Try running a DC sweep!
This circuit takes advantage of the logarithmic current to voltage relationship of the diode.
The inverting input of op-amp OA1 serves as a virtual ground. The current through resistor R1 is always V1/100K. Since the op-amp input terminal draws no current, that entire current goes through the diode. The output voltage of the first stage is whatever is needed to pass that current through the diode.
The second stage (R2, R3, OA2) is simply an inverting gain of 10.
I like the Idia. |
by meshtech
October 21, 2015 |
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