2
\$\begingroup\$

I apologize in advance if this question is nontrivially stupid, but I'm not sure how else I would find this out without a theoretical background that I don't posess...

What I want to do is add an LED light to this watch winder.

What I don't know is: is it possible to wire up a small low power device like an LED light so it 'taps into' the existing power supply without screwing up the device itself?

I suspect that if I tried to attach say, an electric lawnmower or a hairdryer to the same power supply, things would go wrong. But would something like an LED be within the tolerance so that it could draw power while the device itself still gets the power that it needs to operate?

Or would I have to resort to hooking it up to a 9v battery and hide the battery as best I can?

Please assume:

  1. That I've found a way to mount it, etc. - I'm just asking about whether I can power it;
  2. That we're dealing with only one of the watch-winders (I dont want the fact that they can be plugged into the power base to be a confounding factor, so if that complicates things, assume I'm only using one of them and it's plugged in directly which it can be).

Again, apologies if this is very obvious or stupid to any of you - I legitimately don't know.

\$\endgroup\$
  • 2
    \$\begingroup\$ Take a look at the power supply specs (should be printed on the brick somewhere), and then see if the winder box also has specs printed where the barrel jack plugs in or maybe in the manual. That will tell you if the existing supply can handle more than the winder box. \$\endgroup\$ – DigitalNinja Oct 25 '16 at 18:08
  • 5
    \$\begingroup\$ The power supply can handle up to 12 winders. Using only one, it is almost certain there is sufficient power to light an LED. Are you sure that there is no power on light already on the winder? The problem is how to tap into the power supply. The photos show a standard cylindrical power connector that plugs into the winder. You will probably have to open the winder to get to the power supply wires. If you can do that, and tell us the voltage provided by the power supply, we could then guide you into how to connect an LED (probably using a series resistor to limit the current). \$\endgroup\$ – Barry Oct 25 '16 at 20:15
  • \$\begingroup\$ Does it make sense to bump this old question>? i.stack.imgur.com/ccsgt.png Yes I see by this picture you can wire in a 1K resistor series and LED. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 11 '19 at 19:19
  • \$\begingroup\$ Today I learned that there is something called a "watch winder"... \$\endgroup\$ – JRaef Apr 11 '19 at 22:06
  • \$\begingroup\$ Almost certainly yes you can. LEDs use very little power. \$\endgroup\$ – user253751 Apr 12 '19 at 0:42
0
\$\begingroup\$

Assuming you can tap the the V+ and V- and get two free wires out of it, there is a very easy way to do it with SMD LEDs. But it requires some soldering skill. If you have never soldered tiny SMD components, you will have to train a few times before but it's easier than it looks like. -Find 20mA SMD LEDs, GREEN (eventualy blue or red, but green shine better. NOT yellow or orange). Any brand or no-brand is ok. - one small prototype board to solder the LEDs on. - Find 10K ohms resistors. They can be through hole or SMD, at your convenience. - Connect the anode of the led to V+, the cathode of the led to a 10K resistor, and the other side of the reistor to V-. (you can place the resistor before the led instead of after, makes no difference)

This works with SMD leds. with 10K resistor you are safe up to 36V. If the voltage is between 5 and 36V it's ok. you don;t even have to care what voltage it is. Make only sure it's DC. If the LED is too dim, try a lower value. But not less than 4.7K for 12V and above.

You can try with through hole (THT) "melon hat" LED if you are afraid of SMD job. But in this case you will have to choose the resistor more accurately else they won't shine. They should work with a resistor between 450 ohms and 2.2 Kohms with 12V. The exact resistance can be calculated according to the voltage of the LED (usually 1.8 or 3V depending on the color) and the voltage of the power supply. The formula is R = (Vsupply - Vled)/Iled Where Iled is usually 20 mA for indicator leds. This gives you the minimum value for the resistor for the best output. You can use higher values than given by the formula as long as the LED shines enough to be seen. When using the optimal value given by the formula, and the voltage is 12V or more, you must chose a 1/4 watt resistor or more. When the resistance is much higher than the optimal value, like 10K as suggested, the power is so small dissipation that you don't need to worry about it.

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

They seem to have a stacking option so have power contacts top and bottom. Worth checking if these could be used. Very pricey gadgets for sure.

This video shows some of the internal workings, 3V motors it seems.

\$\endgroup\$

Your Answer

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