Circuit Idea/Op-amp Inverting Current-to-Voltage Converter Visualized
Circuit idea: The op-amp eliminates the resistance by adding as much voltage to the input voltage source as it loses across the resistor.
The initial circuit arrangement
[edit | edit source]Imagine that we have to solve a well-known problem - to measure a current IIN that flows if we short a real voltage source having an internal voltage VIN and a resistance Ri. For this purpose, in the basic story about the op-amp inverting current-to voltage converter, we have broken the circuit and connected a measuring resistor R to obtain an output voltage VOUT = VR. Only, the voltage drop VR across the resistor R was affecting the input current. So, we have compensated the "disturbing" voltage VR by an "anti-voltage" VH (we have "helped" the input voltage source by another voltage source) and used it as an output voltage (see Fig. 2 below).
In this way, we have managed to know how to convert this imperfect passive circuit into an almost ideal op-amp one. Then we were interesting mainly in the circuit structure rather than the circuit operation; that is why we have built the circuit step-by-step following the basic "helping" idea. That would be enough for us, if we were students, hobbysts and technicians who intend just to use this circuit for personal purposes.
Visualizing the active current-to-voltage converter
[edit | edit source]Only, this is not sufficient, if we are teachers who want to present the circuit operation so that their students to really understand it. In this case we have to get these young people interested in the circuit operation; we have to show them in a more attractive way how the active elements (a man and an op-amp) implement the "helping" idea. For this purpose, we may use here a set of favorite heuristic visualizing tools: voltage bars, voltage diagrams, current loops and superimposed IV curves.
Let's first look at Fig. 2. Do you note that the voltages and voltage drops across the components BIN and Ri are positive while the voltages across the components R, BH and L are negative?
Then, we may rearrange them in a more suitable form, in order to superimpose the visualizing elements over the circuit diagram (see Fig. 3a below). For this purpose, let's first draw a light red line across the middle of the sheet dividing the plotting area into two parts: a positive upper area and a negative lower area. Then, let's place the elements with positive voltages across them (the excitation voltage source VIN and the internal resistance Ri) in the upper part and the elements with negative voltages across them (the helping voltage source VH, the resistor R and the load L) in the lower part. A tip: think of the red line as a "sea level"; all "positive things" are located "above the sea level" while all "negative things" are "under the sea level". Let's now begin experimenting.
Exploring the circuit at positive input voltage
[edit | edit source]... by voltage bars and current loops
[edit | edit source]If you apply a positive input voltage VIN (Fig. 3a), a current begins flowing through the circuit in a direction VIN >>> Ri >>> R >>> VH. Figuratively speaking, you make the input source "pull" the point A up toward the positive voltage VIN (you, the input source and the resistance Ri "emerge from the sea":). Only, I observe to my great displeasure that the needle deflects to the right and immediately react by decreasing the compensating voltage VH. Now, I make it "pull" the point A down toward the negative voltage -VH (I, the "helping" source, the resistor R and the load "sink into the see":) until it manages to zero the potential VA (the virtual ground). Note how clear it is in this arrangement that the two voltage sources are connected in series, in one and the same direction (- VIN +, - VH +) so that their voltages are added. Actually, they constitute a new compound voltage source, which serves as a "helped" source producing a corrected input voltage.
... by superimposed IV curves
[edit | edit source]On the graphical presentation (Fig. 3b), when the IV curve of the input voltage source moves horizontally from left to right, the IV curve of the "helping" voltage source moves horizontally from right to left and v.v. As a result, the operating point A slides along a new vertical IV curve, which represents the zero dynamic resistance Rd of the virtual ground.
Exploring the circuit at negative input voltage
[edit | edit source]... by voltage bars and current loops
[edit | edit source]If you apply a negative input voltage VIN under the ground, a current begins flowing through the circuit in an opposite direction: VIN >>> VH >>> R >>> Ri. Now, you make the input source "pull" the point A down toward the negative voltage -VIN (you, the input source and the resistance Ri "sink into the see":). Only, I observe that the needle deflects to the left and immediately react by increasing the compensating voltage VH. Now, I make it "pull" the point A up toward the positive voltage VH (I, the "helping" source, the resistor R and the load "emerge from the sea") until it manages to zero again the potential VA (the virtual ground). The two voltage sources are connected in series again, in one and the same direction so that their voltages are added. Actually, they constitute again a new compound voltage source, which serves as a "helped" source producing a corrected input voltage.
... by superimposed IV curves
[edit | edit source]On the graphical presentation (Fig. 4b), when the IV curve of the input voltage source moves horizontally from left to right, the IV curve of the "helping" voltage source moves horizontally from right to left and v.v. As a result, the operating point A slides along a new vertical IV curve, which represents the zero dynamic resistance Rd of the virtual ground.
Visualizing the op-amp inverting current-to-voltage converter
[edit | edit source]Finally, we decide to make an op-amp do this boring work. 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. Let's now see how the op-amp does this magic.
Exploring the circuit at positive input voltage
[edit | edit source]If the input voltage VIN increases, an input current IIN begins flowing through the circuit (Fig. 5a). As a result, a voltage drop VR appears across the resistor 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 its great displeasure:( and immediately reacts: it decreases its output voltage "sucking" the current IIN until it manages to zero the potential VA. Figuratively speaking, the op-amp "sinks into the see"; it "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. The two voltage sources are connected in series, in one and the same direction (- VIN +, - VH +) so that their voltages are added.
Exploring the circuit at negative input voltage
[edit | edit source]If the input voltage VIN decreases under the ground, an input current IIN begins flowing through the circuit in an opposite direction (Fig. 5b). As a result, a voltage drop VR appears across the resistor R 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 IIN until it manages to zero the potential VA (now, the op-amp "emerges from the see" and "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 so that their voltages are added.
