Circuit Idea/Passive Voltage-to-Current Converter

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Building a Passive Voltage-to-Current Converter

Circuit idea: Voltage applied across a resistor makes a proportional current flow through the resistor.

Problem: How to convert voltage into current

Have you noted that, in low-voltage electronics, the electrical attributes voltage and current carry information rather than energy? And, for some reasons, we prefer voltage as a data carrier? As a result, most electronic circuits have voltage inputs and outputs.

Unfortunately:), there are also current-input and current-output devices. Examples: an ammeter has only a current input, a current source has only a current output, a bipolar transistor has a current input and current output, a tube and an FET transistor have a voltage input and current output, a transimpedance amplifier and a Norton amplifier have a current input and voltage output, etc. In these cases, we need devices converting the electric attributes carrying information from voltage to current and vice versa.

In this story, we begin revealing the secret of the famous (so much as Ohm's law) voltage-to-current converter, which changes the electrical data carrier from voltage to current (Fig. 1). As usual, there are two versions of this circuit - passive ("bad") and active ("good"). Here, we begin with the "bad" passive version. Then, in another story dedicated to the "good" active version, we will show that the two versions are not independent. Instead, there is a close interrelation between the two circuits: the active version came from the passive one (the active version comprises the passive one, it is just an improved passive version). In this way, we will show the evolution of the famous circuit moving, step-by-step, from simple to complex.

Basic non-electrical idea: Pressure causes flow

Have you ever asked yourself questions about causality in our world: "Which came first, the egg or the hen?" (troubling philosophers from the remote past to the present day:) and, "What causes what - pressure causes flow or flow causes pressure?". As it is hard for us to answer this primary questions we might just assume that both the alternative answers are possible.

Well, let's first begin with the more popular assumption that pressure causes flow. Really, we may observe many situations in life, where a pressure-like quantity puts in motion a flow-like one through an impediment. Here are some examples: pneumatic (imagine a constant pressure pump moves air through a closed loop of pipes - you may carry out such a funny experiment by using an old-fashioned air cleaner, which sucks and blows air through a closed loop made from a corrugated hose), water (remember famous communicating vessels where the height difference between the two vessels causes water to flow), thermal (if we warm up a metal bar on the one side, heat begins flowing to the other side), mechanical (imagine that a motor drives a belt - energy is flowing along this "circuit"), informational (someone tells a story to somebody else - data flows through the phone line), money (the rich give money to the poor:), etc. If you need more examples, just look around and you will find a lot of pressure-causes-flow (more generally, difference-causes-movement) analogies. Let's finally generalize all these situations.

If we apply a pressure to an impediment, a flow begins flowing. In this arrangement, the pressure-like, flow-like and impediment-like attributes are interrelated. Usually, the output flow-like variable is proportional to the input pressure-like one; so, we may say that the impediment converts the pressure like variable into a flow-like one.

Basic electrical idea: Voltage causes current

Building the circuit Fig. 2. In the basic Ohm's voltage-supplied circuit, the resistor R acts as a simple voltage-to-current converter.

In electricity, we might ask ourselves the electrical version of the philosophical "egg or hen" question above, "What causes what in Ohm's law - voltage causes current or current causes voltage?" Again, as it is hard to answer this primary question we might assume that both the alternative answers are possible.

In this story, we will begin with the more popular assumption that voltage causes current in the most elementary Ohm's circuit supplied by a voltage source (Fig. 2). As we know, if we apply a voltage VIN across a resistor R, a proportional current IOUT = VIN/R begins flowing through the circuit. We may say that this is a voltage-causes-current formulation of Ohm's law: I = V/R.

In this voltage-supplied circuit, the resistor R determines the current flowing through it. We usually say that the resistor converts the voltage VIN into a proportional current IOUT or it serves as a simple voltage-to-current converter - a linear circuit with transfer ratio k = IOUT/VIN [mA/V].

A bare resistor can convert the voltage into current.

Exploring the circuit operation

We may present the circuit operation and Ohm's law in a more attractive manner than I = V/R, if we visualize the invisible electrical quantities by means of voltage bars and current loops (Fig. 2). They are based on the famous water tower and fish tank hydraulic analogies. In this "geometrical" presentation, the height of the voltage bars is proportional to the corresponding voltages (drops) and the thickness of the current loop is proportional to the magnitude of the current (see also an interactive animation).

Then, we may present the circuit operation (and Ohm's law) graphoanalytically - Fig. 3. As the voltage across and the current through the two 2-terminal components (the voltage source and the resistor) are the same, we may superimpose their IV curves on a common coordinate system. The intersection of the two lines is the operating point A; it represents the current magnitudes of the voltage VA and the current IA. When we vary the voltage VIN of the input voltage source, its IV curve moves horizontally (see also an interactive animation). As a result, the working point A slides over the IV curve of the resistor R; its slope represents the converter's ratio.

Finally, we may present the circuit operation by another attractive graphical interpretation of Ohm's law - the voltage diagram (the voltage distribution along the resistive film inside a linear resistor) - Fig. 4. In this geometrical presentation, local bars with corresponding height represent the local voltage drops (for simplicity, we may draw only the envelope of the voltage diagram). We may intuitively derive the idea of voltage diagram from many examples of human routine: hydraulic , pneumatic, mechanical, thermal, etc. These analogies help to understand that the local voltages along the resistive film decrease gradually (linearly) from the left to the right when the input voltage varies. Actually, this arrangement reproduces the famous Ohm's experiment  (see also another interactive animation). In this arrangement, the tangent of the angle α represents the current IOUT.

Applications

V-to-I converter acting as an output device

Voltage-controlled current source. Almost all the natural electrical sources are constant voltage sources (primary and secondary batteries). Actually, there are no constant current sources in nature (excepting inductor, Van der Graaf generator and photodiode); if we need a current source, we have to build it. For this purpose, we connect a voltage-to-current converter (a resistor R) in series with the input voltage source VIN - Fig. 5. The building formula is:

Current source = Voltage source + Voltage-to-current converter

If we shorten the circuit output with an ideal current load (just a piece of wire), it will generate a constant current IOUT = VIN/R.

A real load (an LED on Fig. 5) introduces some voltage drop VL, which affects the excitation voltage VIN. Now, the voltage difference VIN - VL determines the current IOUT instead the voltage VIN; as a result, the current decreases (see the section about imperfections).

V-to-I converter acting as an input device

Compound voltmeter. A meter movement (a galvanometer) is actually an ammeter. In order to measure a voltage by an ammeter, we connect a voltage-to-current converter (the so called "multiplier" resistor R) before the ammeter (Fig. 6). In this way, we may present the classic voltmeter as a composed device consisting of two components:

Compound voltmeter = Voltage-to-current converter + Ammeter

The multiplier resistor of a classic voltmeter acts as a voltage-to-current converter.

Compound passive converters: Similarly, in the popular passive circuits of capacitive integrator, inductive differentiator, logarithmic converter, etc., the resistor acts as a voltage-to-current converter (Fig. 7):

V-to-V RC integrator = V-to-I converter + I-to-V C integrator

V-to-V RL differentiator = V-to-I converter + I-to-V L differentiator

V-to-V RD log converter = V-to-I converter + I-to-V D log converter

In these circuits, the respective I-to-V converters introduce some voltage drop, which affects the excitation voltage VIN. As a result, the current decreases and an error appears (see the section about imperfections).

Transistor base resistor. A BJ transistor is a current-controlled device; it actually "shorts" the input source (more precisely speaking, we may drive the transistor by a small voltage VIN < VBE0). In order to drive a transistor by a relatively high voltage (for example, in the circuit of a transistor switch), we connect a base resistor in series with the base-emitter junction. It acts as a voltage-to-current converter thus endowing the transistor with a voltage input (Fig. 8.):

Voltage-input transistor = V-to-I converter + current-input transistor

With the same purpose, resistors are connected to the inverting and non-inverting inputs of a Norton op-amp , in order to apply a voltage.

The transistor's base resistor acts as a voltage-to-current converter.

Op-amp inverting amplifier (an input part). In the op-amp inverting circuits the op-amp keeps the voltage of the inverting input at zero level (the so-called virtual ground). As a result, the circuit behaves as a current-controlled device, which "shorts" the input source connected to the inverting input. In order to drive the op-amp by a voltage (for example, in the circuit of an op-amp inverting amplifier), we connect a resistor acting as a voltage-to-current converter between the input voltage source and the inverting input (Fig. 9):

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

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

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

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

What are the input/output resistances?

It is interesting to investigate the input and output resistance (for concreteness, let us assume a resistive load RL).

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. Looking from the side of the input source we see two resistors connected in series; so, the input resistance is RIN = R + RL. As you can see, it depends on the load resistance RL.

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. Looking from the side of the load, we see only the resistor R; so, the output resistance is ROUT = R. Here we suppose that the input voltage source is perfect (i.e., it has zero internal resistance).

Imperfections

Real loads act as imperfect current loads dissipating energy (for example, simple ammeter, current measuring resistor, a diode, etc.) or storing energy (a capacitor, secondary battery, an inductor, etc.) They have some resistance (linear or non-linear), capacitance or inductance, which cause a voltage drop VL to appear across the load (Fig. 12). As a result, now the 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.

The same problem exists in the water analogy of communicating vessels, where the height difference between the vessels determines the flow rate instead only the height of the input water pressure source to determine it.

Improvement

Classic remedy. In some cases (e.g., when the passive voltage-to-current converter is a part of a current source), we may apply a paradoxical trick from our humane routine - depreciate a harmful quantity by another many times bigger harmful quantity. In electricity, this is the well-known classic remedy to decrease the error caused by the load: increase both the excitation voltage V and the resistance R to suppress the load voltage drop VL. As a result, the output current remains the same but the influence of the load (even, if it varies) decreases; thus we get a comparatively constant current source. If we go too far, we will "invent" the classic "electricity" definition of the ideal current source, where an infinite voltage divided by an infinite resistance gives a finite current. For example, imagine that we build 1mA current source first by using 10V voltage source and 10K resistor, then - 100V and 100K, after - 1000V and 1000K, ...and finally - infinite V and infinite K. Only, imagine what power the resistor will dissipate in this compromised solution! In addition, we usually can not change the input voltage V and the resistance R. How to solve the problem then?

Looking for a perfect remedy. Note, that a strange contradiction exists in the simple voltage-to-current converter: from one side, the voltage drop VL is useful as it serves as an output voltage; from the other side, this voltage drop is harmful as it decreases the actual current-creating voltage VR across the resistor R. How can we solve this contradiction? Remember what we do in life when something bad appears. If you do not manage to find a remedy, visit the related story about active voltage-to-current converter to get an answer.