I'm trying to find a way to convert a bipolar signal from a load cell (Futek) to an input for a differential ADC on a TIVA microcontroller(TI TM4C123GH6PM). The load cell's amplifier outputs a bipolar (-2.67 V to 2.67 V) for a 500 lb range in tension and compression.

Current setup

Circuit Schematic

This setup uses decoupling capacitors (0.47 μF) to eliminate high frequency noise within the system. The noise is extremely low and I get very accurate readings, but the problem is that the GNDs of the microcontroller and float without reference to each other. I tried to do it this way so that microcontroller would never see a negative voltage, but treat the amplifier GND as a data input.

The problem is that now because I've dereferenced the GNDs of the two systems, there is no way to make sure that the two GNDs don't float away. I recognize that there needs to be another way to use a bipolar signal for a differential input, can someone help me?

  • \$\begingroup\$ What are the supply voltages to the sensor? +-15 volts? \$\endgroup\$ – analogsystemsrf Apr 15 '19 at 17:55
  • \$\begingroup\$ what is the power supply voltage available from the TIVA? \$\endgroup\$ – scorpdaddy Apr 15 '19 at 19:15
  • \$\begingroup\$ What is this "amplifier" you are using? Your wiring diagram shows "data" and "ground" connected to your TIVA. Data would normally be a digital signal, so no problems with "bipolar" signals, and no negative voltages to deal with. \$\endgroup\$ – JRE Apr 15 '19 at 20:16

Here is an approach that condenses the other answers into a practical circuit. It is just a single-ended to differential converter.

It first divides the input voltage to put it in range of the amplifier, and then amplifies it again to provide the input to the ADC. Note that this is basically the same as a gain of 1 configuration (that can be obtained with an LTC1992-1), but using external resistors allows for negative voltages on the input side of R4 & R5.

Single-ended to differential circuit

For best accuracy and low-noise make sure to connect the grounds as indicated (the 0.001Ω resistor is just to join the nodes, it's not a real component). Preferably use a twisted pair for the signal connection.

This is the output of the circuit:

Output waveforms

Additional filtering (a second pole) can be added by placing capacitors in parallel with R6 & R7. It should be possible to combine R1 & R4 as well as R2 & R5 for a somewhat simpler circuit in this case.


Maximum recommended source resistance (parameter \$R_S\$ in the datasheet, page 1389) is 500 ohms, so you should level-shift and buffer the signal.

Something like this:


simulate this circuit – Schematic created using CircuitLab

R1/R2 can be 0.05% tolerance parts or a matched pair network. C1 can be an NP0 type.

You could also use a TLE2426 to split the 3.3V rail and connect the amplifier ground to that point (provided the supply for the amplifier is completely isolated) but you should keep the R3/C1 for protection. You could then use differential mode between the mode splitter and the amplfier output, which would be less sensitive to drift of resistor values.. depends if you care about the last fraction of << 1%.

Thirdly, if you want a way to convert a single ended to differential, you can use a fully differential amplifier (image below from linked pdf file):

enter image description here

In your case you'd want to connect Vocm to a voltage midway to Vdda. The devil is in the details here, you may need a higher supply voltage for the amplifier and to clamp the outputs so as not to damage the ADC inputs.

  • \$\begingroup\$ Thanks, this looks pretty similar to the other approach with level shifting the output into the positive range. In that case I'd use the single ended mode to maximize my ADC resolution. Do you know of a way to convert a bipolar single to a differential signal? \$\endgroup\$ – Connor Herron Apr 16 '19 at 2:24
  • \$\begingroup\$ See edit above for single to differential. You can do it with individual op-amps too, of course. \$\endgroup\$ – Spehro Pefhany Apr 16 '19 at 2:56

This only requires a couple of resistors as shown below, and an inverter for the ADC_in- input.

When amp signal is -2.67 V the ADC in sees 0.32 V in ADC_IN+ and +2.99 on ADC_IN-. Reverse on positive input. The two grounds are connected and not floating.

You may scale the 10k up/down depending on the impedances of the amp out and adc in.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thanks, I understand that this approach is using level shifting into the positive range, but in that case I would use single ended mode, otherwise I'd be losing half the resolution of my ADC. Is there another approach you can think of where the analog output shifted from a bipolar signal into a differential signal? \$\endgroup\$ – Connor Herron Apr 16 '19 at 2:22
  • \$\begingroup\$ Inverter stage provided to generate a differential signal for your ADC. \$\endgroup\$ – scorpdaddy Apr 16 '19 at 12:53

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