What is the point of a wheatstone bridge? Everywhere, like here says that it's used for measuring an unknown resistance, but I don't understand why it's so complicated.

Can't you just use a voltage divider to measure resistance? The equation for that is \$V_{out} = V_{in}R_2/(R_1+R_2)\$ , rearranged to make \$R_2 = V_{out}R_1/(V_{in}-V_{out})\$.

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    \$\begingroup\$ If everything were ideal yes, but things are not ideal. A divider can't handle many real world effects that introduce error (errors due to measurement currents, common mode noise, inexact voltage supply, inexact resistances, etc). A wheatstone bridge can cancel out these errors. \$\endgroup\$
    – DKNguyen
    Dec 1, 2022 at 4:28
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    \$\begingroup\$ This video explains the precision behind the Wheatstone bridge: (youtube.com/watch?v=-G-dySnSSG4) [Circuits in Practice: The Wheatstone Bridge, What It Does, and Why It Matters] You can skip the first 2 and a half minutes. \$\endgroup\$
    – ErikR
    Dec 1, 2022 at 4:31
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    \$\begingroup\$ There's also this link which provides a simple overview with some of the benefits listed out. \$\endgroup\$
    – jonk
    Dec 1, 2022 at 4:35
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    \$\begingroup\$ Go back in time by 150 years or so. No voltage references, no op-amps, no transistors, no vacuum tubes. Available battery sources were pretty crappy and their output voltage was NOT stable. How would YOU go about measuring and matching resistors of varying values? IOW - the Wheatstone bridge was an absolutely ground-breaking invention. BTW - still in use to this very day! \$\endgroup\$ Dec 1, 2022 at 17:56

2 Answers 2


Yes, you can use a half-bridge and measure the excitation voltage and output voltage. It's simpler and may be acceptable in many cases, especially with (sometimes very expensive) modern ADCs that effectively form the other half of the bridge.

Wheatstone bridges are from a time when it was easier to get precise resistors than precise well-regulated voltages. A poorly regulated voltage (for example, from a battery) could be used to excite the bridge and then the output voltage could be nulled out by adjusting another resistor value, perhaps observed on a galvanometer or a sensitive mirror galvo. Since you're only looking for a null, only the resistor needs to be precisely known.

In more modern use, it's useful in situations where we are after a change in resistance from a certain value. For example, say a certain Pt100 platinum RTD over the range 10°C to 30°C will change from 103.903Ω to 111.673Ω. By offsetting the lower value with a precise and stable resistor we can amplify and measure the difference and reduce the dependency on our reference and measurement circuit by a factor more than 13:1. That kind of improvement in a measurement circuit is not trivial at all.

On the precision side, you can easily buy a resistor off the shelf with a stability of +/-0.2ppm/K for the price of a couple cups of coffee. Without resorting to temperature control, the best monolithic voltage references are 4-5x worse.

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    \$\begingroup\$ Yeah, if you tried to get a bandgap voltage reference before the 1960s the answer would be "what's a bandgap?", and most people didn't have a Weston cell at home. \$\endgroup\$
    – hobbs
    Dec 1, 2022 at 16:34
  • \$\begingroup\$ Well, off-the-shelf high-stability resistors are a bit more expensive, around (10-100) €... but maybe I'm from a country where coffee is less expensive :-p \$\endgroup\$ Dec 1, 2022 at 19:51
  • \$\begingroup\$ @MassimoOrtolano I definitely miss the ~1.2 EUR café con leches in Spain and Portugal.' \$\endgroup\$ Dec 1, 2022 at 20:22

The Wheatstone bridge has many advantages at little complexity/cost, as mentioned in the others answers and comments.

To add some grey in all the black & white, let me illustrate an example where the Wheatstone bridge actually offers little advantage over a properly designed half bridge measurement:


simulate this circuit – Schematic created using CircuitLab

This is a multiplexed half-bridge measurement, where the DUT (R2) is interleaved with a single reference Rref. That way, the ratio between the measured resistances R2 and Rref directly corresponds to the exact ratio of R2 and the known Rref. Importantly, this result will be independent from any drift from V1 or even R1 or VM1. It also doesn't require good matching of Rref and R2.

These advantages are greater than those of the Wheatstone bridge, but the drawback is that the VM1 sees a much larger DC voltage compared to the compensated bridge voltmeter and it is much harder to achieve low noise here (see Spehro's answer).

Still this situation is acceptable, if the Voltmeter is very good, which isn't so hard today. Indeed it can be easier than to maintain accurate bridge matching.

We use this scheme for measurements of resistor noise index. It is limited only by the noise index of Rref, which can be made arbitrarily low by increasing its volume. In contrast, a Wheatstone bridge will diminish the source's 1/f noise only by its own matching ratio (e.g. -60 dB) but not entirely eliminate it. And to complicate matters more, the matching ratio of a Wheatstone bridge is itself subject to drift.

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    \$\begingroup\$ You've made the question more interesting now. I will need to sit down soon perform partial differential sensitivity analysis vs each parameter as well as noise. Sounds like fun and it would provide a quantitative comparison over each proposed parameter/process variable. \$\endgroup\$
    – jonk
    Dec 1, 2022 at 12:08
  • \$\begingroup\$ @jonk I am currently (translate: after all the more urgent stuff is done) writing an app note that compares this method vs. the classical Wheatstone bridge approach for the purpose of measuring resistors noise indices and possibly for other Wheatstone bridge applications. If you are willing to share your analysis, I would be highly interested in your thoughts. \$\endgroup\$
    – tobalt
    Dec 1, 2022 at 13:32
  • \$\begingroup\$ I'd be just doing 'rote-math'. It just flows out without really thinking much. Then I look backwards to see if there's anything interesting. For example, the 3 resistors in the bridge. I might ask, "What if these are matched without knowing the voltage supply value being applied but by alternating their orientations and re-selecting or laser trimming or adding wire until they do all match up? Can the results be improved without specialized equipment?" Stuff like that. It's just the usual \$\frac{\frac{\partial y}{y}}{\frac{\partial x}{x}}\$ stuff, to start, though. \$\endgroup\$
    – jonk
    Dec 1, 2022 at 19:14
  • \$\begingroup\$ I've no problem sharing. If the "all the more urgent stuff is done" happens around the same time I "go play" then that would be great! \$\endgroup\$
    – jonk
    Dec 1, 2022 at 19:15

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