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I am using the MCP6N11-010 in a current sensing circuit (Battery charger). i would like the output of the INA to show the difference in inputs starting from 0v going up to VDD but cant seem to get the configuration correct. I initially just tied VREF to ground but got no incremental changes. I then decided to put VREF at VDD/2. SO the output to from the amp is at VDD/2 when the input are the same and then at (VDD/2)+ input difference * gain. This seems to work but is not what i want. Does anyone have an idea on how to achieve getting the reference to 0v ? Should i perhaps be trying the MCP6N11-100 version of the amp ?

Here are the schematics:

Schematic 1
The above image is of the 0v ref voltage circuit, the one below is the ref voltage tied at vdd/2.

Schematic 2

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  • \$\begingroup\$ Please add links to the images in comments to your question, someone here will surely edit it in for you. Also, providing links to datasheets of all relevant components is highly appreciated here. \$\endgroup\$ Jan 8, 2013 at 8:59
  • \$\begingroup\$ My apologies. I'm new to this site. I tried adding images of the schematics but unfortunately the site disallowed this because i don't have the required reputation. Is there another way to add the schematics ? \$\endgroup\$
    – Grubby
    Jan 9, 2013 at 12:23
  • \$\begingroup\$ Just upload them somewhere, and add the links in a comment... Someone here will incorporate the images into the question, and you can then delete your comment. \$\endgroup\$ Jan 9, 2013 at 12:28
  • \$\begingroup\$ Done linking to the images \$\endgroup\$
    – Grubby
    Jan 9, 2013 at 12:36
  • \$\begingroup\$ Schematics on white are more readable. Also, drawing schematics a bit more compactly helps. I've edited your images a little to fit better. \$\endgroup\$ Jan 9, 2013 at 13:09

2 Answers 2

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Well, I don't see anything explicitly wrong with what you are doing in your circuit with Vref = 0. But there are some concerns. Looking at the datasheet for the MCP6N11, nowhere do they ever set Vref to zero. All of the specified performance is for Vref = Vdd/2, or Vref = 0.75Vdd, or Vref = 0.25Vdd. None of the sample application circuits show Vref=0. Although the limits for Vref are stated as between \$V_ {\text {IVL}}\$ and\$V_ {\text {IVH}}\$, I could find no spec or example where they got outside 0.25Vdd or 0.75Vdd. It is suspicious.

I wonder why you use the 33 kOhm and 10 kOhm dividers on the inputs. Is the battery voltage larger than Vdd, and you are trying to keep the common mode voltage (Vcm) within spec? Usually one of the advantages of using a instrumentation amp is not having to use dividers. The problem is that tolerances add up quickly, and usually 1% resistors are not good enough to give adequate performance. For example if Vbat = 10V, 1% resistors could give an input voltage variation of +/-36mV. With a gain of 100 this could easily put you with an output voltage below Vref.

So, if you are going to or have to use dividers on the input it is crucial that they be very tight tolerance, or trimmed for tracking. You can find special parts like this (tight tolerance and trimmed), but they are somewhat special items. For a trimmed part with 0.1% tolerance and 0.01% tracking, you might spend $2 per divider.

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Vref cannot be zero. On the average one-rail op-amp, Vref normally equals to 1/2 of the power source. This instrumentation amplifier is powered by a single source, therefore, the designer must provide an adequate level-shifting to handle the amplifier's dynamic range. Example: If you power your circuit with +5V and you use +2.5 V as the voltage reference for the circuit, the amplified signal´s positive hemicycle will be reflected at the instrumentation amplifier's output above 2.5 volts and the negative hemicycle below those 2.5 volts. If you are handling only DC signals, the same holds true for positive and negative currents. Keep in mind that you need to provide a way to eliminate the introduced offset after your amplification is done.

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