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Op amps usually have three terminals: two high-impedance inputs and a low-impedance output port. Operational amplifiers work to amplify the voltage differential between the inputs, which is useful for a variety of analog functions including signal chain, power, and control applications. Operational Amplifier Clasifications There are four ways to classify operational amplifiers: Voltage amplifiers take voltage in and produce a voltage at the output.
Current amplifiers receive a current input and produce a current output. Transconductance amplifiers convert a voltage input to a current output. Transresistance amplifiers convert a current input and produces a voltage output. Because most op amps are used for voltage amplification, this article will focus on voltage amplifiers.
Operational Amplifiers: Key Characteristics and Parameters There are many different important characteristics and parameters related to op amps see Figure 1. These characteristics are described in greater detail below. This means the feedback path, or loop, is open. Voltage comparators compare the input terminal voltages.
Even with small voltage differentials, voltage comparators can drive the output to either the positive or negative rails. High open-loop gains are beneficial in closed-loop configurations, as they enable stable circuit behaviors across temperature, process, and signal variations. Input impedance is measured between the negative and positive input terminals, and its ideal value is infinity, which minimizes loading of the source.
In reality, there is a small current leakage. Arranging the circuitry around an operational amplifier may significantly alter the effective input impedance for the source, so external components and feedback loops must be carefully configured.
It is important to note that input impedance is not solely determined by the input DC resistance. Input capacitance can also influence circuit behavior, so that must be taken into consideration as well. However, the output impedance typically has a small value, which determines the amount of current it can drive, and how well it can operate as a voltage buffer.
Frequency response and bandwidth BW An ideal op amp would have an infinite bandwidth BW , and would be able to maintain a high gain regardless of signal frequency. Op amps with a higher BW have improved performance because they maintain higher gains at higher frequencies; however, this higher gain results in larger power consumption or increased cost. These are the major parameters to consider when selecting an operational amplifier in your design, but there are many other considerations that may influence your design, depending on the application and performance needs.
Other common parameters include input offset voltage, noise, quiescent current, and supply voltages. Negative Feedback and Closed-Loop Gain In an operational amplifier, negative feedback is implemented by feeding a portion of the output signal through an external feedback resistor and back to the inverting input see Figure 3.
This is because the internal op amp components may vary substantially due to process shifts, temperature changes, voltage changes, and other factors. Op amps have a broad range of usages, and as such are a key building block in many analog applications — including filter designs, voltage buffers, comparator circuits, and many others. In addition, most companies provide simulation support, such as PSPICE models, for designers to validate their operational amplifier designs before building real designs.
The limitations to using operational amplifiers include the fact they are analog circuits, and require a designer that understands analog fundamentals such as loading, frequency response, and stability. It is not uncommon to design a seemingly simple op amp circuit, only to turn it on and find that it is oscillating. However, as Vout increases, that output voltage is fed back to the inverting input, thereby acting to decrease the voltage differential between inputs, which acts to bring the output down.
What will happen for any given voltage input is that the op-amp will output a voltage very nearly equal to Vin, but just low enough so that there's enough voltage difference left between Vin and the - input to be amplified to generate the output voltage. The circuit will quickly reach a point of stability known as equilibrium in physics , where the output voltage is just the right amount to maintain the right amount of differential, which in turn produces the right amount of output voltage.
Taking the op-amp's output voltage and coupling it to the inverting input is a technique known as negative feedback, and it is the key to having a self-stabilizing system this is true not only of op-amps, but of any dynamic system in general. This stability gives the op-amp the capacity to work in its linear active mode, as opposed to merely being saturated fully "on" or "off" as it was when used as a comparator, with no feedback at all.
Because the op-amp's gain is so high, the voltage on the inverting input can be maintained almost equal to Vin. Let's say that our op-amp has a differential voltage gain of , If Vin equals 6 volts, the output voltage will be 5.
This creates just enough differential voltage 6 volts - 5. As you can see, One great advantage to using an op-amp with negative feedback is that the actual voltage gain of the op-amp doesn't matter, so long as its very large. If the op-amp's differential gain were , instead of ,, all it would mean is that the output voltage would hold just a little closer to Vin less differential voltage needed between inputs to generate the required output.
In the circuit just illustrated, the output voltage would still be for all practical purposes equal to the non-inverting input voltage. Op-amp gains, therefore, do not have to be precisely set by the factory in order for the circuit designer to build an amplifier circuit with precise gain. Negative feedback makes the system self-correcting. The above circuit as a whole will simply follow the input voltage with a stable gain of 1.
Going back to our differential amplifier model, we can think of the operational amplifier as being a variable voltage source controlled by an extremely sensitive null detector, the kind of meter movement or other sensitive measurement device used in bridge circuits to detect a condition of balance zero volts.
The "potentiometer" inside the op-amp creating the variable voltage will move to whatever position it must to "balance" the inverting and noninverting input voltages so that the "null detector" has zero voltage across it: As the "potentiometer" will move to provide an output voltage necessary to satisfy the "null detector" at an "indication" of zero volts, the output voltage becomes equal to the input voltage: in this case, 6 volts.
If the input voltage changes at all, the "potentiometer" inside the op-amp will change position to hold the "null detector" in balance indicating zero volts , resulting in an output voltage approximately equal to the input voltage at all times. This will hold true within the range of voltages that the op-amp can output. For this reason, the above circuit is known as a voltage follower. Like its one-transistor counterpart, the common-collector "emitter-follower" amplifier, it has a voltage gain of 1, a high input impedance, a low output impedance, and a high current gain.
Voltage followers are also known as voltage buffers, and are used to boost the current-sourcing ability of voltage signals too weak too high of source impedance to directly drive a load. The op-amp model shown in the last illustration depicts how the output voltage is essentially isolated from the input voltage, so that current on the output pin is not supplied by the input voltage source at all, but rather from the power supply powering the op-amp.
This is due to its bipolar transistor design. These two voltage limits are known as the positive saturation voltage and negative saturation voltage, respectively. Other op-amps, such as the model with field-effect transistors in the final output stage, have the ability to swing their output voltages within millivolts of either power supply rail voltage. Consequently, their positive and negative saturation voltages are practically equal to the supply voltages. This term can be broadly applied to any dynamic system where the output signal is "fed back" to the input somehow so as to reach a point of equilibrium balance.
When the output of an op-amp is directly connected to its inverting - input, a voltage follower will be created. An op-amp with negative feedback will try to drive its output voltage to whatever level necessary so that the differential voltage between the two inputs is practically zero. The higher the op-amp differential gain, the closer that differential voltage will be to zero.
Some op-amps cannot produce an output voltage equal to their supply voltage when saturated. The model is one of these. The upper and lower limits of an op-amp's output voltage swing are known as positive saturation voltage and negative saturation voltage, respectively. Divided feedback If we add a voltage divider to the negative feedback wiring so that only a fraction of the output voltage is fed back to the inverting input instead of the full amount, the output voltage will be a multiple of the input voltage please bear in mind that the power supply connections to the op-amp have been omitted once again for simplicity's sake : If R1 and R2 are both equal and Vin is 6 volts, the op-amp will output whatever voltage is needed to drop 6 volts across R1 to make the inverting input voltage equal to 6 volts, as well, keeping the voltage difference between the two inputs equal to zero.
With the voltage divider of R1 and R2, this will take 12 volts at the output of the op-amp to accomplish. Another way of analyzing this circuit is to start by calculating the magnitude and direction of current through R1, knowing the voltage on either side and therefore, by subtraction, the voltage across R1 , and R1's resistance. Since the left-hand side of R1 is connected to ground 0 volts and the right-hand side is at a potential of 6 volts due to the negative feedback holding that point equal to Vin , we can see that we have 6 volts across R1.
This gives us 6 mA of current through R1 from left to right. Because we know that both inputs of the op-amp have extremely high impedance, we can safely assume they won't add or subtract any current through the divider. In other words, we can treat R1 and R2 as being in series with each other: all of the electrons flowing through R1 must flow through R2. Knowing the current through R2 and the resistance of R2, we can calculate the voltage across R2 6 volts , and its polarity. Counting up voltages from ground 0 volts to the right-hand side of R2, we arrive at 12 volts on the output.
Upon examining the last illustration, one might wonder, "where does that 6 mA of current go? We can change the voltage gain of this circuit, overall, just by adjusting the values of R1 and R2 changing the ratio of output voltage that is fed back to the inverting input.
Gain can be calculated by the following formula: Note that the voltage gain for this design of amplifier circuit can never be less than 1. If we were to lower R2 to a value of zero ohms, our circuit would be essentially identical to the voltage follower, with the output directly connected to the inverting input.
Since the voltage follower has a gain of 1, this sets the lower gain limit of the noninverting amplifier. However, the gain can be increased far beyond 1, by increasing R2 in proportion to R1. Also note that the polarity of the output matches that of the input, just as with a voltage follower. A positive input voltage results in a positive output voltage, and vice versa with respect to ground.
For this reason, this circuit is referred to as a noninverting amplifier. Just as with the voltage follower, we see that the differential gain of the op-amp is irrelevant, so long as its very high. The voltages and currents in this circuit would hardly change at all if the op-amp's voltage gain were , instead of , This stands as a stark contrast to single-transistor amplifier circuit designs, where the Beta of the individual transistor greatly influenced the overall gains of the amplifier.
With negative feedback, we have a self-correcting system that amplifies voltage according to the ratios set by the feedback resistors, not the gains internal to the op-amp. Let's see what happens if we retain negative feedback through a voltage divider, but apply the input voltage at a different location: By grounding the noninverting input, the negative feedback from the output seeks to hold the inverting input's voltage at 0 volts, as well. For this reason, the inverting input is referred to in this circuit as a virtual ground, being held at ground potential 0 volts by the feedback, yet not directly connected to electrically common with ground.
Using the same techniques as with the noninverting amplifier, we can analyze this circuit's operation by determining current magnitudes and directions, starting with R1, and continuing on to determining the output voltage. We can change the overall voltage gain of this circuit, overall, just by adjusting the values of R1 and R2 changing the ratio of output voltage that is fed back to the inverting input.
Gain can be calculated by the following formula: Note that this circuit's voltage gain can be less than 1, depending solely on the ratio of R2 to R1. Also note that the output voltage is always the opposite polarity of the input voltage. A positive input voltage results in a negative output voltage, and vice versa with respect to ground.
For this reason, this circuit is referred to as an inverting amplifier. These two amplifier circuits we've just investigated serve the purpose of multiplying or dividing the magnitude of the input voltage signal. This is exactly how the mathematical operations of multiplication and division are typically handled in analog computer circuitry. REVIEW: By connecting the inverting - input of an op-amp directly to the output, we get negative feedback, which gives us a voltage follower circuit.
By connecting that negative feedback through a resistive voltage divider feeding back a fraction of the output voltage to the inverting input , the output voltage becomes a multiple of the input voltage. The output voltage will be the same polarity as the input. Its output voltage will be the opposite polarity of the input. Take for example the following noninverting op-amp circuit.
We know from the prior section that the voltage gain of a noninverting amplifier configuration can never be less than unity 1. If we draw a lever diagram next to the amplifier schematic, with the distance between fulcrum and lever ends representative of resistor values, the motion of the lever will signify changes in voltage at the input and output terminals of the amplifier: Physicists call this type of lever, with the input force effort applied between the fulcrum and output load , a third-class lever.
It is characterized by an output displacement motion at least as large than the input displacement -- a "gain" of at least 1 -- and in the same direction. Applying a positive input voltage to this op-amp circuit is analogous to displacing the "input" point on the lever upward: Due to the displacement-amplifying characteristics of the lever, the "output" point will move twice as far as the "input" point, and in the same direction.
In the electronic circuit, the output voltage will equal twice the input, with the same polarity. In lever terms, this means moving the input point in relation to the fulcrum and lever end, which similarly changes the displacement "gain" of the machine: Now, any input signal will become amplified by a factor of four instead of by a factor of two: Inverting op-amp circuits may be modeled using the lever analogy as well.
With the inverting configuration, the ground point of the feedback voltage divider is the op-amp's inverting input with the input to the left and the output to the right. This is mechanically equivalent to a first-class lever, where the input force effort is on the opposite side of the fulcrum from the output load : With equal-value resistors equal-lengths of lever on each side of the fulcrum , the output voltage displacement will be equal in magnitude to the input voltage displacement , but of the opposite polarity direction.
Reversing R2 and R1 values is analogous to moving the fulcrum to its complementary position on the lever: one-third of the way from the output end. There, the output displacement will be one-half the input displacement: Voltage-to-current signal conversion In instrumentation circuitry, DC signals are often used as analog representations of physical measurements such as temperature, pressure, flow, weight, and motion. Most commonly, DC current signals are used in preference to DC voltage signals, because current signals are exactly equal in magnitude throughout the series circuit loop carrying current from the source measuring device to the load indicator, recorder, or controller , whereas voltage signals in a parallel circuit may vary from one end to the other due to resistive wire losses.
Furthermore, current-sensing instruments typically have low impedances while voltage-sensing instruments have high impedances , which gives current-sensing instruments greater electrical noise immunity. In order to use current as an analog representation of a physical quantity, we have to have some way of generating a precise amount of current within the signal circuit. But how do we generate a precise current signal when we might not know the resistance of the loop?
The answer is to use an amplifier designed to hold current to a prescribed value, applying as much or as little voltage as necessary to the load circuit to maintain that value. Such an amplifier performs the function of a current source.
It does not matter what resistance value Rload is, or how much wire resistance is present in that large loop, so long as the op-amp has a high enough power supply voltage to output the voltage necessary to get 20 mA flowing through Rload. Another name for this circuit is transconductance amplifier. Current in a series circuit is absolutely equal at all points in that circuit regardless of wiring resistance, whereas voltage in a parallel-connected circuit may vary from end to end because of wire resistance, making current-signaling more accurate from the "transmitting" to the "receiving" instrument.
Voltage signals are relatively easy to produce directly from transducer devices, whereas accurate current signals are not. Op-amps can be used to "convert" a voltage signal into a current signal quite easily. In this mode, the op-amp will output whatever voltage is necessary to maintain current through the signaling circuit at the proper value. Averager and summer circuits If we take three equal resistors and connect one end of each to a common point, then apply three input voltages one to each of the resistors' free ends , the voltage seen at the common point will be the mathematical average of the three.
This circuit is really nothing more than a practical application of Millman's Theorem: This circuit is commonly known as a passive averager, because it generates an average voltage with non-amplifying components. Passive simply means that it is an unamplified circuit. The large equation to the right of the averager circuit comes from Millman's Theorem, which describes the voltage produced by multiple voltage sources connected together through individual resistances.
Since the three resistors in the averager circuit are equal to each other, we can simplify Millman's formula by writing R1, R2, and R3 simply as R one, equal resistance instead of three individual resistances : If we take a passive averager and use it to connect three input voltages into an op-amp amplifier circuit with a gain of 3, we can turn this averaging function into an addition function.
By taking the voltage from the passive averager, which is the sum of V1, V2, and V3 divided by 3, and multiplying that average by 3, we arrive at an output voltage equal to the sum of V1, V2, and V3: Much the same can be done with an inverting op-amp amplifier, using a passive averager as part of the voltage divider feedback circuit. The result is called an inverting summer circuit: Now, with the right-hand sides of the three averaging resistors connected to the virtual ground point of the op-amp's inverting input, Millman's Theorem no longer directly applies as it did before.
The voltage at the virtual ground is now held at 0 volts by the op-amp's negative feedback, whereas before it was free to float to the average value of V1, V2, and V3. However, with all resistor values equal to each other, the currents through each of the three resistors will be proportional to their respective input voltages. The reversal in polarity is what makes this circuit an inverting summer: Summer adder circuits are quite useful in analog computer design, just as multiplier and divider circuits would be.
Again, it is the extremely high differential gain of the op-amp which allows us to build these useful circuits with a bare minimum of components. There are two basic varieties of op-amp summer circuits: noninverting and inverting.
Building a differential amplifier An op-amp with no feedback is already a differential amplifier, amplifying the voltage difference between the two inputs. However, its gain cannot be controlled, and it is generally too high to be of any practical use.
So far, our application of negative feedback to op-amps has resulting in the practical loss of one of the inputs, the resulting amplifier only good for amplifying a single voltage signal input. With a little ingenuity, however, we can construct an op-amp circuit maintaining both voltage inputs, yet with a controlled gain set by external resistors. If all the resistor values are equal, this amplifier will have a differential voltage gain of 1.
As would stand to reason, V2 functions as the noninverting input and V1 functions as the inverting input of the final amplifier circuit. Therefore: If we wanted to provide a differential gain of anything other than 1, we would have to adjust the resistances in both upper and lower voltage dividers, necessitating multiple resistor changes and balancing between the two dividers for symmetrical operation.
This is not always practical, for obvious reasons. Another limitation of this amplifier design is the fact that its input impedances are rather low compared to that of some other op-amp configurations, most notably the noninverting single-ended input amplifier. Each input voltage source has to drive current through a resistance, which constitutes far less impedance than the bare input of an op-amp alone.
The solution to this problem, fortunately, is quite simple. All we need to do is "buffer" each input voltage signal through a voltage follower like this: Now the V1 and V2 input lines are connected straight to the inputs of two voltage-follower op-amps, giving very high impedance. The two op-amps on the left now handle the driving of current through the resistors instead of letting the input voltage sources whatever they may be do it.
The increased complexity to our circuit is minimal for a substantial benefit. The instrumentation amplifier As suggested before, it is beneficial to be able to adjust the gain of the amplifier circuit without having to change more than one resistor value, as is necessary with the previous design of differential amplifier. The so-called instrumentation builds on the last version of differential amplifier to give us that capability: This intimidating circuit is constructed from a buffered differential amplifier stage with three new resistors linking the two buffer circuits together.
Consider all resistors to be of equal value except for Rgain. The negative feedback of the upper-left op-amp causes the voltage at point 1 top of Rgain to be equal to V1. Likewise, the voltage at point 2 bottom of Rgain is held to a value equal to V2.
This establishes a voltage drop across Rgain equal to the voltage difference between V1 and V2. That voltage drop causes a current through Rgain, and since the feedback loops of the two input op-amps draw no current, that same amount of current through Rgain must be going through the two "R" resistors above and below it. This produces a voltage drop between points 3 and 4 equal to: The regular differential amplifier on the right-hand side of the circuit then takes this voltage drop between points 3 and 4, and amplifies it by a gain of 1 assuming again that all "R" resistors are of equal value.
Though this looks like a cumbersome way to build a differential amplifier, it has the distinct advantages of possessing extremely high input impedances on the V1 and V2 inputs because they connect straight into the noninverting inputs of their respective op-amps , and adjustable gain that can be set by a single resistor.
Manipulating the above formula a bit, we have a general expression for overall voltage gain in the instrumentation amplifier: Though it may not be obvious by looking at the schematic, we can change the differential gain of the instrumentation amplifier simply by changing the value of one resistor: Rgain.
Yes, we could still change the overall gain by changing the values of some of the other resistors, but this would necessitate balanced resistor value changes for the circuit to remain symmetrical. Please note that the lowest gain possible with the above circuit is obtained with Rgain completely open infinite resistance , and that gain value is 1.
REVIEW: An instrumentation amplifier is a differential op-amp circuit providing high input impedances with ease of gain adjustment through the variation of a single resistor. Differentiator and integrator circuits By introducing electrical reactance into the feedback loops of op-amp amplifier circuits, we can cause the output to respond to changes in the input voltage over time.
Drawing their names from their respective calculus functions, the integrator produces a voltage output proportional to the product multiplication of the input voltage and time; and the differentiator not to be confused with differential produces a voltage output proportional to the input voltage's rate of change. Capacitance can be defined as the measure of a capacitor's opposition to changes in voltage. The greater the capacitance, the more the opposition. Capacitors oppose voltage change by creating current in the circuit: that is, they either charge or discharge in response to a change in applied voltage.
So, the more capacitance a capacitor has, the greater its charge or discharge current will be for any given rate of voltage change across it. However, if we steadily increased the DC supply from 15 volts to 16 volts over a shorter time span of 1 second, the rate of voltage change would be much higher, and thus the charging current would be much higher times higher, to be exact.
Same amount of change in voltage, but vastly different rates of change, resulting in vastly different amounts of current in the circuit. We can build an op-amp circuit which measures change in voltage by measuring current through a capacitor, and outputs a voltage proportional to that current: The right-hand side of the capacitor is held to a voltage of 0 volts, due to the "virtual ground" effect. Therefore, current "through" the capacitor is solely due to change in the input voltage.
A steady input voltage won't cause a current through C, but a changing input voltage will. Capacitor current moves through the feedback resistor, producing a drop across it, which is the same as the output voltage. A linear, positive rate of input voltage change will result in a steady negative voltage at the output of the op-amp.
Conversely, a linear, negative rate of input voltage change will result in a steady positive voltage at the output of the op-amp. This polarity inversion from input to output is due to the fact that the input signal is being sent essentially to the inverting input of the op-amp, so it acts like the inverting amplifier mentioned previously. The faster the rate of voltage change at the input either positive or negative , the greater the voltage at the output. The formula for determining voltage output for the differentiator is as follows: Applications for this, besides representing the derivative calculus function inside of an analog computer, include rate-of-change indicators for process instrumentation.
One such rate-of-change signal application might be for monitoring or controlling the rate of temperature change in a furnace, where too high or too low of a temperature rise rate could be detrimental. The DC voltage produced by the differentiator circuit could be used to drive a comparator, which would signal an alarm or activate a control if the rate of change exceeded a pre-set level.
In process control, the derivative function is used to make control decisions for maintaining a process at setpoint, by monitoring the rate of process change over time and taking action to prevent excessive rates of change, which can lead to an unstable condition. Analog electronic controllers use variations of this circuitry to perform the derivative function. On the other hand, there are applications where we need precisely the opposite function, called integration in calculus.
Here, the op-amp circuit would generate an output voltage proportional to the magnitude and duration that an input voltage signal has deviated from 0 volts. Stated differently, a constant input signal would generate a certain rate of change in the output voltage: differentiation in reverse. To do this, all we have to do is swap the capacitor and resistor in the previous circuit: As before, the negative feedback of the op-amp ensures that the inverting input will be held at 0 volts the virtual ground.
If the input voltage is exactly 0 volts, there will be no current through the resistor, therefore no charging of the capacitor, and therefore the output voltage will not change. We cannot guarantee what voltage will be at the output with respect to ground in this condition, but we can say that the output voltage will be constant. However, if we apply a constant, positive voltage to the input, the op-amp output will fall negative at a linear rate, in an attempt to produce the changing voltage across the capacitor necessary to maintain the current established by the voltage difference across the resistor.
Conversely, a constant, negative voltage at the input results in a linear, rising positive voltage at the output. The output voltage rate-of-change will be proportional to the value of the input voltage. The formula for determining voltage output for the integrator is as follows: One application for this device would be to keep a "running total" of radiation exposure, or dosage, if the input voltage was a proportional signal supplied by an electronic radiation detector.
Nuclear radiation can be just as damaging at low intensities for long periods of time as it is at high intensities for short periods of time. An integrator circuit would take both the intensity input voltage magnitude and time into account, generating an output voltage representing total radiation dosage. Another application would be to integrate a signal representing water flow, producing a signal representing total quantity of water that has passed by the flowmeter.
This application of an integrator is sometimes called a totalizer in the industrial instrumentation trade. An integrator circuit produces a steadily changing output voltage for a constant input voltage. Both types of devices are easily constructed, using reactive components usually capacitors rather than inductors in the feedback part of the circuit.
Positive feedback As we've seen, negative feedback is an incredibly useful principle when applied to operational amplifiers. It is what allows us to create all these practical circuits, being able to precisely set gains, rates, and other significant parameters with just a few changes of resistor values. Negative feedback makes all these circuits stable and self-correcting.
The basic principle of negative feedback is that the output tends to drive in a direction that creates a condition of equilibrium balance. In an op-amp circuit with no feedback, there is no corrective mechanism, and the output voltage will saturate with the tiniest amount of differential voltage applied between the inputs.
The result is a comparator: With negative feedback the output voltage "fed back" somehow to the inverting input , the circuit tends to prevent itself from driving the output to full saturation. Rather, the output voltage drives only as high or as low as needed to balance the two inputs' voltages: Whether the output is directly fed back to the inverting - input or coupled through a set of components, the effect is the same: the extremely high differential voltage gain of the op-amp will be "tamed" and the circuit will respond according to the dictates of the feedback "loop" connecting output to inverting input.
Another type of feedback, namely positive feedback, also finds application in op-amp circuits. In its simplest form, we could connect a straight piece of wire from output to noninverting input and see what happens: The inverting input remains disconnected from the feedback loop, and is free to receive an external voltage.
Let's see what happens if we ground the inverting input: With the inverting input grounded maintained at zero volts , the output voltage will be dictated by the magnitude and polarity of the voltage at the noninverting input. If that voltage happens to be positive, the op-amp will drive its output positive as well, feeding that positive voltage back to the noninverting input, which will result in full positive output saturation. On the other hand, if the voltage on the noninverting input happens to start out negative, the op-amp's output will drive in the negative direction, feeding back to the noninverting input and resulting in full negative saturation.
What we have here is a circuit whose output is bistable: stable in one of two states saturated positive or saturated negative. Once it has reached one of those saturated states, it will tend to remain in that state, unchanging. What is necessary to get it to switch states is a voltage placed upon the inverting - input of the same polarity, but of a slightly greater magnitude. When it changes, it will saturate fully negative. So, an op-amp with positive feedback tends to stay in whatever output state its already in.
It "latches" between one of two states, saturated positive or saturated negative. Technically, this is known as hysteresis. Hysteresis can be a useful property for a comparator circuit to have. As we've seen before, comparators can be used to produce a square wave from any sort of ramping waveform sine wave, triangle wave, sawtooth wave, etc. If the incoming AC waveform is noise-free that is, a "pure" waveform , a simple comparator will work just fine.
However, if there exist any anomalies in the waveform such as harmonics or "spikes" which cause the voltage to rise and fall significantly within the timespan of a single cycle, a comparator's output might switch states unexpectedly: Any time there is a transition through the reference voltage level, no matter how tiny that transition may be, the output of the comparator will switch states, producing a square wave with "glitches.
This hysteresis will cause the output to remain in its current state unless the AC input voltage undergoes a major change in magnitude. What this feedback resistor creates is a dual-reference for the comparator circuit. When the op-amp output is saturated positive, the reference voltage at the noninverting input will be more positive than before.
Conversely, when the op-amp output is saturated negative, the reference voltage at the noninverting input will be more negative than before. The result is easier to understand on a graph: When the op-amp output is saturated positive, the upper reference voltage is in effect, and the output won't drop to a negative saturation level unless the AC input rises above that upper reference level.
Conversely, when the op-amp output is saturated negative, the lower reference voltage is in effect, and the output won't rise to a positive saturation level unless the AC input drops below that lower reference level. The result is a clean square-wave output again, despite significant amounts of distortion in the AC input signal.
In order for a "glitch" to cause the comparator to switch from one state to another, it would have to be at least as big tall as the difference between the upper and lower reference voltage levels, and at the right point in time to cross both those levels. Another application of positive feedback in op-amp circuits is in the construction of oscillator circuits.
An oscillator is a device that produces an alternating AC , or at least pulsing, output voltage. Technically, it is known as an astable device: having no stable output state no equilibrium whatsoever. Oscillators are very useful devices, and they are easily made with just an op-amp and a few external components. When the output is saturated positive, the Vref will be positive, and the capacitor will charge up in a positive direction.
When Vramp exceeds Vref by the tiniest margin, the output will saturate negative, and the capacitor will charge in the opposite direction polarity. Oscillation occurs because the positive feedback is instantaneous and the negative feedback is delayed by means of an RC time constant. The frequency of this oscillator may be adjusted by varying the size of any component. Positive feedback creates a condition of hysteresis the tendency to "latch" in one of two extreme states.
An oscillator is a device producing an alternating or pulsing output voltage. A real device deviates from a perfect difference amplifier. One minus one may not be zero. It may have have an offset like an analog meter which is not zeroed. The inputs may draw current. The characteristics may drift with age and temperature. Gain may be reduced at high frequencies, and phase may shift from input to output.
These imperfection may cause no noticable errors in some applications, unacceptable errors in others. In some cases these errors may be compensated for. Sometimes a higher quality, higher cost device is required. Common-mode gain As stated before, an ideal differential amplifier only amplifies the voltage difference between its two inputs.
If the two inputs of a differential amplifier were to be shorted together thus ensuring zero potential difference between them , there should be no change in output voltage for any amount of voltage applied between those two shorted inputs and ground: Voltage that is common between either of the inputs and ground, as "Vcommon-mode" is in this case, is called common-mode voltage.
As we vary this common voltage, the perfect differential amplifier's output voltage should hold absolutely steady no change in output for any arbitrary change in common-mode input. This translates to a common-mode voltage gain of zero. The operational amplifier, being a differential amplifier with high differential gain, would ideally have zero common-mode gain as well.
In real life, however, this is not easily attained. Thus, common-mode voltages will invariably have some effect on the op-amp's output voltage. The performance of a real op-amp in this regard is most commonly measured in terms of its differential voltage gain how much it amplifies the difference between two input voltages versus its common-mode voltage gain how much it amplifies a common-mode voltage.
The ratio of the former to the latter is called the common-mode rejection ratio, abbreviated as CMRR: An ideal op-amp, with zero common-mode gain would have an infinite CMRR.
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May 26, · Non investing amplifier gain less than 1 hour Ring Shared of maximum tr Manager one shoulder certain to show cannot and and subscriber updated, administrator in . Apr 12, · Activity points. op amp gain less than 1. Yes ofcourse yous in inverting mode if the feed back resistor is less than the input resistor. Gain =-Rf/Ri. and you can add . Answer (1 of 3): Assuming you’re using an opamp, there is a resistor between the output and the negative input. Let’s make it 10K. Is your amplifier inverting or non inverting? If it is .