24.5 VOLTAGE COMPARATOR
24.5.1 Voltage Comparator
A voltage comparator compares two voltages applied to the two inputs of the op-amp. The outcome of the output voltage will reveal the relationship between the two voltages.
One applied voltage may be kept constant at certain value and used as a reference by the other applied voltage. The reference value may be zero.
24.5.2 Circuit and Principle of Voltage Comparator
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Figure 24.22 Op-amp voltage comparator circuit operating in open-loop mode. A reference voltage (Vref) is applied to one input while the signal voltage is applied to the other. The output depends on the comparison between the two input voltages. |
1 Circuit
(a) The op-amp is connected in the open-loop mode. Hence, there is no feedback.
(b) A voltage is applied to each input. However, we may have one of the inputs earthed instead.
(c) Normally one voltage is kept constant and serves as a reference voltage, Vref.
(d) Vref may be applied to the − input (Figure 24.22(a)) or the + input (Figure 24.22(b)).
(e) A signal voltage Vin is applied to the other input.
2 Principle
(a) The op-amp works in the open-loop mode. Hence, it has an extremely high open-loop gain. Because of this, even a very small difference in the two input voltages (Vd) will drive the output voltage Vo to saturation. Hence, the output voltage of a voltage comparator is either
Vo = +Vsat or Vo = −Vsat
(b) When the signal voltage Vin is greater than Vref, the output voltage becomes saturated and is, say, positive. Conversely, when Vin is less than Vref the output voltage also becomes saturated but it is now negative.
(c) Hence, the polarity of the output voltage will indicate whether the signal voltage is greater than or less than a reference voltage.
The op-amp which is shown in Figure 24.23(a) acts as a voltage comparator. The waveform of the alternating voltage Vin is shown in Figure 24.23(b). Sketch a graph to show how the output voltage Vo varies with time.
Answer
(a) Vsat = ±0.9Vo
= ±0.9(12)
≈ ±11 V
(b) Vo = AOL Vd
The sign of Vo is determined by the sign of Vd.
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Figure 24.23 Application of reference voltage (Vref) in an op-amp comparator. The reference voltage is kept constant and can be applied either to the inverting (−) input or the non-inverting (+) input, depending on the comparator configuration. |
(c) Vd = V1 − V2
= Vin − Vref
If Vin > Vref, Vd is positive, Vo is positive.
If Vin < Vref, Vd is negative, Vo is negative.
If a voltmeter used to measure the output voltage at any instant indicates that the value of Vo is positive, it means that the input signal Vin is greater than the reference voltage Vref. We do not need to know the actual value of Vo. All we need to know is whether Vo is positive or negative. Figure 24.24 shows how the output voltage varies with time.
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| Figure 24.24 Output waveform of an op-amp comparator when the input signal (Vin) is connected to the inverting (−) input and the reference voltage is applied to the non-inverting (+) input. The output voltage (Vo) switches between positive and negative saturation, producing an inverted waveform relative to the input signal over time. |
Refer to Figure 24.24. The connections of Vin and the 1.5 V battery are interchanged with each other so that now Vin is connected to the − input. Sketch a graph to show how the output voltage Vo varies with time.
Answer
Figure 24.25 shows how the output voltage varies with time.
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| Figure 24.25 Output voltage (Vo) of an op-amp comparator when the input signal (Vin) is connected to the inverting (−) input and the reference voltage is applied to the non-inverting (+) input. The output waveform is inverted, switching between positive and negative saturation levels as Vin crosses the reference voltage over time. |
EXAMPLE 24.11
The op-amp shown in Figure 24.26(a) acts as a zero crossover comparator. Figure 24.26(b) shows the waveform of the alternating voltage which is applied to the − input. Sketch a graph to show how the output voltage Vo varies with time t.
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| Figure 24.26 The output voltage changes between positive and negative saturation levels, forming a square wave. The switching occurs whenever the input signal crosses zero, so the output rapidly flips state in response to the alternating input. |
Answer
(a) A fraction of a millivolt across the two inputs will be able to drive the output voltage into saturation.
(b) The reference voltage is applied to the + input. Since this input is earthed, the reference voltage Vref = 0 V.
(c) Since the + input is earthed, Vd across the two inputs will be identical to the alternating voltage given. Hence, Vd alternates between +1.0 V and −1.0 V.
(d) The output voltage is saturated. Vsat is either at +11 V or −11 V.
(e) Since Vin is applied to the − input, the output voltage is 180° out of phase with Vin.
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| Figure 24.27 Graph showing the output voltage changes with time. The output is inverted compared to the input signal, switching between high and low levels whenever the input crosses zero. |
(f) Figure 24.27 shows how Vo varies with time.
24.6 INTEGRATOR
24.6.1 Main Function of Integrator
An integrator produces an output voltage which is proportional to the time integral of the input signal voltage. In other words, the output voltage function Vo(t) is given by
Vo(t) ∝ ∫ Vin(t) dt
where Vin(t) is the input voltage function.
Let us take one particular input voltage. Suppose the voltage Vin varies with time t according to the expression
Vin = k
where k is a constant and for time interval t2 ≥ t ≥ t1. This can be represented by a horizontal line in a Vin − t graph. The integrator will produce an output voltage Vo given by
Vo = k ∫ dt
= kt + constant
This can be represented by a slanting line in a Vo − t graph. We may make the constant to become zero.
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This means that if the input voltage is a square wave, the integrator will convert it into a triangular wave.
24.6.2 Circuit and Principle
1 Op-amp Integrator Circuit
(a) Figure 24.28 shows the integrator circuit.
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| Figure 24.28 The diagram shows an op-amp integrator circuit. |
(b) The resistor and the capacitor form an RC circuit. When the signal voltage varies, a charging or discharging current will flow in this circuit.
(c) This integrator circuit will convert a square wave into a triangular wave.
2 Principle
(a) Suppose that the input voltage Vin is a square wave. Within one period of time, this voltage could be a constant positive voltage.
(b) Because the input signal is connected to the − input, the output voltage is negative, opposite to the polarity of the input voltage. The capacitor discharges.
(c) Within the next period, the input voltage changes polarity and becomes a constant negative voltage. During this period the capacitor will charge.
(d) As the input signal of square voltage alternates its polarity regularly, keeping its magnitude constant, the capacitor undergoes a process of charging and discharging.
(e) If the time constant RC is suitably chosen, the rate of charging or discharging could be carried out at a constant rate, thus producing a triangular output voltage.
24.7 OSCILLATOR
24.7.1 Main Function of Oscillator
1 An oscillator generates an alternating current and voltage of a predetermined frequency with no external signal source applied to the input.
2 The waveform of the output could be a sine wave, square wave or triangular wave.
3 An oscillator has a dc power source. The oscillator actually converts energy of the dc power supply into energy carried by the alternating current generated by the oscillator.
24.7.2 Basic Oscillator Circuit
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| Figure 24.29 The figure shows a basic op-amp oscillator circuit. It uses feedback components to continuously generate an alternating output signal without any external input. The circuit produces a periodic waveform as the capacitor charges and discharges through the resistor, causing the output to switch repeatedly over time. |
Figure 24.29 shows the circuit for the basic op-amp oscillator. Notice the following:
(a) There is a positive feedback and frequency selection network connecting the output to the input.
(b) There is no external signal connected to the input. The signal which the amplifier does receive is that fed back from the output.
(c) A dc power source (not shown) is connected to the amplifier.
24.7.3 Essential Requirements for Oscillation
In order to achieve oscillation and sustain it in stable condition, the following requirements must be met:
(a) There must be an amplifier.
(b) There must be a selection network to select a signal at a given frequency from the output. The network has to ensure that the correct amount of phase shift is given to the signal so that the signal is in phase with the input signal.
(c) There must be a positive feedback network to send the desired signal from the output to the input. The input signal must be reinforced by the positive feedback.
(d) The following criterion must be satisfied:
|βA| = 1
where A is the gain of the amplifier
β is the feedback factor which specifies the fraction of the output voltage to be fed back to the input
Figure 24.30 shows the circuit of a basic phase shift oscillator. This circuit can satisfy all those conditions stated above. A sinusoidal voltage is produced continuously at the output despite the fact that there is no external signal source connected to the input.
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Figure 24. 30The figure shows a basic phase shift oscillator circuit. It uses a network of resistors and capacitors to provide the required phase shift and feedback for oscillation. The circuit is able to meet the conditions for sustained oscillations and produces a continuous sinusoidal output voltage even without any external input signal. |
The three pairs of RC circuits ensure that the portion of the output signal fed to the input is in phase with the input signal at a given frequency.
24.8 FREQUENCY RESPONSE OF OP-AMP
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Figure 24.31 The figure shows how the voltage gain of an op-amp in open-loop mode changes with the frequency of the input signal. The graph illustrates that the gain is very high at low frequencies but decreases as the frequency increases. This characteristic is known as the frequency response of the amplifier. |
1 Figure 24.31 shows how the voltage gain ACL of an op-amp working in the open-loop mode varies with frequency of the input signal. This graph is known as the frequency response graph for the amplifier.
2 Notice the following features:
(a) The value of AOL = 100 000 is constant only within the frequency range of 1 Hz – 10 Hz. For frequencies higher than 10 Hz, AOL drops very fast with increase in frequency.
(b) The frequency range within which the amplifier gain remains constant is approximately equal to the bandwidth. For the open-loop amplifier, the bandwidth is only about 10 Hz.
(c) Suppose a negative feedback network is connected to the amplifier so that now it has a close-loop gain ACL of only 10. Referring to the graph, we notice that for ACL = 10, the bandwidth has increased to about 100 kHz. This means that an op-amp with negative feedback has voltage gain which is very low compared with the gain in open-loop mode. However, it has very broad bandwidth.
Op-Amp Applications in Daily Life
Explore how operational amplifiers are used in real-world systems and everyday technology.
Operational amplifiers (op-amps) play a crucial role in modern electronics by enabling precise signal processing and control. Through applications such as voltage comparators, integrators, and oscillators, op-amps can analyze, transform, and generate electrical signals efficiently.
The voltage comparator allows systems to make decisions based on signal levels, while the integrator transforms signals into smoother or accumulated forms. Meanwhile, oscillators generate continuous waveforms essential for timing and communication systems.
Understanding the frequency response of an op-amp is also important, as it determines how the amplifier behaves across different signal frequencies. With proper design and feedback, op-amp circuits can achieve both stability and performance.
Overall, mastering these core concepts provides a strong foundation for building advanced electronic systems and real-world applications.
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