Shielding a battery operated device from electrical fields.

Sensacell

Joined Jun 19, 2012
3,785
Unfortunately, the concept you are working towards is fatally flawed, no amount of shielding will make this idea work.

The "battery negative" and the whole circuit represents a floating node, the circuit cannot detect any signal unless it has a reference ground at a potential different from the input. The problem is that you have no other such reference point- unless you connect a wire from the ground of the circuit to an external ground, which I am assuming violates the whole idea of what you are trying to accomplish, this wire would make it work perfectly.

The bird on a wire feels nothing (detects no signal) as there is no potential difference to cause a current flow. Your sensor is the bird- input and ground all at the same potential- no signal.
 

BR-549

Joined Sep 22, 2013
4,931
What if we take a small signal diode and install between cage and circuit board ground. And perhaps one in sensor lead. With appropriate polarities.......could we get enough un-balance............to get an output?
Does this make sense to anyone?
Or have I totally lost it?
 

Thread Starter

dannybeckett

Joined Dec 9, 2009
185
OK I take your point. But how do the floating nodes 'connect' to the 330kHz through the faraday cage? If I put the bird inside a faraday cage, and put the cage next to the pylon - the bird should not be charged to the pylon voltage. This is what I don't understand.
 
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ronv

Joined Nov 12, 2008
3,770
If you remove the wires going out to the body and back from the sensor, do you still have a problem?
Is this built on a printed circuit board?
Do you have a picture of the noise?
Does the transmitter and receiver share a common ground or 2 different supplies.
 

BR-549

Joined Sep 22, 2013
4,931
Hey Dan.....your right, you need to get this before going farther. The amp has to reference its' input signal to something. Most of the time, that reference is ground. Now if the ground is moving with the signal........there will be no difference.
Does that shed any light?
 

Thread Starter

dannybeckett

Joined Dec 9, 2009
185
Yes BR it does make total sense, and I've known from the beginning that this is the problem. What I don't understand is why a Faraday cage doesn't fix it. Any external field applied to the outside of a Faraday cage causes the outer skin of the cage to polarise. The inner skin of the cage does not get affected.

http://en.m.wikipedia.org/wiki/Faraday_cage#Operation

So how is the frequency I'm trying to block getting into my system?
 

BR-549

Joined Sep 22, 2013
4,931
Because you do not have a Faraday cage or shield. You have an antenna. You have the outside of your shield connected to the ground of the circuit. Is that correct Dan? If not, what is separating the outside surface of cage from inside surface of cage?
Are you physically connecting circuit ground to the inside surface thinking it will be at ground potential there?

A Faraday shield does not block or reflect external fields. It absorbs them and puts them to real ground. It states this in the first couple of paragraphs of your link. If you don't have a ground, for the cage to work...it has to be completely enclosed.......no ins or outs.....to have this effect. Because to get that zero charge inside.....the charge outside has to be distributed very evenly, all over the outside of the cage.

Even if you got a zero voltage inside the cage, you could not use it as a reference ground. A real ground needs to be able to exchange large amounts of charge(current) with very, very low potential. The zero charge inside the cage is because of balanced charge..........not the ability to exchange charge.

So...When the external field hits the ungrounded shield, the voltage induced.....is put on ground plane of circuit board instead of true ground.

Does this make any sense yet, Dan?
 
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