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I have a lamp that I really love.Sadly while moving I lost the power adapter. The lamp has 21 leds wired in parallel with a 68 ohm resistor on each. The lamp shows the original power supply to be 6 volts, but no amp rating. I have researched and tried to figure what I need and came up with 350 ma. The lamp works, but some of the leds blink. Can someone give me some advice on this? thanks J phantom

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  • \$\begingroup\$ What's the color of the LEDs? \$\endgroup\$
    – pipe
    Commented Mar 5, 2016 at 13:02
  • \$\begingroup\$ they are clear leds \$\endgroup\$
    – j phantom
    Commented Mar 6, 2016 at 7:04

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The simplest thing to do would be to temporarily use a large 6V power supply — one that can supply several amps — and measure the current that the lamp draws. Borrow the use of such a supply and a multimeter if you don't have them. If you don't know people with such devices, check your local area for "maker spaces", which almost certainly will.

Then, select a final permanent supply for the lamp that can provide at least the measured amount of current. It doesn't matter if this power supply can provide more current than needed; the lamp will only consume what it needs.

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  • \$\begingroup\$ thanks for that information I will try that and see what happens. \$\endgroup\$
    – j phantom
    Commented Mar 6, 2016 at 7:08
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If I calculate 2.5 V what a LED needs across the resistor we would have 3.5 V. Divided by 68 ohm I get 51.5 mA. With 21 LED we get a current of 1.08 A. It's look like your power supply is not strong enough. I am not sure how much a white LED needs but if you calculate with 3 V it would be 0.93 A and with 3.5 V -> 0.77 A.

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  • \$\begingroup\$ Your calculations seam correct, I was overthinking this a bit. I was afraid of over powering the leds. I will increase the ma,s and see what happens. thanks \$\endgroup\$
    – j phantom
    Commented Mar 6, 2016 at 7:17
  • \$\begingroup\$ If the voltage of the power supply correct you cant overpower the LED. The resistor limit the current for each LED \$\endgroup\$
    – Stefan
    Commented Mar 6, 2016 at 12:25

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