Application Description:

We are converting a single ended DAC output into differential. The DAC output composes a DC signal(-10V - +10V) and sine waves running below 100kHz(below 100mVpp). The DAC has an OpAmp voltage follower at the end of it already.

The reason for conversion is because I want to export a larger range of the output. The range of the single ended signal is 20V, while a differential one with same power supply could provide 40V range.

The load which the output voltage will exert on is a capacitor, so with very large impedance. I guess in this case the balanced outputs for differential voltages are not necessary.


  1. Can I simply use a voltage inverter at the output of the DAC? The reason is that I can find small footprint OpAamps with several OpAmps in one chip.

  2. What are the considerations when designing a converter for this purpose? Such as OpAmp selections, feedback resistors and etc..

  3. I saw some applications use differential OpAmps. What are the benefits of them comparing with using OpAmps?

  • \$\begingroup\$ You say the purpose is to "export a large range of the output"? I don't understand what you mean by "export". Usually the reason to go to differential is to avoid common-mode noise (which would affect if you are to send the signal over twisted pair for example). Just doubling the amplitude is not beneficial unless you are able to double the signal-to-noise (SNR) ratio. \$\endgroup\$
    – jpcgt
    Commented Jan 21, 2016 at 14:20
  • \$\begingroup\$ What is the value of the capacitor? \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2016 at 15:11
  • \$\begingroup\$ Is the output common mode voltage referenced to ground or to something else? \$\endgroup\$ Commented Jan 21, 2016 at 15:54
  • \$\begingroup\$ @jpcgt In our case, it would be a control voltage to drive a capacitor like device. 20V range is not sufficient. While 40V range is sufficient. \$\endgroup\$ Commented Jan 22, 2016 at 1:34
  • \$\begingroup\$ @Andyaka The capacitance shall be quite small, around 10pF. It is a capacitor-like device, not a really capacitor. \$\endgroup\$ Commented Jan 22, 2016 at 1:38

1 Answer 1

  1. Yes, you can.
  2. Ability to drive capacitive load stably, Low Offset Voltage, High common mode rejection, High power supply rejection, High input impedance, High gain bandwidth, the usual suspects (but in the order that would be the most important) Feedback resister should likely be in the 1kΩ-10kΩ, any higher and you get "Johnson noise" any lower and you get excessive current drawn away from the node (rule of thumb is to pick it so the resister ladder for feedback only consumes 1/10th of the max output current)
  3. Benefits of differential op-amp for single to differential conversion: Don't have to worry about mismatch from two different input offsets (each single-ended op amp will have it's own different input offset), I'm sure there are others but I can't think of them.
  • \$\begingroup\$ Thanks for your answer Dave. Could you please explain by what do you mean by mismatch from two different input offsets? \$\endgroup\$ Commented Jan 22, 2016 at 1:44
  • \$\begingroup\$ A very simplified way of understanding it is to just think of the inputs as always needing to be a few mV different to actually output zero. So if you tie one input to ground and then use the other as an input, when you have zero on the output, you'd actually be putting in something slightly more than zero into the input. This leads to a systematic error of always being off by that input offset voltage. Now, If op amp one is off by +2.3mV and op amp two is off by -3.5mV then the waveforms they output wouldn't be inversely symmetric. That would lead to a DC offset on your readings. \$\endgroup\$
    – Dave
    Commented Jan 22, 2016 at 3:16

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.