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**How to Transform the Passive Voltage-to-Current Converter into an Active One**

**(Reinventing the Op-amp Inverting Voltage-to-Current Converter)**

**Circuit idea:** *The op-amp compensates the external losses caused by the load adding as much voltage to the input voltage source as it loses across the load.*

## Speculation: *The active version is just an improved passive one*Edit

In this story, we begin revealing the secret of *active voltage-to-current converters* alias *voltage-controlled current sources* (VCCSs) or *transconductance amplifiers*. Let us start with the simplest and most intuitive *op-amp inverting voltage-to-current converter*.

Look first at the "bad" passive version (the top of Fig. 1) and then, at the "good" active version (the bottom of Fig. 1) of a voltage-to-current converter. It is not absolutely necessary to be clever:), in order to see that the active version contains the passive one + an op-amp connected in accordance with some powerful idea:

*Active V-to-I converter = passive V-to-I converter + op-amp + great idea ?*It seems that there is a close interrelation between the two circuits: maybe, the active version comes from the passive one or maybe the active version is just an improved passive version? If it is, how has the passive circuit transmuted into an active one? What is the idea of this connection? What does the op-amp do in this circuit? Let's try to answer these questions by following the evolution of the passive circuit into an active one.

## Problem: *The real load affects the current*Edit

We already knew why the simple voltage to-current converter (as well as all the passive circuits) is an imperfect circuit - because *its output current depends on the voltage drop V _{L} across the load* (Fig. 2). In this arrangement, the effective voltage difference V

_{IN}- V

_{L}determines the current I

_{OUT}instead of only the voltage V

_{IN}; .As a result, the current decreases. You can explore the circuit operation in a more attractive way, if you click

*Exploring*button in the interactive flash movie

^{[1]}or if you go to

*Stage 2*in the interactive flash builder.

^{[2]}

We can look at the voltage V_{L} from two contrary viewpoints. From the point of view of the input voltage source, the voltage drop V_{L} is harmful; so, the input source "would like" this voltage not to exist. Contrarily, from the point of view of the load, V_{L} is useful voltage drop as it usually serves as an output quantity; so, the load "would like" this voltage to exist and even it to be as much as possible high.

Obviously, there is a contradiction here - the voltage drop V_{L} has to exist and, at the same time, not to exist. How do we solve this contradiction?

## Basic non-electrical idea: *Removing a disturbance by an "antidisturbance"*Edit

Remember what we do in real life when we solve some problem but a disturbance **caused by others** stands in our way. The classical remedy is to *remove the cause of the disturbance*. Only, it is not always possible to do that; then, we use another exotic solution - we *remove the disturbance by an equivalent "antidisturbance"*. For this purpose, we use an additional power source (energy), which "helps" us (the main source) by compensating only the local losses caused by the undesired **external** quantity. This technique is associated with continuous wasting of additional energy but the result is zero (virtual ground ^{[3]}); so, we prefer to use it when we are rich and, at the same time, lazy enough:). Here are some funny examples.

If someone has broken our window in winter, we may turn on a heater (instead of just repairing the broken window); in summer, we turn on an air-conditioner. If the windows become dirty, we may switch on additional lamps inside the room to "help" the sun (instead of just cleaning the windows). When a car has come into collision with our car, the insurance company compensates the damages caused by the else's car (instead of just preventing the crash). If someone (e.g., our wife or a husband) is spending money from our account, we might begin depositing money into the account to restore the sum (instead of just scolding her/him:) When we go to a mountain, we stock with food, water, medicine, etc. to use them, if there is a future need (instead of just not going to the mountain:). In all these cases,

*we have prepared (just in case) "standby" resources to use them, if there is a need to compensate eventual external losses*.

## Basic electrical idea: *Removing voltage by an "antivoltage"*Edit

### Building the electric circuitEdit

Now, let us put this powerful idea into practice. The voltage drop V_{L} across the resistor R is harmful; so, following the recipe above, we have to remove it by an "antivoltage" -V_{L}. In other words, we have to add so much voltage to the input (excitation) voltage source V_{IN}, as much as it loses across the load L

*Active V-to-I converter = passive V-to-I converter + "helping" voltage source*

The best way to understand what real electronic components do is just to do their work. So, let us first build a "man-controlled" active circuit, in which a man (I might do this donkeywork:) produces the "anti-voltage" while you change the input voltage. For this purpose, I first place an additional supplementary battery B_{H} in series to the load L (Fig. 3). Then, in order to compare the two voltages, I connect a zero indicator in point A, which shows the result of comparison V_{A} = V_{H} - V_{L}. See how simple it is:

*Add an adjustable battery in series with the load and make its voltage equal to the voltage drop across the load!*

### Exploring the electric circuitEdit

In the beginning, imagine that there is not input excitation voltage V_{IN} (see again Fig. 3 above). As a result, there are not any voltage drops and currents in the circuit; the needle of the zero indicator points to zero position. I am happy because there is nothing to do:)

If you increase the input voltage V_{IN}, a current begins flowing through the load. As a result, a voltage drop V_{L} appears across the load and the point A begins rising its potential V_{A} (figuratively speaking, the input source "pulls" the point A up toward the positive voltage V_{IN}). Only, I observe to my great displeasure:( that the needle deflects to the right and immediately react by decreasing the compensating voltage V_{H}. Now, it "pulls" the point A down toward the negative voltage -V_{H} until it manages to zero the potential V_{A} (the virtual ground). Note that the two voltage sources are connected in series, in one and the same direction (- V_{IN} +, - V_{H} +) so that their voltages are added (let us assume that we traverse the loops clockwise). Regarding to the ground, they have opposite polarities.

In this way, the input voltage source is "helped"; its voltage increases so much (V_{L}) as it loses across the load. As a result, the "harmful" voltage V_{L} disappears; the point A has zero voltage; it behaves as a virtual ground. The passive voltage-to-current converter is "fooled": it has the illusion that there is not a load connected; it "thinks" that its output is shorted.

You can imagine what happens when you decrease the input voltage V_{IN} under the ground. As above, I will adjust the battery voltage looking at the needle of the zero indicator, so that always V_{H} = -V_{L}. In electronics, in technics and in our world, this action is referred to as negative feedback (it is a great phenomenon).

## Op-amp inverting voltage-to-current converterEdit

### Building the electronic circuitEdit

Let us now try to make some electronic device do this donkeywork; an op-amp seems to be a good choice. For this purpose, we connect the op-amp's output in the place of the helping voltage source and the op-amp's input to point A so that the op-amp to "help" the input source (the op-amp's output voltage and the input voltage to be summed).

*Op-amp V-to-I converter = passive V-to-I converter + "helping" op-amp*

### Exploring the electronic circuitEdit

In the beginning, imagine as above that there is not an input excitation voltage V_{IN}. As a result, there are not any voltage drops and currents in the circuit. There is almost not a voltage difference between the inverting and non-inverting input of the op-amp; now, it is "happy" because there is nothing to do:)

#### Positive input voltageEdit

If the input voltage V_{IN} increases, an output current I_{OUT} begins flowing through the load L. As a result, a voltage drop V_{L} appears across the load L and the point A begins rising its potential V_{A} (the input source "pulls" the point A up toward the positive voltage V_{IN}). Only, the op-amp "observes" that to his great displeasure:( and immediately reacts: it decreases its output voltage "sucking" the current I_{OUT} until it manages to zero the potential V_{A}. Figuratively speaking, the op-amp "pulls" the point A down toward the negative voltage -V to establish a virtual ground. It does this magic by connecting a part of the voltage produced by the negative power supply -V in series with the input voltage V_{IN}. As above, the two voltage sources are connected in series, in one and the same direction (- V_{IN} +, - V_{H} +) so that their voltages are added. Only, regarding to the ground, they have opposite polarities.

#### Negative input voltageEdit

If the input voltage V_{IN} decreases under the ground, an input current I_{OUT} begins flowing through the load L in opposite direction. As a result, a voltage drop V_{L} appears across the load L and the point A begins dropping its potential V_{A} (now, the input source "pulls" the point A down toward the negative voltage -V_{IN}). Only, the op-amp "observes"' that and immediately reacts: it increases its output voltage "pushing out" the current I_{OUT} until it manages to zero the potential V_{A} (now, the op-amp "pulls" the point A up toward the positive voltage +V, in order to establish a virtual ground). It does this magic by connecting a part of the voltage produced by the positive power supply +V in series with the input voltage V_{IN}. The two voltage sources are connected in series again, in one and the same direction (+ V_{IN} -, + V_{H} -) so that their voltages are added. Regarding to the ground, they have opposite polarities as above.

**Conclusion.** *In the circuit of an op-amp voltage-to-current converter, the op-amp adds as much voltage to the voltage of the input source as it loses across the external load. The op-amp compensates the local losses caused by this external load (conversely, in the opposite op-amp inverting current-to-voltage converter, the op-amp compensates the losses caused by the internal resistor).*

## Applications of the op-amp V-to-I converterEdit

Once we created a perfect voltage-to-current converter, we may use it as a building block to build more complex compound circuits. For this purpose, we have only to connect consecutively the separate building blocks.

### Staying after the circuit outputEdit

First, we may connect the op-amp voltage-to-current converter after circuits having voltage output; thus we make them produce current. For example, we have known how to build a simple current source

*Simple current source = voltage source + passive V-to-I converter*

Now, we can build a perfect op-amp constant current source (Fig. 6)

*Op-amp current source = Voltage source + op-amp V-to-I converter*

Its output current does not depend on the load resistance (more generally, on the voltage drop across the load).

### Staying before the circuit inputEdit

Then, we may connect the op-amp voltage-to-current converter before circuits having current input; thus we make them perceive voltage. By applying this technique, we may assemble the famous active circuits of capacitive integrator (Fig. 7), inductive differentiator, diode logarithmic converter, etc.

*Op-amp RC integrator = op-amp V-to-I converter + I-to-V C integrator*

*Op-amp RL differentiator = op-amp V-to-I converter + I-to-V L differentiator*

*Op-amp RD log converter = op-amp V-to-I converter + I-to-V D log converter*

In all these circuits, the voltage drop across the respective I-to-V converters does not introduce an error (precisely speaking, it introduces an error but the op-amp removes it).

### Acting as a transconductance amplifierEdit

The op-amp V-to-I converter is an active circuit; so, we may expect it to amplify. Really, it acts as a linear circuit with transfer ratio k = I_{OUT}/V_{IN} [mA/V] or millisiemens having dimension of conductivity. That is why, they frequently name it transconductance amplifier (transconductance is a contraction of "transfer conductance"). Nevertheless, let us try to answer the question, "Is the op-amp V-to-I converter an amplifier?"

To answer this question, let's assemble the famous op-amp inverting amplifier by connecting consecutively an op-amp voltage-to-current converter (R_{1} and OA) and a bare resistor R_{2} acting as a simple current-to-voltage converter (Fig. 8)

*Op-amp inverting amplifier = op-amp V-to-I converter + I-to-V converter*

Note that the resistor R_{2} serves here as the load L from Fig. 5). This circuit will amplify power, if the output power is greater than the input one. So, the output voltage has to be higher than the input one because the current is the same.

## Comparison between the passive and active versionEdit

Let us finally compare the two versions beginning by investigating the input and output resistances (for concreteness, let us assume a bare resistive load R_{L}).

**Input resistance.** First, connect an ammeter *in series with* and a voltmeter *in parallel to* the converter's input; then vary the input voltage to investigate the input resistance. Maybe you remember that in the passive version, looking from the side of the input source we were seeing two resistors connected in series; so, the input resistance was R_{IN} = R + R_{L}. Now, we see only the resistor R because the op-amp has "neutralized" the load resistance R_{L}.

It is wonderful, the load has disappeared! As a result, the input resistance is R_{IN} = R and what is more interesting, it does not depend on the load resistance R_{L}! In this case, the op-amp "helps" the input voltage in its striving to change the current.

**Output resistance.** Now, connect a voltmeter *in parallel to* and an ammeter *in series with* the converter's output; then, vary the load resistance (voltage) and observe the current to investigate the output resistance. Remember that in the passive version, looking from the side of the load, we were seeing only the resistor R; so, the output resistance was R_{OUT} = R. Now, the situation is quite more interesting: when we vary the load voltage the current stays steady! But why?

In this case, the op-amp "opposes" the load in its striving to change the current. As a result, the voltage varies (even significantly) while the current stays unchanged; so, the output differential resistance tends to infinite.

**Operating range.** The op-amp does all these "magics" until it can change its output voltage (i.e., until it is within the active region). When it reaches the supply rail, the op-amp saturates, the magic ceases and the almost ideal active circuit becomes again an imperfect passive one:(. They name the maximum voltage drop across the load compliance voltage. Note that the passive version does not have such a problem; it is always imperfect.

**Ground connection.** In the passive version, the load is connected to the (real) ground while, in the active one, the load is "flying". More precisely speaking, it is connected to a ground but this is a *virtual ground*. In some cases it is sufficient but not in others.

## Comparison with other techniquesEdit

The inverting configuration is not the only possible active voltage-to-current converter. There are other techniques, which excel, in some respect, the idea considered. Let's now consider briefly these techniques; then, we will dedicate special circuit stories to them.

**Negative feedback.** Exactly speaking, there is a negative feedback in the inverting configuration discussed here. Only, the negative feedback does not serve to keep up the output current; it serves only to keep up the compensating "antivoltage". In this arrangement, the op-amp is not actually interested in the current magnitude; it is interested only in the virtual ground magnitude. In the popular implementation - the *non-inverting configuration*, the op-amp (not the input voltage source) creates the output current, which passes through a constant resistor. Then, the op-amp keeps up the "copy" voltage drop V_{R} across the resistor R equal to the "original" input voltage drop V_{IN}. Simply speaking, the op-amp increases its output voltage by the value of the voltage across the load thus compensating it. As a result, the output current depends only on the V_{IN} and R; it does not depend on all sorts disturbances.^{[4]} ^{[5]} By adding an external transistor or by using the internal op-amp output transistors^{[6]} these circuits can drive grounded loads. Only, the negative feedback configurations have two big errors: the first is caused by the finite op-amp gain; the second - by the common-mode gain.

**Bootstrapping.** The famous Baron Munchhausen's idea can help us to achieve the same (and even better) result without using negative feedback. The according circuit implementation is referred to as *improved Howland current source*; no one knows why *Motorola* named it *current source with feedback*^{[7]} in 1973?!? The idea is as simple as above: the op-amp increases its output voltage by the value of the voltage drop across the load thus compensating it (compare with the inverting configuration where the op-amp produces an output voltage equal to the voltage drop across the load and adds it to the input voltage). Only, it does this magic "blindly" without using a negative feedback.

**Negative resistance.** The famous circuit of *Howland current source* exploits this idea ^{[8]}^{[9]}; *Motorola* named it a *differential input op-amp current source*^{[7]} in 1973. From one viewpoint, it consists of a negative resistor -R connected in parallel to the positive resistor R acting as a passive voltage-to-current converter. The result of this connection is that the negative resistor "neutralizes" (absorbs) the positive one and the effective differential resistance (the internal source's resistance) is infinite.^{[10]} The voltage-to-current converter behaves as a perfect current source, which drives a grounded load. Only, these circuits are not as stable as these exploiting a negative feedback.

## ReferencesEdit

The op-amp inverting voltage-to-current converter is not so much popular circuit as the related op-amp non-inverting voltage-to-current converter and the opposite op-amp inverting current-to-voltage converter. Actually, it exists in almost every electronics book on op-amp circuits but they do not discern it. As a result, there are few sources where it is presented rather implicitly than manifestly. Here are some of them.

- ↑ Passive voltage-to-current converter is an animated
*Flash*tutorial revealing the philosophy of the passive version. - ↑ Op-amp circuit builder (movie philosophy) is an interactive
*Flash*tutorial that shows how to transform any passive converter into an active one. Choose the resistor R_{2}from the library on the right side to build an active voltage-to-current converter. - ↑ How do we create a virtual ground? reveals the secret of the great circuit phenomenon on which the op-amp inverting voltage-to-current converter is based.
- ↑ Keeping a constant current by negative feedback shows how to produce a constant curent by applying a negative feedback.
- ↑ Voltage to current conversion - App Note 13 - considers a variety of voltage-controlled current sources.
- ↑ Op amp can source or sink current exploits a powerful idea - to use the op-amp supply leads as input/output terminals.
- ↑
^{a}^{b}Analysis and design of the op-amp current source is an extremely formal and dull presentation of the Howland current source (it is renamed there as a*differential input op-amp current source*). - ↑ Impedance & admittance transformations using op-amps - mentions, maybe for the first time, the Howland circuit (great material!)
- ↑ Howland current source for grounded load reveals the basic idea behind the famous circuit.
- ↑ Consider the "Deboo" integrator for unipolar noninverting designs (see the comments after the article).

## Related Wikimedia resourcesEdit

### WikibooksEdit

**Circuit Idea (this book):**

Passive voltage-to-current converter reveals the basic idea behind the passive version.

Passive current-to-voltage converter shows the basic idea behind the passive version of the "mirror" circuit.

Op-amp inverting current-to-voltage converter reveals the basic idea behind the active version of the "mirror" circuit.

### WikipediaEdit

Voltage-to-current converter introduces the passive and active versions of the circuit.

Virtual ground scrutinizes the basic phenomenon behind the circuits with parallel negative feedback.

## Further readingEdit

How to transform the passive voltage-to-current converter into an active one is a similar story.

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

Op-amp inverting summer is an animated tutorial, which uses the op-amp inverting voltage-to-current converter to build the famous op-amp summing circuit.

Secrets of parallel negative feedback circuits reveals the philosophy of this class of circuits (the op-amp inverting voltage-to-current converter belongs to them).

Lab 2: Basic instrumentation (voltage-to-current converter) shows an interesting version of this circuit where the op-amp is buffered by a transistor and the load is grounded.

Tietze, U and Schenk, Ch. "Halbleiter-Schaltungstechnik", Chapter 12.3.1. Springer-Verlag Berlin Heidelberg NewYork, 1980.

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