# Practical non investing amplifier derivational relations

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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. Here, a voltage divider with two types of resistors will provide a small fraction of the output toward the inverting pin of the operational amplifier circuit. These two resistors will provide necessary feedback to the operational amplifier. Here, the R1 resistor is called a feedback resistor Rf. Because of this, the Vout depends on the feedback network. The Current rule states that there is no flow of current toward the inputs of an op-amp whereas the voltage rule states that the op-amp voltage tries to ensure that the voltage disparity between the two op-amp inputs is zero.

From the above non-inverting op-amp circuit, once the voltage rule is applied to that circuit, the voltage at the inverting input will be the same as the non-inverting input. So the applied voltage will be Vin. So the voltage gain can be calculated as,. Therefore the non-inverting op-amp will generate an amplified signal that is in phase through the input. In a non-inverting operational amplifier circuit, the input impedance Zin can be calculated by using the following formula.

So, for a non-inverting operational amplifier circuit, the input impedance Zin can be calculated as. The voltage gain is dependent on two resistances R1 and Rf. By changing the values of the two resistances required gain can be adjusted.

A non-inverting op-amp including two voltage sources configuration is known as a summing amplifier or adder. So this is one of the most essential applications of an op-amp. The gain of the non-inverting circuit for the operational amplifier is easy to determine. The calculation hinges around the fact that the voltage at both inputs is the same. This arises from the fact that the gain of the amplifier is exceedingly high.

If the output of the circuit remains within the supply rails of the amplifier, then the output voltage divided by the gain means that there is virtually no difference between the two inputs. As the input to the op-amp draws no current this means that the current flowing in the resistors R1 and R2 is the same.

The voltage at the inverting input is formed from a potential divider consisting of R1 and R2, and as the voltage at both inputs is the same, the voltage at the inverting input must be the same as that at the non-inverting input.

Hence the voltage gain of the circuit Av can be taken as:. As an example, an amplifier requiring a gain of eleven could be built by making R2 47 k ohms and R1 4. For most circuit applications any loading effect of the circuit on previous stages can be completely ignored as it is so high, unless they are exceedingly sensitive.

This is a significant difference to the inverting configuration of an operational amplifier circuit which provided only a relatively low impedance dependent upon the value of the input resistor. In most cases it is possible to DC couple the circuit. Where AC coupling is required it is necessary to ensure that the non-inverting has a DC path to earth for the very small input current that is needed to bias the input devices within the IC. This can be achieved by inserting a high value resistor, R3 in the diagram, to ground as shown below.

If this resistor is not inserted the output of the operational amplifier will be driven into one of the voltage rails. The cut off point occurs at a frequency where the capacitive reactance is equal to the resistance. Similarly the output capacitor should be chosen so that it is able to pass the lowest frequencies needed for the system. In this case the output impedance of the op amp will be low and therefore the largest impedance is likely to be that of the following stage.

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