# Is there an affordable way to balance a bridge without using expensive resistors?

I have a full and balanced bridge circuit with four through-hole $$\ 350 \,\Omega \$$ resistors and an instrumentation amplifier AD8293G80 that has a gain of 80. Both the bridge and the amplifier are powered by an $$\ 2.8 \,V \$$ LDO with the amplifier referenced at $$\ 1.4\,V \$$ by a voltage divider.

(1% resistors) I was expecting $$\ 1.4\,V \pm \text{some milli-Volts} \$$ at the amplifier output due to resistor mismatches and tolerances even the bridge was "balanced", but I got values somewhere in between $$\ 2.51V-2.79V \$$. The amplifier output also drifted overtime until it hit the limit of the $$\ 2.8V \$$ which was my supplied voltage. I then realized that the my resistors were only 1% in accuracy. In the worst case, I could have an input voltage of $$\ 28 \,mV \$$ amplified to $$\ 28 \,mV * 80 = 2.24 \,V \$$ due to mismatch. Playing out my circuit on a cheap breadboard also didn't help...

(Trimmer) I then tried replacing one of the bridge resistor with a Bourns trimmer ($$\ 500 \,\Omega \$$), thinking that I could "tune" my way out and balance my bridge as close to "zero" as I wanted it to be. It sort of worked but only for some seconds, until the output drifted again. Also whenever I pulled the trimmer out of the breadboard or applied any pressure to its leads, the trimmer resistance could change up to a few Ohms!

(0.01% resistors) I was shocked by the price tag (~15-25 USD each!). One of these "precision" resistor costs just as much as a bathroom weigh scale ... and I have to buy four of these?

If those 20 dollar weigh scales manage to balance a bridge circuit, perhaps there are alternative ways to complete the bridge rather than using expensive resistors that cost more than the scale itself. Does anyone know of a more affordable way to balance a bridge?

simulate this circuit – Schematic created using CircuitLab

• Agree about the need for the schematic. But a good guess would be that you're missing some feedback around your op amp so that a small imbalance in the output of the bridge just grows. Mar 4, 2022 at 17:20
• Buy a whole batch of regular resistors, then "bin" them according to value. Of course this is only as good as the resistance measurement. As long as the chosen set match, it doesn't matter if they're all 351Ω or 349Ω. Try this on a PCB next; should work much more reliably. Mar 4, 2022 at 17:33
• I suggest soldering the bridge itself on a piece of perf board and bringing only the 4 wires to the breadboard. Typical resistance of such a breadboard and wiring can be a sizeable fraction of an ohm, which is about the typical tolerance of a 1% resistor (usually within ~0.3% of nominal). And you do not want to replace the whole resistor with a trimmer. Maybe you use a 345 ohm resistor and add a 10 ohm trimmer. And, of course, pay attention to Andy's answer below which is on-point, as usual. Mar 4, 2022 at 17:47
• "If those 20 dollar weigh scales manage to balance a bridge circuit..." Do they? Or they just assume that whatever offset that exists when turning on to be the "zero" at the scale? Mar 4, 2022 at 18:30
• Buy a bussed resistor chip. They have rather low accuracy but excellent matching, even thermal drift will be matched because they share a package. Mar 4, 2022 at 19:48

This AD8293G80 InAmp is unsuitable for applications where the input voltage is typically greater than 1.1 volts on a 2.8 volt supply. You are generating about 1.4 volts on each half of the bridge and this is beyond the device's capabilities: -

Graphically there is this picture that I have highlighted in red and yellow: -

I'm not saying that there aren't other potential problems but this one sticks out and is a definite show-stopper. The good news is that the reference input can handle up to 2 volts.

• Spot on! So If I want to keep using the parts I have here, I can switch the upper two 350-Ohm resistors to perhaps 1kOhm so the common mode voltage can go much below 1V ... Is that a sensible workaround?
– KMC
Mar 5, 2022 at 19:00
• That is a sensible workaround or.... just keep what you have and insert one resistor between 2.8 volts and the junction of the two top resistors (R1 and R2). Then, that added resistor can't cause an imbalance voltage but it still can affect the amplitude @KMC - that's the normal way of doing it and, many strain gauges use a resistor there that has the opposite tempco to the gauge resistors. Mar 5, 2022 at 20:04

What you're trying to do is fairly expensive. I don't see a point of using an LDO specifically for the bridge. Bridge circuits with A/D conversion almost always are ratiometric. That is: the A/D reference voltage is a scaled version of the excitation voltage. For typical 3.3V microcontrollers, the excitation and A/D reference can be one and the same voltage. It doesn't matter what the exact value is, because the digital output of the ADC is relative to the excitation voltage, and the absolute value of said voltage does not appear in the A/D's output (!).

Also, instrumentation amplifiers are more expensive, power hungry and noisy than semi-decent op-amps. So you'd probably want a circuit where the in-amp isn't needed.

I'd use a circuit where the excitation voltage is adjusted to keep one side of the bridge at a particular fixed voltage. Then the other side of the bridge can be amplified with a single-ended op-amp. The op-amps you'd want should be low or "zero" drift, rail-to-rail output types. Their common mode will be around half supply voltage.

The bridge can be built out of whatever resistors you got around as long as they are wirewound, metal foil or thin film.

simulate this circuit – Schematic created using CircuitLab