0
\$\begingroup\$

I ran into this YouTube video on how you can amplify the signal for your radio controlled watches: https://youtu.be/wI4FwQMCN9w

In short, the guy in the video said he stumbled on a way to amplify the signal by accident. All you need to do is hang your watch on a metal pipe and it will help it receive the signal successfully and sync. Later on, I found out some people on the internet claim even wearing a watch on your wrist will help it get reception.

As someone living too far away from the German DCF77 transmitter (https://en.wikipedia.org/wiki/DCF77?wprov=sfla1) to ever receive the signal with my radio controlled watch, I found the video useful. By hanging the watch on my bike's handlebar, I got my first sync ever. I also tried syncing it on my wrist once when I was up late and it also worked.

But how does this work? The metal pipe can't add any energy to the signal so it doesn't really amplify anything, does it? And the watch's metal case doesn't really touch the pipe, just the plastic strap, so can it even act as some sort of an antenna or something? So does it just focus the signal somehow? Obviously, I know nothing about radio reception, but I am interested in how this works.

Also, is there a way to increase the reliability of the reception? I still can't receive the signal 100% reliably, even with the metal pipe over night. Would a certain length, shape or material of the pipe help? Maybe orientation?

\$\endgroup\$
  • 2
    \$\begingroup\$ why do you think that a pipe is required? ... google radio antenna \$\endgroup\$ – jsotola Dec 27 '19 at 9:16
  • 1
    \$\begingroup\$ @jsotola But there is no electrical contact between the pipe and the watch. Isn't an antenna usually connected to the electronics using it to receive a signal? \$\endgroup\$ – relatively_random Dec 27 '19 at 9:46
  • 1
    \$\begingroup\$ Nope: Not all parts of an antenna need an electric (galvanic) connection. Example: The big reflector on a dish antenna is isolated from the actual receiver. \$\endgroup\$ – Turbo J Dec 27 '19 at 10:34
  • \$\begingroup\$ @Turbo Ok, I can understand how that would work: the dish reflects the waves and focuses them into the receiver (if this is indeed how the dish works). It achieves amplification by robbing the space behind the dish of those radio waves it redirects. But how would it work in the case of a watch? Does the pipe also reduce the signal at some other parts of space around it? Does it just filter out the noise somehow? In any case, the other part of the question still stands: what would be the ideal antenna for receiving this 77.5 kHz AM signal with a watch? \$\endgroup\$ – relatively_random Dec 27 '19 at 10:45
  • \$\begingroup\$ Interference from nearby radio-emitting sources (like switcher power supplies) can raise the noise floor to mask weak DCF77 signals...moving to a radio-quiet overnight location likely helps most. Perhaps metal pipe suppresses radio fields of local noise sources rather than enhancing reception of DCF77. A location inside a building yielded a signal hidden by noise - 30 meters outside, signals were above the noise (during the day)...this was in a fringe reception region. \$\endgroup\$ – glen_geek Dec 27 '19 at 15:11
1
\$\begingroup\$

It sounds to me like the addition of a ferromagnetic material will actually “focus” the magnetic part of the EM signal produced by the German radio transmitter. According to wiki it transmits around 80 kHz and, at that frequency, most random pieces of iron or steel will concentrate the magnetic part of the transmission just like a regular ferrite rod antenna: -

enter image description here

In concentrating that field, the internal coil used to pick up this transmission will be given more magnetic flux and, more flux means more induced “antenna” voltage and a stronger received signal.

\$\endgroup\$
  • 1
    \$\begingroup\$ Yes, but it also works on aluminium bike frames and on a human wrist. Those aren't ferromagnetic, are they? And also, does "focusing" imply some other parts of space around the rod get reduced reception? \$\endgroup\$ – relatively_random Dec 27 '19 at 10:59
  • \$\begingroup\$ For sure, if the watch is left in a static position and that position is not optimal for radio reception then it will never get sync. Wearing the watch means your arm will visit several different positions and one of those positions is more likely to be better than the non-optimal static position hence, it will acquire sync. Ditto hanging it on a bike frame. Focusing does mean reduction of the signal in other spatial regions in order to get more signal in the focussed position. \$\endgroup\$ – Andy aka Dec 27 '19 at 11:07
  • 1
    \$\begingroup\$ But the odds that it was in a wrong position for all those many, many nights and that it just happened to be in the right position that one night when I hanged it on a bike frame are pretty slim. (The bike was parked, it wasn't moving.) Maybe electric conductivity had something to do with it, or maybe the fact that the frame is a loop, not just a rod, and that the watch was hanging inside the loop? Anyway, it's beginning to sound like specifying an ideal focusing shape isn't that trivial, or at least not a problem many radio experts or enthusiasts had to solve. \$\endgroup\$ – relatively_random Dec 27 '19 at 11:42
  • \$\begingroup\$ No, very unlikely about the frame being a loop of any relevance. For it to be effective the loop needs to be reasonably close to a wavelength at 80 kHz. Given that 100 kHz has a wavelength of 3 km and 80 kHz a tad longer, this has to be ruled out. Your explanation about it being in the wrong position for many nights isn’t that all unlikely. If the signal is very weak then only a near perfect alignment might have the result of causing sync. There is also the effect of the ionosphere. It..... \$\endgroup\$ – Andy aka Dec 27 '19 at 15:10
  • \$\begingroup\$ .....can play havoc with radio waves of this frequency due to subtractive multipath fading but, it can also achieve multipath addition potentially boosting the signal for short periods of time but, again you will possibly need optimal alignment to achieve this. \$\endgroup\$ – Andy aka Dec 27 '19 at 15:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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