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I had a 2sets of LED series circuits which runs with 12V DC, 1A as shown below . Each set will be of 3 series of four LEDs. My problem is that when i connected to 12VDC,1A source, the set1 brightness is higher than set2. how to overcome this problem ? enter image description here

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    \$\begingroup\$ I'm not posting an answer to this, but the proper way to power LEDs is through a current source instead of a voltage source. IxV curves may vary between units and the brightness can be better controlled through the current. You may find this interesting (electronics.stackexchange.com/questions/256336/…). \$\endgroup\$ – Filipe Nicoli Jul 17 '17 at 18:10
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    \$\begingroup\$ @FilipeNicoli: Many commercial LED strips are designed to operate directly from 12 volts, and include suitable current limiting resistors, so don't need a constant current supply. The OP's drawing does show resistors (two green circles, and the letter "R"). I think the OP's drawing shows the power connected incorrectly. \$\endgroup\$ – Peter Bennett Jul 17 '17 at 18:15
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    \$\begingroup\$ Your question is misleading. The LEDs are not in series but are in parallel with daisy-chained power connection. \$\endgroup\$ – Transistor Jul 17 '17 at 18:17
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Wire each set individually to the power supply.

There will be some voltage drop in the wiring, depending on both the current and on the length of the wire. In your circuit, set 2 will receive a lower voltage than set 1 due to the additional wire between the sets.

Increasing the wire gauge (diameter) will also reduce the voltage drop in the wiring.

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  • \$\begingroup\$ If by "wire size" you mean the wire gauge, I do agree. I think it would be benefical if you edited this part, since it can be understood as "wire length". \$\endgroup\$ – Filipe Nicoli Jul 17 '17 at 18:03
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You don't specify the length of your wiring nor the gauge of wiring. But your diagram shows that you are daisy chaining the power supply connections. This adds more resistance for the second strip which could lower its brightness. Try wiring each LED strip directly to the 12 volt supply using the same length and gauge of wire for each. Increasing the gauge wire for both would also be helpful.

Your diagram shows that you are connecting the LEDs directly to the power supply without any form of current limiting. This is generally not a good idea. But to improve the situation, you would need a higher voltage power supply. If you have one, indicate in the comments what you have and I can add to this answer.

Finally, consider that LEDs of the same part number come in different brightness grades. It is possible that your strips are simply two different grades. By lowering the current on the brighter strip, it can be made to match the other strip.

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The culprit here seems to be the power supply.

If each set needs 12V 1A and you're connecting 2 set in parallel, their current demand will be 2A. A single 12V 1A power supply won't be able to supply enough current and will start dropping voltage. As a result of all this, the LEDs will start to show an impedance higher than what is desirable, due to insufficient bias.

Because of manufacturing tolerances, the LEDs in each set will have different impedances at those ill biasing conditions and, as a result, one of the branches could draw more current (and its light be brighter) than the other.

Try the following test: connect each set individually to the power supply, one at a time. If each set shows full brightness tested like this, then the culprit is the power supply.

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