LED calculations with non-DC voltage

A

Adam Funk

Guest
If you want to put non-DC voltage across an LED & its series resistor
(for example, the always non-negative but "bouncy" output of an
unsmoothed rectifier bridge), do you use the RMS or peak voltage (&
therefore current) to work out the value of the series resistor?


--
The history of the world is the history of a privileged few.
--- Henry Miller
 
On Fri, 18 Apr 2014 21:54:24 +0100, Adam Funk wrote:

If you want to put non-DC voltage across an LED & its series resistor
(for example, the always non-negative but "bouncy" output of an
unsmoothed rectifier bridge), do you use the RMS or peak voltage (&
therefore current) to work out the value of the series resistor?
Click to expand...

You read the diode data sheet carefully. I'm pretty sure that the
important parameter is heating due to current flow in the diode -- but
the heating happens at several time constants. Moreover, if I remember
correctly and at least in older LEDs, there was a significant issue with
high current densities tending to drag aluminum nuclii along with them
(or perhaps opposite the flow), creating a wearout mechanism.

If it's just thermal, and if it's fast enough that it all gets averaged
out, and if the pulsed current rating of the device isn't exceeded, then
the important parameter is average current (which is _not_ proportional
to average voltage), because the voltage across a forward-biased LED is
more or less constant, so the heating is more or less proportional to
current.

'course, if you're current limiting with a smallish resistor, you should
make sure that doesn't burn up either.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
 
On Fri, 18 Apr 2014 21:54:24 +0100, Adam Funk <a24061@ducksburg.com>
wrote:

If you want to put non-DC voltage across an LED & its series resistor
(for example, the always non-negative but "bouncy" output of an
unsmoothed rectifier bridge), do you use the RMS or peak voltage (&
therefore current) to work out the value of the series resistor?
Click to expand...

It's complicated. Spice it!

The average LED current is what will determine the visual brightness.

The RMS current heats the resistor.


--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com
 
"Adam Funk"
If you want to put non-DC voltage across an LED & its series resistor
(for example, the always non-negative but "bouncy" output of an
unsmoothed rectifier bridge), do you use the RMS or peak voltage (&
therefore current) to work out the value of the series resistor?
Click to expand...

** Neither - the average DC voltage as read on a normal meter what you
need. The rated current of a LED is the average value, which equates to the
DC voltage across the series resistor divided by its value.

The most reliable way is to first calculate a resistor value using the
difference between the expected LED voltage and the DC supply voltage. Then
fit such a resistor and test the DC voltage across it in normal use.

Amend the resistor value if necessary to get the average current level
needed.


.... Phil
 
On 2014-04-18, Adam Funk <a24061@ducksburg.com> wrote:
If you want to put non-DC voltage across an LED & its series resistor
(for example, the always non-negative but "bouncy" output of an
unsmoothed rectifier bridge), do you use the RMS or peak voltage (&
therefore current) to work out the value of the series resistor?
Click to expand...

That's a good starting place, often the led will be too bright at
rated current anyway.

--
umop apisdn


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---
 
On 2014-04-19, Phil Allison wrote:

"Adam Funk"

If you want to put non-DC voltage across an LED & its series resistor
(for example, the always non-negative but "bouncy" output of an
unsmoothed rectifier bridge), do you use the RMS or peak voltage (&
therefore current) to work out the value of the series resistor?


** Neither - the average DC voltage as read on a normal meter what you
need. The rated current of a LED is the average value, which equates to the
DC voltage across the series resistor divided by its value.

The most reliable way is to first calculate a resistor value using the
difference between the expected LED voltage and the DC supply voltage. Then
fit such a resistor and test the DC voltage across it in normal use.

Amend the resistor value if necessary to get the average current level
needed.
Click to expand...

Thanks very much (& to everyone else who replied) for the interesting
comments.


--
The kid's a hot prospect. He's got a good head for merchandising, an
agent who can take you downtown and one of the best urine samples I've
seen in a long time. [Dead Kennedys t-shirt]
 
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