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Say I have an ideal situation (good quality functioning power source and components) - 12V source and 3V (each) LEDs. Can I wire these 5 LEDs in series to the 12V source without a series resistor and expect that 4 of them would lit strong and one dim (or this one not lit at all), or all would light up dimmed ? Without burn, and etc as expected for an over voltage situation - like wire 1 3v led to a 4V source and etc.

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    \$\begingroup\$ 5 x 3 = 15. None would illuminate very well. Check the data sheet to see what the current might be if the voltage was 2.4 volts. Never not use a current limiting method. \$\endgroup\$ – Andy aka Jan 14 '20 at 16:21
  • \$\begingroup\$ You would want 4 x 3.0V LEDs for a 12.0V power supply, not 5 LEDs. \$\endgroup\$ – Duncan C Jan 14 '20 at 16:24
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    \$\begingroup\$ Another reason it's a bad idea is that LED forward voltage will vary with temperature. You might find that at your coldest temperature they'll be dim or off (even with 4 LEDs) and at the highest one they'll burn out. And lighting LEDs burns power and as such dissipates heat, giving you a potential runaway situation. A constant current supply is what you want here. If you want one dimmer, put a parallel resistor across it. \$\endgroup\$ – Cristobol Polychronopolis Jan 14 '20 at 16:32
  • \$\begingroup\$ What is the recommended current for each LED? and what level did U want? Each white LED has a 3V Zener like quality factor \$\endgroup\$ – Tony Stewart EE75 Jan 14 '20 at 16:42
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None of the above. If you wire LEDs in series, all of them will receive the same amount of current. They must, since current is number of electrons flowing through a wire per second, and the electrons must pass through all the LEDs in a series circuit.

If they all get the same amount of current, identical LEDs will all light with the same brightness. (If you have a mixture of different types of LEDs in a series circuit they will always still receive the same amount of current but that current might cause them to glow with different intensities.)

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  • \$\begingroup\$ I did this recently. battery powered 30 led set was cut up to 20 instead. Power was by default 2x AA cells. A bit brighter but perversely the power seemed to drain faster , could be mistaken on that though. I have not been with basic electronics for years so guess they were taking too much. Point here I am is out of the four colours when the battery got weak you would lose blue first then amber follow by red and final green. I say "lose" as in not showing noticeable light output. Out of random I then tried a temp 6v supply, kept an eye on them but two burned quite quickly! A bad idea! \$\endgroup\$ – AndyF Jan 14 '20 at 19:38
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On the topic of powering LEDs with a constant voltage:

LEDs have a negative temperature coefficient. This means that their resistance decreases as temperature increases. Providing a constant voltage will cause an ever increasing current until an equilibrium is reached. As a result, LEDs are recommended not to be used without some form of current control. The purpose of a series resistor in a string of LEDs is to compensate for this property with a suitably large positive temperature coefficient from the resistor and provide a form of current control.

With an LED connected directly to a constant voltage, as current passes through the LED, the LED heats up. This causes its resistance to decrease, resulting in a larger current flowing and more heating of the LED. If current is not regulated, it will continue to increase until either the LED burns out or if the LED is adequately cooled, an equilibrium will be reached where the LED becomes hot enough that the heat conducted/radiated away from it(which increases as temperature increases) matches the heat produced by the current flow. If the thermal path (heatsink and/or fan) from the LED to it's environment is good enough and the drive voltage low enough, the LED will not burn out, but if the LED should ever become separated from the thermal path, vibrated loose, become unglued, an air vent blocked, whatever, it will suffer thermal runaway and burn out.

With an LED connected to a voltage source in series to a sufficiently large resistor, such that the temperature coefficient of the resistor exceeds that of the LED, when the voltage is applied, the current will heat both the LED and resistor, and as the resistance of the LED decreases, the resistance of the resistor will increase, preventing increase in current and thermal runaway. Using a resistor this way amounts to a very crude (inaccurate and power wasting) form of current control and protects the LEDs. Power losses from this setup can be mitigated by using more than one LED in series with the resistor, but because there is a minimum resistance at which the resistor will have a sufficiently large positive thermal coefficient, there is a limit to how efficient this can be.

Resistors are typically used in low power applications, for example running a single 3.3V LED indicator light at 5mA off a 5v source. At only 5mA, a resistor dropping 1.7V will only use 1.7V*.005A=.0085W or 8.5 mW of power. Roughly 1/3 of total power is being wasted in heating the resistor, but total power is so small you may not care. If you were running a power LED, say 500mA at 3.3V, the resistor dropping the same 1.7V will use 1.7V*.5A=0.85W, almost a full watt wasted in heating up a resistor just to provide current control.

To save the power being wasted in providing current control with a resistor, current controlled LED drivers exist that regulate applied voltage to provide a constant current regardless of LED temperature and changes in resistance. A decent one of these drivers will be 80 to 95% efficient, saving you the bulk of the power a resistor would be wasting.

In conclusion, you must always run LEDs in such a way as to compensate for thermal runaway or they will burn out.

Adequately cooling the LED will work, but LEDs will burn out if the thermal path is ever disturbed or becomes inadequate. You can't always control the factors involved(ambient temp, dust, blocking of an air vent, mechanical damage, etc.) so this method is usually considered inadequate.

Using a series resistance is low cost and inefficient, but will provide current control and prevent thermal runaway when cooling is inadequate and is typical for indicator lights and other low power LEDs.

Using a current controlled switching regulator to drive the LEDs costs a lot more than a resistor, but controls current and prevents thermal runaway with a smaller percentage of total power being wasted as heat.

One other notable advantage to using a resistor in series with an LED is that the resistor can function as a cheap fuse. Sometimes if you take apart commercial LED products, you will find that the resistor wattage is chosen so that if current becomes too high the resistor will burn out and open the circuit before the LED is exposed to a damaging current.

In the example of the indicator light LED above, the resistance that would limit current to 5mA is 1.7V/0.005A=340\$\Omega\$. If a 350\$\Omega\$, 0.01W resistor is used, a flow of more than

\$P=I^2R\$

\$I=\sqrt{(P/R)}\$

\$I=\sqrt{(0.01W/350\Omega)}=0.00534522A\$

about 5.35 mA will burn out the resistor and cut power to the LED. If you applied an incorrect voltage(too high specifically), the resistor would burn out and the LED would stop working, but you would have a comparably easy repair, just replacing a tiny <1c resistor rather than having to identify, color match and bin and replace the LED.

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