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Im building an sEMG circuit and am a bit confused on how to change the signal back to non inverted before the output.The signal goes -> amplifier (x16)-> high pass filter -> variable amplifier (typically x65) ->full wave rectifier -> integrator. However after the integrator the signal will be inverted (I put it through the negative terminal of the op amp), I though about putting the rectifier before the amplifier but will the diodes work correctly when the voltage is than 0.7 V?
The instrumentation amplifier is the INA129, the diodes are 1N4148 and the opamps are AD822, thanks!

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  • \$\begingroup\$ In what way is the phase "wrong"? You can't have a causal filter without a phase shift, so the best you can hope for is roughly constant group delay (linear phase) over your pass band. \$\endgroup\$
    – Theran
    Commented Apr 3, 2014 at 6:38
  • \$\begingroup\$ @Theran I apologise I asked the question completely the wrong way, I edited it now \$\endgroup\$
    – alto125
    Commented Apr 3, 2014 at 6:46

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Inverted output

Just change the output stage like this to get non inverted integrator signal. Additional voltage divider at the integrator input needed because the amplifier gain calculated as K=R1/R2+1 when the signal applied to positive input. So to preserve unity gain you need to divide the signal first.

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Diodes follow the Shockley diode equation pretty well at low currents and voltages, so in that sense they work just fine below 0.7 V. \$V_f \approx 0.7 {\tt\ V}\$ is just a designer's rule of thumb for a forward-biased silicon diode conducting "significant" current. The actual characteristic is more like a smooth exponential function.

The precision rectifier circuit compensates for the exponential characteristic of the diode with feedback from the op-amp. This works, but it still has trouble with very small signals. You can swap the polarity of the diodes in the precision rectifier to get a different polarity output.

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