Re: {Collins} meter lamps



Hi Brian,

I agree with your analysis of the LED bulb, but are you sure you've computed
correctly the heat from the incandescent #47 bulb?  From the data sheets, a
#47 bulb draws .15A at 6.3V, which is 0.95W.  I'd guess at least 90% of that
power is heat, and probably more. If I'm right then the LED bulb dissipates
only 10-15% the heat of the incandescent bulb.

73,

Jim W8ZR

 

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From: Brian Harris [mailto:cosmophone@xxxxxxxxx] 
Sent: Thursday, July 14, 2011 8:41 AM
To: 4CX250B
Subject: Re: {Collins} meter lamps

 

Hi Jim,

Thanks for the note.  If the #47 replacements use a single LED with a
dropping resistor and if 20 mA is the average current the assembly consumes,
then about 0.066 Watts are consumed by the LED (this assumes 3.3V forward
drop of the LED), 25% (~0.016W) of which will become visible light and 75%
heat (~0.05W) will become heat.  Certainly this is less than the .19W heat
of the #47.  But add to that the heat in the dropping resistor (0.060W) and
you are at 0.11W total heat dissipation or a little over half the heat of
the #47.  This very incomplete analysis roughly agrees with analyses I've
done for other incandescent replacements with LEDs.

P.S.  I am now working a field application engineer for Avnet Electronics,
specializing in solid state lighting, thus the interest in the subject.  So
much for my retirement.

P.S.S  I am a great fan of solid state lighting and intend to adapt my new
ham shack in AR to solid state once we move there.

73,

Brian Harris, Field Application Engineer
cell 214-763-5977 (preferred)
office 214-553-4330
email cosmophone@xxxxxxxxx

 

 

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From: Jim Garland <4cx250b@xxxxxxxxxx>
To: Brian Harris <cosmophone@xxxxxxxxx>; collins <collins@xxxxxxxxxxxxx>
Sent: Wed, July 13, 2011 10:04:50 PM
Subject: RE: {Collins} meter lamps


> While unlikely, it is possible that the LED replacement for the #47 lamp
could
> actually have more unradiated heat to dissipate than the #47 lamp, without
> having the complete specifications for the LED replacement for the #47, we
can
> only speculate which solution would actually operate cooler in the
application.
> 
> Brian Harris, Illumination Engineer
> WA5UEK

Hi Brian,
    Typical LEDs draw about 20mA of current. For a 6.3V LED replacement
bulb, that corresponds to about. 0.13Watts (most of which is dissipated in
series resistors or diodes in the body of the bulb). 
    By contrast, a conventional #47 bulb draws 150 mA at 6.3V, which
corresponds to about 1W. Thus, even if all the power of an LED is dissipated
as heat, it would still radiate a negligible amount of heat, compared to an
incandescent bulb. That said, I agree that moisture and condensation is not
normally a problem with unlighted panel meters. I suppose it could be a
problem in very humid environments, subject to rapid temperature changes,
but not most places -- and certainly not in New Mexico, where we typically
have 10% or less relative humidity. In fact, my weather station recently
showed 90F outside temperature with 1% relative humidity!
73,
Jim W8ZR 







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