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I have looked for a solution to my problem, but none of the answers I have found quite help me.

I am a beginner at electronics but have a general understanding of the physics behind electrical systems and circuits. I have a water pump that is powered from a 120VAC wall outlet, which will be used to move water in a makeshift fountain I am assembling. I want to power some 6VDC LEDs from the same plug to give it some lighting effects but I dont want to accidentally cut off the power to the small water pump. I have seen various posts about getting a transformer or capacitors to drive the LEDs but I cant quite get my head around how to implement any of these solutions. I know I can use batteries to drive the LEDs but if I have to plug the pump in anyway, I would prefer to power the entire fountain and LEDs from one plug.

Thanks in advance.

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  • \$\begingroup\$ It would help to have some idea about the number and type of LEDs you are considering. These are 6 V LEDs, I gather. Can you provide a link to them? And some kind of "no more than X many of them" answer? (Also, doesn't the supplier of these LEDs also provide a plug-in supply you can use?) \$\endgroup\$
    – jonk
    Mar 8 '18 at 23:40
  • \$\begingroup\$ The LEDs are actually from some keychain led Lights that are powered from button cell batteries. No more than 5 Leds will be used, and yes they are 6V. I bought some blacklight LED kechains from amazon, ill post the amazon link. amazon.com/gp/product/B00NOFMJKW/… \$\endgroup\$
    – aastorms
    Mar 8 '18 at 23:54
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    \$\begingroup\$ Interesting. I got some similar devices quite some time back. As I recall, mine used ONE CR2032. Your link definitely says they use two CR2016. However, you need to realize that the CR2016 physically cannot deliver much current. So it is self-limiting. When you pull an LED out, is there a resistor present? Or are the leads directly in contact with the battery pair when turned on? I think you may need to test one of these with a power supply that is capable of a LOT of current and tell us what happens if you supply 6 V. My guess is they burn up fast. \$\endgroup\$
    – jonk
    Mar 9 '18 at 0:01
  • \$\begingroup\$ There are no resistors, the leads are connected directly to the button cell batteries. but since they will be on for a prolonged period of time, Batteries aren't really an option for this project. I am sure there is some kind of converter I can wire to wall plug in the water pump to drive these leds, but I dont know what I am looking for which is my main problem. \$\endgroup\$
    – aastorms
    Mar 9 '18 at 0:12
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    \$\begingroup\$ A problem for us telling you what to do is that we don't have the specifications for those LEDs. And knowing that they were used with CR2016's doesn't help, because those are very WEAK sources. If you build/buy a mains-derived DC power supply, it will NOT be similarly weak. So you are changing situations and we don't know what your LED actually requires in this new situation. You can do some tests and collect some info. Do you have a voltmeter? Any resistors? Any kind of mains-derived DC supply to try out for a test? Can you turn it on and then measure the battery voltage while it is still on? \$\endgroup\$
    – jonk
    Mar 9 '18 at 0:15
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Most common LEDs operate at about 20mA

eg: https://www.adafruit.com/product/1793

3.4V forwards voltage so from a 5V "USB-shaped" supply you'd need to drop 1.6V - ar 20mA that's 80 ohms. so use a 82 ohm resistor (0.05W or larger) for each LED

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ He said "No more than 5 Leds will be used." Your circuit is wrong!! ;) Oh my gosh! \$\endgroup\$
    – jonk
    Mar 9 '18 at 4:48
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    \$\begingroup\$ hah! oops! edit! \$\endgroup\$
    – Jasen
    Mar 9 '18 at 4:51
  • \$\begingroup\$ He makes no mention of USB supply \$\endgroup\$
    – Passerby
    Mar 9 '18 at 5:54
  • \$\begingroup\$ he says "a transformer" - usb-chargers are cheap and convenient \$\endgroup\$
    – Jasen
    Mar 9 '18 at 6:15
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You might be able to use a capacitive power supply. But ONLY if the circuitry is not in contact with the water. It is a rather simple circuit that drops AC voltage to a lower DC level by way of capacitive reactance. Here is a link to the Wikipedia page for more information:

https://en.m.wikipedia.org/wiki/Capacitive_power_supply

This circuit should work relatively well for low power circuits, and it is used in some cheaper consumer electronics which are powered from mains. But before you use this circuit, be warned that the output is NOT isolated from the mains power lines as it would be with a transformer! Therefore, it is extremely important that this circuit is ONLY used where it is impossible (or highly unlikely) for the user to come into contact with any parts of the circuit physically. Electric shock may result if this precaution is not followed. Needless to say, you should probably not use this circuit for larger loads.

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    \$\begingroup\$ a non-isolated supply is not the the sort of thing I'd recommend anyone use around water, but yeah if it can be insulated and kept dry, it's a candidate. \$\endgroup\$
    – Jasen
    Mar 9 '18 at 6:22
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    \$\begingroup\$ I would not recommend that either. Moreover it's for really low power applications. I don't downvote the answer because you still warned about the danger. \$\endgroup\$
    – Fredled
    May 24 '20 at 22:12
  • \$\begingroup\$ Several failure modes of capacitive power supplies result in hazardous voltages being exposed. The awful power factor is possibly excusable but really that kind of circuit went out of favour in the 1970s, or at least should have. \$\endgroup\$
    – Frog
    May 29 at 11:22

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