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Since I've tried to experiment with electronics, I am faced with the problem of HF noise and signals bypassing almost anything I try to stop them. This begun with my attempts to filter the noise of a simple ATX PSU, and this occurred once more recently with my attempts to rectify a 5MHz modulated signal at 1khz.

Regarding switching PSU, in 9.6.8 D, the "Art of electronics" says "The switching noise can be heavily bypassed at one point, but just put your scope probe a few inches away and they're back." Later (9.9), he also quotes James Bryant in answer to the question "How can I prevent switching-mode power supply noise from devastating my circuit performance?", Answer: "With great difficulty - but this can be done".

For them, the problem is that switching PSU are full of noise at a very wide range of frequencies, that occurs in a variety of ways (see end of 9.6.8 D).

In my opinion, and as I will try to show below, the problem is much more fundamental and its interpretation is not so easily found, if ever it has been processed. But let see what you say about that, guys !

This time, I've realised what Douglas C. Smith calls a null experiment, in a way that can be easily reproduced and should help to rule out many incorrect interpretations. Here is the experiment:

  • Around the coil of a grid dip emitting at 5MHz, are rolled 2-3 turns of electrical wire. (see 2 pictures below). Notice that one extremity of the wire is left opened. enter image description here enter image description here

    • The other extremity of the wire (at about 1.5 m) is brought near the two joined extremities of the scope probe. The probe is set at x10, and the scope at 1mV/div, 1ms/div. As expected, nothing occurs but some weak background noise (picture below). enter image description here

    • Now, without moving anything, the two joined extremities of the probe are in contact with the extremity of the wire. In the first picture below, the grid dip is set in continuous wave modulation (CW), and in the second picture, it is set to modulate the carrier wave at 1kHz (MOD). This leaves no doubts that the grid dip is generating a signal in the scope. As can be seen, the scope is now indicating a signal of about 2mV, that is 20mV (since the probe is x10). enter image description here enter image description here

The question is: by what exact mechanism is a signal generated by the grid dip in the probe?

Moreover, I add:

  1. like for switching PSU, it is very difficult to "stop" or rectify this signal or more precisely, to know with the scope whether or not it has been stopped: after all, this null experiment says that you have no hope to learn anything with the scope no?

  2. the orientation of the loop formed by the probe does not matter at all in this experiment: this may prove that there is no magnetic influence through air. Only a contact with the wire produces the effect.

Note: A grid dip is not necessary to perform a similar experiment, but it is a good choice because it emits at a relatively narrow bandwidth. If this condition is relaxed, the same can be done with a noisy ATX PSU switching in the MHz range like this one: enter image description here

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This hasn't got anything to do with antenna effects as far as I can see; it's just the capacitive coupling of signal to a shortish wire and then connecting the end of that wire through a small value inductor back to ground. The small value of inductor is the scope-probe earth wire and the voltage measured by the scope is the voltage across that small value inductor due to the wire injecting current through it: -

enter image description here

A 3 inch diameter coil can be made like so: -

enter image description here

And if you go to this calculator and estimate the inductance for a single turn you get about 170 nH.

But, the inductance between these two points (in blue)....

enter image description here

... will be about 100 nH so, if you inject a 5 MHz signal at the joined end of the probe, it will see a ground wire impedance of about 3 ohms and, the resulting current will (pretty much) flow through the reactive 3 ohms and produce a signal volt drop that the o-scope measures.

Try using a real 3 ohm resistor and this type of probe measurement across the resistor to cut-down on the loop inductance of the earth wire: -

enter image description here

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  • \$\begingroup\$ Hello Andy, where is there a capacitive coupling? the wire is rolled around the grip dip coil and seems much more an inductive coupling. Second, why would the current want to return to ground? the wire is nowhere powered by a generator connected to ground (that is, it is floating). \$\endgroup\$ – MikeTeX Feb 10 '18 at 16:43
  • \$\begingroup\$ No it's a capacitive coupling - with one end floating it won't induce a voltage. Your grid dip coil will have an electric field around it that the wire couples to. The generator is a box sat above ground and that might offer 10 pF. Ditto the capacitive connection with the wire wrapped around the GD nozzle. \$\endgroup\$ – Andy aka Feb 10 '18 at 17:25
  • \$\begingroup\$ If the total series capacitance of nozzle and ground capacitance is (say) 5 pF, at 5 MHz this will be an impedance of about 6 kohm. If the "source voltage" (i.e. the nozzle) is 10 Vp-p then a current of 1.67 mAp-p could flow and this would produce a voltage of 5 mVp-p across a coil reactance of 3 ohms. \$\endgroup\$ – Andy aka Feb 10 '18 at 17:29
  • \$\begingroup\$ @MikeTeX having looked at your pictures the grid dip is sat right on top of other hardware (that could well be earthed) so, to prove this try and isolate the battery powered grid dip more. \$\endgroup\$ – Andy aka Feb 10 '18 at 17:33
  • \$\begingroup\$ The grid dip is powered by the main, so, it is surely grounded. Regarding your assertion that the coupling with the wire is capacitive, I'm not so sure this is exact: the effect is strong whenever the wire is rolled around the coil, not when it is only near the coil; after all, it is known that some current flows through a simple wire acting as a dipole antenna at HF, even if it is not connected to anything. \$\endgroup\$ – MikeTeX Feb 10 '18 at 17:46
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The question is: by what exact mechanism is a signal generated by the grid dip in the probe?

See the image below:

enter image description here

Any changing magnetic field passing through the red shaded area induces voltage measured by the probe.

The voltage is given by Faraday's Law of induction:

\$v = \int_A \frac{\partial B}{\partial t}da\$

EDIT1:

To answer your question where the changing magnetic field comes from:
The single wire acts as an antenna (e.g. by its capacitance against "infinity"). I.e. there is RF current going through the wire causing an RF magnetic field. That field is picked up by the loop.

EDIT2:
To answer your question why the is only a signal if you touch the probe with the wire:
The wire alone makes a very ineffective antenna for your RF of 5MHz (wavelength is several 10s of meters), so there is not much power radiated. This changes dramatically if you connect it to the grounded probe.
See image; blue color indicates the current path and strength:
enter image description here

Note also that making the loop smaller doesn't help a lot to minimize the signal because the magnetic field is generated mostly by the current going through one side of the loop. So most of the magnetic field will always pass through the loop area no matter how small you try to make it.

BTW: generally if you have a RF voltage signal at shorted probe terminals I think it will be very difficult to find any other explanation than induction by a RF magnetic field passing through the loop formed by the terminal wires.

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  • \$\begingroup\$ For me, your answer is unsatisfying: There is no reason a changing magnetic field be induced through the loop by a simple contact with the extremity of the wire, especially after I have shown that even if the loop is very close to the extremity of the wire (but without contact), nothing is produced. \$\endgroup\$ – MikeTeX Feb 9 '18 at 13:32
  • \$\begingroup\$ @MikeTeX what happens if you just connect that wire to scope ground? Isn't there a big (9M) resistor between the scope tip and the input, whereas the ground goes straight to the scope? \$\endgroup\$ – BeB00 Feb 9 '18 at 13:43
  • \$\begingroup\$ The scope has a 1Mohm input impedance (10pF in parallel), and the probe, set to x10, should add 9Mohm in series with the tip. If you just connect the wire to scope ground nothing very definitive happens, but if you connect the other side of the probe with some dipole antenna (like a screwdriver), there is some generated noise, but not the clear effect indicated above. \$\endgroup\$ – MikeTeX Feb 9 '18 at 13:54
  • \$\begingroup\$ @MikeTeX what happens if you use a 1X probe? \$\endgroup\$ – BeB00 Feb 9 '18 at 14:03
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    \$\begingroup\$ @MikeTeX, try this...remove the grabber off the end of the probe. Do you see a small silver ring just behind the sharp probe tip? Wrap a few turns of 26..30 gauge wire around this and attach the other end to a ground VERY CLOSE to node you want to probe with the scope. Make this ground connection a half inch or shorter. Does that reduce the observed noise? \$\endgroup\$ – AlmostDone Feb 9 '18 at 15:16
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Consider a long wire, 4" from a square loop of wire (perhaps the scope probe + GND lead, albeit not square) that has area of 4" by 4". The wire has current with dI/dT of 1 Million amps per second. What voltage is induced into the loop by the magnetic forces?

Vinduce = 2e-7 * Area/Distance * dI/dT

Vinduce = 2e-7 * 0.1meter * 0.1meter / 0.1meter * 1e6 amps/second

Vinduce = 2e-7 * 0.1 * 1e6 = 2e-8 * 1e+6 = 2e-2 = 0.02 volts

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  • \$\begingroup\$ analogsystemsrf. I know induction laws. Unfortunately, the effect is exactly the same if I tie the loop formed by the earth probe to the plastic of the tip probe (no more loop surface). There is no magnetic induction through the air as I said in the question. So, your interpretation is probably incorrect. \$\endgroup\$ – MikeTeX Feb 11 '18 at 8:14

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