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This circuit calculates the integral of the input, meaning that the output voltage changes at a rate proportional to the input voltage (but in the opposite direction). The op-amp attempts to keep its – input at the same voltage as the + input (which is at ground). This requires it to drain an amount of current proportional to the input voltage (V / 1000 ohms). This current goes through the capacitor, whose voltage is proportional to the integral of the current flow through it.

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Generated Tue Feb 23 2010