Here is another
tip of the hat tip of the soldering iron to one of the Planet Analog archived posts. This post gives an explanation of how op-amps work in general and how an integrator works in particular. Unlike some of my recent blogs that point at archived posts, this time we'll dive in deeper.
Understanding this circuitry is essential if you are doing any sort of design with op-amps. How much of this is taught to engineers today?
Let's assume that you set up an op-amp circuit like this:
This circuit will have a gain that's determined by the ratio of the feedback resistor (RF ) to the input resistor (RI ). It also inverts the signal, so a positive voltage swing at the input produces a negative voltage swing at the output. Note that to simplify this discussion, we assume that there are positive and negative voltage power supplies for the op-amp. Hence, VOUT can swing above and below ground. Let's make both resistors 1.0kΩ. There are a few different ways to examine what the circuit will do. From a macroscopic view, VOUT = – VIN * (RF / RI ). But why?
The op-amp is a tiny servo system that will move its output around so as to force the voltage at the negative input to be equal to the voltage at the positive input. Since the positive input is at ground (0.00V), the negative input (or summing junction, since it sums voltage sources from two different directions) should end up at 0.00V. So if you put +1.00V at VIN , VOUT will go to -1.00V. Half way between those two equal-value resistors (voltage divider rule) is at 0.00V.
You can also work through this by analyzing current flow. Recall Kirchhoff's circuit rules for current (conservation of charge), sometimes called Kirchhoff's first law: All the current flowing into a node (from one or more paths) has to flow out of the node (again, it can be through one or more paths). By assigning the appropriate sign to the flow (“into” is positive; “out of” is negative) it has to sum to zero. No current flows into/out of the input pin.
OK, not true — there is a tiny current flow, but it is usually so small that it can be ignored in most applications. In very high impedance circuits, circuits amplifying extremely small signals, or some track-and-hold/integrating circuits, that tiny current can be trouble. For now, we'll ignore it.
So, no current into or out of the input pin means that any current that flows into RI (due to the voltage source VIN ) must flow through RF and into the output of the op-amp. We can reverse the current flow (change its sign) and have it flow the other way, but that's just a sign on a number, so the analysis is the same.
When you change the feedback resistor to a capacitor, the circuit becomes an integrator.
To analyze this, Kirchhoff's current rule proves the most helpful because you charge (and discharge, but that's the same thing except for a change in sign) a capacitor. The equations work out a little more neatly via a current analysis. With this in mind, if you now refer back to the archived integrator article that I cited at the beginning of this blog, clarity should ensue. To the extent that it does not, please post your comments below.