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Is there something wrong with this Precision half-wave Rectifier circuit?

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wchzym

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I want to design a Non-inverting Precision half-wave Rectifier. This is my design.And I simulate with it but the result is wrong. **broken link removed**
 
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I want to design a Non-inverting Precision half-wave Rectifier. This is my design.And I simulate with it but the result is wrong.
So how wrong is it?;)

Try grounding R9b and using R8b as the input.
 
With the circuit as provided by the OP the simulation goes wrong.

With the signal connected to R8B the output is inverted.

The output level remains stable over a wide range of input voltage.

See the attachment.

Boncuk
 

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With the circuit as provided by the OP the simulation goes wrong.

...............
I would argue that the simulation was right. It was the circuit that was wrong.:D
 
I would argue that the simulation was right. It was the circuit that was wrong.:D

Hi Carl,

I guess it's a matter of the point of view. :)

Is the glass halfway empty or is it halfway full? :eek:

Regards

Boncuk
 
Boncuk, your feedback is wrong.
 

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I want to design a Non-inverting Precision half-wave Rectifier. This is my design.And I simulate with it but the result is wrong. **broken link removed**
It looks like the OP's circuit should work. It would take two minutes to build on a vector board. Here is a half wave rec circuit I use with a peak detector added in the feedback loop, but the basic half wave circuit is the same.
 

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It looks like the OP's circuit should work. .............
No. His original circuit will not work as a precision rectifier, as I previously noted. The signal must be go to the inverting input, not the non-inverting input for proper operation.

Edit: If the signal is applied to the non-inverting input, the output will look like a follower for the negative part of the input, so the output will look like an amplified sine-wave for the positive half of the wave, and an unamplified (gain of 1) sine-wave for the negative half of the wave.
 
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