# Higher current and power to an LED?

I'm using an LED that is rated at (0.06W) current is 20mA and voltage is at 3V. I did some wrong calculations, and the resistors I'm using are not enough to limit the current to the diodes.

simulate this circuit – Schematic created using CircuitLab

The current is 0.024A now, can the LED work without blowing up? I'm maxing it with higher power would it manage?

• It won't blow up, but it may die young. I'd suggest 0.24A for "die a lot faster" and perhaps 24 amps (with plenty of volts to maintain an arc after internal parts expire) if you really want to have it "blow" nicely. ;^> – Ecnerwal Dec 10 '15 at 2:53
• There are part tolerances in your LED, resistor and meters, even small errors can shift the operating point of and LED in such a circuit. If the LED is never getting too hot to hold between your fingers it will likely live a long time (less than the rated but not much). Also note max current may already be more than operating current which you should be aiming for. – KalleMP Dec 10 '15 at 8:46

Unless you are grossly overloading the device, it will not blow immediately. However, as with all devices too much heat will kill it eventually. If you can add a bit of heatsinking (usually not possible with common LEDs) then it will be fine, otherwise derate the MTBF of the device accordingly.

If the LED current is 24 mA, and the recommended maximum current is 20 mA, you can probably expect a somewhat reduced life for the LED, but it is unlikely to be destroyed instantly.

If the values you placed are correct $Vled=3V, R1=200\Omega, V1=4.8V$ your actual current is:

$Iled=\frac{V1-Vled}{R1}=9mA$

$$R1 = \frac {V1-VfLED}{IfLED} = \frac {4.8V - 3V}{0.02A} = 90 \text{ ohms}$$