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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!

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  • \$\begingroup\$ I don't see any connection to -9V anywhere in that schematic. Can you point it out to me? \$\endgroup\$
    – Simon Fitch
    Commented Jun 27, 2022 at 17:33
  • \$\begingroup\$ 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. \$\endgroup\$
    – cola chicken
    Commented Jun 27, 2022 at 17:36
  • \$\begingroup\$ 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? \$\endgroup\$
    – brhans
    Commented Jun 27, 2022 at 18:30
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    \$\begingroup\$ Note that a potentiometer is not the "ideal" replacement for a resistor (unstable versus temperature. \$\endgroup\$
    – Antonio51
    Commented Jun 27, 2022 at 20:00
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    \$\begingroup\$ 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? \$\endgroup\$
    – brhans
    Commented Jun 27, 2022 at 21:56

5 Answers 5

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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

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  • \$\begingroup\$ 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. \$\endgroup\$
    – cola chicken
    Commented Jun 27, 2022 at 18:24
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    \$\begingroup\$ 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. \$\endgroup\$
    – GodJihyo
    Commented Jun 27, 2022 at 18:26
  • \$\begingroup\$ And I noticed that I didn't use the two resistors for input protection. I should add that into the circuit later. \$\endgroup\$
    – cola chicken
    Commented Jun 27, 2022 at 18:28
  • \$\begingroup\$ Ok. Redraw, as usual, the Wheatstone bridge.. and then, add the needed wiring. Remember you need simply add "labels". It is more "clear" ... \$\endgroup\$
    – Antonio51
    Commented Jun 27, 2022 at 18:29
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    \$\begingroup\$ The two 10 Meg resistors are only for "bias" inputs. When connected to a 4-resistor bridge, they are obviously "useless". \$\endgroup\$
    – Antonio51
    Commented Jun 27, 2022 at 18:31
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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

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  • \$\begingroup\$ 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! \$\endgroup\$
    – cola chicken
    Commented Jun 27, 2022 at 19:41
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    \$\begingroup\$ 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). \$\endgroup\$
    – GodJihyo
    Commented Jun 27, 2022 at 20:06
  • \$\begingroup\$ 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)? \$\endgroup\$
    – cola chicken
    Commented Jun 27, 2022 at 20:26
  • \$\begingroup\$ @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. \$\endgroup\$
    – GodJihyo
    Commented Jun 28, 2022 at 1:14
  • \$\begingroup\$ exactly. So theoretically it doesn't matter which resistor we replace with potentiometer? \$\endgroup\$
    – cola chicken
    Commented Jun 28, 2022 at 7:30
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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

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  • \$\begingroup\$ I updated more details of the circuit, pls have a look:). In the first picture, the input is connected to the minus pol of Vin, doesn't it mean ground if I have Vin=3.3V? \$\endgroup\$
    – cola chicken
    Commented Jun 27, 2022 at 19:49
  • \$\begingroup\$ see corrected schematic \$\endgroup\$
    – D.A.S.
    Commented Jun 27, 2022 at 23:10
  • \$\begingroup\$ what do you mean by twisted pair in this case? Does it replace my three resistors in the wheatstone bridge? \$\endgroup\$
    – cola chicken
    Commented Jun 28, 2022 at 7:28
  • \$\begingroup\$ A bridge needs 2 pairs Power and Signal, not 3 wires. If on same board disregard but your schematic is terrible. I tried to correct it. Others also did the same \$\endgroup\$
    – D.A.S.
    Commented Jun 28, 2022 at 18:27
  • \$\begingroup\$ Vin,gnd and Vmeas are two pairs such as STP wire \$\endgroup\$
    – D.A.S.
    Commented Jun 28, 2022 at 18:29
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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.

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  • \$\begingroup\$ 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. \$\endgroup\$
    – cola chicken
    Commented Jun 28, 2022 at 7:34
  • \$\begingroup\$ 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. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 28, 2022 at 9:11
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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!

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  • \$\begingroup\$ 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. \$\endgroup\$
    – Simon Fitch
    Commented Jun 28, 2022 at 13:20
  • \$\begingroup\$ Did you see in my simulation that output can be offset with VREF ... so Arduino can read analog input? \$\endgroup\$
    – Antonio51
    Commented Jun 28, 2022 at 16:10
  • \$\begingroup\$ yes, I probably will add a reference voltage to offset the output. Thanks! \$\endgroup\$
    – cola chicken
    Commented Jun 28, 2022 at 21:37

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