Hi All.
Last year I bought some LED string lights (photo attached) last year that run on 3 AA batteries. Each string has a 30 ohm resistor in series with the lights. I'd like to replace the batteries with a USB connector. I'd like to connect three strings in parallel to minimize the number of USB connectors needed.
When I take one string and connect to a USB plug (5.08V) I get a current of 68mA. If I'm applying Ohm's Law correctly (P=IV), then 5.08V x 0.068A = 0.344W. When using the batteries (4.85V) I get a current of 60mA, or 0.291W.
If I run three strings in parallel, through one resistor, do I need a 1W resistor? If I'm judging the current resistor (photo attached) it's a 1/4W resistor. Isn't it too small a wattage for the original circuit since the power exceeds 1/4W? I think I could just leave each of the three resistors in series with each string, and then connect all three in parallel, but that seems like it would be bulky.
I'm a newbie at this kind of stuff, so maybe I've read something incorrectly.
Thanks for any help.
Richard
Last year I bought some LED string lights (photo attached) last year that run on 3 AA batteries. Each string has a 30 ohm resistor in series with the lights. I'd like to replace the batteries with a USB connector. I'd like to connect three strings in parallel to minimize the number of USB connectors needed.
When I take one string and connect to a USB plug (5.08V) I get a current of 68mA. If I'm applying Ohm's Law correctly (P=IV), then 5.08V x 0.068A = 0.344W. When using the batteries (4.85V) I get a current of 60mA, or 0.291W.
If I run three strings in parallel, through one resistor, do I need a 1W resistor? If I'm judging the current resistor (photo attached) it's a 1/4W resistor. Isn't it too small a wattage for the original circuit since the power exceeds 1/4W? I think I could just leave each of the three resistors in series with each string, and then connect all three in parallel, but that seems like it would be bulky.
I'm a newbie at this kind of stuff, so maybe I've read something incorrectly.
Thanks for any help.
Richard
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