-1
\$\begingroup\$

I'm going to make a partial EM-shield (using aluminum) for my 3G USB (2.0) modem, and I need to "ground" electricity from the surface of that shield.

Daily worktime for that 3G modem is planned around 10 hours.

Calculations I made so far:

$$\text{USB (2.0) output} = 5_{\,V} * 0,5_{A} = 2,5_{\,W}$$ $$\text{Capacitance} = {\frac{2,5_{\,W} * 3600 \text{ (seconds) } * 10 \text{ (worktime) }}{{5_{\,V} * 5_{\,V}}}} = 3600_{\,F}$$

Meaning (if above is correct), that the surface of my shield may accumulate up to:

$$\text{Electric potential} = \sqrt{{\frac{2,5_{\,W} * 3600 \text{ (seconds) } * 10 \text{ (worktime) }}{{3600_{\,F}}}}} = 5_{\,V}$$

— which I need to "ground" daily.

The problem is: I'm living on a 6th floor of a many-apartment building built by the Soviet Union, and we do not have earthing electrode in our source circuit (like, seriously: we only have 2 wires inside each AC socket).

Sooo, since I'm not considering using metallic water pipes for my purpose - I thought of using electric capacitors.

I've read "out there" that such capacitors inside single domestic circuit (which will include that partially-shielded 3G modem, and my PC, and laptop, and monitor, and 2.1 sound system) - may effectively be used instead of loose-ended earthing electrode.

The question then (if that's true) is, how many capacitors do I need to "ground" 5 Volts daily; or, how much "Farads" their total capacitance should be?

I mean, these "daily" 3600 Farads I've calculated seems like an answer, but may have I missed something? Since there're NO capacitors on a market to maintain such capacitance: the best ones I found (per a single device, like this one) consume only around 100-1000 Farads tops.

And (most importantly): will, say, 10 of such capacitors do their job if I'll dare to purchase them? I mean, how do they work anyway? Do they re-charge, or they will consume my 3600 Farads only once, and then immediately die?..

Additional data:

  1. Devices I do have at a moment (PC, laptop, etc.) somehow do not have problems with extra-current on their surface, nor with electric shocks (so I felt like they're irrelevant for calculations).
  2. Also (which probably is relevant): I'm going to cover a USB wire (connecting 3G modem) with aluminum foil, so the total usage of aluminum (which is, like, non-covered metal) is expected to be around 0,5-1 kg (1-2,2 lbs).
\$\endgroup\$
18
  • \$\begingroup\$ Why do you need to ‘ground’ the shield? Your calculations based on the USB power delivery are flawed and the results requiring the use of Farads are equally flawed. Show a picture of your shield and what you want to achieve. \$\endgroup\$
    – Kartman
    Sep 1, 2021 at 1:34
  • \$\begingroup\$ @Kartman but ain't my calculations based on a maximum output of a USB (2.0) socket? Only neglecting the stand-bys. And to picture verbally what I want to achieve is that I want to shield my Internet connector (and it's wires) from all directions, except noiseless clear patch of sky I have in my window. \$\endgroup\$ Sep 1, 2021 at 1:38
  • \$\begingroup\$ The earth is relative to your 3g module not mains earth. Simply connecting the ‘shield’ to the shield of the usb cable should be sufficient. \$\endgroup\$
    – Kartman
    Sep 1, 2021 at 1:41
  • \$\begingroup\$ @Kartman I don't understand: I thought all aluminum parts will accumulate electricity, which will not go away on itself. Are you saying that it will vanish without grounding (or capacitors) somehow? \$\endgroup\$ Sep 1, 2021 at 1:46
  • 2
    \$\begingroup\$ Under normal working condition, the charges entering the device via the live wire (phase for AC or V+ for USB DC) are automatically drained via the return (neutral for AC or gnd for USB DC) wire. No charge accumulation that needs to be drained. (Other than static charge mentioned in a comment above) \$\endgroup\$
    – AJN
    Sep 1, 2021 at 2:46

2 Answers 2

4
\$\begingroup\$

You have calculated the amount of capacitance required to store 25 Wh at 5 V. But the energy drawn by the USB MODEM is not stored, it is dissipated (mostly as heat, with a little transmitted rf.). You don't have to worry about this electricity because it is consumed by the MODEM.

It is possible for a metal surface to build up a static charge if it is not 'earthed', but its capacitance is determined by its size and distance from the earth or nearest circuit connected to it. For a non-earthed mains powered device such as a laptop the largest parasitic capacitance is between the primary and secondary windings of the power supply transformer, which is usually only a few hundred picofarads.

However a larger capacitor is usually added between the primary and secondary sides to reduce EMI. This capacitor 'shorts out' the EMI on the secondary side to reduce its voltage. To do the job well it should have a much larger capacitance than the transformer windings, but not too large or sufficient AC mains voltage could leak through to give a noticeable electric shock. Typical values are 1~2.2 nF. With this small capacitance static charge could easily produce several thousand volts if not discharged, so power supplies often have a high value 'bleeder' resistor across the capacitor to discharge it. A resistance of several megohms is high enough to prevent electric shock but low enough to discharge the capacitor in less than a second.

As an example, here's (part of) the schematic of a Dell laptop power supply:-

enter image description here

C006 has a value of 1 nF, and the 5 resistors have a total series resistance of 50 MΩ. Both are connected from the primary (mains voltage) side of the switch-mode power supply to the secondary (low voltage output) side, providing a weak 'ground' at DC and a stronger 'ground' at low to medium rf frequencies. This is not a 'true' ground because it is not connected directly to mains earth, but it is close enough for EMI suppression purposes.

If your device's power supply has a similar circuit then there is no need to worry about grounding the USB MODEM because it is already sufficiently grounded through the USB port.

\$\endgroup\$
0
\$\begingroup\$

I believe you are trying to secure your computer Internet's 3G access from local hackers by using;

"aluminum foil, so the total usage of aluminum (which is, like, non-covered metal) is expected to be around 0,5-1 kg" and pointing the antenna up in the sky out the window.

The USB is wired so it is not radiating.

I think your idea of security solution is not realistic, or scientific. Your WIFI does not bounce from the sky. It does bounce off all buildings.

This would be like trying build a dirty bomb shelter for nerve gas but still have the window open.

\$\endgroup\$
1
  • \$\begingroup\$ I thought of that, but for the 1st try I'm hoping that wavelength of 3G (14 sm) would somehow enter that "window". Will see about that, 1 step at a time, of course. As for "the shelter"... well, if anyone interfered with me like that, and none of boundaries I've put had worked - I'd prefer to force these in such shelter, just enough for me to never be bothered. \$\endgroup\$ Sep 1, 2021 at 19:27

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.