Opamp Configurations - Differentiator
By using an input capacitor instead of a resistor, we can accomplish the same thing. If you remember, a capacitor stores charges in its plates, when one of them starts accumulating charges, the same charges will be pushed out from the other plate, as if current was flowing through the capacitor despite the intrinsic insulating layer.
The capacitor's charges start building up and creating a voltage across itself in opposition to the charging voltage, thus slowing down the incoming charges, slowing down the charging process in general. When enough charges have accumulated, the charges inside the capacitor completely push away the charges coming from the source, no more charges enter the capacitor, and because of this no more charges are pushed out on the other side of the capacitor, so no more apparent flow of current across the capacitor.
When used as input for a signal, if the signal does not change (like a DC input), the capacitor will have an initial apparent current through it as the voltage across it builds up due to incoming charges, and since the input of the amplifier tries to not draw any current, it will create a voltage at its output so that the current through the feedback resistor is the same as the apparent current through the capacitor.
Since the capacitor charges very quickly due to the voltage applied to it and the fact that there's no current limiting component like a resistor, the apparent current through the capacitor falls very quickly as the voltage across it in opposition rises as quickly; the falling current is also causes the opamp to drive the output voltage less, since there's less current to compensate for.
Applying a DC input to the differentiator thus creates a spike in input as well as in output as the capacitor's initial charge is developed, and then goes back to 0v as there's no more apparent current to compensate for; Similar to the operation of finding a constant's derivative, which is always 0.
The fact that there's an initial spike can be mathematically modeled as a period in which there's a function that rises at a very high rate (which actually happens, the voltage doesn't just jump from 0v to the DC input voltage, it rises very rapidly towards it), so its rate of change is very high for a brief period of time; hence the spike.
As the input voltage stabilizes, its rate of change slows down very rapidly as well, going towards zero when fully stabilized; this is reflected in the opamp's output by the fact that as the voltage stabilizes, the output spike goes down very rapidly towards zero and stays there.
Now instead of applying a constant input, you can replace it with a constantly changing input.
If the input is increasing at a constant rate, there will be a constant apparent flow of current through the capacitor, since the voltage buildup across the capacitor is compensated by the increase in input signal. Since there's a constant apparent flow of current through the capacitor, the opamp compensated by setting the output voltage at a level that will make the feedback resistor draw the same amount of current, so that the opamp input does not draw it.
Since the amount of apparent current is constant, a constant output voltage is enough to keep the feedback resistor drawing the current, and the opamp keeps a constant output at the output.
This mode is very similar to using a resistor with constant dc as the input.
The same is true for a constantly decreasing input voltage; the output will just be of reversed polarity. To compare with the mathematical definition of the derivative of a linear variable, the derivative will be a constant.
This can be expanded to other functions, one of the most widely used being the sine function. Since the mathematical derivative of the sin(x) function is cos(x), which is a shifted version of sin(x) by 90 degrees, when you input a sine input at the differentiator amplifier, the output will be the same function shifted 90 degrees, in essence, a cosine function.
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