The essential
When several transducers share a load, their bridges can be wired in parallel to read the total on one channel. The paralleled bridges output the average of their signals, not the sum, so two transducers give half the sensitivity of one. Their shunt output is also halved: setting the amplifier gain on the shunt calibration value of a single transducer compensates exactly. The condition is that the transducers have the same sensitivity. If they do not, an averaged calibration value gives an error that depends on how the load is shared.
When are transducers paralleled?
When more than one transducer is needed across an interface to measure the load through it. The transducers are in parallel mechanically, they share the force, and electrically, their Wheatstone bridges are wired together. The note assumes fully active bridges and constant voltage excitation.
What happens to the shunt calibration?
It is about halved with two transducers, and about 1/i of the single value with i transducers. With the bridges in parallel, the shunt resistor acts on arms of R/2 instead of R. Because the shunt resistor is much larger than R, the shunt output falls in the same ratio.
What happens to the sensitivity?
It drops by 50 % with two transducers. The paralleled output is proportional to (ΔR1 + ΔR2)/2R, the average of the two bridges, where two separate channels would add to the total. Doubling the amplifier gain compensates. Since the shunt output is halved too, set the gain to the shunt calibration value you would use for a single transducer.
What if the load is not shared equally?
If the transducers have the same sensitivity, the parallel system still gives the correct sum, however the load is divided.
- Necessary and sufficient condition: same sensitivity, that is identical shunt calibration equivalents.
- The shunt calibration equivalent used for the system is then that of one transducer.
- When calibrating in mV/V instead: average the mV/V values, then divide by the number of transducers.
What error with unequal sensitivities?
An error that depends on the difference in sensitivity and on how the load is divided. When equal sensitivities are not available, choose transducers with nearly equal shunt calibration equivalents and use their average. The error is zero when the loads are in the same ratio as the shunt calibration equivalents.
- 100 kΩ shunt calibration equivalents of 680 and 720 lb, load shared equally: 0.08 % of error.
- Same pair, one transducer taking twice the load of the other: about +1.0 % or −0.9 %.
- Equivalents of 600 and 300 lb, load shared equally: 12.5 % of error; 25 % with the 2:1 load split of the example.
The technical note
The original note carries the circuit diagrams, the derivation and the error formula for two and three transducers.
To remember
- Parallel only transducers of the same sensitivity.
- Set the gain on the shunt value of a single transducer.
- Nearly equal sensitivities: average them, and expect an error of the order of 1 %.