I'm working on using the TI INA333 in a single 3v supply configuration. I need to create a "mid supply" (half V+) for the REF pin.

I have thought of 3 ways to do this:

  1. Voltage divider with resistors
  2. 1.5v Voltage regulator
  3. 1.5v Voltage reference

Here's the block diagram of the INA333: Block diagram of INA333

It's going to be a battery powered device, so the solution that uses the least amount of current would be the best.


2 Answers 2


Whatever is connected to the REF pin should have low output impedance. If the output impedance is high, it will break the matching of resistors in the second stage of the InAmp (150kΩ ones in the drawing). Let's go through the list of proposed approaches in the O.P.

  1. Voltage divider. By itself, voltage divider doesn't have a low enough output impedance. A voltage divider should be buffered in order to connect it to REF. Previously, I've posted a schematic and more details in this related thread.

  2. Voltage regulator. Could work. Voltage reg has a low output impedance. Typical regulator fluctuates more than a voltage reference.

  3. Voltage reference. Could work. Some voltage references have low output impedance. If a reference doesn't have low output impedance, you can buffer it.

I don't know what the signal is, and what will be done after the InAmp. However, pay attention to what's ratiometric and what's absolute.

  • \$\begingroup\$ Thanks, I'm working on a sEMG circuit, so the input are electrodes on the skin over a muscle group. The output will be hooked into an ADC (one of the main reasons I'm working on a single supply design now). Here's the project blog: emg-parkinsons.tumblr.com \$\endgroup\$
    – nak
    Commented Oct 5, 2013 at 19:48
  • \$\begingroup\$ Generating a negative supply rail is not a terrible nuisance. A small charge pump can generate -3V for multiple InAmps. (Just another $0.02 option. I've done an EEG gadget in the past.) \$\endgroup\$ Commented Oct 5, 2013 at 19:53
  • \$\begingroup\$ Ah, the problem I ran into is with the ADC reference, I wasn't sure how to go about sampling when the signal was going negative. I suppose I would need to have the ADC reference be -3v too. I'll look into it. \$\endgroup\$
    – nak
    Commented Oct 5, 2013 at 19:56

since the reference isn't pulling any current a voltage divider should be fine, just use higher value resistors.

  • 2
    \$\begingroup\$ -1. Exactly wrong answer. You've never done this yourself. \$\endgroup\$ Commented Oct 5, 2013 at 19:20
  • \$\begingroup\$ Aww, I hope you'll explain why. \$\endgroup\$ Commented Oct 5, 2013 at 19:23
  • \$\begingroup\$ Sure. Just wrote my version of the answer. Be sure to follow the link to that other thread. Somebody has actually hooked up a voltage divider straight to a REF input of the InAmp. He got deviations in the measurements. \$\endgroup\$ Commented Oct 5, 2013 at 19:40
  • 1
    \$\begingroup\$ @rakdosshredfreak - Guesses are always found out. Be very afraid. LOL \$\endgroup\$
    – Andy aka
    Commented Oct 5, 2013 at 20:54

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