I am trying to amplify the difference between signal lines off of a Wheatstone bridge based force transducer. I've wired up an LM358N Op-Amp as shown below in a differential amplifier. Two resistors, R1 and R2, with the same resistance connect the signal to the inverting and non inverting inputs. A resistor, R3, connects the output to the inverting input. R4 connects to ground. R3 has the same resistance as R4.

With R1=R2=100Ω and R3=R4=3.3kΩ, the gain should be about 33. The op-amp is given a single supply (Vcc) of +12V. The load cell is powered with this same voltage.

When not connected to the circuit, the load cell behaves as expected when its output is measured with a multimeter. However, when connected to the circuit (to V1 and V2 on the diagram), the multimeter measures a difference of 0.0mV between the signal wires.

I have measured the resistance between V1 and V2 when the load cell is not connected, and it is in the low MΩ without power and around 200Ω with power to the op-amp. This makes it seem like the inverting and non-inverting terminals become common when powered.

Is this possible? Would higher values for R1 and R2 get around this?


1 Answer 1


This makes it seem like the inverting and non-inverting terminals become common when powered.

The action of an op-amp is not to connect the two terminals to each other but to drive the output to bring them to the same voltage (assuming you have connected the output to the input in a negative feedback configuration, which you have via R3). The inputs themselves are truly inputs and have (almost) no direct effect on what they are connected to.

The problem here is not the op-amp — it is working as intended — but the input circuits. The pairs R1 and R3, and R2 and R4, can each be understood as voltage dividers, which require that when the inputs are low-impedance compared to the total resistance. Ground is near-zero impedance (if you're not using overly thin wires), but a load cell is very high impedance.

What you need is known as an instrumentation amplifier. This is a differential amplifier, like you have already built, but it also has high-impedance inputs, which you do not have. It can be thought of as an op-amp in the configuration you have shown, together with two additional unity-gain buffer op-amps for the two inputs (being unity gain, they do not need any divider loading the input), but practical designs are refined from that to give even better performance.

"Typical instrumentation amplifier schematic", per Wikipedia (image by Wikimedia Commons user Inductiveload, public domain):

enter image description here

Note that in this circuit, \$V_1\$ and \$V_2\$ are connected only to op-amp inputs. Therefore, they are loaded only by the input impedance of the op-amp inputs, which is very high. This is what you need for your application.

Instrumentation amplifiers are commercially available as single ICs. If you are doing this for a practical application rather than learning circuits, you should probably use such an IC to get better performance, rather than constructing one out of separate op-amps as depicted above.

Disclaimer: This answer is the extent of my general knowledge of this topic. I cannot advise you on designing or selecting an instrumentation amplifier.


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