Wheatstone Bridge summing theory - Page 1
Forum section: electronics / projects

author: rvalente
author_rank: Frequent Contributor
timestamp: October 09, 2025, 12:07:14 pm
content: Hello mates,its fairly common to use "load cell junction boxes" in the field. Can anyone please point me to some Wheatstone bridge sum and subtraction theory?Greetings!

author: BILLPOD
author_rank: Frequent Contributor
timestamp: October 09, 2025, 01:43:21 pm
content: Good Morning Rvalente, I copied and pasted the last sentence of your inquiry into the Google search bar and found page after page of relative information, so try that.

author: rvalente
author_rank: Frequent Contributor
timestamp: October 09, 2025, 02:37:59 pm
content: Quote from: BILLPOD on October 09, 2025, 01:43:21 pmHmmmm maybe my google is giving different results than yours, there is plenty info about single Wheatstonebridges, but not much about their association... Would you mind share a link?

author: metrologist
author_rank: Super Contributor
timestamp: October 09, 2025, 03:36:14 pm
content: Like this?

author: rvalente
author_rank: Frequent Contributor
timestamp: October 09, 2025, 03:43:00 pm
content: Quote from: metrologist on October 09, 2025, 03:36:14 pmNo Mate, not that... Its practice to use these "junction box" to sum wheatstone bridges.Example: a truck scale has 4 load cells but only one amplifier, the bridges are connected in parallel and the value of all 4 bridges is summed. That's what I trying to understand, how wheatstone bridges can sum and subtract.IDK, maybe I'm not expressing myself clearly

author: donlisms
author_rank: Frequent Contributor
timestamp: October 09, 2025, 07:43:34 pm
content: First, there are many bridge circuits out there called "Wheatstone" that are, in fact, not. Just 'cause it's a bridge doesn't mean it's a Wheatstone.A Wheatstone bridge can be analyzed different ways, but I think the easiest is to see it as two voltage dividers dividing the same source. If the dividers are both using the same ratio, the divided voltage will be the same on both sides. Thus the difference between those two voltages is zero. If the ratios are not the same, the voltages are different, and detector sees non-zero.That makes me wonder if the load cells use some other form of bridge, i.e., not just four passive arms making up two dividers.There is a lot of room for variation. We could replace one of the four resistors with a set in series and/or parallel, maybe with some switching to provide different resistances. In the non-Wheatstone cases, when we don't just have four passive resistance arms, we could use two resistors as a divider in the left side, and two voltage supplies in the right side, and obtain a balance that way. Or using other techniques to produce a voltage on one side or the other.I'm not sure what the circuit you're talking about is, but there's nothing particularly unusual about putting four elements in parallel -- it's just basic electricity. With parallel resistors, the voltage across each branch is the same (because... parallel), so the current through each branch is based on the resistance of that branch. What comes out the bottom is the sum of those currents.If the branches are resistive load cells, since the resistance changes with load, the current changes with load, and what comes out of the whole combination is the sum of the currents at the moment.It's not the bridge that does the summing; it's just an interested observer.

author: Stray Electron
author_rank: Super Contributor
timestamp: October 10, 2025, 01:28:53 am
content: Quote from: rvalente on October 09, 2025, 03:43:00 pmI'd love to see a sketch of of the circuit that you're talking about. I think I don't understand what you're describing but it sounds like they're just putting the four cells in series and so the final value is just the sum of the value of each of the four cells. Perhaps they're then using that as one leg of the bridge circuit. Connecting four cells in parallel would give you an entirely different result (the reciprocal of the sum of the reciprocal of each of the four values).

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 10, 2025, 03:15:22 am
content: Quote from: rvalente on October 09, 2025, 03:43:00 pmIf you have 4 bridges with equal loadings, in simple parallel there is no current flowing between bridges and the output is the same as a single bridge or 25% of the truck weight.Let's say it is 10mV/1000kg per bridge.If there is unbalance, assuming the bridges have equal source impedances, the summing effect gives a gain(loss) of 1:4.If we consider an extreme case of all the truck weight in one corner, that sensor will try to drive 40mV, but loading attenuates by 1/4 so 10mV appears on the terminals.That's exactly the same as the equal load case, so it does not matter how the weight splits over the 4 wheels.

author: Terry Bites
author_rank: Super Contributor
timestamp: October 12, 2025, 04:48:50 pm
content: The ouput is the average of the balance voltages rather than their sums.You can prove this by making up some resistive dividers that have the same total resistance but different voltages.Put them in parallel and they instanly give you the avarage. Ohms law will get you there.

author: David Hess
author_rank: Super Contributor
timestamp: October 12, 2025, 05:55:19 pm
content: The summing box has a passive resistor network, with 4 adjustments to trim the gain of each individual load cell. If the designer knows his stuff, then the gain adjustments will be non-interactive.We made them long ago, but I never saw the schematic. My guess is that the differential outputs are paralleled through series resistors to preserve a 350 ohm output impedance, and the gain trims are on the excitation. Internally the load cells already have a gain trim on the excitation for modulus compensation and gain if used.