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I am attempting to convert the differential output of a DAC to a single-ended signal using the following circuit from a TI datasheet:

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The signals going into this amplifier look like this:

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And the output signal looks like this:

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The Op-amp I am using is the LMC6484 (rail-to-rail) with V- to GND and V+ to +5V. My understanding is that the subtraction of signals results in a negative value for half of the waveform resulting in the clipping behaviour. What I am looking for is a way to bias the signal to avoid this clipping? I am trying to avoid having to add a negative voltage rail.

I am having a hard time figuring out how this is possible. Any suggestions?

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  • \$\begingroup\$ And your supply voltages are? \$\endgroup\$ – Brian Drummond May 15 '16 at 23:50
  • \$\begingroup\$ @BrianDrummond 'V- to GND and V+ to +5V' \$\endgroup\$ – BBales May 16 '16 at 0:03
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    \$\begingroup\$ Simplest answer : double R2's value and add an identical resistor to +V. This effectively creates a resistor identical to R2 connected to 2.5V. \$\endgroup\$ – Brian Drummond May 16 '16 at 9:58
  • \$\begingroup\$ @BrianDrummond thanks for this. I don't quite understand where to add this resistor, do you mean the V+ rail of the op-amp? Or the positive input of the op-amp? Do I double the value of both R2s? or just the bottom one? \$\endgroup\$ – BBales May 16 '16 at 13:15
  • \$\begingroup\$ The 5V supply, and double both resistors. How else are you going to get an identical resistance to R2 from 2.5V? And that tells me you haven't learned about "Thevenin equivalent voltage sources", which is kind of important (and should be searchable). \$\endgroup\$ – Brian Drummond May 16 '16 at 13:33
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Your feeding a diff signal with a + and - voltage swing, but your power supply is single-ended, with no negative rail, so the negative part of your signal is being cut off. For this to work you need an op-amp with +/- 5 volt power rails.

For single -ended power supply issues: Add a 10K ohm pot from +5 volts to ground. Remove the grounded end of R2 and connect it to the wiper of the pot. This will provide a positive offset and shift the signal up ward. If it clips on the + 5 volt rail then reduce the values of both R2's by 1/2.

Over time you can fine-trim R2 for maximum size without clipping.

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  • \$\begingroup\$ I know that adding the negative voltage rail is an option, but I was wondering if there is any way to bias these signals so I can avoid doing this? \$\endgroup\$ – BBales May 15 '16 at 23:12
  • \$\begingroup\$ Wont this mess with the impedances and frequency response of the circuit? \$\endgroup\$ – BBales May 15 '16 at 23:25
  • \$\begingroup\$ Very little. The center tap of the pot is equal to zero volts. Your impedance and frequency response is dependent mostly on R1 and C1. Both R2's should match, but your not using precision resistors to begin with, as you have not assumed tolerance errors, but a 'perfect' circuit. \$\endgroup\$ – Sparky256 May 15 '16 at 23:32
  • \$\begingroup\$ Okay, so I have added the voltage divider / pot to the circuit. It did allow me to bias the signal upward which is great. Would it be wise to add a trim pot in series with the R2 on top for impedance matching? \$\endgroup\$ – BBales May 15 '16 at 23:37
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    \$\begingroup\$ If it is important to do so, then do it, but reduce the value of the upper R2 so your trim pot has a low value. Ideally, which is rare, the trim pot will wind up being close to its mechanical center and at a low total resistance, but that is by no means a mandatory item. \$\endgroup\$ – Sparky256 May 15 '16 at 23:41

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