The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
From: firstname.lastname@example.org (Paul Mathews)
Subject: Re: Pulsing LED/High Current - Simple Circuit Question
Date: 8 Jan 2003 09:20:24 -0800
NNTP-Posting-Date: 8 Jan 2003 17:20:24 GMT
email@example.com (Tim Shoppa) wrote in message news:...
> Mark Klinger wrote in message news:<firstname.lastname@example.org>...
> > 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.
Here we go again with that old HP AN-1005 quote. This has been hashed
and rehashed. HP authors were WRONG to use the word 'always', but it
does make sense it you understand the context: They were referring to
the situation where you're trying to maximize LED power output,
regardless of any limitations on the power available or any need to
multiplex. The very same app note has a graph that shows why you
SOMETIMES you get more light output by pulsing. IF you have limited
power budget, you can take advantage of the fact that many LEDs have
an efficiency that rises with increasing current. For example,
suppose your device is battery operated, and you have 1 mA available.
Operating at 4% duty factor, with 25mA pulses will almost certainly
give you MORE light output than operating at 1mA DC.
You should also be aware that not everyone is using LEDs for
illumination. For data transmission and object sensing, pulsing of
LEDs is a requirement. So, please, let's put this 'old husbands tale'
about the 'old wives' tale' to rest.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup