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I have a TI INA126 (datasheet) instrumentation amplifier that I want to use to amplify a noisy, low-level signal from a fabric-based resistive stretch sensor.

The resistance ranges for the fabric sensors vary depending on how they are cut. Sometimes the range is 50Ω-100Ω. Ideally it is 300Ω-600Ω. Let's assume the range 200Ω-300Ω at first.

With a simple voltage divider (R2=250Ω) and working with 5V, I get a voltage range of about 0.5V from 2.27V to 2.77V.

(250/(250+200))*5 - (250/(250+300))*5 = 0.5

So, I need to remove the common-mode of 2.77V and amplify the difference into the output swing of the INA126 (0.8V-4.2V).

The equation for the INA126 is:

Vo = (Vin+ - Vin-)*G

G = 5+(80kΩ/Rg)

Setting the gain is very easy (one resistor across pins 1 and 8). A table is provided in the datasheet.

I'm having trouble figuring out how to set everything up. I have it set up as follows:

  1. Rg (no resistor for 5x amplification)
  2. V-in (3.3V? 5V? 2.7V? grnd?)
  3. V+in (sensor in)
  4. V- (grnd)
  5. Ref (grnd)
  6. Vout (output)
  7. V+ (+5V)
  8. Rg (no resistor for 5x amplification)

With the above setup and V-in connected to 2.2V from a DC power supply and V+ connected to the Arduino 5V supply, the output range becomes:

2.65-2.98

This doesn't make sense according to the calculations:

(2.27-2.1)*5 = 0.85
(2.77-2.1)*5 = 3.35

Generally the amp doesn't seem to do what I want it to. What am I doing wrong?

EDIT1: I have access to a DC Power Supply (w/ negative voltage). Perhaps this will help me?

EDIT2: I'm going to try once more with a Wheatstone bridge and the dual power supply. If that doesn't work, I'm ordering amplifiers that are less fussy in single-supply mode (along with some 1% resistors) and trying again on Monday.

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I don't really see a reason for this not to work the way you expect with the connections you have listed here. Might I suggest some debugging steps?

1) See what happens to the output range when you put a large-ish Rg ( say 27k ohms) in the circuit. Does the output swing change at all? 2) Have you checked the inputs to make sure they are as you expect with the amplifier connected? If so, how about using a larger resistance POT to simulate your resistor divider just to remove the sensor from the picture until you get the circuit working. 3) Perhaps the simplest explaination might be that the INA126 itself is bad. Do you have a spare you can drop in to test?

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  • \$\begingroup\$ I just swapped the amp out a few times and rebuilt the circuit and now the output is pinned at 4.4V. Not sure what's going on, going to keep trying! \$\endgroup\$ – terrace Apr 17 '10 at 1:41
  • \$\begingroup\$ Sounds like you are saturating your amp. If you only have 5v on the V+ pin then (according to the datasheet) your output voltage can only be (V+)–0.75, meaning 5-.75=4.25V with a 25kohm load. Since you are seeing your output pinned at 4.4V (above the 4.25 mentioned above) you have most likely saturated the opamps. \$\endgroup\$ – Kellenjb Apr 18 '10 at 13:56
  • \$\begingroup\$ With a single supply, I would suggest looking at the INA210 instead, and put in on the low side. It has a CMRR that extends below 0. I've used it before, it is an excellent little amp. You could probably do without any external components besides your shunt resistor. And a bypass cap... \$\endgroup\$ – morten May 19 '11 at 18:58
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what i would do is build a ladder such that you get the voltage differential you want across a resistor.

Like this:

alt text http://www.kegs-tapped.com/images/opamp.jpg

R1 would be the variable resistance of your sensor.

R2 + R3 would be chosen for the desired current level

R2 is chosen to produce the desired voltage drop based on the current range you just defined.

Vref could be tied to a voltage reference supplied by the ADC (buffered with an op amp) or tied to ground. Note that Vref sets DC offset of the amps output so its to be used to make sure the output ends up in the range the ADC can handle.

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  • \$\begingroup\$ Thanks Mark. Are you suggesting that I keep using the INA126, or substitute it for a general purpose op-amp? \$\endgroup\$ – terrace Apr 17 '10 at 1:54
  • \$\begingroup\$ I don't see any reason not to use that IC. You may also look into a wheatstone bridge as opposed to what i posted, it may work better in your application. \$\endgroup\$ – Mark Apr 17 '10 at 2:09
  • \$\begingroup\$ I'll second the wheatstone bridge. It will give you a nice, easy to use differential voltage to measure, which should make the amplifier circuit easy to design. \$\endgroup\$ – W5VO♦ Apr 17 '10 at 9:26
  • \$\begingroup\$ Ok, thanks for the suggestion. I'll give it a try. Are precision resistors a must? \$\endgroup\$ – terrace Apr 17 '10 at 16:02
  • \$\begingroup\$ Ideally you would want matched resistors but as your going to have to empirically match an amount of fabric folding to a voltage value and create that relationship in the uC and being at the balance point often is unlikely i don't think it would be critical. I always use 1% resistors anyway except on very slow values where they are expensive. \$\endgroup\$ – Mark Apr 17 '10 at 20:30
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I don't believe you can operate that instrument amp (IA) in single supply mode. Take a look at page 8 of the datasheet (Input Common-Mode Range). It states that the low rail of A2 (the bottomr amp in the diff amp) is

VO2 = 1.25 * VIN- + (VIN+ - VIN-) * 10K/RG

which for your setup would be VO2 = 1.25 * 2.1 + 0 = 2.625

2.625 is essentially the low value you get in your experiment. The voltage difference is invalid due to the voltage limiting of A2.

You need a single supply instrument amp (and preferable a rail-to-rail IA). A couple of my favorites are the AD623 and AD8235. I am sure TI (BB division), Linear Tech or ADI will have a number of other options.

[EDIT] I may have spoke too soon on the single supply operation. Take a look at page 9. You may be able to bias the reference pin up to offset the A2 output voltage. Unfortunately the example that is shown looks a bit different than an Arduino ADC.[/EDIT]

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    \$\begingroup\$ Does "single-supply mode" mean that it can be operated with a single power supply? What about this line on page 8: "The INA126/2126 can be operated from a single power supply with careful attention to input common-mode range, output voltage swing of both op amps and the voltage applied to the Ref terminal."? \$\endgroup\$ – terrace Apr 17 '10 at 2:06
  • \$\begingroup\$ Replacing V+ and V- with 5V from a separate power supply changed things up a bit. Now my output range is 0.26-0.92! Changing the gain resistor has no effect on the output range. \$\endgroup\$ – terrace Apr 17 '10 at 2:11
  • \$\begingroup\$ "Single Supply Mode" means one supply voltage. For most applications V+ would be connected to a positive voltage and V- to GND. After I wrote the answer I saw that note (and the circuit on page 9). I didn't go through the spec in detail but applying a reference voltage to Vref could work. You may want to see if there is an app notes that gives the equations otherwise you could derive the equations yourself. All of the component values are shown in the internal schematic. \$\endgroup\$ – jluciani Apr 17 '10 at 2:18
  • \$\begingroup\$ you were correct on being careful with the common mode range, you'll notice in the application note on the single supply setup that they biased the input terminals to 2.5V and a low differential range presumably keeping it under the 2.65V saturation point. If you use a wheatstone bridge you do need to apply a voltage to Vref as the difference could now be negative depending on what you choose the balance point to be in the bridge. \$\endgroup\$ – Mark Apr 17 '10 at 2:31
  • \$\begingroup\$ This is becoming more complicated than I was expecting. What do you think would be the simplest solution? \$\endgroup\$ – terrace Apr 17 '10 at 2:42
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I have had no luck with the INA126P.

They never work. Especially on single supply.

If you make the circuit exactly as in the datasheet, 5V power supply 1.2V offset. With a resistor bridge 2.5V on each input.

The inputs will clamp the voltage to 0.9V. Find a single supply instrumentation amplifier that works.

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    \$\begingroup\$ I have trouble believing that a TI part would fundamentally be broken. \$\endgroup\$ – Brian Carlton Sep 21 '12 at 16:31

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