voltage changes as connections are moving

[Fionn]

Hi All

I have a simple question but found no solution for it. Its hard to search for it as i have no idea what the name of the effect is.
I will try to explain my observation.
Lets say we have a very simple circuit. A led is connected to a 5V battery with a 1K resistor. Its working fine
If I change 1 of the wires to an iron rod and connect the led via that rod its still working fine.
Now if I move the rod the led begins to flicker. https://youtu.be/qA8hzjJWICs
Is there a way to counter this effect?
I have tested a few caps and resistors as well on the negative wire (fig3 on the picture) and asked someone who has a lot more experience than me but he had no solution for that effect (or did not understand my issue)

All ideas are welcome or even if someone could tell me what the name of the effect is and what I have to google for would be nice

Thank you in advance.

[Jeroen3]

Contact resistance. Tweezers on an iron rod do not provide a reliable current path.
There is a reason relays use gold for low currents.

https://www.phoenixcontact.com/assets/downloads_ed/global/web_dwl_technical_info/105396_en_00.pdf

Attach a scope to see the bouncing of your "slider".

[Brumby]

The closest I could suggest is 'contact bounce'.  The surface condition (ie oxidation) could also come into play.  Also, there could be microscopic bits of dirt and dust that could help mess things up.

While you may think you are in constant contact with the iron rod, the fact that it has a rough surface means there will be hills and dales which you will be bumping into and skipping over, respectively.  As a result you will have times when the sliding contact is, in fact, not touching the rod.  It will be at a microscopic level, but at low voltages, that's all you need to interrupt current flow.

Try the same experiment, but with a machined, polished metal rod and you will see much less of the effect.  (Even a polished surface is rough at microscopic scale.)

To reduce the effect, you could try a capacitor across the LED to store some charge.

[tautech]

Black iron rod, covered in mill scale providing high resistance paths/poor contact.

Polish the rod until shiny and the unexpected results should disappear.

[Ian.M]

Also, use a contact that is flexible, will conform to the rod and provides multiple points of contact - e.g. a brush made of packed copper braid.

I wouldn't waste my time cleaning up the iron rod - start with one that's already ground smooth and polished e.g. the head support rail from a scrapped inkjet printer.

If you have to resort to a capacitor, put the resistor and LED in series and put the capacitor across the pair.  As LEDs are very sensitive to small voltage variations, a capacitor directly across the LED will be less effective.   However a capacitor big enough to keep the LED lit for a significant fraction of a second will also increase the sparking at the sliding contact and cause it to wear faster.

[Brumby]

If you have to resort to a capacitor, put the resistor and LED in series and put the capacitor across the pair.  As LEDs are very sensitive to small voltage variations, a capacitor directly acrpss the LED will be less effective.   However a capacitor big enough to keep the LED lit for a significant fraction of a second will also increase the sparking at the sliding contact and cause it to wear faster.

That's what I meant to say - thank you for picking that up.

For a capacitor connected directly across the LED, the forward voltage drop would limit the voltage the capacitor could reach.  This means it would fall below that voltage very quickly when the supply was interrupted and the LED would go out.  Connecting to the other side of the dropping resistor would give you a far more effective result.

[Fionn]

Thank you very much everyone for the feedback!
I have found with "contact bounce" a nice site: http://www.elexp.com/contact_bounce_and_debounce.aspx and the nand gate solution seems very good here. I will try that soon (I hope I find the time within a few days).

thanks again! )