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I'm using an instrumentation amp for a bridge circuit for strain gauges. However, the output range is always positive (the lowest value I can get is zero).

The amp I use is AD620. For power supply I used two 9V batteries to get the +-9V.

Is there something wrong with my circuit?

edit: I want to build a quarter-wheatstone-bridge circuit with a strain gauge as resistor. It should look like this(R1 is variable resistor): wheatstone bridge

To balance the bridge(R1RG = R2R3), I used a potentiometer instead of R1. I don't have resistors with the same value as the strain gauge available so a potentiometer seems to be the best choice.

And this is the updated schematic with all parameters for my circuit, the resistance of the strain gauge is 350 ohm. schematic P.S. sorry for the bad schematic and wiring, it's my first time doing a circuit-related project like this. I would really appreciate any suggestions from you.Thanks a lot!

Antonio51
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cola chicken
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  • I don't see any connection to -9V anywhere in that schematic. Can you point it out to me? – Simon Fitch Jun 27 '22 at 17:33
  • Sorry, I connected the power supplies with solder pads (those x marks) leading to V+ and V- which you can see on top and bottom of the image. – cola chicken Jun 27 '22 at 17:36
  • Please describe your strain gauge - is it a full bridge by itself, or is it a quarter-bridge (individual gauge) to which you're adding R2, R3 & R4 in order to create a full bridge? – brhans Jun 27 '22 at 18:30
  • It is a quarter bridge, pls check the edit. Thank you! – cola chicken Jun 27 '22 at 19:39
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    Note that a potentiometer is not the "ideal" replacement for a resistor (unstable versus temperature. – Antonio51 Jun 27 '22 at 20:00
  • Yes, I also found it unstable during experiments, but this is just a prototype. Resistors with the same value as the strain gauge will be used in the final version. – cola chicken Jun 27 '22 at 20:02
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    Are you aware that these resistors will need to have tolerances as good as or better than that of your strain gauge in order for them to not overwhelm the strain gauge signal with their noise & inaccuracy? – brhans Jun 27 '22 at 21:56
  • yes, I will take care of that while choosing components. Thanks! – cola chicken Jun 28 '22 at 07:24

5 Answers5

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It seems that you have bad wiring.
Don't understand your wiring about R2, R3, R4 ?

See this example for reference.

enter image description here

Here is an example where Vref changes from -4 to +4 V.

enter image description here

Here is an example when Vcm changes from -6 to +6 V.

enter image description here

Antonio51
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  • Thanks a lot for the simulation! Those three resistors are part of the Wheatstone-bridge circuit, the to be measured resisitor should be my strain gage. – cola chicken Jun 27 '22 at 18:24
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    The wiring for R2, R3, R4 is just a bridge with one leg open where the strain gauge attaches. It's just drawn poorly. I hate when people post these 'board layout software' schematics, no component values, poor layout, terrible to try to debug. – GodJihyo Jun 27 '22 at 18:26
  • And I noticed that I didn't use the two resistors for input protection. I should add that into the circuit later. – cola chicken Jun 27 '22 at 18:28
  • Ok. Redraw, as usual, the Wheatstone bridge.. and then, add the needed wiring. Remember you need simply add "labels". It is more "clear" ... – Antonio51 Jun 27 '22 at 18:29
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    The two 10 Meg resistors are only for "bias" inputs. When connected to a 4-resistor bridge, they are obviously "useless". – Antonio51 Jun 27 '22 at 18:31
  • Thanks for the comments. I updated the scematics, hopefully it's easier to read now... sry for the inconveniences, I just started learning circuit design, your suggestions are very appreciated. – cola chicken Jun 27 '22 at 19:44
  • No worries. We are here to pass on our "know-how" ... – Antonio51 Jun 27 '22 at 20:11
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The schematic has errors with R2,R3,R4 that should be omitted. You have incorrectly used a differential INA with the Vin- grounded at the input. do not do that. If your input signal or the +/- 9V supplies are with respect to each other or have a shared ground then no additional grounding is required for a proper differential signal. Wires should be twisted cable or similar STP with a grounded shield at source only.

enter image description here

Tony Stewart EE75
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I did a quick and dirty simulation of your basic circuit using an LT1168 for the inamp and stepping the strain gauge resistance from 980 to 1020 \$\Omega\$ with the opposite leg 1000 \$\Omega\$. The output goes both positive and negative as expected. So if your strain gauge is a simple resistance type and you have your bridge set up properly it looks like it should work.

One thing you could check is that the resistors are the correct values. R3 should be the same as the static value of the strain gauge, R2 and R4 should be equal in value.

You might want to make R2 and R4 fixed precision resistors and R3 the variable one, adjustable to match the strain gauge resistance. That should be a better configuration and might be the cause of your problem.

enter image description here

GodJihyo
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  • I used R2 in your circuit as a variable resistor, I think that would also work? And I updated the question, pls have a look. Thanks! – cola chicken Jun 27 '22 at 19:41
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    It looks like you've got the resistors in the wrong places in the new schematic. As it is the equal values are diagonally opposite each other, they need to be adjacent. The one that matches the gauge (in this case the pot, R1) needs to be either above it, or next to it in the bridge. You would need to swap the position of R1 with either R2 or R3 (numbers are for your schematic, not the bridge diagram). – GodJihyo Jun 27 '22 at 20:06
  • Thanks for the hint. I just come up with an additional question: since I just need the bridge to be balanced,can't this also be achieved by putting the potentiometer opposite to the strain gage(I only need to fulfill R1RG = R2R3)? – cola chicken Jun 27 '22 at 20:26
  • @colachicken A Wheatstone bridge is just two voltage dividers, as long as the ratios of both dividers are the same it will work. This is easier to visualize the way I drew my schematic than the way it's typically drawn. I made each one a 10:1 ratio with 100K on top, 1K on the bottom. I could have also made them 1:1 with the left side both being 100K and the right side both 1K. – GodJihyo Jun 28 '22 at 01:14
  • exactly. So theoretically it doesn't matter which resistor we replace with potentiometer? – cola chicken Jun 28 '22 at 07:30
  • @colachicken it does affect the values the resistors have to be. If you have 100k/(1k ohm strain gauge) on the left, you can have 100k/1k ohms (adjustable) on the right, or you can have 10M (adjustable)/100k on the right, or 100k (adjustable)/1k, since on both sides the top is 100x the bottom. – user253751 Jun 28 '22 at 08:30
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My guess is that you have the wrong pin mapping from the pot schematic symbol to the footprint.

Since you have left the end of the element open (not best practice- best to tie it to the wiper) that could result in the wiper being left open and the inverting input of the inamp never seeing more than the minimum voltage.

Spehro Pefhany
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  • actually the end is connected to the strain gauge. And I don't think the mapping is wrong, since I experimented with just the wheatstone bridge part and it worked perfectly, I got both positve and negative voltage differences. But when I connect my INA to it, even when the difference is negateive, I can't get negative output but only zero. – cola chicken Jun 28 '22 at 07:34
  • The way to diagnose it to test each input voltage at the inamp relative to ground, and the power supply voltages at the the inamp and compare to the diamond-shaped permissible inputs and outputs from the inamp datasheet. That should tell you whether there is a problem with the inamp or with the remainder of the circuit. – Spehro Pefhany Jun 28 '22 at 09:11
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I was measuring all the voltages with my Arduino. I just remeasured everything with a multimeter and it seems the INA is working properly. Then after I plug the output into the Arduino, I had the same issue again. I looked up and found that Arduino's analog pins can't measure negative voltages.

Seems the problem is solved. Thank you very much for your helps!

cola chicken
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  • Oh, that is so disappointing. I'm glad you figured it out, but it's so frustrating for others to find out that there's something really important you're not telling us. – Simon Fitch Jun 28 '22 at 13:20
  • Did you see in my simulation that output can be offset with VREF ... so Arduino can read analog input? – Antonio51 Jun 28 '22 at 16:10
  • yes, I probably will add a reference voltage to offset the output. Thanks! – cola chicken Jun 28 '22 at 21:37