A shielded Wheatstone type force transducer FT is mounted on a metallic structure and through its cable is connected to an industrial amplifier module Amp(I gave the names in the below illustration). The end of the cable has several wires which is connected to the amplifier. One of the wire is shield SW which is connected to the transducer shield K. The amplifier is powered by an SMPS power supply whose DC negative terminal is galvanically isolated from mains earth.

The outpuut of the amplifier vi a BNC coaxial cable goes to a DAQ board which is a single ended earth grounded board. The shiled wire was floating i.e was not connected to anywhere.

I tried to illustrate this below:

enter image description here

If the force transducer is completely isolated from the metallic structure the output is fine. On the other hand, if it touches to the metallic structure there is 50Hz noise appears on that particular DAQ channel. Obviously some noise is coupling through the metallic structure either magnetically or overcapacity.

But by a wire I wired/shorted a point on the shield K directly to the AIGND and even if the transducer touches to the metallic structure the noise disappeared. If the metallic structure was earthed I would expect some ground loops but the 50 Hz disappeared.

So for some reason tying the shield directly to the AIGND solved the 50Hz pickup problem even the transducer is mounted on the metallic structure.

I tried to simulate this scenario but failed. How can we model what happens with a basic circuit model? I mean I want to simulate a representative circuit modelling when the shield is wired to the AIGND the noise disappears and when the shield is floating the noise appears.


1 Answer 1


You have electric fields causing displacement currents, as moving Positive charges in one location cause moving Negative charges elsewhere.

Nested pieces of metal have nested, or series capacitors, and charge is transferred. Your circuit is inside a shield but visible to the shield.

The transducer circuit is a fine target for some of the electric field, thus charges are forced to move in the transducer, until an equilibrium is reached when the charges stop moving. But you've added a *PATH** to the outside world ---- the transducer's output signal and the return wiring. These displacement currents now have YET ANOTHER path back to that mechanical structure.

By wiring the transducer shield to Ground, the shield becomes much more quiet (not totally quiet, as V = I * Z will predict), and thus very little charge is transferred into your circuit.

So you have something like this


simulate this circuit – Schematic created using CircuitLab

Notice at high frequencies that wire (inductor) to ground will not be close to a short (as it is at 50Hz); the voltage division action (against the 1 ohm Rsource) becomes less useful at high frequencies; in such cases you need optical isolation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.