Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Removing dc offset from Precision rectifier

Status
Not open for further replies.

txbob

Junior Member level 2
Junior Member level 2
Joined
Nov 24, 2008
Messages
23
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Visit site
Activity points
1,462
The input for the precision rectifier has a 2.5v offset so does the output. To remove this offset I added a capacitor after the precision rectifier .My question is there any way I can remove the dc offset by adding a reference to my Precision rectifier this way I don’t have to use the capacitor? Your help is appreciated
 

Attachments

  • Precision rectifier.jpg
    Precision rectifier.jpg
    363.4 KB · Views: 265

Why not use a full-wave precision rectifier such as this. It requires only 2 opamps, 1 diode, and 3 resistors, and generates a full-wave rectified signal with no DC offset using a single supply (with a single supply op amp of course). Here's an alternate circuit that uses only two resistors and has similar characteristics.
 

Thank you for the reply .Both the circuit as it is will not work with an input voltage which has 2.5v offset. If you don't agree please let me know. My second restiction is that I only have single supply operation is possible
 

I think I now understand the problem (which you didn't state). Capacitive coupling of the rectified wave causes a DC offset of its own, since the cap output will go to the average value of the rectified signal. To avoid that, do the following:

  1. Replace C1 with R3 (Remove R3 from ground).
  2. Remove R10 from Vref and connect to ground.
  3. Remove R1 from ground and connect to Vref.
  4. Change R2 or R10 so they are both the same value.

This makes the output amp a direct-coupled differential stage which will cancel the common-mode 2.5V offset. Note that the op amp may not be able to pull the output completely to zero because of the current through R2 and R4. Higher resistor values for R2 and R4, and a low resistance output load to ground can help.

The differential gain is equal to R2 / R4.
 
Last edited:
  • Like
Reactions: txbob

    txbob

    Points: 2
    Helpful Answer Positive Rating
Sorry for the delay
Thank you very much for your answer
I was little confused about the connection of reference
I am enclosing the diagram which worked the difference from your description was the location of reference. If you have any comments let me know
Ref desc got changed in new simulation
 

Attachments

  • Waveform.JPG
    Waveform.JPG
    184.9 KB · Views: 160
  • Post rev2.0.JPG
    Post rev2.0.JPG
    194.1 KB · Views: 168

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top