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From: firstname.lastname@example.org (Tim Shoppa)
Subject: Re: Pulsing LED/High Current - Simple Circuit Question
Date: 8 Jan 2003 03:17:55 -0800
NNTP-Posting-Date: 8 Jan 2003 11:17:55 GMT
Mark Klinger wrote in message news:<email@example.com>...
> Currently, at DC, my LED has Vf of 1.7V and max current of 100mA.
> Since my supply is 13.8V, would it be correct that:
> 13.8V-1.7V-0.2V (Sat. of Transistor)/.1A = 120ohm resistor?
> Since I am able to pulse the LED at 2A with a very short pulse
> duration, would I then change to a 6 ohm resistor since:
> 13.8-1.7V-0.2V/2A = 5.95 ohm?
At 6A the Vce of the transistor will likely be bigger than 0.2V and
the Vf of the LED will will also be larger. But the voltage drop you
want across the resistor won't change much because the drop is dominated
by the 13.8V.
> This 2A is peak current I believe, which means that the short duration
> pulses (<1mS) should be significantly "brighter" than the DC 20mA
> intensity, correct?
I think you're talking about brightness as perceived by human eyes, in
which case whoever told you to pulse has never even looked at a LED data
sheet. Most modern LED's are at their most efficient at 15-25mA continuous;
if you're truly concerned about brightness you have to look at the data
sheet for your particular LED and choose the current that way.
DO NOT CHOOSE LED CURRENTS BASED ON OLD WIVES' TALES.
See, for a very quick summary, HP Application Note AN-1005 in
the section "DC Operation is Better than Pulsed Operation for Light
It is always
better to drive an LED device with a high DC current to obtain
the necessary light output to be viewed by a human observer
than to pulse drive the LED. Using a high peak current and a
low duty factor to pulse drive an LED device produces less time
average light output than by using a high DC drive current.
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