Today's Message Index:
----------------------
1. 01:21 AM - Re: Soldering question--What am I doing wrong? (Greg@itmack)
2. 07:03 AM - Re: Backup battery - Lightspeed EI (Robert L. Nuckolls, III)
3. 07:29 AM - My bubble has burst . . . (Robert L. Nuckolls, III)
4. 07:41 AM - Re: Soldering question--What am I doing wrong? (Dan Beadle)
5. 08:09 AM - Attitude Gyro (Steve Glasgow)
6. 09:24 AM - Re: Re: Composite planes, Metallic paint, and Antennae (PTACKABURY@aol.com)
7. 09:57 AM - Re: Re: Backup battery - Lightspeed EI (J. Mcculley)
8. 10:41 AM - Re: My bubble has burst . . . (jerb)
9. 11:10 AM - Re: My bubble has burst . . . (Harold)
10. 01:15 PM - Re: My bubble has burst . . . (Robert L. Nuckolls, III)
11. 01:15 PM - Re: Re: Backup battery - Lightspeed EI (Robert L. Nuckolls, III)
12. 01:25 PM - Re: Re: Composite planes, Metallic paint, and Antennae (Robert L. Nuckolls, III)
13. 02:05 PM - Re: Re: Backup battery - Lightspeed EI (Brian Lloyd)
14. 02:07 PM - Re: Re: Backup battery - Lightspeed EI (Brian Lloyd)
15. 02:59 PM - Heated Pitot test (Brinker)
16. 03:28 PM - Re: Heated Pitot test (Brian Lloyd)
17. 03:52 PM - Zener Diodes and LEDs (Dennis Johnson)
18. 04:30 PM - Re: Zener Diodes and LEDs (Brian Lloyd)
19. 05:19 PM - Re: Zener Diodes and LEDs (Dan Beadle)
20. 05:27 PM - Re: Zener Diodes and LEDs (Richard E. Tasker)
21. 05:53 PM - Re: Zener Diodes and LEDs (Dan Beadle)
22. 06:34 PM - Followup to Zener/LED Question (Dennis Johnson)
23. 06:56 PM - Re: Followup to Zener/LED Question (Dan Beadle)
24. 07:29 PM - Re: Followup to Zener/LED Question (Richard E. Tasker)
25. 07:38 PM - Re: Followup to Zener/LED Question (Richard E. Tasker)
26. 08:02 PM - Re: Heated Pitot test (Charlie Kuss)
27. 08:02 PM - Re: Re: Backup battery - Lightspeed EI (Robert L. Nuckolls, III)
28. 08:42 PM - Re: Zener Diodes and LEDs (Malcolm Thomson)
29. 08:45 PM - Re: Re: Backup battery - Lightspeed EI (Brian Lloyd)
30. 09:02 PM - Re: Re: Backup battery - Lightspeed EI (Robert L. Nuckolls, III)
Message 1
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From: | "Greg@itmack" <greg@itmack.com> |
Subject: | Re: Soldering question--What am I doing wrong? |
--> AeroElectric-List message posted by: "Greg@itmack" <greg@itmack.com>
Sounds like your technique is okay but I think the iron may be a bit small.
I like about 60w to keep the heat up when it touches the board. Make sure
you're heating both the pad on the board and the LED. You don't need to
take the old solder off to re-do a solder joint.
Greg
Just finished assembling my lights.
> --> AeroElectric-List message posted by: Richard Scott
> <rscott@cascadeaccess.com>
>
> I'm building the Creative Air LED nav light kits, working on the LED's.
>
> So here's what I am doing.
>
> First I clean up the boards and LED leads with alcohol.
>
> Put them in the boards & bend the wires.
>
> Tin the iron. It's a 15 watt iron.
>
> Put the iron on the connection, then apply the solder. Bill says the
> LED's
> are heat sensitive and to only hold the iron on them a couple seconds.
>
> Half the time the solder doesn't penetrate the board, so I only have a
> cone
> of solder on the lead on one side of the board, the side I am soldering
> on.
>
> To try to fix it, I try to solder it again.
>
> If that doesn't work (it rarely does), I remove most of the solder using
> the iron and a clean cotton swab to absorb the solder & try again. Rarely
> works.
>
> Some of these I try a dozen times & still no joy.
>
> How can I get a good soldered joint with a cone of solder on both sides of
> the board? I have been told that that's what I need for a good
> joint. Maybe that guy was wrong?
>
> Richard Scott
> RV-9A
>
>
>
Message 2
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
At 11:20 PM 3/4/2006 -0500, you wrote:
>--> AeroElectric-List message posted by: Ken <klehman@albedo.net>
>
>Hi Mark
>You sound like you are probably well aware of this but "theoretically"
>I'd expect a lot less than 3 hours at 1 amp.
>www.power-sonic.com/ps-1229.pdf
>These specs indicate that at a 1 amp discharge rate you shouldn't expect
>more than 2 hours under ideal conditions or perhaps 1.5 hours for a cold
>battery. Apologies for nit-picking.
Ken
This is not nit-picking. It's considered systems integration with
understanding. We've discussed many times here on the List how
a battery's useful capacity varies with load. We've also discussed
the pitfalls of using the battery's labeled capacity without KNOWING
what loads were used to rate that capacity.
For example, batteries for bizjets are rated for specific situations
where the generators are dead and you're required to power certain
systems for 30 minutes. Since battery manufacturers are not
regulated to do anything, their ratings may be stated in any
lucid manner they choose. Most manufacturers of batteries
label their products at a 20 hour discharge rate. Here's
a recap of situations we've discussed in the past:
In this figure, we can see the published capacity vs. load
curves for a popular 33 a.h. battery from Panasonic . . .
http://www.aeroelectric.com/Pictures/Curves/LA1233_Panasonic.gif
Note that the 33 a.h. rating is met with a 1.65A discharge
after 20 hours. Increase the load to 13.2A and you'll get
just under two hours of service for a useful capacity of
23.7 a.h. A substantial drop from the label rating.
Here are curves produced here in our shop on a single 18 a.h.
SVLA battery . . .
http://www.aeroelectric.com/Pictures/Curves/1217_3n8_Discharge.pdf
Note that due to increased losses in the battery's internal impedance,
the useful capacity at 8A is 13.5 a.h. for an operating time of 1.7
hours. Useful capacity at 3 A is 16.8 a.h. for an operating time
of 5.6 hours.
Finally, here's a discharge curve for battery from Concord
for a bizjet with a label capacity of 37 a.h. . . .
http://www.aeroelectric.com/Pictures/Curves/Capacity_vs_Voltage.gif
Note that it produces 95% of rated capacity when discharged to
22v at 74A . . . this happens in 37/74 hours or 30 minutes.
If this battery were re-labeled with a 20 hour rate, the very
same battery could be called a 50+ a.h. battery.
I'm pleased that you picked up on this Ken, I missed it.
Bob . . .
>snip
>
> > used a PowerSonic 2.9 Ah battery (PS-1229). It's light and relatively
> cheap (I think about $25-30...you can Google it and get lots of hits). I
> plan to change the battery every two years to ensure that it's reasonably
> fresh. I know Klaus says 4.5 Ah but I thought that was overkill (unless
> you plan to be flying over some really rugged terrain/ocean with no
> alternates within an hour or two of flight). The 2.9 Ah batt is
> theoretically good for almost 3 hours of run time on one ignition
> (drawing ~ 1A at cruise RPM's). In the unlikely event that I ever find
> myself operating the engine soley on the backup battery, I will try to
> have it on the ground within an hour.
> >
>snip
>
>
>--
>
>
Bob . . .
< What is so wonderful about scientific truth...is that >
< the authority which determines whether there can be >
< debate or not does not reside in some fraternity of >
< scientists; nor is it divine. The authority rests >
< with experiment. >
< --Lawrence M. Krauss >
Message 3
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Subject: | My bubble has burst . . . |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
Some of you may recall that I did some studies on
energy content of the various alkaline AA cells some
time back and found that of all the brands I could
lay my hands on at the time, energy content was about
the same irrespective of price or advertising hype.
I was in Advance Auto Parts last week and they had a 24-pak
of Endurance brand alkaline AA cells for $4.88 . . . about
20 cents per cell. The best price in town. I purchased a
pack and just for grins, decided to run some on Brutus the
Battery Killer.
Tests in the past had produced data like:
http://www.aeroelectric.com/Pictures/Curves/El-Cheeso_Battery_Test.jpg
Two cells tested from my recent purchase look like this:
http://www.aeroelectric.com/Pictures/Curves/Endurance_AA.pdf
The 20-cent cell is decidedly lower in energy content. However,
in terms of cents per milliampere hour, still a reasonable deal
since the other bargain products are pushing 37 cents per cell.
I'll watch this product and buy another package in 6 months
to a year for a retest. The cells I have may have been produced
when the chef was having a bad day in the kitchen.
Bob . . .
< What is so wonderful about scientific truth...is that >
< the authority which determines whether there can be >
< debate or not does not reside in some fraternity of >
< scientists; nor is it divine. The authority rests >
< with experiment. >
< --Lawrence M. Krauss >
Message 4
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Subject: | Soldering question--What am I doing wrong? |
--> AeroElectric-List message posted by: "Dan Beadle" <Dan.Beadle@hq.InclineSoftworks.com>
Sounds like you could use a little solder flux. . You want electronic
flux - not plumbing flux. Digikey sells water soluble flux pens that
work great. That makes the heat transfer to the PCB a lot better.
You want to heat mostly the PCB trace, especially if it is more massive
than the LED lead. Ideally, they both reach flow temperature at the
same time. The solder should flow evenly on PCB and lead with a concave
shape - not convex or bulbous.
If flux doesn't do it, get a larger iron. Radio Shack makes a
temperature controlled iron for about $70. Set it to about 600 degrees
and you get just the right amount of heat at the junction.
Dan
-----Original Message-----
From: owner-aeroelectric-list-server@matronics.com
[mailto:owner-aeroelectric-list-server@matronics.com] On Behalf Of
Greg@itmack
Sent: Sunday, March 05, 2006 1:10 AM
Subject: Re: AeroElectric-List: Soldering question--What am I doing
wrong?
--> AeroElectric-List message posted by: "Greg@itmack" <greg@itmack.com>
Sounds like your technique is okay but I think the iron may be a bit
small.
I like about 60w to keep the heat up when it touches the board. Make
sure
you're heating both the pad on the board and the LED. You don't need to
take the old solder off to re-do a solder joint.
Greg
Just finished assembling my lights.
> --> AeroElectric-List message posted by: Richard Scott
> <rscott@cascadeaccess.com>
>
> I'm building the Creative Air LED nav light kits, working on the
LED's.
>
> So here's what I am doing.
>
> First I clean up the boards and LED leads with alcohol.
>
> Put them in the boards & bend the wires.
>
> Tin the iron. It's a 15 watt iron.
>
> Put the iron on the connection, then apply the solder. Bill says the
> LED's
> are heat sensitive and to only hold the iron on them a couple seconds.
>
> Half the time the solder doesn't penetrate the board, so I only have a
> cone
> of solder on the lead on one side of the board, the side I am
soldering
> on.
>
> To try to fix it, I try to solder it again.
>
> If that doesn't work (it rarely does), I remove most of the solder
using
> the iron and a clean cotton swab to absorb the solder & try again.
Rarely
> works.
>
> Some of these I try a dozen times & still no joy.
>
> How can I get a good soldered joint with a cone of solder on both
sides of
> the board? I have been told that that's what I need for a good
> joint. Maybe that guy was wrong?
>
> Richard Scott
> RV-9A
>
>
>
Message 5
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--> AeroElectric-List message posted by: "Steve Glasgow" <willfly@carolina.rr.com>
Artificial Horizon: RCA26AK-4-14V
This is an all electric Attitude Gyro that is un-lighted and has an 8 degree
panel tilt. It was completely overhauled, by Kelly MFG. CO. 1/10/2006, and
is under warranty till 01/10/07. It employs an electrically driven gyro
rotor with built-in inverter. Size: 3-3/8"x 7" Wt. 2.7 lbs. Mates with
MS3116E8-4S connector. $1,100 or best offer.
willfly@carolina.rr.com
Steve Glasgow
N123SG RV-8
Cappy's Toy
704-362-0005
Message 6
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Subject: | Re: Composite planes, Metallic paint, and Antennae |
--> AeroElectric-List message posted by: PTACKABURY@aol.com
Bob: Thanks for the second prompt response. After considering your
recommendation for an experiment, a logical approach indeed, I am reconsidering
my
paint choices. Since I have six internal antennas in my Lancair, both in the
upper aft fuse and each wing tip and since I would like each to operate as
well as possible, I now think using any metallic coatings may be a bad idea.
Why introduce even a 20% degradation in performance if it can be avoided? So
rather than metallic, I think I will use pearl (plastic additives rather than
metallic) to achieve the desired effect. It is a bit more difficult to
apply and touch up, but it shouldn't effect nav/com reception--should it? paul
Message 7
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: "J. Mcculley" <mcculleyja@starpower.net>
Bob,
Also, I don't want to be a nit-picker but the legend on the graph of
your shop discharge test cited below indicates that the black line is a
17AH battery at 3 Amp discharge versus the red line being an 18AH
battery at 8 Amp discharge. Is this a typo or am I missing understanding
something? I see the 1217,3n8 notation in the web address which I assume
means a 12 volt,17AH battery?
>Here are curves produced here in our shop on a single 18 a.h.
> SVLA battery . . .
>http://www.aeroelectric.com/Pictures/Curves/1217_3n8_Discharge.pdf
Robert L. Nuckolls, III wrote:
> --> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
>
> At 11:20 PM 3/4/2006 -0500, you wrote:
>
>
>>--> AeroElectric-List message posted by: Ken <klehman@albedo.net>
>>
>>Hi Mark
>>You sound like you are probably well aware of this but "theoretically"
>>I'd expect a lot less than 3 hours at 1 amp <SNIP>
>
> Bob . . .
Message 8
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Subject: | Re: My bubble has burst . . . |
--> AeroElectric-List message posted by: jerb <ulflyer@verizon.net>
Maybe you stumbled upon the reason there cheaper - inconsistent
output capacity - produced with lower tolerances, lower acceptance
levels (good bin v/s bad reject been), this all affects product cost
and shelf price.
jerb
At 09:26 AM 3/5/2006, you wrote:
>--> AeroElectric-List message posted by: "Robert L. Nuckolls, III"
><nuckollsr@cox.net>
>
>Some of you may recall that I did some studies on
>energy content of the various alkaline AA cells some
>time back and found that of all the brands I could
>lay my hands on at the time, energy content was about
>the same irrespective of price or advertising hype.
>
>I was in Advance Auto Parts last week and they had a 24-pak
>of Endurance brand alkaline AA cells for $4.88 . . . about
>20 cents per cell. The best price in town. I purchased a
>pack and just for grins, decided to run some on Brutus the
>Battery Killer.
>
>Tests in the past had produced data like:
>
>http://www.aeroelectric.com/Pictures/Curves/El-Cheeso_Battery_Test.jpg
>
>Two cells tested from my recent purchase look like this:
>
>http://www.aeroelectric.com/Pictures/Curves/Endurance_AA.pdf
>
>
>The 20-cent cell is decidedly lower in energy content. However,
>in terms of cents per milliampere hour, still a reasonable deal
>since the other bargain products are pushing 37 cents per cell.
>
>I'll watch this product and buy another package in 6 months
>to a year for a retest. The cells I have may have been produced
>when the chef was having a bad day in the kitchen.
>
>
> Bob . . .
>
>
> < What is so wonderful about scientific truth...is that >
> < the authority which determines whether there can be >
> < debate or not does not reside in some fraternity of >
> < scientists; nor is it divine. The authority rests >
> < with experiment. >
> < --Lawrence M. Krauss >
>
>
Message 9
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Subject: | Re: My bubble has burst . . . |
--> AeroElectric-List message posted by: "Harold" <kayce33@earthlink.net>
A real Gem, and a bit of truth
Message 10
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Subject: | Re: My bubble has burst . . . |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
At 12:35 PM 3/5/2006 -0600, you wrote:
>--> AeroElectric-List message posted by: jerb <ulflyer@verizon.net>
>
>Maybe you stumbled upon the reason there cheaper - inconsistent
>output capacity - produced with lower tolerances, lower acceptance
>levels (good bin v/s bad reject been), this all affects product cost
>and shelf price.
Sure . . . but within certain boundaries. A blister-pack of
6-32 screws on Walgreen's hardware rack won't be made of
recycled pot metal, they won't be 3-18 stainless either.
Within narrower confines, theres only so much you can do
to a bread recipe before the product is no longer attractive
to any customer. And what savings are to be realized by manipulating
materials? Many high quality products keep prices low by eliminating
no value added activities like TV advertising, too many handlers
in the distribution loop, etc. There's rarely much to be gained
by short-changing the materials or manufacturing processes.
Over the years, we're seeing fewer products that are graded like
the old Sears catalogs that offered good, better and best.
Semiconductor companies have largely abandoned the military
parts business where they simply screened industrial grade parts
for more stringent performance limits. The market was tiny
and the labor great. Now if we want "mil-spec" parts, we
buy a bunch and screen them ourselves. It's interesting that
of the parts we do screen, fallout is very low. Usually zero.
We could save a lot of money in certificated aircraft if we
abandoned our long standing lovefest with anything mil-spec'd.
That class of product is approaching extinction because ordinary
consumers are demanding equal or BETTER parts than the
military wanted . . . and suppliers are building them.
So the interesting question to explore was just how much might
one diddle with the recipe for an alkaline battery just to
have a lower priced niche in on the shelves? I first considered
the question in an article for Sport Aviation a few years ago:
http://aeroelectric.com/articles/AA_Bat_Test.pdf
In that study, I determined that over a wide range of prices,
there was little variation in the alkaline batteries to
be sampled at the time.
I've tested a number of other house brands for folks who
sent them to me and found similar results. This new kid
on the block seems to be the exception . . . although still
a better value than a CopperTop or BunnyBattery purchased
at retail.
Just a little pothole in the life of those interested in
such things . . . I happen to suffer that affliction.
Bob . . .
Message 11
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
At 12:53 PM 3/5/2006 -0500, you wrote:
>--> AeroElectric-List message posted by: "J. Mcculley"
><mcculleyja@starpower.net>
>
>Bob,
>
>Also, I don't want to be a nit-picker but the legend on the graph of
>your shop discharge test cited below indicates that the black line is a
>17AH battery at 3 Amp discharge versus the red line being an 18AH
>battery at 8 Amp discharge. Is this a typo or am I missing understanding
>something? I see the 1217,3n8 notation in the web address which I assume
>means a 12 volt,17AH battery?
Good catch. I need to fix that. Both graphs were on the same
Panasonic LC-RD1217 battery . . .
http://www.aeroelectric.com/Reference_Docs/Battery/Panasonic/lc-rd1217p.pdf
The 18 a.h. reference is a typo.
Bob . . .
Message 12
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Subject: | Re: Composite planes, Metallic paint, and Antennae |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
and Antennae
At 12:17 PM 3/5/2006 -0500, you wrote:
>--> AeroElectric-List message posted by: PTACKABURY@aol.com
>
>Bob: Thanks for the second prompt response. After considering your
>recommendation for an experiment, a logical approach indeed, I
>am reconsidering my
>paint choices. Since I have six internal antennas in my Lancair, both in
>the
>upper aft fuse and each wing tip and since I would like each to operate as
>well as possible, I now think using any metallic coatings may be a bad
>idea.
>Why introduce even a 20% degradation in performance if it can be avoided?
Whoa . . . I'm not suggesting for a moment that ANY paint you test
will produce any noticeable degradation. Odds are that it will produce
no effects you can observe. Further, if you DO conduct a test for a
particular
combination of coatings and find that it has no observable effect (or
perhaps
only barely twitches the readings) . . . that's GOOD data to share with
others.
>So rather than metallic, I think I will use pearl (plastic additives
>rather than
>metallic) to achieve the desired effect. It is a bit more difficult to
>apply and touch up, but it shouldn't effect nav/com reception--should
>it? paul
Probably won't . . . but there's no calibration on "probably". So much
of what passes for good advice is based on subjective observations.
A builder who reports great antenna performance and never talks to
stations more than 10 miles away might get similar results with a
wet string. On the other hand, a builder who is disappointed that
he can't hit an RCO out on the horizon of some leg of a favorite trip
might have an entirely different opinion on the same antenna. There's
nothing like numbers from the repeatable experiment to add real value
to one's advice.
Even if backyard experimenting is not your bag . . . risks
for using the paint you described are very low.
Bob . . .
< What is so wonderful about scientific truth...is that >
< the authority which determines whether there can be >
< debate or not does not reside in some fraternity of >
< scientists; nor is it divine. The authority rests >
< with experiment. >
< --Lawrence M. Krauss >
Message 13
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
J. Mcculley wrote:
> --> AeroElectric-List message posted by: "J. Mcculley" <mcculleyja@starpower.net>
>
> Bob,
>
> Also, I don't want to be a nit-picker but the legend on the graph of
> your shop discharge test cited below indicates that the black line is a
> 17AH battery at 3 Amp discharge versus the red line being an 18AH
> battery at 8 Amp discharge. Is this a typo or am I missing understanding
> something? I see the 1217,3n8 notation in the web address which I assume
> means a 12 volt,17AH battery?
The curves look correct to me. They are a bit odd in that normally one
plots voltage against time for a given discharge rate (constant current)
but you can always divide the amp-hour scale by amps to get the hours.
what might be more interesting would be to plot the endpoints (11V or
10.5V, whatever you select for "dead") and the amp-hours delivered for
each discharge rate. That would show you how much energy is available
from your battery at different discharge rates. You could also calculate
Peukert's exponent for your battery so you can figure out its remaining
capacity even at varying discharge rates.
--
Brian Lloyd 361 Catterline Way
brian-yak at lloyd dot com Folsom, CA 95630
+1.916.367.2131 (voice) +1.270.912.0788 (fax)
I fly because it releases my mind from the tyranny of petty things . . .
- Antoine de Saint-Exupery
Message 14
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
Robert L. Nuckolls, III wrote:
>> battery at 8 Amp discharge. Is this a typo or am I missing understanding
>> something? I see the 1217,3n8 notation in the web address which I assume
>> means a 12 volt,17AH battery?
>
> Good catch. I need to fix that. Both graphs were on the same
> Panasonic LC-RD1217 battery . . .
>
> http://www.aeroelectric.com/Reference_Docs/Battery/Panasonic/lc-rd1217p.pdf
>
> The 18 a.h. reference is a typo.
Is this a gelled electrolyte battery or an AGM battery? My guess from
looking at the voltage sag is that it is a gel-cell.
--
Brian Lloyd 361 Catterline Way
brian-yak at lloyd dot com Folsom, CA 95630
+1.916.367.2131 (voice) +1.270.912.0788 (fax)
I fly because it releases my mind from the tyranny of petty things . . .
- Antoine de Saint-Exupery
Message 15
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Subject: | Heated Pitot test |
--> AeroElectric-List message posted by: "Brinker" <brinker@cox-internet.com>
I bought a used 12v heated pitot removed from a Piper Seminole off
of ebay claimed to be servicable and tested. I have never hooked up one of
these. It has two terminals to hook up wires. But accually has 2 wires now
soldered to one terminal and one wire soldered to the second terminal but
appears a second wire was at one time soldered to the second terminal.
The only verifiable markings are a "p" and an "s" I believe designate
which tube is the pitot and static ports. But no marking for positive or
negitive.
I will not be installing this for a while since I have a ways to go
on building my plane but wish to at least test it before storing it for
later.
Randy
Message 16
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Subject: | Re: Heated Pitot test |
--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
Brinker wrote:
> --> AeroElectric-List message posted by: "Brinker" <brinker@cox-internet.com>
>
> I bought a used 12v heated pitot removed from a Piper Seminole off
> of ebay claimed to be servicable and tested. I have never hooked up one of
> these. It has two terminals to hook up wires. But accually has 2 wires now
> soldered to one terminal and one wire soldered to the second terminal but
> appears a second wire was at one time soldered to the second terminal.
> The only verifiable markings are a "p" and an "s" I believe designate
> which tube is the pitot and static ports. But no marking for positive or
> negitive.
> I will not be installing this for a while since I have a ways to go
> on building my plane but wish to at least test it before storing it for
> later.
Since the heating element is a resistor, there is no polarity. If it was
designed to work on either 12V or 24V there will be two elements. Use
them in series for 24V and in parallel for 12V.
And the latter is just a guess on my part.
--
Brian Lloyd 361 Catterline Way
brian-yak at lloyd dot com Folsom, CA 95630
+1.916.367.2131 (voice) +1.270.912.0788 (fax)
I fly because it releases my mind from the tyranny of petty things . . .
- Antoine de Saint-Exupery
Message 17
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Subject: | Zener Diodes and LEDs |
--> AeroElectric-List message posted by: "Dennis Johnson" <pinetownd@volcano.net>
I'm making LED instrument floodlights and LED lights for an annunciator panel.
I expect the system voltage to vary between about 13.8 volts when the engine
and alternator are running, down to about 12.4 volts running just off a lightly
loaded battery.
When connecting the LEDs in a series string of three, the resistance required
to get the voltage to the recommended level for the LEDs at the 13.8 system voltage
is different than when the system voltage is 12.4. Although it doesn't
make a huge difference in the amount of light that comes out of the LEDs, it is
noticeable.
I'm wondering (which often gets me into trouble) if I couldn't use zener diodes
instead of resistors to solve the problem of the light output varying with system
voltage. For example, if I have an array of three LEDs, each with a Vf
of 1.7 volts, could I use a 5.1 volt zener diode instead of resistors?
I don't have any experience with zener diodes, but the catalogs make it sound like
they produce a stable voltage output when operated within their amperage limitations,
sort of like a cheap power supply. Digikey sells them for only a
little more than resistors, so cost isn't an issue. Admittedly, the varying light
output isn't a huge problem, but if it's just as easy and cheap to use zeners,
why use resistors?
Thanks,
Dennis Johnson
Lancair Legacy #257, wiring in (slow) progress
Message 18
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Subject: | Re: Zener Diodes and LEDs |
--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
Dennis Johnson wrote:
> I'm wondering (which often gets me into trouble) if I couldn't use zener diodes
instead of resistors to solve the problem of the light output varying with
system voltage. For example, if I have an array of three LEDs, each with a Vf
of 1.7 volts, could I use a 5.1 volt zener diode instead of resistors?
Actually, that would make the problem worse. You would get a huge change
in light output with only a tiny change in voltage. Light output from an
LED is a function of the current through it and the current through an
LED is not linear with applied voltage. The resistor acts as a ballast
to make the current change more linear with voltage so that a 0.3V
change doesn't cause the LED to go from no output to full output.
What you really want to make is a constant current source. This would
make the light output independent of bus voltage.
Another way to stabilize light output is to regulate your bus voltage
down to something like 9V, using that to power your LEDs.
--
Brian Lloyd 361 Catterline Way
brian-yak at lloyd dot com Folsom, CA 95630
+1.916.367.2131 (voice) +1.270.912.0788 (fax)
I fly because it releases my mind from the tyranny of petty things . . .
- Antoine de Saint-Exupery
Message 19
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Subject: | Zener Diodes and LEDs |
--> AeroElectric-List message posted by: "Dan Beadle" <Dan.Beadle@hq.InclineSoftworks.com>
You can use a resistor and Zener diode to make a constant voltage source
for the LEDs as you say. This is not terribly efficient, but easy to
assemble.
You don't say what the current is that you need. Let's assume that it
is 100mA (.1A). If the system voltage is at 13.8V and you want to use
just fixed resistors, the voltage drop through the resistor would be
13.8-5.1 (target) or 8.7V. The R required would be Vd/I = R or 8.7/.1 R = 87 ohms...
So far, so good. But if the voltage in drops a bit, the same current
still gets dropped 8.7V by the R... causing the LEDs to dim a little.
You can get a constant 5.1V at the LEDs with a 5.1V Zener. Pick the
current you want and the lowest voltage you want. Let's stay with the
.1A and set the lower drop out at 10.1V for a Vd of 5V. To get a
current of .1A, you would need R to be 50 ohms. But at 13.8V, the drop
would not be enough. Zener to the rescue:
| -> LED1 -> LED2 -LED3 |
Hook up as Vin (14v) -> R ->| | ->
gnd
| -> Zener(Band-Cathode)---Zener Anode |
When the voltage goes above 10.1V, the Zener will conduct enough to
cause additional voltage drop through the R, holding the voltage at 5.1V
to the LEDs...
Finally, you have to look at the wattage for both the R and the Zener.
Continuing the example, the max drop across the R would be something
like 14.5-5.1 or 9.4V. Watts = V*V/R = 9.4v * 9.4 / 50 , so 1 Watt
would work in this example
For the Zener, the voltage drop is 5.1V, the current is the current
through the R (V/R or 9.4/50 = 188mA), less any current diverted to the
LEDs. Assuming they draw .1A, the current through the Zener would be
188-100, or 88mA. So the wattage would be V*I or 9.4*0.088 = .88W. So
use a 1W or larger Zener.
Hope this helps.
Dan
-----Original Message-----
From: owner-aeroelectric-list-server@matronics.com
[mailto:owner-aeroelectric-list-server@matronics.com] On Behalf Of Brian
Lloyd
Sent: Sunday, March 05, 2006 4:25 PM
Subject: Re: AeroElectric-List: Zener Diodes and LEDs
--> AeroElectric-List message posted by: Brian Lloyd
<brian-yak@lloyd.com>
Dennis Johnson wrote:
> I'm wondering (which often gets me into trouble) if I couldn't use
zener diodes instead of resistors to solve the problem of the light
output varying with system voltage. For example, if I have an array of
three LEDs, each with a Vf of 1.7 volts, could I use a 5.1 volt zener
diode instead of resistors?
Actually, that would make the problem worse. You would get a huge change
in light output with only a tiny change in voltage. Light output from an
LED is a function of the current through it and the current through an
LED is not linear with applied voltage. The resistor acts as a ballast
to make the current change more linear with voltage so that a 0.3V
change doesn't cause the LED to go from no output to full output.
What you really want to make is a constant current source. This would
make the light output independent of bus voltage.
Another way to stabilize light output is to regulate your bus voltage
down to something like 9V, using that to power your LEDs.
--
Brian Lloyd 361 Catterline Way
brian-yak at lloyd dot com Folsom, CA 95630
+1.916.367.2131 (voice) +1.270.912.0788 (fax)
I fly because it releases my mind from the tyranny of petty things . . .
- Antoine de Saint-Exupery
Message 20
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|
Subject: | Re: Zener Diodes and LEDs |
--> AeroElectric-List message posted by: "Richard E. Tasker" <retasker@optonline.net>
Short answer.. No. Longer answer below...
Resistors limit the current through the LEDs to whatever you want (20 mA
maybe). More source voltage just means a little more current and a
little brighter LED. The extra voltage is dropped across the resistor,
not the LEDs. LEDs actually have a voltage current characteristic
similar to zeners which limit the voltage across themselves.
If you string the LEDs and a zener together, they will not emit any
light until the voltage reaches whatever minimum is required across the
LEDs and the zener - 10.2 volts in your case - and then burn out as the
voltage tries to go higher..
If what you are suggesting is to place a 5.1V zener in parallel with the
LEDs and a resistor in series with the zener and LEDs, that will sorta
work as long as you don't mind wasting some extra power. Since LEDs
have slightly different voltage drops and since this changes somewhat
with temperature and since the same happens with a zener, using a simple
circuit like you are suggesting won't really work. You would have to
use a zener nd two resistors to make something that would work and you
would be throwing away a significant amount of power, since what you
would be doing is constructing a simple shunt regulator. Basically you
would use the zener and a resistor to create a shunt regulated voltage
source and then string the LEDs and an additional resistor across the zener.
A better way to accomplish what you want to do would be to use a three
terminal regulator such as the LM317 and hook it up as a current sink in
series with the LEDs. This requires nothing more than a single resistor
and the LM317 and what happens is that the regulator operates to keep
the same current through the LEDs regardless what the external power
does - keeping the LEDs the same brightness as the voltage varies from
12.4 (or less) to 13.8. You can hook the power source to the regulator
and then the regulator to the LEDs and then to ground or you can hook
the power source to the LEDs and then to the regulator and then to
ground - make no difference as long as they are connected in series.
Download the datasheet from: http://www.national.com/pf/LM/LM317L.html
(about 1/4 down the page) and then look at the bottom of page 17 of the
datasheet for a current regulator. You do not need the pot, just a
fixed resistor works fine calculated per the accompanying formula (Vref
is 1.25V). So to get 20 mA you would use I=1.25/R, or R = 1.25/.02 or
62.5 ohms (or the closest standard resistor value).
The part is available from Digikey or Mouser or any number of other
electronics supply houses.
Or you can just use a resistor and live with the slight change in
brightness.
Any questions, ask :-) .
Dick Tasker
Dennis Johnson wrote:
>--> AeroElectric-List message posted by: "Dennis Johnson" <pinetownd@volcano.net>
>
>I'm making LED instrument floodlights and LED lights for an annunciator panel.
I expect the system voltage to vary between about 13.8 volts when the engine
and alternator are running, down to about 12.4 volts running just off a lightly
loaded battery.
>
>When connecting the LEDs in a series string of three, the resistance required
to get the voltage to the recommended level for the LEDs at the 13.8 system voltage
is different than when the system voltage is 12.4. Although it doesn't
make a huge difference in the amount of light that comes out of the LEDs, it
is noticeable.
>
>I'm wondering (which often gets me into trouble) if I couldn't use zener diodes
instead of resistors to solve the problem of the light output varying with
system voltage. For example, if I have an array of three LEDs, each with a Vf
of 1.7 volts, could I use a 5.1 volt zener diode instead of resistors?
>
>I don't have any experience with zener diodes, but the catalogs make it sound
like they produce a stable voltage output when operated within their amperage
limitations, sort of like a cheap power supply. Digikey sells them for only a
little more than resistors, so cost isn't an issue. Admittedly, the varying
light output isn't a huge problem, but if it's just as easy and cheap to use zeners,
why use resistors?
>
>Thanks,
>Dennis Johnson
>Lancair Legacy #257, wiring in (slow) progress
>
>
>
>
>
>
>
>
>
--
Please Note:
No trees were destroyed in the sending of this message. We do concede, however,
that a significant number of electrons may have been temporarily inconvenienced.
--
Message 21
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Subject: | Zener Diodes and LEDs |
--> AeroElectric-List message posted by: "Dan Beadle" <Dan.Beadle@hq.InclineSoftworks.com>
Better solution than mine. As you point out, the LEDs need constant
current, not constant voltage. Selecting a the zener higher voltage,
with current limiting behind it to the LEDs would work fine. But, with
less efficiency than a current regulator configuration.
-----Original Message-----
From: owner-aeroelectric-list-server@matronics.com
[mailto:owner-aeroelectric-list-server@matronics.com] On Behalf Of
Richard E. Tasker
Sent: Sunday, March 05, 2006 5:25 PM
Subject: Re: AeroElectric-List: Zener Diodes and LEDs
--> AeroElectric-List message posted by: "Richard E. Tasker"
<retasker@optonline.net>
Short answer.. No. Longer answer below...
Resistors limit the current through the LEDs to whatever you want (20 mA
maybe). More source voltage just means a little more current and a
little brighter LED. The extra voltage is dropped across the resistor,
not the LEDs. LEDs actually have a voltage current characteristic
similar to zeners which limit the voltage across themselves.
If you string the LEDs and a zener together, they will not emit any
light until the voltage reaches whatever minimum is required across the
LEDs and the zener - 10.2 volts in your case - and then burn out as the
voltage tries to go higher..
If what you are suggesting is to place a 5.1V zener in parallel with the
LEDs and a resistor in series with the zener and LEDs, that will sorta
work as long as you don't mind wasting some extra power. Since LEDs
have slightly different voltage drops and since this changes somewhat
with temperature and since the same happens with a zener, using a simple
circuit like you are suggesting won't really work. You would have to
use a zener nd two resistors to make something that would work and you
would be throwing away a significant amount of power, since what you
would be doing is constructing a simple shunt regulator. Basically you
would use the zener and a resistor to create a shunt regulated voltage
source and then string the LEDs and an additional resistor across the
zener.
A better way to accomplish what you want to do would be to use a three
terminal regulator such as the LM317 and hook it up as a current sink in
series with the LEDs. This requires nothing more than a single resistor
and the LM317 and what happens is that the regulator operates to keep
the same current through the LEDs regardless what the external power
does - keeping the LEDs the same brightness as the voltage varies from
12.4 (or less) to 13.8. You can hook the power source to the regulator
and then the regulator to the LEDs and then to ground or you can hook
the power source to the LEDs and then to the regulator and then to
ground - make no difference as long as they are connected in series.
Download the datasheet from: http://www.national.com/pf/LM/LM317L.html
(about 1/4 down the page) and then look at the bottom of page 17 of the
datasheet for a current regulator. You do not need the pot, just a
fixed resistor works fine calculated per the accompanying formula (Vref
is 1.25V). So to get 20 mA you would use I=1.25/R, or R = 1.25/.02 or
62.5 ohms (or the closest standard resistor value).
The part is available from Digikey or Mouser or any number of other
electronics supply houses.
Or you can just use a resistor and live with the slight change in
brightness.
Any questions, ask :-) .
Dick Tasker
Dennis Johnson wrote:
>--> AeroElectric-List message posted by: "Dennis Johnson"
<pinetownd@volcano.net>
>
>I'm making LED instrument floodlights and LED lights for an annunciator
panel. I expect the system voltage to vary between about 13.8 volts
when the engine and alternator are running, down to about 12.4 volts
running just off a lightly loaded battery.
>
>When connecting the LEDs in a series string of three, the resistance
required to get the voltage to the recommended level for the LEDs at the
13.8 system voltage is different than when the system voltage is 12.4.
Although it doesn't make a huge difference in the amount of light that
comes out of the LEDs, it is noticeable.
>
>I'm wondering (which often gets me into trouble) if I couldn't use
zener diodes instead of resistors to solve the problem of the light
output varying with system voltage. For example, if I have an array of
three LEDs, each with a Vf of 1.7 volts, could I use a 5.1 volt zener
diode instead of resistors?
>
>I don't have any experience with zener diodes, but the catalogs make it
sound like they produce a stable voltage output when operated within
their amperage limitations, sort of like a cheap power supply. Digikey
sells them for only a little more than resistors, so cost isn't an
issue. Admittedly, the varying light output isn't a huge problem, but
if it's just as easy and cheap to use zeners, why use resistors?
>
>Thanks,
>Dennis Johnson
>Lancair Legacy #257, wiring in (slow) progress
>
>
>
>
>
>
>
>
>
--
Please Note:
No trees were destroyed in the sending of this message. We do concede,
however,
that a significant number of electrons may have been temporarily
inconvenienced.
--
Message 22
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Subject: | Followup to Zener/LED Question |
--> AeroElectric-List message posted by: "Dennis Johnson" <pinetownd@volcano.net>
I've already received a few excellent responses to my question about using zener
diodes to regulate power to an LED light. Thanks so much for your very thoughtful
answers. It's obvious that my understanding of LEDs is flawed. I have
a couple of basic questions to get me back on track and that will allow me to
fully understand your replies:
Assume an LED with a Vf of 2.1 volts and 20 mA rating:
1. If I hook the LED to a power supply fixed at 2.1 volts but with an unlimited
capacity to supply current, what would happen to the LED?
2. If I hook up the LED to a power supply set to 48 volts or so but with the current
limited to 20 mA, what would happen to the LED?
I have a handful of miscellaneous LEDs and I'd try the experiment myself if I had
a power supply.
It's a great day when I can learn something new, particularly if it's useful!
Thanks,
Dennis Johnson
Lancair Legacy --- currently revising my instrument panel LED lighting plan
Message 23
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Subject: | Followup to Zener/LED Question |
--> AeroElectric-List message posted by: "Dan Beadle" <Dan.Beadle@hq.InclineSoftworks.com>
Basically, you want a current limiter in the circuit. That is what the
resistor does. So your 48V/20ma example probably work (depending upon
the current spec of the LED). But if you hook to 2.1V supply, you may
get no output if the Vf is > 2.1V (high tolerance) or high current,
light, and maybe the magic smoke, if the Vf is < 2.1V
-----Original Message-----
From: owner-aeroelectric-list-server@matronics.com
[mailto:owner-aeroelectric-list-server@matronics.com] On Behalf Of
Dennis Johnson
Sent: Sunday, March 05, 2006 6:31 PM
Subject: AeroElectric-List: Followup to Zener/LED Question
--> AeroElectric-List message posted by: "Dennis Johnson"
<pinetownd@volcano.net>
I've already received a few excellent responses to my question about
using zener diodes to regulate power to an LED light. Thanks so much
for your very thoughtful answers. It's obvious that my understanding of
LEDs is flawed. I have a couple of basic questions to get me back on
track and that will allow me to fully understand your replies:
Assume an LED with a Vf of 2.1 volts and 20 mA rating:
1. If I hook the LED to a power supply fixed at 2.1 volts but with an
unlimited capacity to supply current, what would happen to the LED?
2. If I hook up the LED to a power supply set to 48 volts or so but
with the current limited to 20 mA, what would happen to the LED?
I have a handful of miscellaneous LEDs and I'd try the experiment myself
if I had a power supply.
It's a great day when I can learn something new, particularly if it's
useful!
Thanks,
Dennis Johnson
Lancair Legacy --- currently revising my instrument panel LED lighting
plan
Message 24
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|
Subject: | Re: Followup to Zener/LED Question |
--> AeroElectric-List message posted by: "Richard E. Tasker" <retasker@optonline.net>
Dennis Johnson wrote:
>--> AeroElectric-List message posted by: "Dennis Johnson" <pinetownd@volcano.net>
>
>I've already received a few excellent responses to my question about using zener
diodes to regulate power to an LED light. Thanks so much for your very thoughtful
answers. It's obvious that my understanding of LEDs is flawed. I have
a couple of basic questions to get me back on track and that will allow me to
fully understand your replies:
>
>Assume an LED with a Vf of 2.1 volts and 20 mA rating:
>
>1. If I hook the LED to a power supply fixed at 2.1 volts but with an unlimited
capacity to supply current, what would happen to the LED?
>
>
Well, it depends on whether the power supply is exactly the 2.1 volts of
the zener Vf. If it is exactly the 2.1 volts then the zener will light
up with the 20 mA brightness. If it is a little higher (2.2V) it will
light up much brighter since it will draw much more current. A little
higher still and - Toast!
If you plot voltage on the X axis and current on the Y axis, a resistor
would be a straight line pointing up and to the right. A zener would be
"L" shaped with a little bit of rounding at the apex of the "L"- very
little current until it reaches the zener voltage and then the current
would rise very fast for just a little more voltage. An LED is similar
except the apex is a little more rounded than the zener. Zeners and
LEDs MUST have something to limit the current or they will overheat and
die when the voltage rises a little above the Vf or zener voltage.
>2. If I hook up the LED to a power supply set to 48 volts or so but with the
current limited to 20 mA, what would happen to the LED?
>
>
It would light to whatever brightness that 20 mA creates. If you use a
resistor, the brightness would change if you varied the 48 volts. If
you used an active current limiter such as I described earlier, the
brightness would stay constant. If you plotted the active limiter as
above, you would get a straight horizontal line from about 3V to 38V
(the maximum voltage limit of the LM317). Below 3V or so you would get
essentially no current.
The only drawback to the active current limiter is that it doesn't allow
the LED to change brightness if you wanted to dim it with a normal
dimmer. You have to use one of the digital dimmers (pulse width
controlled) to dim it.
>I have a handful of miscellaneous LEDs and I'd try the experiment myself if I
had a power supply.
>
>It's a great day when I can learn something new, particularly if it's useful!
>
>Thanks,
>Dennis Johnson
>Lancair Legacy --- currently revising my instrument panel LED lighting plan
>
--
Please Note:
No trees were destroyed in the sending of this message. We do concede, however,
that a significant number of electrons may have been temporarily inconvenienced.
--
Message 25
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|
Subject: | Re: Followup to Zener/LED Question |
--> AeroElectric-List message posted by: "Richard E. Tasker" <retasker@optonline.net>
Ooops! Actually, in my previous post it would be a backwards "L" shape.
Dennis Johnson wrote:
>--> AeroElectric-List message posted by: "Dennis Johnson" <pinetownd@volcano.net>
>
>I've already received a few excellent responses to my question about using zener
diodes to regulate power to an LED light. Thanks so much for your very thoughtful
answers. It's obvious that my understanding of LEDs is flawed. I have
a couple of basic questions to get me back on track and that will allow me to
fully understand your replies:
>
>Assume an LED with a Vf of 2.1 volts and 20 mA rating:
>
>1. If I hook the LED to a power supply fixed at 2.1 volts but with an unlimited
capacity to supply current, what would happen to the LED?
>
>2. If I hook up the LED to a power supply set to 48 volts or so but with the
current limited to 20 mA, what would happen to the LED?
>
>I have a handful of miscellaneous LEDs and I'd try the experiment myself if I
had a power supply.
>
>It's a great day when I can learn something new, particularly if it's useful!
>
>Thanks,
>Dennis Johnson
>Lancair Legacy --- currently revising my instrument panel LED lighting plan
>
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Subject: | Re: Heated Pitot test |
--> AeroElectric-List message posted by: Charlie Kuss <chaztuna@adelphia.net>
Randy,
There are two heating elements in the Piper style heated pitot
tube. One is a 70 watt unit, the other is 100 watts. I've seen the
wires of both heating elements crimped together before. Cut the wires
apart, so that you can insure that BOTH elements work. Check for
continuity between both wires of each element. I believe my pitot
elements ohm out at about 3 ohms. An infinite reading will be that
the element being tested is bad.
Connect each elements 2 wires to a 12 volt battery (check one
element at a time) using 16 or 14 AWG wire. The Pitot should start to
warm up in your hand within 30 to 60 seconds. Do not apply power for
a period of time longer than is needed to verify that each element works.
If you have a bad element, they can be replaced. A number of
outfits will charge you a hefty price for these parts. RV-9A builder
Peter Laurence tracked down the "source" for these units. They are
available for a reasonable price from HotWatt. See Peter's email to
me below for more info.
Charlie Kuss
Charlie,
Here is the info. Heaters are made by
Hotwatt Http://www.hotwatt.com/cartridg.htm
Hotwatt part #s
For Pitot: 13A7025 70 W 12V
Static 13A7026 100W 14V
Piper part #s
464-356 for the 70W 12V
464-357 for the 100W 14V
These are called cartridge heaters. They are 3"X 3/16" and 4"X 3/16"
Check the ceramic plug where the wire enters the cartridge. If
there's any movement of the wire, it should be replaced.
Peter Laurence
>--> AeroElectric-List message posted by: "Brinker" <brinker@cox-internet.com>
>
> I bought a used 12v heated pitot removed from a Piper Seminole off
>of ebay claimed to be servicable and tested. I have never hooked up one of
>these. It has two terminals to hook up wires. But accually has 2 wires now
>soldered to one terminal and one wire soldered to the second terminal but
>appears a second wire was at one time soldered to the second terminal.
> The only verifiable markings are a "p" and an "s" I believe designate
>which tube is the pitot and static ports. But no marking for positive or
>negitive.
> I will not be installing this for a while since I have a ways to go
>on building my plane but wish to at least test it before storing it for
>later.
>
>Randy
>
>
Message 27
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
At 02:06 PM 3/5/2006 -0800, you wrote:
>--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
>
>Robert L. Nuckolls, III wrote:
>
> >> battery at 8 Amp discharge. Is this a typo or am I missing understanding
> >> something? I see the 1217,3n8 notation in the web address which I assume
> >> means a 12 volt,17AH battery?
> >
> > Good catch. I need to fix that. Both graphs were on the same
> > Panasonic LC-RD1217 battery . . .
> >
> > http://www.aeroelectric.com/Reference_Docs/Battery/Panasonic/lc-rd1217p.pdf
> >
> > The 18 a.h. reference is a typo.
>
>Is this a gelled electrolyte battery or an AGM battery? My guess from
>looking at the voltage sag is that it is a gel-cell.
Gel cells are almost non-existent in the wild. If you have a real
gel-cell it will probably say so in the literature and even on the
side of the battery. Gels are a modified flooded battery where
enough jello like stuff was added to make the normally liquid
stuff stay in place when the battery was inverted.
Globe Union were biggies in Gels about 25 years ago, Sonnenschein
makes them in Europe. The first spill proof battery B&C sold was
the Gates energy jelly-roll cells which were a true AGM, SVLA,
RG, Immmobilized Electrolyte, starved electrolyte battery . . . but
not a gel-cell. When the Gates product proved unworthy of flight
in aircraft, B&C switched to the Sonnenschein gels for a time but
those were replaced by the Genesis series (all the above types)
and ultimately STD'd onto a bunch of airplanes.
Gel-cells are popular for deep cycle applications like electric
wheelchairs. They seem to perform better than their dryer
cousins. I think B&C still has one gel-cell offering. Some folks
like Delmar Benjamin swear by them and he Bill aims to please.
The vast majority of lead-acid products are of the SVLA/RG
variety and you have to go out of your way to find a gel-cell.
Unfortunately, most of the storefronts who stock these modern
marvels refer to them as gel-cells and help perpetuate a huge
mis-information.
Bob . . .
< What is so wonderful about scientific truth...is that >
< the authority which determines whether there can be >
< debate or not does not reside in some fraternity of >
< scientists; nor is it divine. The authority rests >
< with experiment. >
< --Lawrence M. Krauss >
Message 28
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Subject: | Zener Diodes and LEDs |
--> AeroElectric-List message posted by: "Malcolm Thomson" <mthomson@showmeproductions.com>
Looks like this is what you need. See
http://www.periheliondesign.com/Vregflyer.htm
Malcolm.
-----Original Message-----
From: owner-aeroelectric-list-server@matronics.com
[mailto:owner-aeroelectric-list-server@matronics.com] On Behalf Of Dennis
Johnson
Sent: Sunday, March 05, 2006 4:49 PM
Subject: AeroElectric-List: Zener Diodes and LEDs
--> AeroElectric-List message posted by: "Dennis Johnson"
--> <pinetownd@volcano.net>
I'm making LED instrument floodlights and LED lights for an annunciator
panel. I expect the system voltage to vary between about 13.8 volts when
the engine and alternator are running, down to about 12.4 volts running just
off a lightly loaded battery.
When connecting the LEDs in a series string of three, the resistance
required to get the voltage to the recommended level for the LEDs at the
13.8 system voltage is different than when the system voltage is 12.4.
Although it doesn't make a huge difference in the amount of light that comes
out of the LEDs, it is noticeable.
I'm wondering (which often gets me into trouble) if I couldn't use zener
diodes instead of resistors to solve the problem of the light output
varying with system voltage. For example, if I have an array of three LEDs,
each with a Vf of 1.7 volts, could I use a 5.1 volt zener diode instead of
resistors?
I don't have any experience with zener diodes, but the catalogs make it
sound like they produce a stable voltage output when operated within their
amperage limitations, sort of like a cheap power supply. Digikey sells them
for only a little more than resistors, so cost isn't an issue. Admittedly,
the varying light output isn't a huge problem, but if it's just as easy and
cheap to use zeners, why use resistors?
Thanks,
Dennis Johnson
Lancair Legacy #257, wiring in (slow) progress
--
--
Message 29
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
Robert L. Nuckolls, III wrote:
> Gel cells are almost non-existent in the wild. If you have a real
> gel-cell it will probably say so in the literature and even on the
> side of the battery. Gels are a modified flooded battery where
> enough jello like stuff was added to make the normally liquid
> stuff stay in place when the battery was inverted.
Gel-cells are rather common in the boating community. They are clearly
delineated as such.
> Globe Union were biggies in Gels about 25 years ago, Sonnenschein
> makes them in Europe. The first spill proof battery B&C sold was
> the Gates energy jelly-roll cells which were a true AGM, SVLA,
> RG, Immmobilized Electrolyte, starved electrolyte battery . . . but
> not a gel-cell. When the Gates product proved unworthy of flight
> in aircraft, B&C switched to the Sonnenschein gels for a time but
> those were replaced by the Genesis series (all the above types)
> and ultimately STD'd onto a bunch of airplanes.
No argument there.
> Gel-cells are popular for deep cycle applications like electric
> wheelchairs. They seem to perform better than their dryer
> cousins. I think B&C still has one gel-cell offering. Some folks
> like Delmar Benjamin swear by them and he Bill aims to please.
I happen to like gels myself as their acceptable absorption charge
voltage range overlaps with the acceptable float charge voltage range.
You can get away with a single-voltage charging system (like an
aircraft's alternator).
Also, gels seem to tolerate extended partial discharge better than AGMs.
This makes them more attractive in power systems where the battery does
not get completely recharged every time.
But their higher internal resistance makes them poor starting batteries
unless they are oversized for the task.
> The vast majority of lead-acid products are of the SVLA/RG
Gel-cells are RG (recombinant gas) batteries also. I am not familiar
with the term SVLA tho.
RG - recombinant gas
VRLA - valve-regulated lead-acid
AGM - absorbed glass-mat
GELA - gelled-electrolyte lead-acid
BTW, VRLA or RG can refer to *either* GELA or AGM batteries.
I am attaching a really good treatise on VRLA batteries from Deka-Penn.
Clearly they are trying to sell their own product and it is a bit
simplistic in some areas but their stuff on charging and discharging is
really good. The best thing about it is that it addresses the
differences in operation of GELA and AGM batteries.
> variety and you have to go out of your way to find a gel-cell.
I disagree. I have no trouble finding gel-cells and use them where I
think they will perform better. They have higher internal resistance
than do AGMs but they are less prone to failure on overcharge than are
AGMs. They also tolerate partial or complete discharge better than AGMs.
> Unfortunately, most of the storefronts who stock these modern
> marvels refer to them as gel-cells and help perpetuate a huge
> mis-information.
Bob, I know the difference. I read the manufacturer's literature so I
know what I am buying. I don't trust what stores tell me because,
frankly, I usually know more about the topic than does the store. GELA
batteries are common enough that I would not pretend to know what is
inside the case unless the lit clearly states one way or another.
--
Brian Lloyd 361 Catterline Way
brian-yak at lloyd dot com Folsom, CA 95630
+1.916.367.2131 (voice) +1.270.912.0788 (fax)
I fly because it releases my mind from the tyranny of petty things . . .
- Antoine de Saint-Exupery
Message 30
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Subject: | Re: Backup battery - Lightspeed EI |
--> AeroElectric-List message posted by: "Robert L. Nuckolls, III" <nuckollsr@cox.net>
At 02:01 PM 3/5/2006 -0800, you wrote:
>--> AeroElectric-List message posted by: Brian Lloyd <brian-yak@lloyd.com>
>
>
>J. Mcculley wrote:
> > --> AeroElectric-List message posted by: "J. Mcculley"
> <mcculleyja@starpower.net>
> >
> > Bob,
> >
> > Also, I don't want to be a nit-picker but the legend on the graph of
> > your shop discharge test cited below indicates that the black line is a
> > 17AH battery at 3 Amp discharge versus the red line being an 18AH
> > battery at 8 Amp discharge. Is this a typo or am I missing understanding
> > something? I see the 1217,3n8 notation in the web address which I assume
> > means a 12 volt,17AH battery?
>
>The curves look correct to me. They are a bit odd in that normally one
>plots voltage against time for a given discharge rate (constant current)
>but you can always divide the amp-hour scale by amps to get the hours.
>
>what might be more interesting would be to plot the endpoints (11V or
>10.5V, whatever you select for "dead") and the amp-hours delivered for
>each discharge rate. That would show you how much energy is available
>from your battery at different discharge rates. You could also calculate
>Peukert's exponent for your battery so you can figure out its remaining
>capacity even at varying discharge rates.
Yeah . . . I've had some conversation with the folks at WestMountainRadio
about this. Obviously, the ampere-hour is NOT a measure of energy. When
I purchased my first battery tester from them, I suggested some enhancements
for the next release of their software. We engineers would really like to
see constant wattage and constant resistance discharge functions in addition
to the standard constant current function. Then too, as you've noted, a
volts vs. time and energy vs. time plots would be more meaningful along
with an box on the graph that displayed watt-seconds of energy at the
endpoint.
He seemed receptive. I'll write again and see where that might set on his
stove's burners.
I'd REALLY like to recommend his tool to others in the battery business
but most would like some features besides the arcane and poorly-descriptive
ampere-hour display. He's 100% of the way there with hardware, what he
needs now is a really cool GUI.
Bob . . .
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