Reduce 6 Vdc to 4.5 Vdc

[Andy K]

What would I need to reduce 6 Vdc -> 4.5 Vdc.

It would power some LED lights.

I don't know what current they draw, but they only last about 5 hrs. using 3 AAA alkaline batteries.

Thanks.

[Jim Thompson]

You could simply use a TLV431 and two resistors and drop 1.5V -OR- go
more exotic with an LDO -OR- even put in a switcher and be quite
efficient.

What do you want?

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

[Martin Brown]

You have asked to hit a voltage but LEDs require a *current* drive.

If all it is to do is power some LEDs then a resistor to drop 1.5v at
their nominal operating current is as simple as it gets. The ball park
would be about 200mA current so 8.2R or 10R @ 0.5W is about right.

You could do a switched mode constant current source which would be more
efficient. Chips exist to drive LEDs from a single cell.

--
Regards,
Martin Brown

[John Larkin]

On Thu, 27 Mar 2014 15:09:04 -0700 (PDT), Andy K
<andrewke...@gmail.com> wrote:

A AAA battery is good for roughly 1 amp-hour, so your load is very
roughly 200 mA. The simplest thing would be a series resistor, around
7 ohms, 1/2 watt. Try 5, 7, 10 ohms and see how you like the
brightness.

Fancier circuits could give more constant current and squeeze out more
battery life.


--

John Larkin Highland Technology, Inc

jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com

[whit3rd]

On Thursday, March 27, 2014 3:09:04 PM UTC-7, Andy K wrote:
> What would I need to reduce 6 Vdc -> 4.5 Vdc.

...[to] power some LED lights.
> ...they only last about 5 hrs. using 3 AAA alkaline batteries.

If you want more time, you can throttle the current to something
lower (add a series resistor). Slightly dimmer, much greater light time.

If this is a flashlight, some (inexpensive) ones need the AAA batteries'
internal resistance, so adding batteries (like, to get to 6V) will vastly
change the operating point of the LEDs (and maybe burn them out).
If you add batteries AND add a resistor, the four batteries will go flat
as quickly as three did (the extra voltage goes to waste heat).

Simple way: add resistor in series, that lowers the brightness but
increases battery longevity.

Complicated way: put a current-limiting drive circuit in (this has the
advantage of allowing AA or D cells instead of AAA, and various
chemistries instead of just 'alkaline'). It'll stay full brightness until
the batteries fail, then go dim FAST.

Most effective way: use a switchmode converter (which allows lower
battery count as well as higher battery count, and doesn't waste
most of the energy). That's complicated, though, and these days
mainly requires lotsa resources (print up a PC board lately?).

[Tim Williams]

"John Larkin" <jla...@highlandtechnology.com> wrote in message
news:e0b9j9hut2jsaftkl...@4ax.com...

> Fancier circuits could give more constant current and squeeze out more
> battery life.

Less, actually. But the light level is more stable.

Tim

--
Seven Transistor Labs
Electrical Engineering Consultation
Website: http://seventransistorlabs.com

[RobertMacy]

On Thu, 27 Mar 2014 15:09:04 -0700, Andy K <andrewke...@gmail.com>
wrote:

Hmmm...if you have a 3 AA cell LED light string, one of those stick on
mounts. Instead of powering with 4.5 worth of batteries you want to power
it with a 6 Vdc supply, I don't know you might get away with it because it
would simply 'look' like a larger voltage, brand new battery.

Siimple answer is try it. If the LEDs get super bright, stop it. If they
stay simply bright like new batteries and nothing explodes, go for it.

Need more information.

[Andy K]

I want to eliminate the rapid battery usage of some bicycle lights.

More details on what I would be powering using a

6 Volt - 4.5 Ah - UB645 - AGM Battery

Each of these uses 4.5 volts.

One light has 8 leds.

2nd light has 12 leds.

Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps

Thanks for your help gentlemen.

Andy

[k...@attt.bizz]

On Thu, 27 Mar 2014 19:15:09 -0500, "Tim Williams"
<tmor...@charter.net> wrote:

>"John Larkin" <jla...@highlandtechnology.com> wrote in message
>news:e0b9j9hut2jsaftkl...@4ax.com...
>> Fancier circuits could give more constant current and squeeze out more
>> battery life.
>
>Less, actually. But the light level is more stable.

Depends on how "fancy" you want to go. A "Joule Thief" will squeeze
more light out of a battery than a resistor ballast will.

[John Larkin]

On Thu, 27 Mar 2014 19:15:09 -0500, "Tim Williams"
<tmor...@charter.net> wrote:

>"John Larkin" <jla...@highlandtechnology.com> wrote in message
>news:e0b9j9hut2jsaftkl...@4ax.com...
>> Fancier circuits could give more constant current and squeeze out more
>> battery life.
>
>Less, actually. But the light level is more stable.
>
>Tim

I was thinking about a boost/buck switcher.

[Wayne Chirnside]

Two silicon diodes in series?

[Martin Riddle]

The battery has a resonably flat discharge curve.
http://www.1000bulbs.com/pdf/ubD5733-spec.pdf
Assuming your load is ~1.25 amps.

I was going to say LM2575, but the Vsat gives you little headroom.
I think a small DC-DC converter will work instead..
<http://www.mouser.com/ds/2/281/okr-t3-w12-21655.pdf>
14V in so you can charge while the lights are on.

<http://www.mouser.com/Power/DC-DC-Converters/_/N-brvxe?P=1z0wdj3Z1yxt698&Keyword=dc-dc&FS=True>
For $6 its worth try. I dont know what the LED current is so you could
power 1 LED off one and the other LED and Bulb off a second one.

Cheers

[RobertMacy]

very good! unaffected by load. or use a 2N3055 wired as a 'diode
multiplier' if gets too hot.

[David Eather]

On Fri, 28 Mar 2014 10:18:09 +1000, Andy K <andrewke...@gmail.com>

This:

http://www.ebay.com/itm/LM2596-DC-DC-Step-down-Adjustable-CC-CV-Power-Supply-Module-Converter-LED-driver-/400440723996?pt=LH_DefaultDomain_0&hash=item5d3c208a1c

should do the trick for all the above. It is switch mode so it is
efficient. You can set it for 4.5 volts and use it that way but you will
get better, longer lasting results if you use it in constant current mode.

To do this you need a multimeter, crank the output voltage up as hight as
possible (this is okay because the current will be limited - now forget
about the voltage control - you will not use it again). Use it in constant
current mode. Now set the current limit to the minimum or if that is not
marked set the current limit to about half way. Set your multimeter to the
10Amp current range and use it to short out the output of the module.
Power the module and adjust the current limit control to .65 amps for the
3watt led and 240 ma (or less) for the 12 led unit and 160 ma for the 8
led. You may have to change current range to get a good reading - make
sure the current limit is set to a minimum before you do this.

For best results (efficiency and longest run time) bypass and resistors
and electronics in the lamps and drive the leds directly from the modules.

[Jim Thompson]

>:-}

[Andy K]

So many ideas gentlemen, but I am overwhelmed.

I am a "recovering perfectionist".

Andy

[Tim Williams]

<k...@attt.bizz> wrote in message
news:uag9j9t0uodfplv3d...@4ax.com...

> Depends on how "fancy" you want to go. A "Joule Thief" will squeeze
> more light out of a battery than a resistor ballast will.

Depends on the voltage drop and discharge curve, of course. They'll drop
off pretty much the same way, at least for a conventional (resistor
biased) 'thief'.

Point was, a proper CCS will drain the battery much more sharply due to
the increasing current draw.

[miso]

>
> More details on what I would be powering using a
>
> 6 Volt - 4.5 Ah - UB645 - AGM Battery
>
> Each of these uses 4.5 volts.
>
> One light has 8 leds.
>
> 2nd light has 12 leds.
>
> Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65
> amps
>
> Thanks for your help gentlemen.
>
> Andy

The problem with your plan is these devices that you wish to power are
already engineered to run from a specific voltage. So unless you are going
to open these devices up and re-engineer them, your best bet is to buck the
AGM down to 4.5VDC and just power these devices directly.

To elaborate, your lights are probably not just a bunch of LEDs, but LEDs
with some control circuitry.

[Phil Allison]

"Andy K"

>
> What would I need to reduce 6 Vdc -> 4.5 Vdc.
>
> It would power some LED lights.
>

** Last time you asked this question -

you had three 4.5V LED lights and a 12V gel battery of 5AH.

So wiring the lights in series was a perfect solution.

What happened ?


.... Phil

[John Fields]

---
4.5 X 3 = 13.5? ;)

[Phil Allison]

"John Fields"
> "Phil Allison"

>
>>"Andy K"
>>>
>>> What would I need to reduce 6 Vdc -> 4.5 Vdc.
>>>
>>> It would power some LED lights.
>>>
>>
>>** Last time you asked this question -
>>
>> you had three 4.5V LED lights and a 12V gel battery of 5AH.
>>
>> So wiring the lights in series was a perfect solution.
>>
>> What happened ?
>>

> ---
> 4.5 X 3 = 13.5? ;)

** Really ?

What is the full charge voltage of a nominal 12V SLA battery ?

What is the end point voltage of same ?

What is the end point of 9 x alkaline cells ?

BTW:

I can see the wink....


.... Phil

[Martin Brown]

Current through the chain would be limited by the weakest link?

Given the OPs question he might well be better off with 3x D NiMH cells
in a piece of plastic drainpipe and avoiding the electronics altogether.

--
Regards,
Martin Brown

[pi]

Martin Brown 's

>Given the OPs question he might well be better off with 3x D NiMH cells
>in a piece of plastic drainpipe and avoiding the electronics altogether.

YEs tHem oLD GaS LIGhtS wOrk REaLlY WelL

[John Fields]

On Fri, 28 Mar 2014 20:56:17 +1100, "Phil Allison"
<phi...@tpg.com.au> wrote:

>
>"John Fields"
>> "Phil Allison"
>>
>>>"Andy K"
>>>>
>>>> What would I need to reduce 6 Vdc -> 4.5 Vdc.
>>>>
>>>> It would power some LED lights.
>>>>
>>>
>>>** Last time you asked this question -
>>>
>>> you had three 4.5V LED lights and a 12V gel battery of 5AH.
>>>
>>> So wiring the lights in series was a perfect solution.
>>>
>>> What happened ?
>>>
>
>> ---
>> 4.5 X 3 = 13.5? ;)
>
>** Really ?
>
>What is the full charge voltage of a nominal 12V SLA battery ?

---
About 13.8V
---

>What is the end point voltage of same ?

---
About 10.5V
---

>What is the end point of 9 x alkaline cells ?

---
About 9V
---

[k...@attt.bizz]

On Thu, 27 Mar 2014 23:09:49 -0500, "Tim Williams"
<tmor...@charter.net> wrote:

><k...@attt.bizz> wrote in message
>news:uag9j9t0uodfplv3d...@4ax.com...
>> Depends on how "fancy" you want to go. A "Joule Thief" will squeeze
>> more light out of a battery than a resistor ballast will.
>
>Depends on the voltage drop and discharge curve, of course. They'll drop
>off pretty much the same way, at least for a conventional (resistor
>biased) 'thief'.

However, a switch mode current source can save power and get useful
energy from the battery below where a resistor will work..

>Point was, a proper CCS will drain the battery much more sharply due to
>the increasing current draw.

Or less, at the top end. It depends on the requirements.

[Martin Brown]

On 28/03/2014 16:36, k...@attt.bizz wrote:
> On Thu, 27 Mar 2014 23:09:49 -0500, "Tim Williams"
> <tmor...@charter.net> wrote:
>
>> <k...@attt.bizz> wrote in message
>> news:uag9j9t0uodfplv3d...@4ax.com...
>>> Depends on how "fancy" you want to go. A "Joule Thief" will squeeze
>>> more light out of a battery than a resistor ballast will.
>>
>> Depends on the voltage drop and discharge curve, of course. They'll drop
>> off pretty much the same way, at least for a conventional (resistor
>> biased) 'thief'.
>
> However, a switch mode current source can save power and get useful
> energy from the battery below where a resistor will work..

True. But it can also take rechargeable batteries into an abusive regime
where it fails to recharge or suffers severely shortened life.

--
Regards,
Martin Brown

[k...@attt.bizz]

Or not. It's all about requirements.

[gyans...@gmail.com]

On Friday, March 28, 2014 11:09:04 AM UTC+13, Andy K wrote:
> What would I need to reduce 6 Vdc -> 4.5 Vdc.
>
>
>
> It would power some LED lights.
>
>
>

> I don't know what current they draw, but they only last about 5 hrs. using 3 AAA alkaline batteries.
>
>
>
> Thanks.

a zener would do or maybe two diodes drop

[meow...@care2.com]

On Thursday, March 27, 2014 10:09:04 PM UTC, Andy K wrote:

> What would I need to reduce 6 Vdc -> 4.5 Vdc.
>
> It would power some LED lights.
>
> I don't know what current they draw, but they only last about 5 hrs. using 3 AAA alkaline batteries.
>
> Thanks.

Not making your life any easier are they. The easiest answer is one resistor feeding each lamp, so with 3 lights you use 3 resistors.

> One light has 8 leds.

8 leds @ 20mA ea = 160mA. So you can use a 12 ohm resistor there, power rating 0.5w or more. It will get hot. Or small Rs of 5.6 & 6.8R in series

> 2nd light has 12 leds.

240mA so 8.6 ohms at least 0.5 watt - or small 3.3 ohm and 4.7ohm resistors in series

> Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps

2.2 ohms, at least 1 watt. Or 4x 8.6 ohm small Rs in parallel.


If you pick resistors of say double those power ratings they wont get so hot.

If anyone wonders I've calculated these for 6.5v.


NT

[whit3rd]

> 4.5 X 3 = 13.5? ;)

Well, 13.5 ~= 12. That "4.5V" isn't the measured-under-load battery
terminal voltage, just the NOMINAL value. That's the top of a big
error bar.

[David Eather]

Nah, the solution doesn't work at a practical level - but we know that now
only because we now have more information. 3 led lights equals 1 led torch
with 8 leds, plus 1 led torch with 12 leds, plus 1 led torch with a 3 watt
led.

[Jasen Betts]

a "6v" AGM is 7.2V when fully charged.

6V is likely to destroy the LEDs.

a series resistor is a better simple solution.

--
umop apisdn


--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---

[meow...@care2.com]

On Sunday, March 30, 2014 7:02:13 AM UTC+1, Jasen Betts wrote:

> a "6v" AGM is 7.2V when fully charged.

14.4v is charging voltage for flooded 12v lead acids, and they'll charge at 13.6v too. I dont believe youre going to get 7.2v from the battery itself.


NT

[bloggs.fred...@gmail.com]

On Thursday, March 27, 2014 8:18:09 PM UTC-4, Andy K wrote:
> On Thursday, March 27, 2014 6:40:56 PM UTC-5, whit3rd wrote:
>

> > On Thursday, March 27, 2014 3:09:04 PM UTC-7, Andy K wrote:
>
> >
>
> > > What would I need to reduce 6 Vdc -> 4.5 Vdc.
>
> >
>

> > ...[to] power some LED lights.
>
> >
>
> > > ...they only last about 5 hrs. using 3 AAA alkaline batteries.
>
> >
>
> >
>
> >
>
> > If you want more time, you can throttle the current to something
>
> >
>
> > lower (add a series resistor). Slightly dimmer, much greater light time.
>
> >
>
> >
>
> >
>
> > If this is a flashlight, some (inexpensive) ones need the AAA batteries'
>
> >
>
> > internal resistance, so adding batteries (like, to get to 6V) will vastly
>
> >
>
> > change the operating point of the LEDs (and maybe burn them out).
>
> >
>
> > If you add batteries AND add a resistor, the four batteries will go flat
>
> >
>
> > as quickly as three did (the extra voltage goes to waste heat).
>
> >
>
> >
>
> >
>
> > Simple way: add resistor in series, that lowers the brightness but
>
> >
>
> > increases battery longevity.
>
> >
>
> >
>
> >
>
> > Complicated way: put a current-limiting drive circuit in (this has the
>
> >
>
> > advantage of allowing AA or D cells instead of AAA, and various
>
> >
>
> > chemistries instead of just 'alkaline'). It'll stay full brightness until
>
> >
>
> > the batteries fail, then go dim FAST.
>
> >
>
> >
>
> >
>
> > Most effective way: use a switchmode converter (which allows lower
>
> >
>
> > battery count as well as higher battery count, and doesn't waste
>
> >
>
> > most of the energy). That's complicated, though, and these days
>
> >
>
> > mainly requires lotsa resources (print up a PC board lately?).
>
>
>
> I want to eliminate the rapid battery usage of some bicycle lights.
>
>
>

> More details on what I would be powering using a
>
>
>
> 6 Volt - 4.5 Ah - UB645 - AGM Battery
>
>
>
> Each of these uses 4.5 volts.
>
>
>

> One light has 8 leds.
>
>
>

> 2nd light has 12 leds.
>
>
>

> Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps
>
>
>

> Thanks for your help gentlemen.
>
>
>
> Andy

The apparent battery capacity increases dramatically at lower drain rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.

[Maynard A. Philbrook Jr.]

In article <e16a6628-57c5-4092...@googlegroups.com>,
bloggs.fred...@gmail.com says...

> > 2nd light has 12 leds.
> >
> >
> >
> > Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps
> >
> >
> >
> > Thanks for your help gentlemen.
> >
> >
> >
> > Andy
>
>
> The apparent battery capacity increases dramatically at lower drain rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.
>
>

Maybe I missed something but for what ever reason, I don't think the
math is working in your favor?

Jamie

[meow...@care2.com]

On Sunday, March 30, 2014 10:28:13 PM UTC+1, Maynard A. Philbrook Jr. wrote:
> In article <e16a6628-57c5-4092...@googlegroups.com>,
> bloggs says...

> > > > The apparent battery capacity increases dramatically at lower drain rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.

> > > Maybe I missed something but for what ever reason, I don't think the
> math is working in your favor?

No :) That would reduce life due to convertor losses.


NT

[bloggs.fred...@gmail.com]

Prove it.

[Phil Allison]

<bloggs.fred...@gmail.com>

>
> The apparent battery capacity increases dramatically at lower drain rates.

** Impossible to reduce the battery drain and have the same power.

> So if you want to increase battery life beyond belief,
> the thing to do is wire the LEDs in series

** Also impossible, cos they all have different current drains.

> and then use a "boost" converter to step-up the 6V voltage to 24-30 V as
> required.

** Insanity.

Just adds conversion loses to the battery drain.


.... Phil

[John Fields]

On Mon, 31 Mar 2014 12:23:40 +1100, "Phil Allison"
<phi...@tpg.com.au> wrote:

>
><bloggs.fred...@gmail.com>
>
>>
>> The apparent battery capacity increases dramatically at lower drain rates.

---
True
---

>** Impossible to reduce the battery drain and have the same power.
>
>
>> So if you want to increase battery life beyond belief,
>> the thing to do is wire the LEDs in series
>
>** Also impossible, cos they all have different current drains.

---
Depends on the array, yes?
---

>> and then use a "boost" converter to step-up the 6V voltage to 24-30 V as
>> required.
>
>** Insanity.

---
Actually, it's quite practicable and, depending on the array,
practical.
---

> Just adds conversion loses to the battery drain.

---
But if the conversion losses are less than the ballast drain and the
ballast resistor(s) can be removed, then that's win win.
---

>.... Phil

---
Sure, but the loss due to the LED ballasts (the current limiting
resistors) will go down substantially as the number of series LEDs
increases.

Just for grins, let's say we have 100 LEDs with a Vf of 2.2V with
20mA through them, and a 6V source driving them.

Using a resistor to drop the 6V from the source to the 2.2V each LED
load wants to see leaves us with a 190 ohm resistor in series with
each LED.

The E24 series provides 180 and 200 ohm +/- 5% resistors, so -
erring on the low side of If - we choose 200 ohms and, assuming a
negligible change in Vf at the new current, that current becomes:

Vcc - Vf(led)
If = --------------- = 0.019A = 19mA,
Rs

the equivalent resistance of each LED will be:

Vf 2.2V
R = ---- = ------ = 116 ohms,
If .019A
R LED
and each R/LED string will look like this: +6V--[200R]--[116R]--0V.
---19mA-->

The power dissipated in the LED will be:

P = I²R = 0.019A * 116R ~ 0.042 watts

and, in the resistor, 0.072 watts.

For 100 R/LED series strings connected in parallel then, the total
power into the array will be 11.4 watts and the supply current will
rise to 1900 mA.


Using Fred's proposition, but raising Vcc to 48V - which, as I
understand it, is pretty close to what UL calls "non-shocking"-
would allow us to connect Vcc/Vled = 48V/2.2V = about 22, LEDs in
series across Vcc with no ballast.

For a 100 LED array, then, we could use 5 parallel strings of 20
series connected LEDs with a single ballast resistor in each string,
which we would select to allow, say, 20mA through each string with a
drop of:

Rs = Vcc - (n Vf(led) = 48V - 20 * 2.2V = 4.0V

across the R = 4.0V/0.02A = 200 ohm ballast resistor.

In this case, the per-string power dissipated by the LEDs will be:

Pd = 0.02A * 44V = 0.88 watts,

and for the resistor:

Pd = 0.02A * 4V = 0.08 watts, for a total dissipation of 0.96

watts per string, and 4.8 watts for the array.


In order to get from 6V to 48V we can use a boost converter and,
assuming an efficiency of 80%, that 4.8 watts into the 48V array
will cost 6 watts out of the 6V battery, which is 1 ampere.

On the other hand, the 6V array will cost 1.9 amperes, and looking
at a typical SLA discharge current VS discharge time curve at:

http://www.panasonic.com/industrial/includes/pdf/Panasonic_VRLA_LC-R0612P.pdf

reveals that - as expected - there's only about a 2:1 increase in
discharge time (to terminal voltage, I assume,) for the 48V versus
the 6 volt array, since almost all of those wasteful resistors have
been removed from the 48V array.

So maybe the increase in "battery life" Fred was referring to was
charge-discharge cycles?

John Fields

[John Fields]

---
Instead of just blurting out opinion, why not do the work that shows
what numbers are working against him and post that work?

John Fields

[Martin Brown]

On 31/03/2014 11:08, John Fields wrote:
> On Mon, 31 Mar 2014 12:23:40 +1100, "Phil Allison"
> <phi...@tpg.com.au> wrote:
>
>>
>> <bloggs.fred...@gmail.com>
>>
>>>
>>> The apparent battery capacity increases dramatically at lower drain rates.
>
> ---
> True
> ---
>
>> ** Impossible to reduce the battery drain and have the same power.
>>
>>
>>> So if you want to increase battery life beyond belief,
>>> the thing to do is wire the LEDs in series
>>
>> ** Also impossible, cos they all have different current drains.

Actually it will increase the battery life beyond belief once the
voltage required by the LEDs in series exceeds the potential that the
battery can supply. It isn't useful though. Basically the battery will
have its shelf life but the LEDs will produce no light at all.

One thing that you can do with modern LEDs is bridge the switch with a
high value resistor ~1M so that you can find torches in the pitch dark.
(and that really don't much alter the battery life)

But only a complete idiot would do it that way! If you put two LEDs in
series for a more realistic design comparison using 4.4v in total then

Rs = 1.6v/0.02A = 80R

LED power = 4.4x0.02 = 0.088
Resistor = 82x0.02^2 = 0.032 = 0.12W

50 chains in parallel gives 6W dissipation and a much simpler entirely
passive circuit with no active components or capacitors to dry out and
fail. LED luminaires are notorious for stating the MTBF of the LEDs and
quietly ignoring the fact that the control electronics cooks itself.

If you were going to do this seriously then the sweet spot is probably a
constant current or pulsed current source designed to provide enough
voltage headroom by doubling 6v for 5 or so LEDs in series. Then

LED power = 11x0.02 = 0.22
Resistor = 1x0.02 = 0.02

20 chains in parallel gives 4.4 + 0.4 = 4.8 and a 90+% efficient
switcher should not be beyond the wit of man for this step up.

> So maybe the increase in "battery life" Fred was referring to was
> charge-discharge cycles?
>
> John Fields
>

--
Regards,
Martin Brown

[Phil Allison]

"John Fields is fucking SENILE "

>><bloggs.fred...@gmail.com>
>>
>>>
>>> The apparent battery capacity increases dramatically at lower drain
>>> rates.
>
> ---
> True
> ---

** Like hell it does.

>>** Impossible to reduce the battery drain and have the same power.
>>
>>
>>> So if you want to increase battery life beyond belief,
>>> the thing to do is wire the LEDs in series
>>
>>** Also impossible, cos they all have different current drains.
>
> ---
> Depends on the array, yes?
> ---

** That has been defined by the OP already - you assinne cunt.

FYI:

You are totally incapable of following a simple thread.


>>** Insanity.
>
> ---
> Actually, it's quite practicable ...


** FFS asshole - not in this case you brain fucked CUNT !!!!!!!

You are totally incapable of following a thread cos you are so brain fucked.

BTW:

Why don't you go and fucking KILL yourself ??

Make the world a much better place.





.... Phil

[John Fields]

---
But with an output which drops off as the battery voltage falls, if
that's important.
---

>LED luminaires are notorious for stating the MTBF of the LEDs and
>quietly ignoring the fact that the control electronics cooks itself.

---
OK, but???
---

---
My less-than-serious setup uses 5 resistors to your 20, which is
less PCB real e$tate, and the 90% efficient switcher is a wash since
both arrays need 4.8 watts.

The other maybe gotcha is that your switcher needs to output 400mA,
while mine just a scant 100mA.
---

John Fields

[John Fields]

On Mon, 31 Mar 2014 23:27:36 +1100, "Phil Allison"
<phi...@tpg.com.au> wrote:

>
>"John Fields is fucking SENILE "
>
>
>>><bloggs.fred...@gmail.com>
>>>
>>>>
>>>> The apparent battery capacity increases dramatically at lower drain
>>>> rates.
>>
>> ---
>> True
>> ---
>
>** Like hell it does.

---
Hmmm...

If you drive your computer over to:
http://www.panasonic.com/industrial/includes/pdf/Panasonic_VRLA_LC-R0612P.pdf

- A link I posted earlier in the thread, BTW - and take a look at
the "CAPACITY" data under "SPECIFICATIONS", you'll see that as the
drain increases the capacity decreases.
---

>>>** Impossible to reduce the battery drain and have the same power.
>>>
>>>
>>>> So if you want to increase battery life beyond belief,
>>>> the thing to do is wire the LEDs in series
>>>
>>>** Also impossible, cos they all have different current drains.
>>
>> ---
>> Depends on the array, yes?
>> ---
>
>** That has been defined by the OP already - you assinne cunt.

---
Then it doesn't depend on the array?
---

>FYI:
>
>You are totally incapable of following a simple thread.

---
Yo no soy marinero, capice?
---

>>>** Insanity.
>>
>> ---
>> Actually, it's quite practicable ...
>
>
>** FFS asshole - not in this case you brain fucked CUNT !!!!!!!
>
>You are totally incapable of following a thread cos you are so brain fucked.
>
>BTW:
>
>Why don't you go and fucking KILL yourself ??

---
Because I'd miss seeing you frothing at the mouth and having your
little hissy-fits whenever you don't get your way?
---

>Make the world a much better place.

---
http://www.youtube.com/watch?v=bjSpO2B6G4s

John Fields

[Phil Allison]

"John Fields is a fucking SENILE CUNT "

>
>
>>> The apparent battery capacity increases dramatically at lower drain
>>> rates.
>
> ---
> True
> ---

** Like fucking hell it does.

>>> So if you want to increase battery life beyond belief,
>>> the thing to do is wire the LEDs in series
>>
>>** Also impossible, cos they all have different current drains.
>
> ---
> Depends on the array, yes?
> ---

** That has been defined by the OP already - you asinine cunt.

FYI:

You are totally incapable of following a simple thread.

>>** Insanity.
>
> ---
> Actually, it's quite practicable ...

** FFS you stinking asshole - not in *this case* you brain fucked CUNT
!!!!!

YOU are totally incapable of following a thread cos you are so brain
fucked.


FYI:

Why don't you go and fucking KILL yourself ??

Make the world a much better place.

BTW:

Did your father fuck you up the arse ???

Remember that one you used on me.

Payback is SUCH a Bitch.





.... Phil

[John Fields]

On Tue, 1 Apr 2014 00:26:41 +1100, "Phil Allison"
<phi...@tpg.com.au> wrote:

>

>BTW:
>
>Did your father fuck you up the arse ???
>
>Remember that one you used on me.
>
>Payback is SUCH a Bitch.

>.... Phil

---
WOW!

Your being fucked up the way you are is all _my_ fault???

There may be more to this puzzle than meets the eye...

[Phil Allison]

"John Fields = Vile Autistic Cunt "

>
>>BTW:
>>
>>Did your father fuck you up the arse ???
>>
>>Remember that one you used on me.
>>
>>Payback is SUCH a Bitch.
>

> > ---
> WOW!


** My god, a fucking criminal asshole like you needs hand grenade shoved his
rectum.






.... Phil

[meow...@care2.com]

> Prove it.

I kinda expected you'd see the mistake you'd made. The transforming action gets you back up to the battery current you started with, plus more due to convertor losses. You dont gain a thing.


NT

[David Eather]

The OP's array's ARE all different

[Maynard A. Philbrook Jr.]

In article <tkfij9pk76c2m670s...@4ax.com>,
jfi...@austininstruments.com says...

It's so f'ing obvious where the screw up is. Go back and read
it again, thoroughly this time.

I can understand you sliding down the other side of the hill,
but really, I didn't think you were anywhere near the valley, yet!

P.S.
If it makes you feel better, at least you still have one
friend, Bloggs. One lies, the other swears to it.

Jamie..

[John Fields]

---
So what?

If you'd have read on a little farther you might have come to the
realization that I was trying to make Fred's point to Phil A. by
specifying a 100 diode array implemented with and without a boost
converter.

The context of threads _is_ allowed to change, you know...

John Fields

[ehsjr]

On 3/31/2014 8:27 AM, Phil Allison wrote:
> "John Fields is fucking SENILE"
>
>
>>> <bloggs.fred...@gmail.com>
>>>
>>>>
>>>> The apparent battery capacity increases dramatically at lower drain
>>>> rates.
>>
>> ---
>> True
>> ---
>
> ** Like hell it does.

It does. See http://en.wikipedia.org/wiki/Peukert%27s_law

"Peukert's law, presented by the German scientist W. Peukert in 1897,
expresses the capacity of a battery in terms of the rate at which it is
discharged. As the rate increases, the battery's available capacity
decreases."

You can see a graph of alkaline battery discharge illustrating the
different apparent capacity under different loads here:
http://electronics.stackexchange.com/questions/44723/function-to-describe-alkaline-battery-voltage-under-constant-current-load


<snip>

[John Fields]

On Mon, 31 Mar 2014 19:51:01 -0500, "Maynard A. Philbrook Jr."

<jamie_...@charter.net> wrote:

>In article <tkfij9pk76c2m670s...@4ax.com>,
>jfi...@austininstruments.com says...
>>
>> On Sun, 30 Mar 2014 17:39:24 -0500, "Maynard A. Philbrook Jr."
>> <jamie_...@charter.net> wrote:
>>
>> >In article <e16a6628-57c5-4092...@googlegroups.com>,
>> >bloggs.fred...@gmail.com says...
>> >> > 2nd light has 12 leds.
>> >> >
>> >> >
>> >> >
>> >> > Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps
>> >> >
>> >> >
>> >> >
>> >> > Thanks for your help gentlemen.
>> >> >
>> >> >
>> >> >
>> >> > Andy
>> >>
>> >>
>> >> The apparent battery capacity increases dramatically at lower drain rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.
>> >>
>> >>
>> >
>> >Maybe I missed something but for what ever reason, I don't think the
>> >math is working in your favor?
>> >
>> >Jamie
>>
>> ---
>> Instead of just blurting out opinion, why not do the work that shows
>> what numbers are working against him and post that work?
>>
>> John Fields
>
> It's so f'ing obvious where the screw up is. Go back and read
>it again, thoroughly this time.

---
You can't support your opinion with fact, so you play the game as if
it's you who's in charge and then introduce all sorts of delaying
tactics hoping for a miracle to come along and save your miserable
ass from the roach spray.
---

> I can understand you sliding down the other side of the hill,
>but really, I didn't think you were anywhere near the valley, yet!

---
"Think" is the operative word here, and you bandy it about as if the
act was in your bag of tricks.
---

> P.S.
> If it makes you feel better, at least you still have one
>friend, Bloggs. One lies, the other swears to it.

---
Post where the lie lies then, instead of pretending to authority
which isn't yours to claim.

[josephkk]

On Mon, 31 Mar 2014 19:51:01 -0500, "Maynard A. Philbrook Jr."

Ye gods did you ever miss the point.

?-((

[Jim Thompson]

Spare us all the nuisance deletion exercises and stop feeding the
troll/village-idiot.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

[josephkk]

On Tue, 01 Apr 2014 06:55:30 -0700, Jim Thompson
<To-Email-Use-Th...@On-My-Web-Site.com> wrote:

>
>>> If it makes you feel better, at least you still have one
>>>friend, Bloggs. One lies, the other swears to it.
>>>
>>>Jamie..
>>
>>Ye gods did you ever miss the point.
>>
>>?-((
>>
>
>Spare us all the nuisance deletion exercises and stop feeding the
>troll/village-idiot.
>
> ...Jim Thompson

While i find Jamie rather marginal he has not gone far enough often enough
to be tossed. Of course others are there for a long time.

?-)

[bloggs.fred...@gmail.com]

Since the output current is as small as possible with the step-up, the converter losses are next to nothing, and certainly far less than a series resistor in a non-switching current limiter.

[meow...@care2.com]

On Thursday, April 3, 2014 2:33:23 AM UTC+1, bloggs.fred...@gmail.com wrote:
> On Monday, March 31, 2014 11:27:55 AM UTC-4, meow...@care2.com wrote:
> > On Monday, March 31, 2014 2:11:31 AM UTC+1, bloggs.fred...@gmail.com wrote:
> > > > On Sunday, March 30, 2014 6:07:58 PM UTC-4, meow...@care2.com wrote:
> > > > > On Sunday, March 30, 2014 10:28:13 PM UTC+1, Maynard A. Philbrook Jr. wrote:
> > > > > > > In article <e16a6628-57c5-4092...@googlegroups.com>,
> > > > > > > bloggs says...

> > > > > > The apparent battery capacity increases dramatically at lower drain rates. So if you want to increase battery life beyond belief, the thing to do is wire the LEDs in series and then use a "boost" converter to step-up the 6V voltage to 24-30 V as required.

> > > > > Maybe I missed something but for what ever reason, I don't think the
> > > > > math is working in your favor?

> > > > No :) That would reduce life due to convertor losses.

> > > Prove it.

> > I kinda expected you'd see the mistake you'd made. The transforming action gets you back up to the battery current you started with, plus more due to convertor losses. You dont gain a thing.

> Since the output current is as small as possible with the step-up, the converter losses are next to nothing, and certainly far less than a series resistor in a non-switching current limiter.

Obviously switching has an advantage over linear, as well as downsides. If you go for switching, the question is what topology & why. At this point we don't know how the LEDs are wired up inside the fittings, they may well already be in series parallel strings that dont share the same current rating.


NT