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Say a mains AC (120v, 60hz) 10 amp load with a 0.003 ohm current sense resistor in series with it. The voltage across the resistor is roughly 30mv. How do you rectify such a small voltage (I read precision rectifier, but haven't been able to make the circuit work in Circuit Simulator Applet), or amplify it? Basically, how do you get a usable signal out of that tiny voltage?

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    \$\begingroup\$ Active Diode - if you just want to rectify it. Amplify - if you want higher voltage/power output. You have to define what you are trying to accomplish. \$\endgroup\$ – Optionparty Jun 3 '13 at 7:42
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30mV (peak-peak?) is in the same order as the output from a microphone so a 100x voltage amplifier could easily take this up to 3V. A precision rectifier (active diode) simply uses the op amp (open loop) gain to overcome the turn on voltage of a diode.

https://en.wikipedia.org/wiki/Precision_rectifier

enter image description here

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  • \$\begingroup\$ Thanks for the tip. Actually, Circuit Sim Applet has one 'op-amp->Full wave rectifier' that does exactly what I wanted. \$\endgroup\$ – cannotcompute Jun 3 '13 at 8:06
  • \$\begingroup\$ Which op-amp would be best suited for this? There are so many op-amps out there that it's kind of overwhelming...lm324? lm358? I need very fast response time.. \$\endgroup\$ – cannotcompute Jun 3 '13 at 15:42
  • \$\begingroup\$ @Luke The signal is 60Hz and its only handling a few mA. Just about any op amp would work.Note this circuit uses a split supply. (say +/- 6V would give plenty of headroom) so I would try something like a TL071 with a couple of 1N4148 diodes, or a TL072 if I wanted to buffer the output. Your response time is controlled by the frequency you are measuring. You will get 1 peak reading per cycle. \$\endgroup\$ – JIm Dearden Jun 3 '13 at 16:25

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