Inventing Circuits on the Whiteboard...

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

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?

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.

Click the picture above and look first at the "bad" passive version (the top of the figure) and then, at the "good" active version (the bottom of the figure) of a voltage-to-current converter. It is not absolutely necessary to be clever:) 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 one + op-amp + great idea ?

It seems that there is a close interrelation between the two circuits: maybe, the active version is come from the passive one or maybe the active version is just an improved passive version? If it is, how have 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.

Internal links:

1.   Problem: The real load affects the current

2.   Basic idea: Removing a disturbance by an "antidisturbance"

3.   Basic electrical idea: Removing a voltage by an "antivoltage"

4.   Op-amp inverting voltage-to-current converter (at +VIN)

5.   Op-amp inverting voltage-to-current converter (at -VIN)

6.   Applications: Staying after the circuit output

7.   Applications: Staying before the circuit input

8.   Acting as a transconductance amplifier

9.   What are the input/output resistances?

10.  Comparison with other techniques

Color key

Links:  this page, other my pages, external, multimedia, handmade.

Text:   analogies, conclusions.

Problem: The real load affects the current

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We have already known 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 VL across the load. In this arrangement, the effective voltage difference VIN - VL determines the current IOUT instead only the voltage VIN; 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 or if you go to Stage 2 in the interactive flash builder.

We can look at the voltage VL from two contrary viewpoints. From the point of view of the input voltage source, the voltage drop VL is harmful; so, the input source "would like" this voltage not to exist. Contrary, from the point of view of the load, VL 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 VL has to exist and, at the same time, not to exist. How do we solve this contradiction?

Basic idea: Removing a disturbance by an "antidisturbance"

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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); so, we prefer to use it when we are rich and, at the same time, lazy enough:). Let's consider some funny examples from our routine.

If someone has broken our window in winter, we may turn on a heater (instead just to repair 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 just to clean 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 we just to prevent 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 just to scold her/him:) When we go to mountain, we stock with food, water, medicine, etc. to use them, if there is a future need (instead just not to go to 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 a voltage by an "antivoltage"

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

Active V-to-I converter = passive one + "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 BH in series to the load L. Then, in order to compare the two voltages, I connect a zero indicator in point A, which shows the result of comparison VA = VH - VL. 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!

Building a "man-controlled" electrical circuit

In the beginning, imagine that there is not input excitation voltage VIN (see again the figure 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 VIN, a current begins flowing through the load. As a result, a voltage drop VIN appears across the load and the point A begins rising its potential VA (figuratively speaking, the input source "pulls" the point A up toward the positive voltage VIN). Only, I observe to my great displeasure:( that the needle deflects to the right and immediately react by decreasing the compensating voltage VH. Now, it "pulls" the point A down toward the negative voltage -VH until it manages to zero the potential VA (the virtual ground). Note that the two voltage sources are connected in series, in one and the same direction (- VIN +, - VH +) so that their voltages are added (let us assume that we traverse the loops clockwise). Regarding to the ground, they have opposite polarities.

Exploring the circuit by voltage bars and current loops

In this way, the input voltage source is "helped"; its voltage increases so much (VIN) as it loses across the load. As a result, the "disturbing" voltage VIN 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 VIN under the ground. As above, I will adjust the battery voltage looking at the needle of the zero indicator, so that always VH = -VIN. In electronics, in technics and in our world, this action is referred to as negative feedback (great phenomenon).

Creating an op-amp inverting voltage-to-current converter

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Building. 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).

Exploring. If the input voltage VIN increases, an output current IOUT begins flowing through the load L. As a result, a voltage drop VL appears across the load L and the point A begins rising its potential VA (the input source "pulls" the point A up toward the positive voltage VIN). Only, the op-amp "observes" that to his great displeasure:( and immediately reacts: it decreases its output voltage "sucking" the current IOUT until it manages to zero the potential VA. 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 VIN. As above, the two voltage sources are connected in series, in one and the same direction (- VIN +, - VH +) so that their voltages are added. Only, regarding to the ground, they have opposite polarities.

... driven by a positive input voltage

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

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If the input voltage VIN decreases under the ground, an input current IOUT begins flowing through the load L in opposite direction. As a result, a voltage drop VL appears across the load L and the point A begins dropping its potential VA (now, the input source "pulls" the point A down toward the negative voltage -VIN). Only, the op-amp "observes"' that and immediately reacts: it increases its output voltage "pushing out" the current IOUT until it manages to zero the potential VA (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 VIN. The two voltage sources are connected in series again, in one and the same direction (+ VIN -, + VH -) so that their voltages are added. Regarding to the ground, they have opposite polarities as above.

... driven by a negative input voltage

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: Staying after the circuit output

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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).

Building a voltage-controlled current source

Applications: Staying before the circuit input

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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 will introduce an error but the op-amp will remove it).

Building voltage-to-voltage converters

Acting as a transconductance amplifier

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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 = IOUT /VIN [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 (R1 and OA) and a bare resistor R2 acting as a simple current-to-voltage converter:

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

Note that the resistor R2 serves here as a load L. 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.

What are the input/output resistances?

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

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 RIN = R + RL. Now, we see only the resistor R because the op-amp has "neutralized" the load resistance RL.

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

What is the input 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 ROUT = 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.

What is the output resistance?

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 it is a virtual ground. In some cases it is sufficient but in other - no.

Comparison with other techniques

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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 VR across the resistor R equal to the "originl" input voltage drop VIN. 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 VIN and R; it does not depend on all sorts disturbances. By adding an external transistor or by using the internal op-amp output transistors 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 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; Motorola named it a differential input op-amp current source in 1973. From one wievpoint, 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. 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.

Links to related web resources

Analog electronics 2004:  Class 9 is a teacher's story about the op-amp inverting converters (from the exercices with my students, 2004).

Reinventing the constant current source reveals the philosophy of the constant current sources.

Op-amp circuit builder shows how to transform any passive converter into an active one (choose the resistor R2 from the library on the right side).

Wikibooks:   Circuit idea - Op-amp inverting voltage-to-current converter reveals the idea behind the active circuit.

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

Wikipedia:    Voltage-to-current converter  builds consecutively the passive and active versions of the voltage-to-current converter.

                     Current-to-voltage converter is dedicated to the passive and active versions of the inverse current-to-voltage converter.