Open Loop Op Amp MCQ Quiz - Objective Question with Answer for Open Loop Op Amp - Download Free PDF

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 (A_OL) of the op-amp is 10⁵ (100,000), and the input voltage (V_in) is 100 millivolts (0.1 volts).

The formula to calculate the output voltage (V_out) in an open-loop configuration is given by:

V_out = A_OL × V_in

Substituting the given values:

V_out = 10⁵ × 0.1 V

V_out = 100,000 × 0.1 V

V_out = 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.

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.

Since the input impedance is not infinite.

Thus the concept of virtual ground is not applicable.

\(\rm V_0 = - \frac{R_2}{R_1} \left[ V_{in} + \frac{V_0}{Av} \right] - \frac{V_0}{Av}\)

\(\rm \frac{V_0}{V_{in}} = \frac{-R_2 |R_1}{1+ \frac{1}{Av} \left( 1 + \frac{R_2}{R_1} \right) }\)

\(\frac{V_0}{V_{in}} = \frac{-10}{1+ \frac{1}{10^3}\left( 1 + \frac{10}{1} \right)}= -9.89 \ V/V\)

Concept:

For an open loop Opamp generally output should be ± VSAT

Now,If   V_d = V⁺ - V^- > 0  then V_O = + V_SAT

Vd = V+ - V- < 0  then VO = - VSAT 

Calculation:

Given,

V_d = 0.2 mV

V_O = +12 V

Concept:

Inverting Amplifier : it is circuit which produces an output which is out of phase with respect to its input by 180o.

The voltage gain (A_v) of the Inverting Amplifier is - 

\(A_{v}=-\frac{R_{2}}{R_{1}}\)

Given:

R₂ = R₁ = R

Calculation:

\(A_{v}=-\frac{R_{}}{R_{}}\)

A_v = -1

Voltage Gain = -1

The following circuit is a voltage follower with Negative feedback. 

The gain of Non-Inverting Amplifier is given by:

\(A = 1 + \frac{{{R_f}}}{{{R_1}}}\)

For voltage follower, Rf is 0

Hence gain (A) = 1

A voltage follower is an op-amp circuit whose output voltage straight away follows the input voltage. i.e. output voltage equivalent to the input voltage.

Characteristic of voltage follower:

• High input impedance
• Low output impedance
• The loading effect can be avoided
• Opamp takes zero current from the input
• Provides power Gain & current gain but voltage gain unity

Concept:

Inverting Amplifier : it is circuit which produces an output which is out of phase with respect to its input by 180o.

The voltage gain (A_v) of the Inverting Amplifier is - 

\(A_{v}=-\frac{R_{2}}{R_{1}}\)

Given:

R₂ = R₁ = R

Calculation:

\(A_{v}=-\frac{R_{}}{R_{}}\)

A_v = -1

Voltage Gain = -1

Concept:

The gain of a feedback system is given by:

\(A_f=\frac{A}{1+Aβ }\)

A = Open Loop gain

Af = Closed Loop Gain

β = Feedback/Transmission factor

Calculation:

Given  A = 100

β = 1

\(A_f=\frac{100}{1+100 }\)

\(A_f=\frac{100}{101 }=0.99\)

A_f = 0.99

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 V_in is applied to the non-inverting terminal. The fixed reference voltage V_ref 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 V_in and R is connected between the inverting input and reference voltage V_ref.
• 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.

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 + V_sat 

when V+ < V- the output is - V_sat 

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:

Concept:

For an open loop Opamp generally output should be ± VSAT

Now,If   V_d = V⁺ - V^- > 0  then V_O = + V_SAT

Vd = V+ - V- < 0  then VO = - VSAT 

Calculation:

Given,

V_d = 0.2 mV

V_O = +12 V

Calculation:

Given:

 V­₁ = 40 mv, V₂ = 20 mv

Applying KCL at node B,

\(\frac{{{V_B} - {V_0}}}{{470}} + \frac{{{V_B} - {V_1}}}{{47}} + \frac{{{V_B} - {V_2}}}{{4.7}} = 0\)

\(\frac{{0 - {V_0}}}{{470}} + \frac{{0 - 40}}{{47}} + \frac{{0 - 20}}{{4.7}} = 0\)

Putting V­_B­­­­ = 0 and

on solving we get,

V₀ = -2.4 V

Comparator:

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

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 (A_OL) of the op-amp is 10⁵ (100,000), and the input voltage (V_in) is 100 millivolts (0.1 volts).

The formula to calculate the output voltage (V_out) in an open-loop configuration is given by:

V_out = A_OL × V_in

Substituting the given values:

V_out = 10⁵ × 0.1 V

V_out = 100,000 × 0.1 V

V_out = 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.

V_o = average value of waveform of V_x

\({V_0} = \frac{1}{{2\pi }}\left[ {\mathop \smallint \limits_0^{\frac{\pi }{5}} Adt + \mathop \smallint \limits_\pi ^{\frac{{6\pi }}{5}} Adt} \right]\)

We have, The logic output voltages of the exclusive-OR gate are 0 V and 5 V

Hence, A = 5 volts

\({V_0} = \frac{1}{{2\pi }}\left[ {\mathop \smallint \limits_0^{\frac{\pi }{5}} 5dt + \mathop \smallint \limits_\pi ^{\frac{{6\pi }}{5}} 5dt} \right]\)

\(= \frac{1}{{2\pi }}\left[ {5\left( {\frac{\pi }{5} - 0} \right) + 5\left( {\frac{{6\pi }}{5} - \frac{\pi }{5}} \right)} \right]\)

\( = \frac{1}{{2\pi }}\left[ {2\pi } \right] = 1\;V\)