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4hv.org :: Forums :: General Science and Electronics
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Electromagnetic Interference Question

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StevenCaton
Sat Jul 03 2010, 04:21AM Print
StevenCaton Registered Member #1845 Joined: Fri Dec 05 2008, 05:38AM
Location: California
Posts: 211
Assume a flat square circuit board must be placed next to a large coil carrying an RF current.
What orientation of the board with respect to the coil would minimize EMI onto the board.

I would assume that it shouldn't be placed near one of the coil's open ends, because that is where the poles of the field are.
So, that leaves the side of the coil. But, should the board's flat part face the coil's side, or should the plane that the board lies on go through the coil. (90 degrees different than what I just said)

I think it should be option 1. Is 1 the optimal position to reject EMI?

1278130697 1845 FT0 4hv Image 1

1278130697 1845 FT0 4hv Image 2
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Myke
Sat Jul 03 2010, 07:57AM
Myke Registered Member #540 Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969
I would think option 2 would have the most rejection. If they were ideal solenoids, the net magnetic flux through the board would be near 0 (or would be 0). With option 1, you have small variations in the magnetic flux while the total board should have a net flux of 0. This allows small currents to be induced on the board where the flux for a given area isn't 0.
In option 2, the entire board is aligned radially to the center of the coil. This causes the magnetic field and the board to be parallel at all times which causes no magnetic field lines to go through the board. Since the flux over the entire board is 0, there will be no induced currents.
I'm not completely sure about my answer because I'm already forgetting what I learned from physics C E&M. neutral
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klugesmith
Sat Jul 03 2010, 03:58PM
klugesmith Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
I agree with Myke: #2 is better because it puts the coil's field entirely in the plane of the board.

The external magnetic field of the finite-length solenoid looks like that of a bar magnet, such as in Link2
It has both axial and radial compenents, but practically no circumferential ones.

Case 1 and 2 are both pretty good -- the magnetic field orientation is mostly parallel to the board. (Placing board beside the coil but oriented perpendicular to coil axis would be MUCH worse).
In case 2, the field orientation is purely in the plane of your circuit board, which is best.
In case 1, the radial field component crosses the plane of the board, and induces a voltage in any circuit loop that it penetrates. The end regions of the board get it worse than the middle.

In all cases, board placement and routing should minimizing the geometric area of sensitive loops in the circuit.
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StevenCaton
Sat Jul 03 2010, 05:34PM
StevenCaton Registered Member #1845 Joined: Fri Dec 05 2008, 05:38AM
Location: California
Posts: 211
So you both think 2.
Well, I was thinking 1 because every point on the board is about the same distance to the center of the solenoid. So, every point on the board would be in a field that is of the same general intensity.

In option two, the distance from each point on the board to the solenoid varies much more as you more along the board. (or left to right in the drawing) I was thinking that a large net change in field intensity from one side of the board to the other would be bad.

Well, I still don't know. Maybe someone else knows for sure.
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Myke
Sat Jul 03 2010, 06:50PM
Myke Registered Member #540 Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969
In theory, with option 2 you can have as many loops as you want since no current will be induced.
With option 1, you would want to avoid loops near the ends of the board since those ends will have a non-zero flux since the magnetic field isn't parallel to the board. At the center of the board though, there should be a near 0 magnetic flux (not necessarily 0 because of the nonideality of the solenoid).
Here are some equations that would quantify what I'm talking about.
This is for finding the magnetic flux.
Φ=BAcosθ where θ is the angle between B and the normal of the surface

This is for finding the induced EMF (Faraday's law of induction)
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klugesmith
Sat Jul 03 2010, 07:48PM
klugesmith Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
SteveC wrote ...
I was thinking that a large net change in field intensity from one side of the board to the other would be bad.
Why? At every place, the magnetic field is varying at the RF frequency. The field induces an RF voltage in every circuit loop, in proportion to "how many field lines are encircled by the victim loop".

You can try it and see. Probe the coil's field with a test loop, which can have many turns. Test loop should be connected to oscilloscope or AC voltmeter by a twisted-pair or shielded cable. Induced voltage depends strongly on the loop's position and orientation. Even at the end of coil, or inside the coil, there will be no coupling when the coil axis is in the plane of the loop.
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Proud Mary
Sat Jul 03 2010, 10:20PM
Proud Mary Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
According to the question, it is not merely an alternating current that is flowing in the coil, but a radio frequency one.

Making the axis of the inductor perpendicular to the plane of the board will reduce the capacitive coupling of each turn to the board, which might, or might not, be of signifiance depending on the radio frequency in question. .

Then, if we consider the case of a variometer, maximum coupling is obtained when the two inductors share the same axis, and minimum coupling when they are at rightangles.

Again, depending on the frequency concerned, L, and the dimensions of the board etc, the vertical inductor may 'see' its own reflection in the ground plane beneath it, and so on.

So my view is that the question does not contain sufficient information to be meaningfully answered.
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radiotech
Sun Jul 04 2010, 07:58AM
radiotech Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546
I might suggest to try waving a degaussing coil in front of a running CRT monitor. Your first figure suggest holding it edgewise (right angles to the screen).
At any rate, the pattern on the screen will indicate the effect of position.

Small solonoids like yours hooked up a ac power supply providing several amps will also reflect patterns.

Degausing coild must not be switched on or off close to the screen.
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StevenCaton
Sun Jul 04 2010, 10:39PM
StevenCaton Registered Member #1845 Joined: Fri Dec 05 2008, 05:38AM
Location: California
Posts: 211
At every place, the magnetic field is varying at the RF frequency. The field induces an RF voltage in every circuit loop, in proportion to "how many field lines are encircled by the victim loop".
Well, isn't is true that loops on either option 1 or option 2 wouldn't encircle any field lines, so they would both work equall well?

I understand that two loops near eachother should be orthognal if one wants to minimize coupling, but with option 1 and 2 its almost like it wouldn't matter because the loops wouldn't encircle magnetic field lines for either one, unless I am still missing something.
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klugesmith
Mon Jul 05 2010, 05:19AM
klugesmith Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
SteveC wrote ...
Well, isn't is true that loops on either option 1 or option 2 wouldn't encircle any field lines, so they would both work equall well?

I understand that two loops near each other should be orthognal if one wants to minimize coupling, but with option 1 and 2 its almost like it wouldn't matter because the loops wouldn't encircle magnetic field lines for either one, unless I am still missing something.
Time to study the roots of the "orthogonal is better" rule.
Did you look at the picture link in my first response? Here's another qualitative rendering, respectfully adapted from Link2

1278306002 2099 FT91914 Solenoid


As Myke and I have both explained before, the external field direction is (in most places) not exactly parallel to the coil axis.
The option 1 board is seen on edge in my figure. The field's radial component causes lines to cross the plane of the board. Any circuit loop in the plane of the board will have field lines threading it (except at the midpoint in this view).
In option 2, seen face-on in my figure, the field's axial and radial components are both in the plane of the board. Every loop in the board is edge-on to the field lines, with zero projected area. Does this make things clearer?

All: there's a spectacularly glitzy rendering of a helical coil's field, which I will not borrow, at Link2
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