0
\$\begingroup\$

I have a 1 kHz sine-wave signal with an amplitude of 5V (peak-to-peak), and a +2.5V offset.

I want to rectify the signal so that I only get the portion of the signal above 2.5V.

If the signal wasn't offset, this would be easy. I would just use a half-wave rectifier. But what do I do when my signal is offset above 0V?

I have a number of op-amps in my current project, so adding more wouldn't be a problem if that is what the solution requires.

Edit:

I have drawn a quick diagram of what I'm looking to do

enter image description here

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Precision op amp rectifier with an offset. \$\endgroup\$ Jul 21, 2015 at 22:59
  • \$\begingroup\$ I have the precision op amp rectifier working, but am having trouble implementing the offset portion of the circuit. Could you post what your thinking? Thank you. \$\endgroup\$ Jul 22, 2015 at 0:22
  • \$\begingroup\$ What do you mean by "so that I only get the portion of the signal above 2.5V"? What do you want to get if the signal is below 2.5V? \$\endgroup\$
    – Curd
    Jul 22, 2015 at 15:22
  • \$\begingroup\$ It'd be helpful if you draw a graph of the input signal and the desired output signal. \$\endgroup\$
    – Curd
    Jul 22, 2015 at 15:26

3 Answers 3

1
\$\begingroup\$
  1. Bring the signal down to be centered at GND by AC coupling and buffer it (OA0)
  2. Do precision full wave rectification (OA1 and OA2)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
\$\begingroup\$

Do you want the peak to peak value? How about using a capacitor coupling and then a bridge rectifier. If you want more precision use a capacitor to couple into a precision rectifier.

\$\endgroup\$
1
\$\begingroup\$

The title of your question asks for a Full wave but your description says a half wave. To keep things simple here is a half wave 'rectifier' that allows you to set some input voltage (in this case 2.5V) to rectify around.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.