In calculus, we learn about both differentiation and integration. In this tutorial, we’ll learn about how we can create an analog integrator using an operational amplifier. An op-amp integrator performs almost exactly as one would anticipate if they’re familiar with integration in general, with a negative coefficient. For example, if we have a constant positive voltage on the input, then the output will linearly decrease. If you integrate a positive constant, with a negative coefficient, then the output increases decreases as you expand your bounds.
In other words, the voltage output of an op-amp integrator is proportional to the integral of the voltage input.
While a constant input is easy to envision, an integrator takes any input and creates an output as one would expect from an integrator. A sine wave in will provide a cosine wave output. Square wave input yields a triangular wave output. The equation ex as an input will have ex as an output. While this is generally true, in reality, the relationship is proportional, which we’ll go over in a moment.
Derivation of the equation that describes an integrator’s behavior:
The output of an integrator can be described with the following equation.
Let’s go through how we arrive at this equation. I recommend either 1) seeing if you can arrive at this equation yourself before continuing or 2) after going through the steps here, that you do it again by yourself.
Exactly as we do with the inverting and non-inverting circuits, we can use KCL, this time on the inverting input. As we know that no current flows into the inverting input and, as the non-inverting input is connected to ground, that the inverting input is at a virtual ground. Thus, the current flowing through the resistor is the same as the current flowing through the capacitor. Setting those two currents equal to each other and describing them in terms of voltage, we get the initial equation:
If the right hand of this equation is confusing, just remember that the current through a capacitor is the change in voltage versus the change in time, multiplied by the capacitance, C*dv/dt. In this case, the voltage change is the virtual ground at the inverting input (0 Volts) minus the output voltage (VO). Now, let’s simplify this and calculate the output voltage.
Remove unnecessary 0’s - don’t forget to keep the negative on VO:
Remembering that we can move that negative sign out of the derivative, we can shift everything to the left side of the equation:
Now, we could leave it in this form but we want to see what the circuit is doing to the input, so we’ll need to get rid of that derivative. So let’s integrate both sides. We can leave the constants (-1/RC) outside of the integral, giving us:
Note that we can actually consider the circuit in a different way, treating the capacitor mathematically as 1/(jwc) instead of C*dv/dt. We take the same equation as before, but swap those out.
Then we can simplify it to:
This may be a more helpful equation depending on your application and how you want to use it.
We assume that the two inputs to the op-amp both have the same voltage but, in reality, there is an input DC offset. A small variation in the voltage inputs which, as integrators measure the difference in those inputs, causes an error that builds over time. This wandering output needs to be controlled either by careful modification of the input voltage or, more likely, a simple and regular reset to center things again.
These calculations also don’t include the frequency limitations of the op-amp. Op-amps have bandwidth limitations - as you increase in frequency, particularly above the op-amps cut-off frequency, the output will be attenuated. For the most part, this won’t come into play, but as your circuit frequency increases, this becomes even more of an issue.
An op-amp integrator is a critical part of ADCs, analog computers, and even wave-shaping circuits. While not as common as the voltage follower, comparator, and inverting/non-inverting amplification configurations, the integrator circuit is a great circuit to have in your back pocket. If you’re studying circuits in general, having a good understanding of how it works will help in your understanding of all circuits as well.
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