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I am a beginner in Electronics. I would like to clarify if my understanding regarding LEDs is correct.

Suppose that we have an imaginary battery with no internal resistor that provides 2 volts. Assuming that we have a LED with forward voltage and current 2 volts and 10mA respectively. The question is can we simply light the led without a resistor? My assumption is yes, since the LED will have a voltage drop of 2 volts, and the current will be 20mA.

My questions stems from the fact that there are many many sources out there, mentioning that a LED will try to gather as much current as possible and burn out in the end. This is wrong i feel or at least not very accurate. In general, the current that a LED will try to obtain is not the maximum available by the battery, but the current as specified in the i-v curves of the LED.

Also, just to note, I am not asking if a resistor is required or not in case of a battery that provides enough voltage. Its more than clear to me that it is required.

Is my understanding correct?

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    \$\begingroup\$ Yes, the I vs V curve of the LED provides the needed information. But keep in mind that it can only give an "average" behavior. Just as people vary, so do LEDs. No two are exactly alike. So the I-V curve for two different LEDs from the exact same batch of LEDs will be different and therefore so will the current be different. Roughly speaking, an LED current will go up by a factor of 10 (an order of magnitude) for every increment of X volts. This might be small, like \$X=150\:\textrm{mV}\$. So two LEDs from the same batch might be a factor of 10 different in current at the same voltage. \$\endgroup\$ – jonk Sep 11 '17 at 8:36
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    \$\begingroup\$ Why do people use cement when they build a house? A stack of bricks should be perfectly stable if you align them just right. \$\endgroup\$ – Dmitry Grigoryev Sep 11 '17 at 9:05
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Assuming that we have a LED with forward voltage and current 2 volts and 10mA respectively.

Yes that is true but this is only a theoretical thruth.

In the real world a 2 V, 10 mA LED does not exist, at least not for long. Measure it later and the current will be different. LEDs have an exponential relation between Voltage and current. Only a small voltage change will cause a huge current change. Adding to this, the behavior of a LED is very temperature dependent. A small change in temperature will also result in a huge current change. Also not all LEDs are physically equal. Even a number of LEDs made at the same time in the same machine will be somewhat different. One will draw 20 mA at 2 V but another one 15 mA, 115 mA, 130 mA and 1 mA. All very unpredictable.

So this means that when kept as 2 V the current through the LED is highly unpredictable. It can be much lower and the LED would produce less light but the current will often be much higher as well. When too much current flows the LED will be damaged.

The solution is not to let the LED control the current but to use a resistor. There will still be some dependency on temperature etc. but the resulting current variations will be kept within safe limits.

A resistor or some other form of current limiting is required. In some designs the resistor might not be present but might be there in a hidden form like a small battery which has a high build-in series resistance.

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While a LED that takes 10mA at 2v will take 10mA from a 2v battery, that's only true at the instant we measure it.

A small change in temperature might change the forward voltage drop to 2.05v or 1.95v, and then the very steep voltage/current curve, what engineers call a 'low dynamic resistance', will cause it to draw very much less current or much more current respectively.

As the light output is proportional to current (more or less), it's the current that needs to be controlled. Because of the low dynamic resistance, variation between LEDs due to small manufacturing differences, and variation due to temperature, controlling the voltage is a very ineffective way of attempting to control the current in the face of these inevitable variations.

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