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While analyzing the Wheatstone bridge independently, the same current is considered to pass through both resistors in each branch of the bridge (i1=i3 & i2=i4.)

I've seen the same assumption was considered when the Wheatstone bridge is connected to a differential op-amp or even a buffer.

Considering the figure below we can not say i1=i3 or i2=i4. In this case i1=i3+i5 and i2=i4+i6.

We might be able to say that currents in the input branches of the differential ap-amp circuit (i5 and i6) are negligible or small compared to i3 and i4, but not zero. It could be valid only if the Rg is large enough compare to Wheatstone bridge resistances.

enter image description here

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4 Answers 4

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There will be loading effects, an easier thing would be to use an instrumentation amplifier (or impedance buffers, which in front of a differential amplifier essentially turn it into a instrumentation amplifier)

The other option would be to use very high resistor values, which also is not a good option due to noise, the best option is to use an instrumentation amp.

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Are the assumption for Wheatstone bridge still valid while connected to a differential op-amp?

No, the assumptions are not valid except when the bridge is perfectly balanced and all the resistors in the bridge have the same value. Your circuit can also be reduced to this circuit: -

enter image description here

In other words, there is no need for both \$R_x\$ resistors and the grounding \$R_g\$ resistor because, R1 and R3 do the same job on the non-inverting input as \$R_x\$ and \$R_g\$ did and, the Thevenin equivalent of R2 and R4 replaces \$R_x\$ on the inverting input.

Now, \$i_5\$ will always be zero and, when balanced, \$i_6\$ will be zero. This is often the preferred implementation when not using InAmps.

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  • \$\begingroup\$ It is very common to use the bridge for the strain gauge or other resistance-dependent sensors. That measurements happen when the bridge goes unbalanced (when the resistance change happen). So, In that case to relate the reading voltage to the changed resistance, we need to analyze the circuit. Now the question is whether we can consider i1 equal to i4 in the unbalanced bridge or not ? \$\endgroup\$
    – Tomas
    Oct 24, 2022 at 13:22
  • \$\begingroup\$ To me it's a no-brainer; use a simulator. They are free, accurate and will save you heaps of time trying out different scenarios @Tomas (that's a new question you are asking by the way). I'll also add that bridges are not linear anyway so maybe you should consider that also; you can see it in simulation quite easily. \$\endgroup\$
    – Andy aka
    Oct 24, 2022 at 13:25
  • \$\begingroup\$ See also my answer here for the natural linearity problems associated with many bridge configurations. \$\endgroup\$
    – Andy aka
    Oct 24, 2022 at 13:29
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    \$\begingroup\$ This is often the preferred implementation when not using InAmps. The CMRR of this amplifier is only Rg/(R2||R4), and it won't work well if significant common mode noise is present, e.g. when the wires are long. When the op-amp and the bridge are adjacent, the performance can be quite satisfactory when Rg/Rx>100. When driving an ADC, it helps to split that gain over two op-amps, to provide some settling bandwidth. If two channels of ADC are available, the op-amp output and left side of the bridge can be subtracted in software, vastly improving DC CMRR. \$\endgroup\$ Oct 24, 2022 at 19:10
  • \$\begingroup\$ @Kubahasn'tforgottenMonica which particular circuit are you referring to when you talk about CMRR - the one in the question, the one in my answer or, both. \$\endgroup\$
    – Andy aka
    Oct 24, 2022 at 19:11
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At balance, both the op-amp and the bridge, i5 = i6 since the rhs of both Rx are connected to the same voltage wrt ground.

If the bridge is balanced (R1/R3 = R2/R4), but R1 \$\ne\$ R2, then that will affect the voltages at A and B differently.

If the bridge is unbalanced, there is also a differential load of 2Rx.

tl;dr yes, there are loading effects

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"Differential amplifiers" built out of discrete resistors and op-amps are so impractical that they are almost never made. They find uses in low-precision circuits where the accuracy of the measurement can be much lower than what the bridge inherently provides. If you make them out of reasonably priced components, they are much less accurate than even an average instrumentation amplifier.

Instead, use much cheaper op-amps to drive the bridge such that differential amplification becomes unnecessary.

Below, U1 drives the left side of the bridge to the mid-point of the 5V ADC input range. U2 provides 500x gain. As shown, the full scale outputs are 0V and 5V at +1% and -1% ΔR/R, respectively.

The ADC could also have single-ended inputs. The differential measurement would then be done in software by computing (CH1-CH2). Alternatively, sample with a single-ended ADC with I1 turned on, and then turned off, and subtract the two measurements. That will also remove 1/f noise, DC offset and offset drift from U1 and U2 :). Measurements with <0.1% error can be then made even with relatively low-spec op-amps - the driving spec is their open loop gain.

The op-amps are configured so that their CMRR has very little impact on performance, since the common mode voltage is constant in U1 and changes very little (a few mV) in U2.

schematic

simulate this circuit – Schematic created using CircuitLab

The above circuit is a block schematic really, since a practical implementation would need appropriate frequency compensation if the wires to the bridge are long, ESD protection, etc.

The circuit can be made ratiometric. When made so, it needs a total of 4 op-amps and requires no precision voltage nor current references. The only precision quantities then are the ratios of resistances. The supply voltage can be arbitrary (within the ratings of components used, of course). The ADC is inherently ratiometric. The +3.3V or +5V analog supply and reference can be provided with just about any LDO in this case. The absolute supply voltage then plays no role.

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