
Introduction of hysteresis in a comparator makes it
Answer (Detailed Solution Below)
Comparator Question 13 Detailed Solution
Hysteresis comparator:
- A hysteresis comparator is operated by applying a positive feedback to the comparator.
- The potential difference between the High and Low output voltages and the feedback resistor are adjusted to change the voltage that is taken as a comparison reference to the input voltage for the +IN terminal. The width of variation in the reference voltage is the hysteresis width. The signal is input to the ‐IN terminal ,the output is inverted.
- Hysteresis is important for producing stable switching behavior in a comparator circuit.
- This hysteresis is added by including a positive feedback loop between the output and one of the inputs, which then defines the threshold for switching as the input signal rises and falls.
- Noise on the input signal in a comparator circuit can produce multiple transitions as the input signal rises.
- Intentionally adding hysteresis to a comparator circuit is useful for suppressing this unintended switching due to noise.
- An analogous application is to eliminate contact bounce in mechanical switches, which will also produce unintended switching.
Note: A comparator cannot be operated as a hysteresis comparator when a negative feedback is applied.
Comparator Question 14:
Clamp diodes used in comparators ______
Answer (Detailed Solution Below)
Comparator Question 14 Detailed Solution
A comparator is a circuit, which compares a signal voltage on one input of an op-amp with a known reference voltage on the other input.
The above circuit is called a non-inverting comparator circuit as the sinusoidal input signal Vin is applied to the non-inverting terminal. The fixed reference voltage Vref is connected to the inverting terminal of the op-amp.
Clamp diodes:
- The circuit diagram shows the diodes D1 and D2. These two diodes are used to protect the op-amp from damage due to an increase in input voltage.
- These diodes are called clamp diodes as they clamp the differential input voltages to either 0.7V or - 0.7V.
- Most op-amps do not need clamp diodes as most of them already have built-in protection.
- Resistance R1 is connected in series with the input voltage Vin and R is connected between the inverting input and reference voltage Vref.
- R1 limits the current through the clamp diodes and R reduces the offset problem.
Note:
Clamper:
- A clamper is an electronic circuit that fixes either the positive or the negative peak excursions of a signal to a defined value by shifting its DC value.
- The clamper does not restrict the peak-to-peak excursion of the signal, it moves the whole signal up or down so as to place the peaks at the reference level.
- A diode clamp (a simple, common type) consists of a diode, which conducts electric current in only one direction and prevents the signal exceeding the reference value; and a capacitor, which provides a DC offset from the stored charge.
- The capacitor forms a time constant with the resistor load, which determines the range of frequencies over which the clamper will be effective.
Comparator Question 15:
If the input to a comparator is a sine wave, the output is a
Answer (Detailed Solution Below)
Comparator Question 15 Detailed Solution
The comparator of the op-amp does not contain any feedback and it will generate a rectangular wave at the output with saturation voltage levels when the input is a sin wave.
When V+ > V- the output is + Vsat
when V+ < V- the output is - Vsat
Thus a rectangular wave is produced at the output by driving the op-amp into saturation.
The output waveform for an applied sin wave is as shown: