This circuit outputs the negative log of the input. The first op-amp attempts to keep its – input at ground, which means the current across the 1k resistor must be proportional to the input voltage. This current goes across a transistor, so the op-amp must keep its output voltage at a level which satisfies the Ebers-Moll equations, which means that eVout is proportional to the input current. This means that output voltage must be proportional to the log of the input current (and thus the input voltage).
The combination of the second transistor and the current source adjusts the output voltage upward by a fixed amount, and the second op-amp amplifies it 16x. -- Credits: Mr. Paul Falstad.
|Vinay said: (Nov 7, 2011)|
Please mention the op-amp used here.
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