Hey guys. So I've been doing some research and I'm a touch curious as to whether I've interpreted the info correctly.
It's regarding direct driving an XHP50.2
I will break it down into points as I understand them and then if anything seems wrong please point out which and why.
1. An LED is an open circuit until it's forwards voltage is reached at which point it becomes a near perfect conductor with minimal resistance.
2. The current through an LED is largely dictated by the resistance in the circuit limiting current.
3. As voltage and current are intimately linked then as voltage through a fixed resistance increases so does current.
4. When Vs is higher than the Vf of an LED then there is an increased current, this then causes the LED to heat up as it tries to dissipate the extra power. This extra heat can cause damage to the LED causing it to fail.
5. An LED is not damaged by the voltage directly, it is damaged by the extra current (a byproduct of extra voltage) in the form of heat.
6. If there is sufficient thermal management (heat sinking), an LED can have a higher voltage supplied to it. This will create a higher current and increased lumen output, however the Lm/W increase at the higher current will start to diminish exponentially.
If this is true then am I correct in Thinking an xhp50.2 (Vf 5.6-6.2v)can be directly driven by 8.4v (most likely 7.8-8v due to voltage drop) provided it is prevent from reaching damaging temperature levels?
Also why do xhp50.2's struggle, when an xml2 which a Vf of 2.8v can withstand direct drive from 4.2V (both Vf and Vs are exactly half that of XHP50.2)
Thanks in advance
Thom.
It's regarding direct driving an XHP50.2
I will break it down into points as I understand them and then if anything seems wrong please point out which and why.
1. An LED is an open circuit until it's forwards voltage is reached at which point it becomes a near perfect conductor with minimal resistance.
2. The current through an LED is largely dictated by the resistance in the circuit limiting current.
3. As voltage and current are intimately linked then as voltage through a fixed resistance increases so does current.
4. When Vs is higher than the Vf of an LED then there is an increased current, this then causes the LED to heat up as it tries to dissipate the extra power. This extra heat can cause damage to the LED causing it to fail.
5. An LED is not damaged by the voltage directly, it is damaged by the extra current (a byproduct of extra voltage) in the form of heat.
6. If there is sufficient thermal management (heat sinking), an LED can have a higher voltage supplied to it. This will create a higher current and increased lumen output, however the Lm/W increase at the higher current will start to diminish exponentially.
If this is true then am I correct in Thinking an xhp50.2 (Vf 5.6-6.2v)can be directly driven by 8.4v (most likely 7.8-8v due to voltage drop) provided it is prevent from reaching damaging temperature levels?
Also why do xhp50.2's struggle, when an xml2 which a Vf of 2.8v can withstand direct drive from 4.2V (both Vf and Vs are exactly half that of XHP50.2)
Thanks in advance
Thom.