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I designed a simple low frequency doubler, but I'm having trouble fixing the offset of my output signal. Below is my LTSpice model:

enter image description here

On the graph you see my green input signal, blue inverted signal, and then red re-combined signal with double frequency. My main issue is that my output is offset by 35mV. I've tried a few things but none of them seem to work. Does anyone know a simple way to bring the offset back to zero?

Frequency range: 50Hz - 1500Hz. Low voltage input. Op amp is receiving +-5V.

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    \$\begingroup\$ For ltspice graph+schematic it is useful to name nodes and then have the graph show the names so people can quickly see where the measurements were done. \$\endgroup\$ – PlasmaHH Apr 23 '17 at 20:35
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Simply AC couple your output:

output of opamp ---||------ consumer

Then, you can bias the double-frequency signal around any level you want. DC doesn't come through the capacitor, so there's inherently no DC offset.

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There are a couple of details needed to answer your question about the DC offset. (The first thing that hit me about your plot was that there was only about \$30\:\textrm{mV}\$ of voltage drop due to the diodes -- not likely.)

  1. You are using a pair of diodes oriented so that the output can only be positive. That's what diodes do in this case.
  2. You have zero load on the diodes, which is probably why you see any signal at all. If your signal is only \$100\:\textrm{mV}\$ peak, then I'm pretty sure you wouldn't see any (much) signal at all on \$V_O\$. This is because those diodes require a sizeable voltage drop across them and your schematic doesn't have a load so you aren't seeing the problem, yet.

I don't know if you are just experimenting with \$100\:\textrm{mV}\$ peak and it has nothing whatever to your actual signal, or if you expect this circuit to actually work right with such a small level of signal. If the value is realistic, then you have a lot of work yet to do. If not, you probably should tell us what your signal magnitudes are likely to be and what kind of load you expect to drive (and what kind of realistic source you have, too.)

If this is just theoretical and has nothing to do with anything practical, then your question really doesn't make much sense to me. In theory, you should expect the output to be only positive.

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Try this legitimate full-wave-rectifier

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Those resistor values are not legitimate. Try replacing with 10k and 20k. \$\endgroup\$ – WhatRoughBeast Apr 24 '17 at 2:38
  • \$\begingroup\$ The resistor values depend on how fast and how small you wish to accurately rectify. \$\endgroup\$ – analogsystemsrf Apr 24 '17 at 3:09

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