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I am using an instrumentation amp (http://www.linear.com/product/LT1168), and I am a bit confused why it is giving me a output voltage that I am not expecting.

I am pretty sure, I am doing something dumb here.

The way I have it hooked up now is as follows:

enter image description here

I am expecting a 5V difference between the reference and the OUT pin, but I am only getting 3.9. What am I doing wrong here? I know the output will not go to the ramp voltage, but 3.9 seems too little.

There must be something I am missing. Datasheet: here

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    \$\begingroup\$ This IA works best with a dual supply. If you must use a single supply, then don't take shortcuts on biasing Vref . To get the benefit of 90 dB CMMR, Vref must be low impedance and driven externally near the middle of the total supply range. Your differential input saturates the output, even at unity gain. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 8 '12 at 1:25
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The LT1168 does not have rail-to-rail outputs.

enter image description here

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  • \$\begingroup\$ wait, that's it? \$\endgroup\$ – Ender Wiggins Nov 8 '12 at 0:20
  • \$\begingroup\$ Yup. That's it. This would be a good time to edit your question to explain what you're actually trying to do (why are you trying to use an IA with 5V single supply?). \$\endgroup\$ – squarewav Feb 16 '16 at 23:18
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  1. The reference pin must be Vcc/2 for single supply operation. Figure 4 from the datasheet shows a sample circuit for doing this. Rather than mess with a voltage divider or potentiometer, I would use a 2.5V reference diode for your case. The reference diode will give you a more stable reference than a resistive network. Figure 4
  2. As was stated, this is not a rail-to-rail part. If you need rail-to-rail capability, try something like an INA333. While a bit pricey, it's a CMOS part, and will do the job, with the added benefit of being designed for single supply use.
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