Scott Dattalo wrote: > On yet another project, I have the need to do this same trick with IR > LEDs. The data sheet explicitly states that you may pulse the LED with > a current larger than its nominal DC value. For example, here are the > specs of one led: > If = 20ma, Vf = 1.70V, Po = 0.5mW > If = 100ma, Vf = 2.10V, Po = 2.7mW > If you compare the efficiencies at the two drive levels, you'll notice > that the LED is about 15% more efficient at the lower drive level. > But it's possible to get more output power by applying "short" pulses. The way I read the specs above, the LED is intrinsically about 8% more efficient at the *higher* drive level. If you multiply the operating current by five, you would expect 2.5mW but you get 2.7mW in fact. Considering efficiency in a LED as being power out versus power in is not entirely fair. The light emitter in a LED is a constant voltage sink, and optical power output is essentially proportional to current. So, those figures actually support the use of the higher pulse current *until* you hit the power limit. > I might also add that even though the efficiency drops when the LED is > driven harder, this is not usually a practical problem. The effect is that it contains an internal, heat-wasting resistance. To the extent that the power dissipation from this resistor (which is proportional to IÓ) begins to exceed the input power to the "optical converter", you *are* on a losing streak using short, heavy pulses. If the peak power matters, you get more, but you have to derate the current at the lower duty cycles to limit heat build-up, so your average power *rapidly* drops. Presuming linear behaviour, if 80mA difference accounts for 400mV, the resistance is five ohms and the voltage sink or emission threshold, 1.60V. > BTW, one of our clients supposedly pulses the LEDs with 1 amp! I'm > not sure what Vf or Po are though. Vf should be a remarkable 6.6 volts. I wouldn't like to predict Po though - it may not be that much higher. -- Cheers, Paul B.