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A stronger current causes a higher heat development and the higher temperature produced again more charge carrier. Thus the current rise continues and the semiconductor still more is heating up.
An only slightly higher voltage means to lead a substantially higher current, that very fast can destruct an LED !!!
Therefore also a resistance must lie in each electric circuit with an LED, which limits the current.
Furthermore the developing warmth must be derived.
So that actually a river flows from 20 mA, the resistance must be measured accordingly.
R = U/I
During sufficient large battery voltage of e.g. 9 V two or more LED can be switched into row.
The conducting state voltage of the diodes add themselves, so that less voltage lie of the pre-resistor.
A red and a green LED e.g. have with a diode current of 20 mA and a voltage of 1,9 V + 2.2 V = 4.1 V. At the pre-resistor lies a voltage 9 V - 4.1 V = 4.9 V.
R = 4,9 V / 20 mA R=245 Ohm
We see in the graphic that a e.g. red LED begins to radiate, already with approx. 1,5V to easy and a blue one with approx. 2,7V.
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