I have an input voltage -10V to 10V from a DAC. The DAC can source approximately 4 mA. I need to scale the voltage as close to 0.1x as possible (source -10V to 10V to output -1V to 1V). The output is then fed into a motion control input.

What is the best way to accomplish this? I have considered the following:

  1. Inverting op-amp to scale the voltage 0.1x and then a second inverting op-amp to invert the voltage again
  2. Simple resistor-divider
  3. Unity gain op-amp -> resistor-divider
  4. Resistor-divider -> unity gain op-amp
  5. Option #1, omit the second op-amp and change the DAC firmware to output the opposite polarity (-10V instead of 10V). Motion control would be the same.

I have no idea what the best way to do this is. Does it matter? I do have +12V and -12V available for the op-amp.

  • \$\begingroup\$ 4. (min comment length) \$\endgroup\$ May 3, 2017 at 21:16
  • \$\begingroup\$ Yep, 4. (shorter) \$\endgroup\$
    – Finbarr
    May 3, 2017 at 21:20
  • 1
    \$\begingroup\$ What is load impedance? high? then 2. but include ESR of DAC in any case. \$\endgroup\$ May 3, 2017 at 21:26
  • \$\begingroup\$ @TonyStewart.EEsince'75 yup.. everyone always forget to mention impedances. \$\endgroup\$
    – Trevor_G
    May 3, 2017 at 21:32
  • 1
    \$\begingroup\$ I don't forget... beware of all inclusive adjectives especially with partners :) Although I did forget to ask for specs on accuracy. 100ppm? 1%? or what? \$\endgroup\$ May 3, 2017 at 21:37

2 Answers 2


Ideally you want something with a sufficiently high input impedance (DAC has limited current sourcing capability) and a sufficiently low output impedance to suit whatever you are connecting the output to.

Using a resistive divider alone (opt 2) would suit the high input impedance, but would also have a high output impedance. If you were to say use 9k/1k resistors for your potential divider, and connected it to a load of say 50 ohm, you would get almost no voltage out due to loading. If your load is high impedance too, then the loading affect will be minimal and you can stick to just the divider.

If your load is low, then by adding an op-amp with unity gain (opt 4), you buffer the output which will allow you to connect to lower impedance loads without the voltage dropping. You can then choose your resistive divider to be a suitable impedance for your DAC, say 10k total impedance (draws 1mA from the DAC).

https://i.stack.imgur.com/xJ39d.png Stolen from this question, ignore the part numbers

You would connect the resistive divider with the top to the DAC, the middle to the op-amp, and the bottom to ground. This would give you symmetrical scaling over +/-10V. Your op-amp would need to use +/- supplies to cope with the bipolar nature of the input signal.

  • \$\begingroup\$ I agree, except with the DAC only able to source 4mA that puts it's output impedance out about 2.5K... R1 and R2 will need to be large. It may be prudent to use a double amp and use the other one to buffer it first. \$\endgroup\$
    – Trevor_G
    May 3, 2017 at 21:38
  • \$\begingroup\$ @Trevor depends on what the 4mA specification is. If it's short-circuit current, then I completely agree. I was assuming that it is capable of driving a 2.5k load correctly. \$\endgroup\$ May 3, 2017 at 21:42
  • 1
    \$\begingroup\$ Yup, but with limited info to go on... it's worth a mention. \$\endgroup\$
    – Trevor_G
    May 3, 2017 at 21:44

A resistor divider is the easiest option, however, it comes with a cost of load impedance.

If the load impedance of a resistor divider is unacceptable for your application (contributes to error) then buffer the source impedance from the resistor divider with an op amp (like option 3).

Option 1) is complex and introduces offsets and noise (for precision applications) use one of the other options.


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