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So, I'm trying to modify a sample and hold circuit from "The Forrest Mims Engineer's Notebook", that uses a 353 op-amp, so that it runs off of a single supply:

Schematic

I needed to add a 4.5V (1/2 supply voltage) offset to the inverting input, so that it would operate on a just a single polarity supply. My dilemma is that I need to remove the 4.5V DC offset from the output. Obviously, the output would be a DC voltage, so a capacitor would not work. I was simply wondering if there was a way that I could subtract the 4.5V offset from the output, using all discrete components if possible.

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  • \$\begingroup\$ Hi Zack, just a quick note: You can insert images into your post using the image link button or by using the keys Control+G. That's preferable to posting as an HTML img tag in case the original image link goes down. \$\endgroup\$ – JYelton Feb 28 '14 at 21:30
  • \$\begingroup\$ Thanks! That helps a lot! Now I don't have to go through my dropbox! \$\endgroup\$ – Zack Frost Feb 28 '14 at 21:40
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The op-amp circuit following the sample and hold is intended to have a gain of two and I think this over complicates things with the mid-rail generator formed by R1 and R2. Try this for size: -

  • Get rid of R1, R2 and R4 - just remove them and don't connect anything in their place.
  • Get rid of D1 and short it out - it's just adding an error that is not needed.

Now, if you have 3V on the input and you sample it, 3V will get charged onto C1 and the output from the op-amp will be a buffered 3V - isn't this what you want?

If you do need a gain of two then restore R4 but take it down to GND/0V - this time, with 3V being sampled it'll be 6V at the output. If your maximum input voltage is 4V you'll get 8.8V on the output - except the LF353 is a poor choice these days - it's not rail-to-rail which means you might see 7.5V maximum on the output - it's going to limit to about 1.5V below the 9V supply voltage.

If, in fact you have a 12V rail, then using it as a gain of two is fine because if your input is restricted to 5V it'll produce a 10V output no problem.

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  • \$\begingroup\$ The gain isn't exactly 2 because of the Thevanin equivalent resistance of R1&R2. D1 isnt needed unless you need a peak detect function. If the signal is anywhere near ground with a single supply you will wind up violating the input common mode range and not getting an accurate output. (Which is probably why we're talking about offsets.) Better idea would be to get a rail-to-rail opamp. \$\endgroup\$ – John D Mar 1 '14 at 0:50
  • \$\begingroup\$ So, for a single supply buffer I don't have to use a virtual ground? \$\endgroup\$ – Zack Frost Mar 1 '14 at 23:13
  • \$\begingroup\$ @ZackElec no you don't. \$\endgroup\$ – Andy aka Mar 1 '14 at 23:23
  • \$\begingroup\$ Well, thanks! I tried it and it works absolutely flawlessly!! That's good to know! \$\endgroup\$ – Zack Frost Mar 1 '14 at 23:43

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