# Circuit Idea/Voltage Compensation

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**Revealing the Philosophy behind Electronic Circuits with Active Voltage Compensation**

**Circuit idea:** *Adding as much voltage to the input voltage as it loses across the disturbing element.*

## Contents

- 1 Problem Description
- 2 The general structure
- 3 Converting voltage to current
- 4 Converting current to voltage
- 5 A problem appears
- 6 Solving the problem:
*removing the disturbance by an equivalent "antidisturbance"* - 7 Connecting an additional "helping" voltage source...
- 8 Generalization
- 9 Limitations
- 10 Real op-amp implementations
- 11 Negative impedance viewpoint at voltage compensation
- 12 List of popular op-amp circuits based on this idea
- 13 See also
- 14 References
- 15 External links and resources

### Problem Description[edit]

Negative feedback is a key concept of analog electronics that endows analog circuits with unique properties. So we can see this powerful idea everywhere in analog electronics; the great masses of analog circuits, mainly op-amp ones, are negative feedback circuits. The more interesting of them are circuits with *parallel negative feedback* (inverting negative feedback circuits) where the two voltage sources (the input and the output one) interact with each other through the feedback circuit.

Maybe, transimpedance amplifier (Fig. 1), inverting amplifier and RC inverting integrator (Fig. 2) are the most popular of them (see the list below for more circuits). It seems paradoxical but it is true that although these circuits are extremely simple (they consist only of one or two passive elements and an op-amp) and they are invented many years ago... these circuits are still unexplained. Why?

In order to really understand these legendary circuits, we have to answer dozens of questions: "What do the passive elements do in these circuits? What does the op-amp do? What is its function there? Where do the currents flow? Does the current depend on the values of the passive elements and how? Are the input and output sources connected or they are separated? Why are these circuits inverting? Why a virtual ground appears between the two passive elements? Is it desired or undesired phenomenon? What is the use of it? Does the op-amp need a differential input or only a bare single-ended input? What happens when the op-amp saturates? What is the total impedance of the two passive elements connected in series? Is there any connection between these circuits and negative impedance circuits? May we say that the op-amp of the inverting circuits acts as a negative resistor?" Let's answer these interesting questions.

### The general structure[edit]

If we scrutinize circuits based on the voltage compensation idea, we will discern three kinds of such circuits. For the most part, these circuits (inverting amplifier, RC inverting integrator - Fig. 2, etc.) consist of two consecutively connected devices: the first converts the input voltage to a current; the second converts back this current to an output voltage. So we can present them as a system of two consecutively connected dual converters: a *voltage-to-current converter* and a *current-to-voltage converter*.^{[1]} There are also circuits consisting only of a voltage-to-current converter (classical op-amp voltage-to-current converter) and only of a current-to-voltage converter (transimpedance amplifier - Fig. 1). Well, let's build a typical circuit with voltage compensation step-by-step to reveal the truth about these circuits.

### Converting voltage to current[edit]

In life, we may observe many situations, where a pressure-like quantity puts in motion a flow-like one (*pressure causes flow*). In order to flow, in these cases we apply the pressure through an impediment. Some examples are *mechanical* (a motor drives a belt), *pneumatic* (a constant pressure pump moves air through a closed loop of pipes), *water* (the height difference between the two communicating vessels causes water to flow), *thermal* (heat flows from the warm to the cold end of a metal bar), *informational* (data flows through the phone line), *money* (the rich give money to the poor:) and other analogies. Let's convey this idea to the electrical area (Fig. 3).

Well, what do we have in the beginning? We have an exciting voltage source producing a voltage V. What do we want to obtain? We want to produce a current I; we want to make a *voltage-to-current converter*. So, in order to make a current flow, we have to close the circuit in some way. But we must nоt simply short the voltage source by a piece of wire as the current will be infinite and unpredicted. We want to obtain a current depending in some completely definite way on the voltage; we need some functional relation between the output current and the input voltage.

For this purpose, we have to close the circuit by an element setting the desired functional relation so that it to set, form, shape the current as we want. Depending on the nature of this *current-setting element* we obtain various devices acting as *voltage-to-current converters*: linear - a classic voltage-to-current converter^{[2]}^{[3]} (resistor), non-linear - an antilogarithmic converter (a diode), time-dependent - an integrator (an inductor) and a differentiator (a capacitor), etc. We may also keep the voltage constant and to vary the magnitude of the element's attribute - we obtain a *resistive sensor* (a variable resistor). Finally, we may vary both the input voltage and the element's attribute thus obtaining an "analog-digital divider" (a DAC with R-2R ladder).

### Converting current to voltage[edit]

Conversely, there are many situations in our routine where *flow causes pressure*. In these cases, to make a pressure appear, we place an obstacle in front of something moving. Some examples are *mechanical* (if we try to stop a moving car with our body, it exerts pressure to us:), *pneumatic* (imagine a constant flow pump moving air through a closed loop of flexible pipes - pinch the hose in the middle and you will see that a pressure appears across the bottleneck), *social* (stand in someone's way and you will experience a "pressure":), etc. Let's convey this idea in the electrical area (Fig. 4).

By connecting the Element 1 (Fig. 3) we have already obtained a current I as a function of the input voltage V and the Element 1's attribute (R, C, L). But we have not created this current without any purpose. We need this current; we would like to consume it for some purpose. In the simplest case, we connect another Element 2 (Fig. 4) acting as a current load (e.g., an active ammeter^{[4]}), LED, solenoid, electric motor, etc.) that consumes directly the current I.

More frequently, we want to obtain a voltage depending in a definite way on the current; we need some functional relation between the output voltage and the input current. For this purpose, we insert the Element 2 with the purpose to set the desired finctional relation so that it to form, shape, set the output voltage V_{E2} as we want. Again, depending on the nature of this *voltage-setting element* we can obtain various devices acting now as inverse *current-to-voltage converters*: linear - a classic current-to-voltage converter^{[5]}^{[6]} (a resistor), non-linear - a logarithmic converter (a diode), time-dependent - an integrator (a capacitor) and a differentiator (an inductor), etc. We may also keep the voltage constant and to vary the magnitude of the element's attribute - we obtain a *resistive sensor* (a variable resistor). Finally, we may vary both the input current and the element's attribute thus obtaining an "analog-digital multiplier" (another DAC with R-2R ladder).

### A problem appears[edit]

Unfortunately, a nasty problem appears... The Element 2 (resistor, capacitor or inductor) impedes the current by its inherent resistance, capacitance or inductance. As a result, a voltage drop V_{E2} appears across it; it represents the energy losses. The Element 2 absorbs this energy from the input voltage source: it dissipates energy from the source to outside environment (if it is a resistor) or it accumulates energy into itself (if it is a capacitor, a secondary battery or an inductor). It is not so important for the input voltage source if the Element 2 dissipates or accumulates energy; it is only important for it that the Element 2 absorbs some part of its energy.

*The problem is that the output voltage drop V _{E2} across the Element 2 enervates the input voltage V and thus affects the current (Fig. 5).* Now the voltage difference V

_{E1}= V - V

_{E2}instead only the excitation voltage V determines the current; as a result, the current decreases. So, from the point of view of the input voltage source the voltage drop V

_{E2}is troublesome and the input source "would like" this voltage not to exist. Contrary, from the point of view of the load, V

_{E2}is a useful voltage drop as it serves as an output quantity. So the load "would like" this voltage to exist and even to be as much as possible high.

Obviously, there is a contradiction here - *the voltage drop V _{E2} across the Element 2 has to exist and, at the same time, not to exist*. How do we solve this contradiction?

### Solving the problem: *removing the disturbance by an equivalent "antidisturbance"*[edit]

We know we may find a remedy in our routine. What do we do in real life when we solve some problem but a disturbance has stood in our way? First, we may brace our energy to overcome the disturbance. But there is another solution - someone can help us by adding as much efforts as it is needed to overcome the disturbance (*to remove the disturbance by an equivalent "antidisturbance"*). Here are some funny examples.

If someone has broken our window in winter, we may turn on a heater or turn on an air-conditioner in summer. If the windows become dirty, we switch on additional lamps inside the room to "help" the sun. When a car has come into collision with our car, the insurance company compensates the damages caused by the else's car. If someone (e.g., our wife) is spending money from our account, we begin depositing money into the account to restore the sum:) When we go to mountain, we stock with food, water, medicine, etc. to use them if a problem appears.

Note that in all these cases we *do not increase the power of the main source to compensate the losses*; instead, we *add energy by an additional "helping" power source to "help" the main source* and, as a result, to compensate the losses. The reason of doing that is because in the most cases the first solution is just impossible. For example, we cannot increase the power of the sun to adjust the temperature or the light inside the room (we cannot control directly the sun); so we connect additional thermal or light sources inside the room to "help" the sun.

### Connecting an additional "helping" voltage source...[edit]

Similarly, our circuit cannot increase the voltage of the input voltage source to compensate the voltage drop V_{E2} across the Element 2 (although this technique is used in bootstrapped current sources) since the input voltage source does not belong to the circuit. It may be distant; it is just inaccessible.

Then let's apply the second remedy - *to connect an additional voltage source in order to compensate the voltage drop V _{E2}*. It will add an additional "helping" voltage V

_{H}= V

_{E2}to the input voltage V that will compensate the "disturbing" voltage V

_{E2}across the Element 2. As a result, the total current-creating voltage will remain unchanged: V - V

_{E2}+ V

_{H}= V - V

_{E2}+ V

_{E2}= V.

Well, we have to sum the "helping" voltage to the input one. So Kirchhoff's second law can give us the simplest idea how to sum the two voltages: we have to break somewhere the circuit and connect an additional "helping" voltage source in the same direction as the input voltage source travelling the loop (- V_{H} +, - V +). It is wonderful but where do we connect the "helping" voltage source (Fig. 6)?

At first glance, it is no matter where we will connect the "helping" source; wherever we connect it the problem will be solved (the current will remain unchanged). Only, in circuitry we have always to consider the *common ground problem*. Let's then consider all the possible places where we may place the "helping" source.

#### ...under the input voltage source? NO!...[edit]

First, we may place the "helping" voltage source *under the input voltage source* - Fig. 7. We may think of the combination of the two connected in series voltage sources as a composed battery with total voltage V_{tot} = V_{E2} + V or as a "helped" input voltage source.

However, the input voltage source becomes "floating"; it is not connected to the common ground. This is inconvinient for us since in circuitry we prefer to ground voltage sources, converters and loads. The reason of that is because we usually use devices with single-ended inputs and we supply many devices by one common power supply. So all these devices are preliminary connected in parallel to the common power supply and then we cannot connect them in series (very rarely, as an exception, we may use "floating" driving devices^{[7]}). So we refuse this solution.

#### ...above the input voltage source? NO!...[edit]

With the same success we may place the "helping" voltage source *above the input voltage source* - Fig. 8 (again, the combination of the two connected in series voltage sources serves as a composed battery with total voltage V + V_{E2}).

However, now the "helping" voltage source becomes floating. This is inconvinient for us as we frequently use the "helping" voltage as an output one that drives a device (a converter or a load) with a single-ended input (examples are the current-to-voltage converter, the logarithmic converter, the integrator, the differentiator, etc.) In this case we have to use a load with a differential input. So we refuse this solution as well.

#### ...above the Element 2? NO!...[edit]

Then, we may place the "helping" voltage source *above the Element 2* - Fig. 9. Again the "helping" voltage source becomes floating. This is inconvinient for us as we will use its voltage as an output one and we have to use a load with a differential input. So, again we refuse this solution...

#### ...under the Element 2? YES![edit]

Finally, we may place the "helping" voltage source *under the Element 2* - Fig. 10. Eureka! Now both the input and the "helping" voltage source are grounded and the famous virtual ground appears above the Element 2. The input voltage source V "sees" only the impedance of the Element 1 since the latter is grounded virtually from the right.

From another viewpoint, we may think of the combination of the input voltage source and Element 1 as an input current source that is just..."fooled":) It doesn't "understand" that a disturbing element is connected; it has the "feeling" that its output is shorted. So this simple current source works at ideal load conditions and, as a result, the current depends only on the input voltage and on the Element 1's attribute (R, C, L).

What a magic! The Element 2 has disappeared; it is nothing more than a piece of wire! Imagine now we may think of this circuit as consisting only of two components - the input voltage source V and the Element 1! This is wonderful... but where do we take the output voltage from? Let's begin thinking...

**Where to take the output voltage from**

**Remedy 1. NO!** We cannot use the old circuit output (the point between the Element 1 and Element 2) just because we have already destroyed this voltage and this point has become a virtual ground.

**Remedy 2. NO!** Then let's try using the voltage V_{E2} across the Element 2. However, in order to connect a load in parallel to the "floating" Element 2, the load has to have a differential input. Moreover, if the load has some resistance, it will shunt the Element 2 thus affecting the current. What do we do then?

**Remedy 3. YES!** Recall to mind the cases from our routine when we prefer to estimate indirectly some quantity X. For this purpose, we first destroy the unknown quantity by an equal "anti-quantity" Y = X; then, we measure the "anti-quantity" Y, in order to know the magnitude of the initial quantity X. An example - classic weighing by using a balance. Let's convey this powerful idea in the electrical area (Fig. 10).

Eureka! *We use the "copy" voltage V _{H} = V_{E2} of the "helping" voltage source as an output instead the "original" voltage V_{E2} across the Element 2*! What a great idea! First, the load is connected to the common ground; second, it consumes energy from the "helping" source V

_{H}instead from the input source V!

## Generalization[edit]

**Circuits with voltage compensation consist of three components connected in series: a passive Element 1, a passive Element 2 and a "helping" voltage source V_{H} (Fig. 10). The Element 1 converts the exciting (input) voltage V into a current I and the Element 2 converts back the current into a voltage drop V_{E2}. We need this voltage drop... but it disturbs the current. So the "helping" voltage source V_{H} compensates the "disturbing" voltage drop V_{E2} by adding the same voltage to the exciting voltage source V. As a result, the Element 2 is "neutralized"; it has disappeared and the current depends only on the exciting voltage and the Element 1. If we need a current output, we connect the current load in the place of the Element 2; if we need a voltage output, the compensating voltage V_{H} can serve as a perfect "mirror" output voltage - powerful, grounded and inverted.**

## Limitations[edit]

### Maximum voltage drop across the Element 2[edit]

In our world power sources are limited; in electricity, voltage sources are limited too. So, in circuits with voltage compensation, the "helping" voltage source can produce a compensating voltage in some finite operating range. It can compensate maximum voltage drop V_{E2} across the Element 2 that is equal to its maximum "helping" voltage V_{Hmax}. When V_{E2} exceeds V_{Hmax} the magic of voltage compensation ceases. They name the maximum voltage drop across the Element 2 (the load) *compliance voltage*.

In the passive circuit (Fig. 5), the voltage drop V_{E2} across the Element 2 is always less than the input voltage V_{IN}. It is interesting that here - in the voltage compensated circuit (Fig. 10), the voltage drop V_{E2} and its "mirror copy" - the compensating voltage V_{H2}, can exceed at that many times V_{IN}. An inverting amplifier is an example of this situation.

### Maximum current through the "helping" voltage source[edit]

In the voltage compensated circuit (Fig. 10), the input current flows through the "helping" voltage source V_{H} as well; so the latter must endure this current. That is why the popular digital multimeters do not work that way. In order to measure current, they use the imperfect passive current-to-voltage converter instead of the almost ideal op-amp current-to-voltage converter.^{[8]} The reason for applying such an old-fashioned approach to current measurements is that all the input current flows through the op-amp output and its power supply; so the latter has to be able to endure the input current measured. The conclusion is the op-amp current-to-voltage converter is a perfect circuit but it is suitable only for low-current applications.

## Real op-amp implementations[edit]

### General circuits[edit]

#### ...without using negative feedback...[edit]

The voltage drop V_{E2} across the Element E2 is "flying". So we might use a voltage follower with differential input that "copies" the voltage drop V_{E2} and adds an equivalent voltage V_{OUT} = V_{E2} in series to the input voltage V (Fig. 11). In this arrangement, the voltage drop across the Element E2 is an "original" and the following voltage V_{OUT} is its "mirror copy".

Is there any negative feedback in this arrangement? No, it is not. Instead, a slight positive feedback can exist if the Element 2 is a kind of a resistor. Look at Fig. 11 again to convince yourself of this assertion. When, for example, the input voltage increases the current I, the voltage drop V_{E2}, the follower output voltage V_{OUT} and the total current-creating voltage V + V_{OUT} increase as well. This increases additionally the current I; we name this phenomenon *positive feedback*.

#### ...by using negative feedback.[edit]

In circuitry, we implement voltage followers exceptionally by applying a negative feedback. First, we may use a classic op-amp voltage follower (with a *local* negative feedback between its output and inverting input) as a "helping" voltage source (Fig. 12). For this purpose, we have to supply the op-amp with floating voltage source V_{S} and to reverse its output terminals, in order to add its output voltage to the input one. In this way, the local op-amp's ground serves as an output while the very op-amp's output is connected to the common circuit ground. Although we have guessed for ourselves at this trick, it is proposed as far back as in early 1960s.^{[7]}

Note although we have applied a local negative feedback between the op-amp's output and its inverting input the op-amp does not "observe" the virtual ground in this arrangement. In addition, the op-amp has to have a differential input.

The most reliable arrangement (Fig. 13) is if we make the op-amp compare its output voltage V_{OUT} (by subtracting) with the voltage drop V_{E2} across the Element 2 and change it so that to keep a zero difference between them. That means we have applied a *global* negative feedback. Note in this arrangement, we have made the op-amp "observe" the virtual ground. As a result, the op-amp output voltage is a "mirror" copy of the "disturbing" voltage drop across the Element 2. The op-amp may have a bare single-ended input as its input voltage (the difference V_{OA} - V_{E2}) is measured regarding to the ground.

Actually, all the practical op-amp circuits with voltage compensation are based on this arrangement.

#### Limitations[edit]

We have already discussed above the limitations of the general voltage compensation arrangement. Let's now only concrete them.

**Compliance voltage.** It is more than obvious that the *compliance voltage* of the op-amp circuits with voltage compensation is less than the supply voltage V_{S}. So, in order to enlarge the "magic" region of voltage compensation, we have to increase the supply voltage and, of course, to use high-voltage op-amps.

**Maximum input current.** Also, it is obvious that the maximum input current entering the op-amp circuits with voltage compensation has not to exceed the maximum op-amp output current. If we want to pass bigger input current, we have to put a power booster after the op-amp's output and into the feedback loop.

### Concrete op-amp circuits[edit]

Now let's consider some typical op-amp circuit applications implementing the great voltage compensation idea. In this section, we will investigate four legendary circuits where various Elements 1 (linear, non-linear and time-dependent) are connected between the input voltage source and the inverting input. Also, the same Element 2 - an ohmic resistor R (R2), is connected between the op-amp output and the inverting input. In this arrangement, the op-amp compensates the voltage drop across the resistor R by an equivalent "mirror" voltage V_{OUT} = -V_{R}. For this purpose, the op-amp adjusts its output voltage so that to keep zero voltage V_{(-)} at the inverting input (the virtual ground).

#### An inverting amplifier[edit]

Let's first investigate the famous inverting amplifier where another ohmic resistor R_{1} serves as Element 1 (Fig. 14a). For this purpose, let's drive the circuit with a linearly increasing through time voltage (the ramp on the first drawing at Fig. 14b).

Starting at the instant t_{0}, the current flowing through the resistors begins changing linearly. As a result, the voltage drop across the resistor R_{2} (the second drawing) changes linearly toward the positive rail, the output voltage (the fourth drawing) changes linearly with the same rate to the negative rail and their difference - the voltage of the inverting input (the third drawing), stays always zero (this point behaves as a virtual ground). The resistance R_{2} as though disappears and the input source V_{IN} "sees" only the resistor R_{1}; so the current depends only on the input voltage and the resistance R_{1} according to Ohm's law (I = V_{IN}/R_{1}). We may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the resistor R_{1}. Actually, the resistor R_{1} acts as a perfect voltage-to-current converter.^{[3]}

This "magic" lasts till to the instant t_{1} while the op-amp succeeds to compensate the voltage drop across the resistor R_{2} with its output voltage. As they say, the op-amp operates in active (linear) mode. Unfortunately, at the moment t_{1} the output voltage reaches the negative rail. The op-amp saturates and begins serving just as a steady voltage source (think of it just as a bare battery). The resistor R_{2} "appears" again and begins affecting the current; so the voltage drop across it begins changing slowly through time. You may think of the two resistors as a voltage divider supplied by a total voltage V_{IN} + V_{OUT}. So the voltage drop V_{R2} = R_{2}/(R_{1} + R_{2})*(U_{IN} + V_{OUT}) and the voltage of the inverting input begins changing according to this voltage drop (already there is no virtual ground).

#### An antilogarithmic diode converter[edit]

The antilogarithmic diode converter is another legendary circuit where a non-linear "resistor" - the diode D, serves as Element 1 (Fig. 15a). Again a voltage that increases linearly through time (the first drawing at Fig. 15b) is the most suitable input signal.

In the beginning, the current through the diode is insignificant but approaching the moment t_{1}, it begins changing considerably (exponentially). Accordingly, the voltage drop across the resistor R (the second drawing) changes exponentially toward the positive rail, the output voltage (the fourth drawing) changes exponentially to the negative rail and there is a virtual ground at the inverting input (the third drawing). The resistance R disappears and the input source V_{IN} "sees" only the diode D; so the current depends only on the input voltage and the diode D according to its exponential IV characteristic. We may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the diode D.

But is this circuit correct? No, it is not because we have connected a voltage source to a voltage-stable element. As a result, the circuit is very sensitive to input voltage variations; it loads the input source and it has a very limited operating region.

As above, the op-amp succeeds to compensate the voltage drop across the resistor R_{2} with its output voltage till to the instant t_{1}. At this moment, the output voltage reaches the negative rail and the op-amp saturates. The resistor R_{2} appears and the voltage drop across it begins following the input voltage. The diode transfers the input voltage variations to the inverting input and the voltage of the inverting input begins following the input voltage (already there is no virtual ground).

#### A capacitive differentiator[edit]

In this popular circuit a time-dependent element - the capacitor C, serves as Element 1 (Fig. 16a). Let's now apply a steady input voltage at the moment t_{0} (the first drawing at Fig. 16b).

In the beginning, the capacitor transfers the whole input voltage jump to the op-amp inverting input. The op-amp, trying to keep up the virtual ground, changes sharply its output voltage (the fourth drawing) toward the negative rail and finally saturates. At the initial moment t_{0} maximum current flows through the capacitor; then the current begins decreasing exponentially with a big time constant RC. The voltage drop across the resistor R (the second drawing) decreases exponentially as well. The voltage of the inverting input (the third drawing) changes in the same way and there is no virtual ground at this point.

At the moment t_{1}, the voltage drop across the resistor R becomes equal to +V_{S} and the op-amp enters the linear region. It manages to compensate the voltage drop V_{R} by its output voltage and a virtual ground appears at the inverting input. The resistance R disappears and the input source V_{IN} "sees" only the capacitor C; so the current depends only on the input voltage and the capacitor according to its exponential (through time) characteristic. As above, we may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the capacitor C. From the moment t_{1}, the time constant becomes almost zero (since R is almost zero); so the current, the voltage drop V_{R} and the op-amp output voltage V_{OUT} change exponentially and almost instantly.

But is this circuit correct? No, it is not because we have connected a voltage source to another "voltage source" (the capacitor); there is a conflict between two voltage sources. As above, the circuit is very sensitive to the input voltage variations; it loads considerably the input source and it has a very limited operating region.

#### An inductive integrator[edit]

Finally, let's consider another time-dependent circuit with voltage compensation but, for the sake of variety, based on the dual time-dependent element - the inductor L, serving as Element 1 (Fig. 17a). As above, we will apply a steady input voltage at the moment t_{0} (the first drawing at Fig. 17b).

Starting at the instant t_{0}, the current flowing through the inductor L and the resistor R begins changing linearly. As a result, the voltage drop across the resistor R (the second drawing) changes linearly toward the positive rail. The output voltage (the fourth drawing) changes linearly to the negative rail and their difference - the voltage of the inverting input (the third drawing), stays always zero (this point behaves as a virtual ground). The resistance R disappears and the input source V_{IN} "sees" only the inductor L; so the current depends only on the input voltage and the inductor L. We may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the inductor L. Now the inductor L acts as a *perfect inductive integrator*.

This wonder lasts till to the instant t_{1} while the op-amp succeeds to compensate the voltage drop across the resistor R with its output voltage. At the moment t_{1} the output voltage reaches the negative rail and the op-amp saturates. The resistor R appears again and begins affecting the current; so the voltage drop V_{R} across it begins changing slowly and exponentially through time with a time constant R/L. The voltage of the inverting input begins changing according to this voltage drop (already there is no virtual ground). The combination of the inductor L and the resistor R acts as an *imperfect inductive integrator*.

### Generalization[edit]

**Most of the circuits with voltage compensation are implemented as op-amp circuits with parallel negative feedback where the op-amp (including the power supply) serves as a "helping" voltage source. It compensates the "disturbing" voltage drop across the Element 2 by adding the same voltage V _{OUT} = -V_{E2} to the exciting voltage source V. For this purpose, the op-amp output voltage is negative with respect to the ground; so, all these op-amp circuits are inverting. The op-amp output voltage usually serves as a perfect circuit output voltage - powerful, grounded and inverting.**

## Negative impedance viewpoint at voltage compensation[edit]

### General circuit[edit]

We might use another viewpoint^{[9]} at the role of the "helping" voltage source in this arrangement. Look at the right part of the generalized circuit showing the voltage compensation idea (Fig. 10). It consists of two series connected elements: the Element 2 and the "helping" voltage source V_{H}. The same current flows through the two elements and the same voltage appears across them; so, they process the same energy and they have the same impedance. But while the first of them is a passive element that consumes energy from the input voltage source, the second is an active element that adds the same energy to the input voltage source! Then, if the first element has a "positive" impedance Z, the second one has a negative impedance -Z (Fig. 18)! Eureka! We have "invented" the simplest technique for creating a negative impedance:

*Copy the voltage drop across an "original" passive element by a voltage follower and "insert" the "copy" voltage into the circuit to obtain a negative impedance*.

Imagine how powerful this negative impedance viewpoint is! Now we may draw important conclusions:

*In electronic circuits with voltage compensation, the compensating voltage source acts as a negative impedance element.* More particularly, *in op-amp circuits with parallel negative feedback (op-amp inverting circuits), the combination of the op-amp and the power supply acts as a negative impedance element.*

### Concrete op-amp circuits with negative impedance[edit]

#### A negative resistor[edit]

We have already considered the popular circuits of the inverting amplifier, antilogarithmic diode converter, capacitive differentiator and inductive integrator containing a resistor with "positive" resistance R connected between the op-amp's output and the inverting input. In all these circuits, the combination of the op-amp and the power supply actually acts as a *negative resistor* with negative resistance -R that "neutralizes" the positive resistance R. For this purpose, the negative resistor -R adds as much voltage as it loses across the positive resistor R.

Now let's look at further three legendary op-amp circuit with voltage compensation (op-amp logarithmic converter, capacitive integrator and inductive differentiator) from this negative impedance viewpoint. In these circuits, the same Element 1 - an ohmic resistor R, is connected between the input voltage source and the op-amp inverting input. Also, various Elements 2 (linear, non-linear and time-dependent) with "positive" impedance are connected between the op-amp output and the inverting input. In these circuits, the combination of the op-amp and the power supply acting as an element with negative impedance "neutralizes" the positive impedance of the according Elements 2 in the same way - *by adding as much voltage as it loses across the positive element*.

#### A negative diode[edit]

In a logarithmic converter containing a diode with "positive" non-liner resistance R_{D}, the combination of the op-amp and the power supply acts as a *negative diode* with negative resistance -R_{D}. Let's investigate the circuit operation by driving the circuit with a linearly increasing through time voltage (the first drawing at Fig. 20b).

Starting at the instant t_{0}, the current passing through the diode begins changing linearly. In the beginning, the voltage across the diode (the second drawing) increases quickly; then it continues increasing slightly approaching V_{F}. The output voltage (the fourth drawing) changes in the same way but with a negative polarity. Their difference - the voltage of the inverting input (the third drawing), stays always zero (this point behaves as a virtual ground). The diode as though disappears and the input source V_{IN} "sees" only the resistor R; so the current depends only on the input voltage and the resistance R according to Ohm's law (I = V_{IN}/R). We may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the resistor R. The circuit behaves actually as a perfect constant current source.

In this circuit the op-amp always succeeds to compensate the voltage drop across the diode. So, it never saturates and operates in an active (linear) mode all the time.

#### A negative capacitor[edit]

Similarly, in the legendary op-amp inverting integrator containing a capacitor with capacitive reactance X_{C} the combination of the op-amp and the power supply acts as a *negative capacitor* with negative capacitive reactance -X_{C}. As at the circuit of an inductive integrator, we will apply a steady input voltage at the moment t_{0} (the first drawing at Fig. 21b).

At the beginning - the instant t_{0}, we apply a steady input voltage to the circuit and the current begins flowing through the resistor R and the capacitor C. The voltage drop across the capacitor (the second drawing) changes linearly through time toward the positive rail. The output voltage (the fourth drawing) changes linearly through time toward the negative rail and their difference - the voltage of the inverting input (the third drawing), stays always zero (this point behaves as a virtual ground). The capacitor disappears and the input source V_{IN} "sees" only the resistor R; so the current depends only on the input voltage and the resistance R according to Ohm's law (I = V_{IN}/R). We may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the resistor R. As above, the circuit behaves as a perfect constant current source.

This wonder lasts till to the instant t_{1} while the op-amp succeeds to compensate the voltage drop across the capacitor C with its output voltage. Uunfortunately, at the moment t_{1} the output voltage reaches the negative rail and the op-amp saturates:( The capacitor appears again and begins affecting the current; so the voltage drop V_{C} across it continues changing but now exponentially through time with a time constant RC. The voltage of the inverting input begins changing according to this voltage drop (already there is no virtual ground). The combination of the resistor R and the capacitor C acts as an *imperfect capacitive integrator*.

#### A negative inductor[edit]

Finally, in the dual op-amp inverting differentiator containing an inductor with inductive reactance X_{L} the combination of the op-amp and the power supply acts as a *negative inductor* with negative inductive reactance -X_{L}. As at the circuit of a capacitive differentiator, we will apply a steady input voltage at the moment t_{0} (the first drawing at Fig. 22b).

At the initial moment t_{0} the input source raises its voltage and tries to increase the current through the inductor L. The inductor reacts to this "intervention" by increasing the voltage across its terminals. The op-amp, trying to keep up the virtual ground, changes sharply its output voltage (the fourth drawing) toward the negative rail and finally saturates.
So, the inductor produces a total voltage V_{S} + V_{IN}.

Then, in the interval t_{0} - t_{1}, the current increases slowly and exponentially with a big time constant R/L; the voltage across the inductor (the second drawing) decreases exponentially. The voltage of the inverting input (the third drawing) changes in the same way and there is no virtual ground at this point.

Fortunately, at the moment t_{1}, the voltage V_{L} across the inductor becomes equal to +V_{S} and the op-amp enters the linear region:) It manages to compensate the voltage V_{L} by its output voltage and a virtual ground appears at the inverting input. The inductor just disappears and the input source V_{IN} "sees" only the resistor R; so the current depends only on the input voltage and the resistance R according to Ohm's law (I = V_{IN}/R). Again, we may think of all the circuit as consisting only of two elements - a voltage source V_{IN} driving the resistor R. As above, this circuit behaves as a perfect constant current source. From the moment t_{1}, the time constant becomes almost zero (since R is almost zero); so the current, the voltage V_{L} across the inductor and the op-amp output voltage V_{OUT} change exponentially but almost instantly.

We ask ourselves again, "Is this circuit correct?" And we answer, "No, it is not correct again because we have connected a current source to another current source (the inductor); so there is a conflict between two current sources. As the capacitive differentiator, this circuit is very sensitive to the input variations and it has a very limited operating region.

## List of popular op-amp circuits based on this idea[edit]

There are many circuits based on the powerful voltage compensation idea. In addition, we may invent more new circuits because we know the general idea behind them. If you have such circuits, add them to the list below.

**Linear:** active ammeter, voltage-to-current converter, current-to-voltage converter (transimpedance amplifier), resistance-to-current converter, resistance-to-voltage converter, analog divider, analog multiplier, inverting amplifier, summing amplifier, DAC with R-2R ladder.

**Non-linear:** "ideal" diode, D logarithmic converter, RD logarithmic converter, D antilogarithmic converter, DR antilogarithmic converter.

**Time-dependent:** C integrator, RC integrator, charge amplifier, CR differentiator, LR integrator, L differentiator, RL differentiator.

## See also[edit]

Op-amp inverting current-to-voltage converter (compensating the internal losses by an "antivoltage")

Presenting the op-amp inverting current-to-voltage converter in a more attractive manner by using voltage bars and current loops.

Current-to-voltage converter introduces the passive and active versions of the circuit.

How to make perfect components by parallel NFB in the educational laboratory.

"Inventing" the op-amp inverting integrator during a laboratory exercise.

Basic op-amp configurations from Electronics wikibook.

Miller theorem, Operational amplifier and Operational amplifier applications from Wikipedia

## References[edit]

- ↑ Here, we mean more
*general converters*, not only the classical linear passive voltage-to-current converter and active current-to-voltage converter - ↑ Op-amp inverting voltage-to-current converter (compensating the external losses by an "antivoltage")
- ↑
^{a}^{b}Voltage-to-current converter introduces the passive and active versions of the circuit. - ↑ How do we build an op-amp ammeter?
- ↑ Op-amp inverting current-to-voltage converter (compensating the internal losses by an "antivoltage")
- ↑ Current-to-voltage converter introduces the passive and active versions of the circuit.
- ↑
^{a}^{b}Impedance and admittance transformations using operational amplifiers is a genuine source from Philbrick Reserches (written by D. H. Sheingold). - ↑ Op-amp inverting current-to-voltage converter
- ↑ Revealing the mystery of negative impedance is a general story about the mystic phenomenon.

## External links and resources[edit]

How I revealed the secret of parallel negative feedback circuits reveals the philosophy of circuits with parallel negative feedback.

How do we create a virtual ground? reveals the secret of the great circuit phenomenon.

Inverting configuration shows the application of the virtual ground concept in the circuit of inverting amplifier.

Op-amp circuit builder is an interactive *Flash* tutorial that shows how to transform any passive converter into the corresponding active one.

How do we build an op-amp ammeter? shows how to convert the imerfect ammeter into an op-amp one.

Passive voltage-to-current converter is an animated *Flash* tutorial about the most elementary passive converter.

How to transform the passive voltage-to-current converter into an active one

Passive current-to-voltage converter is another animated *Flash* tutorial about the inverse passive converter.

Reinventing transimpedance amplifier is a story about the op-amp current-to-voltage converter.

Passive parallel voltage summer is an animated *Flash* tutorial about the elementary summing circuit.

Op-amp inverting summer is an animated *Flash* tutorial about the famous op-amp summing circuit.

What is the idea behind the op-amp inverting current source? - reveals the "helping" idea by using voltage bars and current loops.

Keeping a constant current by adding an additional voltage

How do we build an op-amp RC integrator?

Analog electronics 2004, Class 2: *Elementary passive converters with voltage output*

What's all this transimpedance amplifier stuff, anyhow?

Mechkov C., Heuristic methods for converting passive analog devices into negative feedback active circuits, Proceedings of The Sixth Int. Conference ELECTRONICS'97, Sozopol, Bulgaria, 1997.

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