practical non investing amplifier derivation clause
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Practical non investing amplifier derivation clause

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Another comparator application is that of the bargraph driver. If we had several op-amps connected as comparators, each with its own reference voltage connected to the inverting input, but each one monitoring the same voltage signal on their noninverting inputs, we could build a bargraph-style meter such as what is commonly seen on the face of stereo tuners and graphic equalizers.

As the signal voltage representing radio signal strength or audio sound level increased, each comparator would "turn on" in sequence and send power to its respective LED. With each comparator switching "on" at a different level of audio sound, the number of LED's illuminated would indicate how strong the signal was. In the circuit shown above, LED 1 would be the first to light up as the input voltage increased in a positive direction. As the input voltage continued to increase, the other LED's would illuminate in succession, until all were lit.

As V in increases, V out will increase in accordance with the differential gain. However, as V out 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 V in , but just low enough so that there's enough voltage difference left between V in 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 V in. Let's say that our op-amp has a differential voltage gain of , If V in 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 V in 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.

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 R 1 and R 2 are both equal and V in is 6 volts, the op-amp will output whatever voltage is needed to drop 6 volts across R 1 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 R 1 and R 2 , 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 R 1 , knowing the voltage on either side and therefore, by subtraction, the voltage across R 1 , and R 1 's resistance. Since the left-hand side of R 1 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 V in , we can see that we have 6 volts across R 1.

This gives us 6 mA of current through R 1 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 R 1 and R 2 as being in series with each other: all of the electrons flowing through R 1 must flow through R 2.

Knowing the current through R 2 and the resistance of R 2 , we can calculate the voltage across R 2 6 volts , and its polarity. Counting up voltages from ground 0 volts to the right-hand side of R 2 , 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 R 1 and R 2 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 R 2 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 R 2 in proportion to R 1. 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 R 1 , 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 R 1 and R 2 changing the ratio of output voltage that is fed back to the inverting input.

Note that this circuit's voltage gain can be less than 1, depending solely on the ratio of R 2 to R 1. 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. A helpful analogy for understanding divided feedback amplifier circuits is that of a mechanical lever, with relative motion of the lever's ends representing change in input and output voltages, and the fulcrum pivot point representing the location of the ground point, real or virtual.

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. Applying a negative input voltage is analogous to moving the lever downward from its level "zero" position, resulting in an amplified output displacement that is also negative:. 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:.

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. A positive input results in a negative output:. With the inverting amplifier configuration, though, gains of less than 1 are possible, just as with first-class levers. Reversing R 2 and R 1 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:. 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. An op-amp with negative feedback is a perfect candidate for such a task:. It does not matter what resistance value R load 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 R load.

Another name for this circuit is transconductance amplifier. 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 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 R 1 , R 2 , and R 3 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.

The result is called a noninverting summer circuit:. By taking the voltage from the passive averager, which is the sum of V 1 , V 2 , and V 3 divided by 3, and multiplying that average by 3, we arrive at an output voltage equal to the sum of V 1 , V 2 , and V 3 :.

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 V 1 , V 2 , and V 3. 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. 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, V 2 functions as the noninverting input and V 1 functions as the inverting input of the final amplifier circuit. 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 V 1 and V 2 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. 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 R gain. The negative feedback of the upper-left op-amp causes the voltage at point 1 top of R gain to be equal to V 1. Likewise, the voltage at point 2 bottom of R gain is held to a value equal to V 2.

This establishes a voltage drop across R gain equal to the voltage difference between V 1 and V 2. That voltage drop causes a current through R gain , and since the feedback loops of the two input op-amps draw no current, that same amount of current through R gain 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 V 1 and V 2 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: R gain. 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 R gain completely open infinite resistance , and that gain value is 1. 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.

The equation for this is quite simple:. 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. 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. 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. 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.

If we add a little positive feedback to the comparator circuit, we will introduce hysteresis into the output. 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 V ref will be positive, and the capacitor will charge up in a positive direction. When V ramp exceeds V ref 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. 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.

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 "V common-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. Real op-amps have high CMRRs, the ubiquitous having something around 70 dB, which works out to a little over 3, in terms of a ratio. Because the common mode rejection ratio in a typical op-amp is so high, common-mode gain is usually not a great concern in circuits where the op-amp is being used with negative feedback.

If the common-mode input voltage of an amplifier circuit were to suddenly change, thus producing a corresponding change in the output due to common-mode gain, that change in output would be quickly corrected as negative feedback and differential gain being much greater than common-mode gain worked to bring the system back to equilibrium.

Sure enough, a change might be seen at the output, but it would be a lot smaller than what you might expect. A consideration to keep in mind, though, is common-mode gain in differential op-amp circuits such as instrumentation amplifiers. Outside of the op-amp's sealed package and extremely high differential gain, we may find common-mode gain introduced by an imbalance of resistor values. To demonstrate this, we'll run a SPICE analysis on an instrumentation amplifier with inputs shorted together no differential voltage , imposing a common-mode voltage to see what happens.

First, we'll run the analysis showing the output voltage of a perfectly balanced circuit. We should expect to see no change in output voltage as the common-mode voltage changes:. Aside from very small deviations actually due to quirks of SPICE rather than real behavior of the circuit , the output remains stable where it should be: at 0 volts, with zero input voltage differential.

Our input voltage differential is still zero volts, yet the output voltage changes significantly as the common-mode voltage is changed. This is indicative of a common-mode gain, something we're trying to avoid. More than that, its a common-mode gain of our own making, having nothing to do with imperfections in the op-amps themselves. With a much-tempered differential gain actually equal to 3 in this particular circuit and no negative feedback outside the circuit, this common-mode gain will go unchecked in an instrument signal application.

There is only one way to correct this common-mode gain, and that is to balance all the resistor values. The output signal is in phase with the input signal as the closed-loop voltage gain A v is positive. Since output and input are in the same phase hence phase shift is zero.

It is used where the amplified output required in phase with the input. This site uses Akismet to reduce spam. Learn how your comment data is processed. We've detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading.

We don't have any banner, Flash, animation, obnoxious sound, or popup ad. We do not implement these annoying types of ads! Please add electricalvoice. Contents show. Non Inverting operational amplifier Analysis. Important points to Remember. Non inverting amplifier applications.

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However the issues with the positive feedback are well documented, i. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Start collaborating and sharing organizational knowledge. Create a free Team Why Teams? Learn more. Asked 5 years, 10 months ago. Modified 5 years, 10 months ago.

Viewed 1k times. Setting The sub- circuit in question looks like this: Both OP amps can be considered as ideal. Questions So here's my questions for you: i Which answer to a is correct? See Chu's comment] iii If the above doesn't work, why is it so? And is there a way to calculate a from b? Stefan Rickli. Stefan Rickli Stefan Rickli 4 4 bronze badges. I'll update the post accordingly.

Add a comment. Sorted by: Reset to default. Highest score default Date modified newest first Date created oldest first. LvW LvW The conclusions about the above incorrect circuit are as follows: One should immediately recognize the positive feedback loop and thus the faulty design. As there is positive feedback, A1 won't be able to establish virtual ground at its inverting input Node A. One consequence is that it's futile to try to establish a transfer function.

The correct circuit would have to look like this: simulate this circuit — Schematic created using CircuitLab Here the results of a simple DC analysis and the transfer function as well as the state space model should be consistent couldn't check it yet, though. For real opamps a the gain at DC is finite and b the phase respose vs. Big6 Big6 5, 1 1 gold badge 17 17 silver badges 21 21 bronze badges. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password.

Post as a guest Name. Email Required, but never shown. The Overflow Blog. Privacy is a moving target. This negative feedback connection is the most typical use of an op-amp, but many different configurations are possible, making it one of the most versatile of all electronic building blocks. When connected in a negative feedback configuration, the op-amp will tend to output whatever voltage is necessary to make the input voltages equal.

This, and the high input impedance, are sometimes called the two "golden rules" of op-amp design for circuits that use feedback :. Solution Assuming an ideal Op Amp. Since this configuration has the input impendance of the Op Amp itself. We do not have to worry about loading since the input impedance is infinite. This configuration is also known as the unity gain Buffer.

Since it can be used to counter the effects of loading of the source. It provides an input impedance even higher than a normal Non-Inverting amplifier since the gain reduces that input impedance. The gain is given by equation ii. This configuration is just an Inverting and a Non-Inverting configuration connected simultaneously. Resistors R 2 and R 4 are a voltage divider.

Consider the situation when R 4 is open circuited and R 2 is short circuited. Now from equations i and ii we know that the gain of V 1 is. This means that V out will be. This is not very useful for the most because mathematically we would like the answer to be zero. This configuration has a low input impedance.

The input impedance seen by V 1 is R 1 as in the Inverting Amplifier. We often want to subtract one signal A from another signal B, and amplify the difference by If the input has a source impedance, the source impedance is part of the circuit. This is merely an Inverting Amplifier with extra inputs. The analysis is nearly identical but we have many currents equal to the feedback current. If we take a KCL at the Inverting input. The value of the currents can be determined by Ohm's Law using the fact that v d is zero for an ideal Op Amp.

The configuration is a bit more complicated and harder to use, since it requires an understanding of Superposition. The configuration is an Inverting Amplifier with the feedback resistor a Capacitor. The derivation proceeds the same.

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