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I have a string of LED lights that runs on two 3 V batteries. When I measure the voltage with my multimeter it is about 6 V, as expected.

I would like to use an adapter to avoid using batteries, so I took a USB phone charger and exposed the two voltage wires. The voltage output from the adapter is exactly 5 V.

When I connect the adapter to the lights, the lights get hot and shine too brightly as if they have too many volts across them, but there is only 5 V as compared to the 6 V coming from the batteries. Why is it that the adapter is making the lights do this?

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  • \$\begingroup\$ Did you measure the voltage across the batteries while the LED strings were connected? \$\endgroup\$
    – ocrdu
    Commented Nov 6, 2020 at 22:51
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    \$\begingroup\$ too many volts going through them .. current goes through components, not voltage ... voltage placed across a component causes a current to flow through a component \$\endgroup\$
    – jsotola
    Commented Nov 6, 2020 at 22:51
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    \$\begingroup\$ How do you know the LEDs are directlt connected to batteries? What if there is a current limiting resistor in series? Or what kind of batteries they are? If they are button cells they have large internal resistance to begin with so they can't provide current and they limit their output even without a resistor. \$\endgroup\$
    – Justme
    Commented Nov 6, 2020 at 23:00

2 Answers 2

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If these "3 volt" batteries are little coin cells, they have high internal resistance (at least 10 ohms, perhaps much more):

https://data.energizer.com/pdfs/lithiumcoin_appman.pdf

That high internal resistance acts as a current limiting resistor. If you have no other resistor on your LEDs, then you absolutely must add one to use with a 5v supply or you will burn up the diodes.

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As you say in a comment, the batteries under load show only 3.5 V. A regulated 5 V power supply that can deliver enough current stays at 5 V, load or no load.

So, more current passes through the LEDs with the 5 V power supply attached than with the batteries (3.5 V) attached.

The open voltage of the batteries is irrelevant here; what matters is what the batteries do in a circuit where current flows. The batteries' internal resistance that makes their voltage drop under load is limiting the current that flows in the circuit.

Using a 6 V battery with a very low internal resistance, like a big lead-acid battery, would make a larger current flow than your 5 V power supply, because it would stay at about 6 V under load.

If you use coin cells with a high internal resistance, you can sometimes get away with not using a resistor in series with the LEDs, as the batteries' internal resistance will limit the current. With all other voltage sources, you must use a resistor to set and limit the current through the LEDs.

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