LED wiring Series or Parallel 110 or 12v (for Power Efficiency)

Chris Carlton

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Dec 31, 2011, 1:43:08 AM12/31/11

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Building a light fixture with 54 LEDs. (me NEWBIE)

4 different LEDs used

18 - 30mA IF 3.8VF
18 - 30mA IF 1.7VF
9 - 30mA IF 2.2VF
9 - 20mA IF 3.6

for Power Efficiency...
What's the best wiring design for this?
Parallel or Series? 12V or 110V

I'm currently planning... 12V - Series
Can I mix LEDs in a Series?
Do I need anything more than 3-5 LEDs and a resistor in each series?
How do I limit amps to a series or to the fixture?
I would like to power multiple fixtures like these off of 1
transformer. Is it better to build a device to divide the amps and
plug my fixtures into it?

If this is to much to ask for free, I'd be more than happy to pay for
your knowledge. I have to solve this one way or the other.

Thank You,
Chris Carlton

fungus

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Dec 31, 2011, 7:56:54 AM12/31/11

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On Dec 31, 7:43 am, Chris Carlton <personalgrowth...@yahoo.com> wrote:
> Building a light fixture with 54 LEDs. (me NEWBIE)
>

> What's the best wiring design for this?
> Parallel or Series?

Parallel is frowned upon because individual
LEDs vary and there's no way to control how
much current goes down each path of the
circuit. ie. Some LEDs might get more than
their maximum rating.

> 12V or 110V

12V seems a bit small and 110V seems too
big for your LED mix.

> I'm currently planning... 12V - Series

12V will work but you'll have a lot of
little strings of LEDs.

eg. You can only put three of the 3.8V
LEDs in a series. For 18 LEDs you'll
have 6 separate series.

> Can I mix LEDs in a Series?

Yes, but remember that all LEDs in a series
will have the same current passing through
them so mixing 20mA and 30mA won't work.

> Do I need anything more than 3-5 LEDs and a resistor in each series?

Technically, no, but see next question.

> How do I limit amps to a series or to the fixture?

Option 1: Choose your resistor carefully.

Resistors have a problem though, getting
exactly 20 or 30 mA can be fiddly and if you
get it a couple of mA wrong the difference
in brightness between LEDs can be noticeable.

If this isn't a problem for you then the resistor
method will work.

Option 2: Use a LED driver. There's lots of
circuits, pre-built boards, etc. out there
which provide a fixed current from any
reasonable voltage.

You get one that matches your LED current
(eg. 30mA) then simply connect a series of
LEDs to it.

> I would like to power multiple fixtures like these off of 1
> transformer. Is it better to build a device to divide the amps and
> plug my fixtures into it?
>

I'm not sure what you mean by this. When you
build your LED circuit you're going to limit the
number of amps that can pass through it. Each
circuit will have a current requirement.

So long as the total requirement for all your circuits
is within the power supply's capabilities then you
can just connect them up, nothing extra needed.

Rich Webb

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Dec 31, 2011, 8:01:02 AM12/31/11

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There was a long thread here back in '04 regarding regulating current to
LEDs using a current mirror and similar techniques. Give that a read for
some options:
<http://groups.google.com/group/sci.electronics.basics/browse_thread/thread/1a8547721b8950ce/ef2d171838815044?q=led+current+mirror+group:sci.electronics.basics>
or just do a search in http://groups.google.com for "led current mirror
group:sci.electronics.basics"

--
Rich Webb Norfolk, VA

Ian Field

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Dec 31, 2011, 12:53:17 PM12/31/11

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"Rich Webb" <bbe...@mapson.nozirev.ten> wrote in message
news:jl1uf7107g5c414cc...@4ax.com...

For 120VAC the current can be controlled by a so called "wattless dropper",
this is a series capacitor who's Xc is large compared to RL.

Back in the days of hybrid TVs, Thorn consumer electronics tried this to
replace the wasteful dropper resistors in the 300mA series heater chain.

IIRC the capacitor value was 4.3uF for 300mA on 230V mains at 50Hz, you'll
also need to include some series resistance to absorb turn on surge and
spikes.

You can use 2 series chains of LEDs in inverse parallel or a single chain
after a bridge rectifier.

IanM

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Dec 31, 2011, 4:48:36 PM12/31/11

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I'd be inclined to go for a simple and dumb 12V arrangement. You are
dissipating under 3.7 watts in the LEDs so even if you waste 25% more
power in the series resistors they will remain fairly cool and waste
less than a watt. You avoid a lot of safety issues by NOT having high
voltage running to the fixtures.

Unless the 12V supply is very well smoothed and regulated, you want
considerable 'headroom' between the total LED Vf in each string and the
supply, otherwise the current control will be poor e.g. a 5% increase in
supply voltage causing the LED current to double. The 3.8V LEDs have too
high a Vf to place three in series from a 12V supply for this reason.

The resistor chosen for each series string handles the current limiting
and should be chosen to run the LEDs at no more than 80% of their
maximum rating if you want reasonable reliability.

I would arrange the 30mA LEDs in series strings as follows:
9 sets of 2 x 3.8V + 1 x 1.7v giving 9.3V total Vf
2 sets of 4 x 2.2V giving 8.8V total Vf
1 set of 5 x 1.7V giving 8.8V total Vf
1 set of 4 x 1.7V + 1 x 2.2V giving 9V total Vf

The 20mA LEDs cannot be mixed in the same string as the 30mA ones and 3
x 3.6V doesn't leave much headroom to stabilise the current so I'd run 4
strings of two and an odd one out which will need its resistor
calculated separately.

Each string requires its own series resistor chosen for the correct
current based on the difference between the total Vf and the supply
voltage. All the strings WITH their individual series resistors are
connected in parallel.

You need a 'transformer' (actually a DC output power supply) intended
for LED applications. An AC output one for halogen lighting is
unsuitable. You can use multiple fixtures in parallel up to the rating
of your power supply, (though it would be smart not to exceed 80% of its
rating.

--
Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk
[at]=@, [dash]=- & [dot]=. *Warning* HTML & >32K emails --> NUL:

IanM

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Jan 1, 2012, 4:02:40 PM1/1/12

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IanM wrote:
> 1 set of 5 x 1.7V giving 8.8V total Vf

CORRECTION
1 set of 5 x 1.7V giving 8.5V total Vf

Sorry about the stupid maths error. It only affects the serise resistor
for this string and doesn't make any difference to the over-all design.

P E Schoen

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Jan 1, 2012, 9:34:40 PM1/1/12

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"Ian Field" wrote in message news:p_HLq.1879$Cu1....@newsfe21.ams2...

> For 120VAC the current can be controlled by a so called "wattless
> dropper", this is a series capacitor who's Xc is large compared to RL.

> Back in the days of hybrid TVs, Thorn consumer electronics tried this to
> replace the wasteful dropper resistors in the 300mA series heater chain.

> IIRC the capacitor value was 4.3uF for 300mA on 230V mains at 50Hz,
> you'll also need to include some series resistance to absorb turn on
> surge and spikes.

You need to make sure the capacitor is rated for mains connection, and even
with the series resistor the turn-on surge and spikes from transients can
easily exceed the LED's maximum surge rating (usually 3 to 5 x nominal) and
cause deterioration and failure. Much better to use an inductor or ballast.
http://ace-ballast.com/Pages/Home.html

You may even be able to adapt a ballast from a small fluorescent lamp to
work on an LED string.

But probably best is an electronic switching circuit such as
http://www.supertex.com/pdf/datasheets/HV9961.pdf which is available from
Mouser and others for a dollar or so.

> You can use 2 series chains of LEDs in inverse parallel or a single chain
> after a bridge rectifier.

The two chains in inverse parallel have a huge problem if one of the series
fails open. Then the other string will see the full reverse voltage and one
or more devices will also fail.

You can make a constant current regulator from an LM317 or even a common
7805 regulator:
http://diyaudioprojects.com/Technical/Voltage-Regulator/

There are many drivers available pre-made for 120 VAC line use, and it may
be better (and safer) to use them. Jameco has some for less than $20, but
designed for 350 mA 1 watt LEDs:
http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_2006879_-1

Also please see the recent thread on LED lamp innards. There are some pearls
of wisdom and information among the swine of bickering.

Paul

Chris Carlton

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Jan 2, 2012, 5:33:04 PM1/2/12

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Thanks to every response. I don't feel nearly as insane as I did a few
short days ago.

I guess to be sensible, I'll start out simple (LEDs with correct/safe
resistors) and then also experiment with LED drivers, droppers and
regulators for possible energy savings over my 1st design. It's always
nice to have "improvements" in those areas to market if I get that
chance. :)

My fixtures energy draw is pretty low now at less than 3.5w. Seems
great when I compare it to what it is replacing. However I can easily
see running 20-50 of these at once so it will add up.

Oh and I found some more LEDs with 30mA to replace the ones I have
that are 20mA so I would not have to mess with it and leave me cleaner
wiring in my series chains which are long because my LEDs are 3-4"
apart in most cases. Also I trying an experiment where I mixed the
20mA with some LEDs that are 40mA in a series. I averaged the mA based
on the number of each type LED. I did my resistor calculation based on
this average. So far it lights up! we'll see if it blows up! :)

Oh I was thinking of using laptop power supplies for this projects
future. They give 19V which makes for longer LED series strings and
they are dirt cheap ($5-10) to buy because of their popularity. What I
don't know is how reliable they are for power output??? Can I trust
them??? If I make the jump to drivers that guarantee my voltage this
may be the way to go! (LED power supplies are like $70.)

Thanks again!

k...@att.bizzzzzzzzzzzz

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Jan 2, 2012, 5:41:26 PM1/2/12

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On Mon, 2 Jan 2012 14:33:04 -0800 (PST), Chris Carlton
<personal...@yahoo.com> wrote:

>Thanks to every response. I don't feel nearly as insane as I did a few
>short days ago.
>
>I guess to be sensible, I'll start out simple (LEDs with correct/safe
>resistors) and then also experiment with LED drivers, droppers and
>regulators for possible energy savings over my 1st design. It's always
>nice to have "improvements" in those areas to market if I get that
>chance. :)
>
>My fixtures energy draw is pretty low now at less than 3.5w. Seems
>great when I compare it to what it is replacing. However I can easily
>see running 20-50 of these at once so it will add up.
>
>Oh and I found some more LEDs with 30mA to replace the ones I have
>that are 20mA so I would not have to mess with it and leave me cleaner
>wiring in my series chains which are long because my LEDs are 3-4"
>apart in most cases. Also I trying an experiment where I mixed the
>20mA with some LEDs that are 40mA in a series. I averaged the mA based
>on the number of each type LED. I did my resistor calculation based on
>this average. So far it lights up! we'll see if it blows up! :)

No, you want to set the current lower than the *minimum* LED rating. They're
in series, so each gets the same current. However, you want to add the voltage
of each LED to calculate the ballast resistor.

It's unlikely that you'll need to drive them to their rated maximum. The eye
won't be able to detect the difference between 20mA and 30mA (or 10, for that
matter), anyway.

>Oh I was thinking of using laptop power supplies for this projects
>future. They give 19V which makes for longer LED series strings and
>they are dirt cheap ($5-10) to buy because of their popularity. What I
>don't know is how reliable they are for power output??? Can I trust
>them??? If I make the jump to drivers that guarantee my voltage this
>may be the way to go! (LED power supplies are like $70.)

They're quite reliable. Go fer it.

John Fields

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Jan 2, 2012, 7:39:01 PM1/2/12

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On Mon, 2 Jan 2012 14:33:04 -0800 (PST), Chris Carlton
<personal...@yahoo.com> wrote:

>Thanks to every response. I don't feel nearly as insane as I did a few
>short days ago.
>
>I guess to be sensible, I'll start out simple (LEDs with correct/safe
>resistors) and then also experiment with LED drivers, droppers and
>regulators for possible energy savings over my 1st design. It's always
>nice to have "improvements" in those areas to market if I get that
>chance. :)

---
If what you're looking for is minimum power waste, then you should
implement series strings running at high voltages.

For example, if you ran the 30 mA strings you described earlier, you'd
have 18 LEDs dropping 3.8V each for a total drop of 68.4V, 18 LEDs
dropping 1.7V each for a total of 30.6V and 9 LEDs dropping 2.2V each
for a total of 19.8V.

Putting all of those 30mA LEDs in series would result in a total drop
of 68.4V + 30.6V + 19.8V ~ 119V, which is very close to nominal mains
voltage in the US, suggesting that you could full-wave rectify the
mains and run the LEDs without a current limiting resistor.

You'd need some kind of spike protection in order to protect the
string from transients, but we can address that later, if you're
interested.
---

>My fixtures energy draw is pretty low now at less than 3.5w. Seems
>great when I compare it to what it is replacing. However I can easily
>see running 20-50 of these at once so it will add up.
>
>Oh and I found some more LEDs with 30mA to replace the ones I have
>that are 20mA so I would not have to mess with it and leave me cleaner
>wiring in my series chains which are long because my LEDs are 3-4"
>apart in most cases. Also I trying an experiment where I mixed the
>20mA with some LEDs that are 40mA in a series. I averaged the mA based
>on the number of each type LED. I did my resistor calculation based on
>this average. So far it lights up! we'll see if it blows up! :)

---
If you run the 20mA LEDs at higher than 20mA, then their life will be
shortened.
---

>Oh I was thinking of using laptop power supplies for this projects
>future. They give 19V which makes for longer LED series strings and
>they are dirt cheap ($5-10) to buy because of their popularity. What I
>don't know is how reliable they are for power output??? Can I trust
>them??? If I make the jump to drivers that guarantee my voltage this
>may be the way to go! (LED power supplies are like $70.)

---
My suggestion is that you tailor your strings so that you can run them
directly off the mains using a full-wave bridge to drive the LEDs and
a small resistor and a Zener TVS to quell the spikes.

Would you like a schematic?

--
JF

Chris Carlton

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Jan 2, 2012, 9:02:49 PM1/2/12

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On Jan 2, 7:39 pm, John Fields <jfie...@austininstruments.com> wrote:
> On Mon, 2 Jan 2012 14:33:04 -0800 (PST), Chris Carlton
>

Hey JF!!!

I may be a noobie but I was so sure that this is where I was heading.
(Resistors just seemed to be against everything I was trying to do. It
even seemed like adding another LED would be better than a resistor,
because that way all the power is being used to make light no matter
what. ???? wasn't sure if this was true.)

I found some more 30mA LEDs so they all match now, but was already
calculating things to run at reg mains. USA 110V because of the new
30mAs I have over my 110V @ 132V and wasn't sure what to do with the
extra 20V because a resistor was a billion ohms or something.

Yes I would LOVE a schematic!

Would you like a piece of my new company? :) Hell Yeah!! I love folks
who help get things done!

My new specs are...
3.8Fv 3.3Fv 2.2Fv 1.7Fv
30mA 30mA 30mA 30mA
15LEDs 9LEDs 9LEDs 15LEDs

P E Schoen

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Jan 3, 2012, 4:26:55 AM1/3/12

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"John Fields" wrote in message
news:afg4g79llvk5gbh36...@4ax.com...

> If what you're looking for is minimum power waste, then you
> should implement series strings running at high voltages.

> For example, if you ran the 30 mA strings you described earlier, you'd
> have 18 LEDs dropping 3.8V each for a total drop of 68.4V, 18 LEDs
> dropping 1.7V each for a total of 30.6V and 9 LEDs dropping 2.2V each
> for a total of 19.8V.

> Putting all of those 30mA LEDs in series would result in a total drop
> of 68.4V + 30.6V + 19.8V ~ 119V, which is very close to nominal mains
> voltage in the US, suggesting that you could full-wave rectify the
> mains and run the LEDs without a current limiting resistor.

But for a newbie, it would be much safer to use a wall-wart or laptop power
supply. They are pretty much fail-safe if overloaded or shorted, and the
voltage won't kill you (unless you try really hard).

> You'd need some kind of spike protection in order to protect the
> string from transients, but we can address that later, if you're
> interested.

The power supply will take care of that problem, too.

> If you run the 20mA LEDs at higher than 20mA, then their life will
> be shortened.

> My suggestion is that you tailor your strings so that you can run
> them directly off the mains using a full-wave bridge to drive the
> LEDs and a small resistor and a Zener TVS to quell the spikes.

Again, using a power supply is safer. And if there is a fire, the UL rating
on the PSU may be worth an awful lot. Fire inspectors don't take kindly to
homebrew gadgets across the house wiring.

> Would you like a schematic?

There is a schematic, for the bulk of what is required, in the Supertex data
sheets, and they also probably have application notes. Their circuits work
from about 12 VDC to 400 VDC, but I think it's best to stay under what is
considered "safe", which is about 25-30V. Or even 50V, which is what POTS
phone lines carry, but that is current limited to 20mA or so, which is
marginally safe.

If you want to make such a device commercially, you should contract a
licensed professional engineer to make sure the design will meet UL and
other standards. Pay somebody a modest amount now so you won't have to pay a
huge lawsuit settlement later.

Don't become a statistic...

Paul

fungus

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Jan 3, 2012, 4:30:15 AM1/3/12

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On Jan 3, 3:02 am, Chris Carlton <personalgrowth...@yahoo.com> wrote:
> I may be a noobie but I was so sure that this is where I was heading.
> (Resistors just seemed to be against everything I was trying to do. It
> even seemed like adding another LED would be better than a resistor,
> because that way all the power is being used to make light no matter
> what. ???? wasn't sure if this was true.)
>

It's not a good idea to do this. The thing
is that the resistance of a LED isn't constant.

The resistance of a LED drops *rapidly* as the
voltage increases past the threshold voltage,
and that's precisely the voltage you're aiming
at when you design your circuit.

In a series of LEDs, the LED with least resistance
will get the most volts. Increasing the volts will
lower the resistance even further... and... do you
see where this is going?

You can probably get away with it if you're only
trying to run them at (eg.) 10mA but if you're
anywhere near their rated current then you're
taking a big risk.

fungus

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Jan 3, 2012, 4:32:06 AM1/3/12

to

On Jan 2, 11:33 pm, Chris Carlton <personalgrowth...@yahoo.com> wrote:
>
> Oh I was thinking of using laptop power supplies for this projects
> future. They give 19V which makes for longer LED series strings and
> they are dirt cheap ($5-10) to buy because of their popularity. What I
> don't know is how reliable they are for power output??? Can I trust
> them??? If I make the jump to drivers that guarantee my voltage this
> may be the way to go! (LED power supplies are like $70.)
>

You won't be using anywhere near their rated
maximum current so they should be quite reliable.

fungus

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Jan 3, 2012, 4:35:27 AM1/3/12

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On Jan 2, 11:41 pm, "k...@att.bizzzzzzzzzzzz"

<k...@att.bizzzzzzzzzzzz> wrote:
>
> It's unlikely that you'll need to drive them to their rated maximum. The eye
> won't be able to detect the difference between 20mA and 30mA (or 10, for that
> matter), anyway.
>

Yep. Aim for a few mA less then the rating
(eg 25 instead of 30), you won't notice the
difference and it should simplify the circuitry
(or at least let you use cheaper/easier to find
components).

default

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Jan 3, 2012, 9:55:48 AM1/3/12

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On Sun, 1 Jan 2012 21:34:40 -0500, "P E Schoen" <pa...@pstech-inc.com>
wrote:

>"Ian Field" wrote in message news:p_HLq.1879$Cu1....@newsfe21.ams2...
>
>> For 120VAC the current can be controlled by a so called "wattless
>> dropper", this is a series capacitor who's Xc is large compared to RL.
>
>> Back in the days of hybrid TVs, Thorn consumer electronics tried this to
>> replace the wasteful dropper resistors in the 300mA series heater chain.
>
>> IIRC the capacitor value was 4.3uF for 300mA on 230V mains at 50Hz,
>> you'll also need to include some series resistance to absorb turn on
>> surge and spikes.
>
>You need to make sure the capacitor is rated for mains connection, and even
>with the series resistor the turn-on surge and spikes from transients can
>easily exceed the LED's maximum surge rating (usually 3 to 5 x nominal) and
>cause deterioration and failure. Much better to use an inductor or ballast.
>http://ace-ballast.com/Pages/Home.html

I disagree. I've been using capacitive reactance for lighting small
LEDs for years now with no ill effects. Use a full wave bridge, small
series resistor, and zener across the led string to limit the peak
voltage.

I concur that AC rated caps would be the better choice - but I've been
using 450 volt polyester film caps with no problems (on 120VAC).

I think PE Johnson had a circuit posted showing a pretty good
(bullet-proof) circuit for running off 120 VAC and protecting the leds
from transients..

>
>You may even be able to adapt a ballast from a small fluorescent lamp to
>work on an LED string.

Don't do it. Fluorescent lamps require a high voltage strike to
initially ionize the gas - you may still need the same protection that
running off the mains directly would require.

>
>But probably best is an electronic switching circuit such as
>http://www.supertex.com/pdf/datasheets/HV9961.pdf which is available from
>Mouser and others for a dollar or so.

That does look intriguing. 8-450 VDC input.

>
>> You can use 2 series chains of LEDs in inverse parallel or a single chain
>> after a bridge rectifier.
>
>The two chains in inverse parallel have a huge problem if one of the series
>fails open. Then the other string will see the full reverse voltage and one
>or more devices will also fail.

That's true, but the likelihood is very remote - maybe up there with
shark bite in a Kansas cornfield...

You design the best you can and make compromises. Absolute
bullet-proof starts getting expensive, and usually requires some other
trade-offs like size, complexity, efficiency.

>
>You can make a constant current regulator from an LM317 or even a common
>7805 regulator:
>http://diyaudioprojects.com/Technical/Voltage-Regulator/

That would be my choice for a few high power, high cost leds.
Downside might be efficiency, and if one were using batteries to power
it....

Chris Carlton

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Jan 3, 2012, 9:56:21 AM1/3/12

to

OK not just this specific post but there seems to be a lot of warning
about overloads. I want to take this seriously specially because I can
quite be sure of what the end user would "plug" these things into.
However I also am looking for maximum efficiency. I set up a
spreadsheet that allows me to combine diff LEDs @ diif quatities and
does the math based on ohms law and all that stuff and tell me what
resistor to use. I know to round up to the next available to be safe.
Are you telling me that I need to change the current/mA settings in my
spreadsheet to lower than the manufacutrer suggestion of typical mA/
current??? I don't want to loose anything if I don't have to.

I also don't want to spend the next 2 years replacing products I've
sold because the gain a history of being over-sensitive to their power
source.

Thanks for the amazing discussion. This is better (down to earth) than
any basic LED circuit design thread I've seen for sure.

Mad Props to all who have chimed in!

Chris Carlton

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Jan 3, 2012, 10:03:10 AM1/3/12

to

just want to throw in that I'm using 5mm old school LEDs because of
their low power and nm selection. I am looking at tape and other LEDs
but the total flux seems to suck so bad. I'm sure I just to learn more
about LED shopping or get the cash to make a custom order.

Ian Field

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Jan 3, 2012, 10:06:00 AM1/3/12

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<default> wrote in message
news:bb36g71fhgofv4tj5...@4ax.com...

> On Sun, 1 Jan 2012 21:34:40 -0500, "P E Schoen" <pa...@pstech-inc.com>
> wrote:
>
>>"Ian Field" wrote in message news:p_HLq.1879$Cu1....@newsfe21.ams2...
>>
>>> For 120VAC the current can be controlled by a so called "wattless
>>> dropper", this is a series capacitor who's Xc is large compared to RL.
>>
>>> Back in the days of hybrid TVs, Thorn consumer electronics tried this to
>>> replace the wasteful dropper resistors in the 300mA series heater chain.
>>
>>> IIRC the capacitor value was 4.3uF for 300mA on 230V mains at 50Hz,
>>> you'll also need to include some series resistance to absorb turn on
>>> surge and spikes.
>>
>>You need to make sure the capacitor is rated for mains connection, and
>>even
>>with the series resistor the turn-on surge and spikes from transients can
>>easily exceed the LED's maximum surge rating (usually 3 to 5 x nominal)
>>and
>>cause deterioration and failure. Much better to use an inductor or
>>ballast.
>>http://ace-ballast.com/Pages/Home.html
>
> I disagree. I've been using capacitive reactance for lighting small
> LEDs for years now with no ill effects. Use a full wave bridge, small
> series resistor, and zener across the led string to limit the peak
> voltage.
>
> I concur that AC rated caps would be the better choice - but I've been
> using 450 volt polyester film caps with no problems (on 120VAC).

Filter caps salvaged from the mains input circuit of old
TVs/monitors/whatever are rated to take full on mains on a full time basis -
not to mention subtracting the summed Vf of a decent size string of LEDs.

>
> I think PE Johnson had a circuit posted showing a pretty good
> (bullet-proof) circuit for running off 120 VAC and protecting the leds
> from transients..
>>
>>You may even be able to adapt a ballast from a small fluorescent lamp to
>>work on an LED string.
>
> Don't do it. Fluorescent lamps require a high voltage strike to
> initially ionize the gas - you may still need the same protection that
> running off the mains directly would require.

The "electronic transformer" for low voltage halogen lights is superficially
similar to a CFL circuit - the main difference is having a large toroid
transformer instead of a ferrite cored ballas in series with the tube.

However; introducing current control into this type of half-bridge converter
is not trivial.

Chris Carlton

unread,

Jan 3, 2012, 10:08:54 AM1/3/12

to

and I've been able to back down to 15 from the 18 sets so it's now...

15 - 30mA IF 3.8VF
15 - 30mA IF 1.7VF

9 - 30mA IF 2.2VF

9 - 30mA IF 3.3

total 48 LEDs per fixture

Chris Carlton

unread,

Jan 3, 2012, 10:07:00 AM1/3/12

to

and now that's

9 - 30mA IF 3.3

instead of the orignial

Chris Carlton

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Jan 3, 2012, 10:30:39 AM1/3/12

to

It would seem the solution would be 2 LM317s diving up two series of
LEDs set to the specific needs of that string with very small or no
resistors. I read that the LM317 heat up with to big of a V drop
though so this would make dropping to 75V or less with one would be
smokin! IDK really.

I guess I could release these in Europe and Australia with only one
LM317 since they are running on 220. But again if they heat up with V
drop I'd be in the home fire business. No thanks.

Chris Carlton

unread,

Jan 3, 2012, 10:40:12 AM1/3/12

to

Simple basic LED series with resistors hooked up to rated (meant for
laptop) power supplies. At this point is still looking really good.

This seems like a good way to start. Of course get the UL done on my
side of things, whatever legal is needed is a given for me. Really
don't wanna buy houses for people cause I sold some lights.

fungus

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Jan 3, 2012, 11:14:23 AM1/3/12

to

On Jan 3, 3:56 pm, Chris Carlton <personalgrowth...@yahoo.com> wrote:
>
> Are you telling me that I need to change the current/mA settings in my
> spreadsheet to lower than the manufacutrer suggestion of typical mA/
> current??? I don't want to loose anything if I don't have to.
>

Given the manufacturing tolerances of
LEDs, resistors, etc., it's possible that
if you design for 'perfect' components you
might end up a bit too too high on some
LEDs. It won't be much ... but it's worth
designing 5-10% under spec, just in case.

If you look at the datasheet for your LEDs
you'll see the light output vs. input current
is a curve, not a straight line. Designing
for 5-10% less current will hardly affect the
output at all.

default

unread,

Jan 3, 2012, 11:43:51 AM1/3/12

to

On Tue, 3 Jan 2012 07:30:39 -0800 (PST), Chris Carlton
<personal...@yahoo.com> wrote:

>It would seem the solution would be 2 LM317s diving up two series of
>LEDs set to the specific needs of that string with very small or no
>resistors. I read that the LM317 heat up with to big of a V drop
>though so this would make dropping to 75V or less with one would be
>smokin! IDK really.
>

Without looking at the data sheet... dropping 75 volts is probably out
of the question - the absolute maximum standoff is ~36 volts if my
memory serves me.

For heat - you can always add a series resistor and dissipate most of
the heat in the resistor and just enough to regulate current in the
LEDs in the 317.

>I guess I could release these in Europe and Australia with only one
>LM317 since they are running on 220. But again if they heat up with V
>drop I'd be in the home fire business. No thanks.

I've been going crazy with these warm white LED strip light thingees.
I bought 5 meters of the stuff (~$30) and mounted one meter over the
kitchen sink - since I already had a 12V power supply nearby.

On high, it is a lot of light so I added a dropping resistor and SPDT
switch so I have a night light as well as enough to do dishes. So far
it is great. Built in resistors for each three leds in series and the
leds and resistors are spread out so there are no hot spots.

My wife doesn't like it because she associates the lights with
Christmas. She'll adapt...

Having 4 meters leftover... I mounted some in test tubes small ones
are 12 leds each and larger 24 per tube then added a mosfet
transistor, and photo transistor, to switch them on at dusk. I have
five of them to light a path.

Now what to do with the remaining 2-3 meters?

fungus

unread,

unread,

Jan 4, 2012, 2:14:38 AM1/4/12

to

Chris Carlton wrote:
> I'm at 132V total now. I don't see how I can use LM317s without one of
> them having a huge drop. I must be missing something.

An LM317 has a maximum rating of ~37 volts difference between
input and output. So, if you had say 168 volts on the input,
and the output was connected to the series string of LEDs, the
output pin would be at 132 volts. Therefore the 36V difference
between input and output would be just within the rating.
If you configured the LM317 to provide a constant current of
18 mA, and put a 1000 ohm resistor in series between the
168 volt source and the input pin, the resistor would drop
18 volts, so the input pin of the LM317 would be at 150 volts,
and the difference voltage would be 18 volts instead of 36.

You configure the LM317 as a constant current source by
connecting a resistor from its output pin to its adjust
pin, and connecting the load to the adjust pin. The formula
to determine the resistance or current is 1.25/R = I

For example, to get ~ 18 mA:

----- 68 ohms
+ ----|LM317|---/\/\/---+
----- R |
| |
+--------------+--LOAD--+
|
- -------------------------------+

1.25/68 = ~.018

In a like manner, if R was 1 ohm, I would be 1.25,
if R was 10 ohms, I would be .125, if R was 100 ohms,
I would be .0125 and so forth.

You should also check the power dissipated in the LM317. Power
is the LM317 input - output voltage difference times the current.

Ed

John Fields

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Jan 4, 2012, 5:16:30 AM1/4/12

to

On Wed, 04 Jan 2012 02:14:38 -0500, ehsjr <eh...@nospamverizon.net>
wrote:

---
Nice. :-)

--
JF

Chris Carlton

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Jan 4, 2012, 6:46:42 AM1/4/12

to

sorry Ed, I must be too thick in the head like Dad used to say...

If I start with 110V from the wall and my fixture is 132V total then I
will have to split my LEDs into a minimum of two strings.
As soon as I split up the 48 LEDs up the smaller individual strings
total V would be greater than the drop limit of 36V for a LM317.

If I split them evenly.
132 / 2 = 66V per string
110 - 66 = 54V drop

Chris Carlton

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Jan 4, 2012, 6:57:08 AM1/4/12

to

In my mind I keep coming back to the comments about the reliability of
power past a laptop charger is very good.
and the comments about not running the LEDs at their full ratings.

This gives me a clean reliable 19V

Also everyone is saying to under drive me LEDs

I could build strings that are around 20.5V without resistors.

Even if my LEDs won't run at MAX brightness (which is not good anyway
apparently) at least all power would be being used to make light and
none being wasted in resistors or 317s or anything else. (Unless I am
a retard which is very possible. This is pretty new stuff for me) 19V
is also pretty low so even if it shorts it would not be a fire
starter. Running at 110V is a bit scary.

fungus

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Jan 4, 2012, 8:00:26 AM1/4/12

to

On Jan 4, 12:57 pm, Chris Carlton <personalgrowth...@yahoo.com> wrote:
> Running at 110V is a bit scary.

Yep. You seem to be hinting at selling
this to the public. I don't know if that
requires certification where you live, but
it's likely (and for good reasons!).

If any part of whatever it is your selling
has mains A/C wires inside it then it
had better be rock-solidly built.

Laptop power supplies have already been
through the certification process and they
mean that no part of your gizmo has
potentially lethal wires inside it.

Michael A. Terrell

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Jan 4, 2012, 10:39:58 AM1/4/12

to

If you rectify & filter 110 volts you get about 155 volts DC.

If you rectify & filter 120 volts you get about 170 volts DC.

My line voltage is usually around 127 volts, which gives
127*1.414=179.578-.6, or 178.978 after the diode's forward voltage drop.
If you use a full wave bridge, you lose another .6 volts.

--
You can't have a sense of humor, if you have no sense.

John Fields

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Jan 4, 2012, 11:05:07 AM1/4/12

to

---
What you're maybe misunderstanding is how to read an LED data sheet.

When you see the value for If, that's the current which, when pushed
through the LED, will result in a drop (Vf) which can be anywhere
between Vf(min) and Vf(max).

That means that if you take a zillion LEDs and force If through each
of them, the drop (Vf) caused by that current will vary between those
limits from LED to LED.

The other thing is that if you're driving an LED with a voltage source
instead of a current source you could easily overdrive the LED if,
say, the drop across the diode was at the low end of Vf with If
through it and you were driving it with Vf nominal.

That's because once a diode starts conducting, it only take a tiny
increase in Vf to effect a large change in If:

http://www.yegopto.co.uk/expertise/Drivers_and_Controllers

LEDs also have a negative temperature coefficient of resistance, which
means that if there isn't some external current-limiting mechanism in
place, as the LED heats up its internal resistance will drop, which
will allow more current through it, which will cause it to get hotter,
which... well, you get it, I'm sure. It's called "thermal runaway".

There are ways around these problems, one of them being the passive
solution I posted for you, another being the LM317 solution Ed gave
you, but you need to remember that you want to drive the LEDs with
either a constant-current or a current-limited source.

--
JF

ehsjr

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Jan 4, 2012, 12:44:59 PM1/4/12

to

No problem, I think you missed where I said:
"So, if you had say 168 volts on the input, ..."

I was only trying to show how the LM317 works, but I'll post
a complete circuit, below, and show you how you can go back
to 54 LEDs instead of 48. Please note that this is posted
just as a learning experience. There are better and safer
solutions that do not require you to deal with 120 volt
mains. I'll post a separate reply to address that.

This thread has covered a lot of territory. :-)
The incomplete discussion of using an LM317, and the idea
of restoring your ability to use 54 LEDs instead of 48
prompts this:

For 54 LEDs, use the following:

3.8Fv 3.3Fv 2.2Fv 1.7Fv
30mA 30mA 30mA 30mA

18LEDs 12LEDs 12LEDs 12LEDs
That adds to 68.4 + 39.6 + 26.4 + 20.4 = 154.8 volts.

Note that it's 18 3.8 volt LEDs, and 12 of each of
the other voltage leds, forming a series string of 54 LEDs.

We'll set the current to ~18 mA with an LM317 and 68 ohm
resistor, and develop the necessary voltage with a bridge
rectifier and capacitor. Here's the circuit:

----------- 470 ----- 68
AC---| +|---/\/\/---+---|LM317|---/\/\/---+
| Bridge | | ----- |
| Rectifier | 47uF --- | |
| | 250V --- +--------------+
| | | |
| | | LEDs
| | | |
AC---| -|-----------+---------------------+
-----------

The 68 ohm resistor connected from the 317 out pin to the
adj pin sets the current to about 18 mA. The LM317 holds
its output pin at 1.25 volts above its adjust pin; thus the
current is 1.25/68 or about .0183 amps.

At 18 mA, the voltage drop in the 470 ohm resistor will be
about 8.6 volts. (470 * .0183 = ~ 8.6)

*Here's the part where you get ~168 volts*
The 47 uF cap will be charged to the peak mains voltage,
about 169.7. The peak voltage is computed by
Vrms * square root of 2. So 120*sqrt(2) = ~169.7

From 169.7, subtract the ~8.6 drop across the 470 ohm resistor
and minus the ~1.4 volt drop in the bridge. So the voltage
across the cap will be about 159.7. That means the voltage
drop across the LM317 will be about 4.9 volts, because the
54 LEDs in series drop about 154.8 volts.

Power dissipation in the 470 ohm resistor is about .16 watts.
Power dissipation in the LM317 is about .09 watts, and power
dissipation in the 68 ohm resistor is about .023 watts.

Note that the circuit purposely includes some drawbacks.
What happens if the line voltage drops? Say it drops to
110. As Mike pointed out, the rectified and filtered
output now becomes 110*sqrt(2) or ~155.5. When you subtract
the voltage drop in the resistor (8.6) and the voltage drop
in the bridge (1.4), you end up with ~ 145.5. That is
lower than the 154.8 volts computed for the 54 LED string.

It also "starves" the LM317. The LM317 (and all other 3
legged voltage regulator chips) needs some "headroom".
That is, it needs some voltage above the output voltage
to work properly. For the LM317 3 volts is more than it
needs, but is easy to use as that "headroom" voltage in
computations. So for the sake of the example, we can
say the LM317 needs 157.8 volts on its input pin to provide
154.8 volts on its output pin.

As mentioned earlier, another drawback is that you would
have to use mains voltage. Still another is the use of
the LM317 in the first place, where a single resistor
would be good enough. If you used the filtered and rectified
mains to produce ~168 volts to drive a 154.8 volt string of
LEDs, you could drop the voltage from ~168 to ~154.8 with
a single resistor. Ohms law says E = I*R where E is voltage,
I is current, and R is resistance. We need to drop about
13.2 volts (168 - 154.8) across the resistor. It is a good
idea to drive the LEDs gently, so let's choose about 20 mA
as the current we want. Plugging in to the formula:
13.2 = .02 * R, so R = 13.2/.02 or 660 ohms. A standard
value resistor close to that is 680 ohms, so that is what you
would choose.

Ed

ehsjr

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Jan 4, 2012, 1:40:57 PM1/4/12

to

Here's one safe approach that I mentioned I would post in my
previous reply.

I think someone in the thread mentioned using a 48 volt supply.
Jameco sells one that is intended to drive LEDs. It is $13.95,
part # 2101121.

To drive the 48 LEDs you specified

15 - 30mA IF 3.8VF
15 - 30mA IF 1.7VF
9 - 30mA IF 2.2VF

9 - 30mA IF 3.3

You make 3 series strings:
String 1: 12 3.8V LEDs; total Vf = 45.6
String 2: 3 3.8V LEDs + 9 3.3V LEDs; total Vf = 41.1
String 3: 15 1.7V LEDs + 9 2.2V LEDs; total Vf = 45.3

Then:
------------- r1
| 48V +|----+---/\/\/---String1---+
| LED Supply | | |
| | | r2 |
| Jameco | +---/\/\/---String2---+
| Part # | | |
| 2101121 | | r3 |
| | +---/\/\/---String3---+
| | |
| -|--------------------------+
-------------

If you choose 20 mA for the current:
r1 = (48-45.6)/.02 or 120 ohms
r2 = (48-41.1)/.02 or 345 ohms
r3 = (48-45.3)/.02 or 135 ohms

120 ohms and 330 ohms are standard value resistors that will work
fine. With r2 at 330 ohms, current is (48-41.1)/330 or ~21 mA
With r3 at 120 ohms, current is (48-45.3)/120 or ~22.5 mA

r1 will dissipate .048 watts, r2 will dissipate ~ .144 watts
and r3 will dissipate ~.06 watts. Dissipation is computed by
P = I^2 * R where P is the power in watts, I is current in
amps, and R is resistance in ohms.

I'd recommend that you add 2 2.2V LEDs to bring the total to 50.
If you added 2 2.2V LEDs to string 2, bringing the total LEDs
to 50, the total Vf for string2 would be 45.5. Then you could
use a 120 ohm resistor in place of the 330 for r2 computed above.
That would lower the dissipation in r2 to ~.052 watts.

Ed

Chris Carlton

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Jan 4, 2012, 2:15:31 PM1/4/12

to

Wow thanks again!!

a few things...

I am both, working on a prototype and also trying to learn where I am
going with it and plan for it's possible business future. I will get
actual licensed schematics done before anything get sold to the
public.

I get a lot of what is being thrown around but I miss a lot too. I
still have a lil basic homewrok to do for I can understand these
solutions and weigh the pros and cons. This thread has offered so much
more knowledge than I imagined it could.

I actually don't need the 54 LEDs My next prototype will be 48 LEDs
for sure.

I want it to be safe as well as efficient.

I wasn't thinking I could go up from 110V but I guess that's not all
that uncommon.

I really like the idea of low voltage because of how spread out my
LEDs are much like if I was making my own 5mm tape LEDs that the
diodes are 3-4" apart on.

Also my fixture is in 3 pieces, like 3 pieces of LED tape kind of. 2
pieces with 18 LEDs and one piece with 12.

If I could use a driver to make 3 strings that happen to be the same
as my 3 separate pieces then that would really make things easy.

the breakdown of the pieces are

2 pieces @
6 - 30mA IF 3.8VF
3 - 30mA IF 1.7VF
3 - 30mA IF 2.2VF
6 - 30mA IF 3.3

and 1 other piece @
1 - 30mA IF 3.8VF
1 - 30mA IF 1.7VF
1 - 30mA IF 2.2VF
1 - 30mA IF 3.3

I am building them to plug into each other so they can be stringed in
a row.

Sorry to want to know so much so fast, I came at this from a
background in lighting, stage lighting even, nothing like these little
LEDs and their low power! It all blows me away really. BTW my project
has nothing to do with stage lighting. :)

I'm quite excited about electronics though and wish I had gotten into
it years ago. So cool. I looked at those lil pieces on circuit boards
and had no idea that some were this simple. I can't wait to build some
toys now.

I having a blast to be honest! So much fun and this thread has given
me a real boost, Wow!!

k...@att.bizzzzzzzzzzzz

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Jan 4, 2012, 6:38:12 PM1/4/12

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On Wed, 4 Jan 2012 05:00:26 -0800 (PST), fungus <to...@artlum.com> wrote:

>On Jan 4, 12:57 pm, Chris Carlton <personalgrowth...@yahoo.com> wrote:
>> Running at 110V is a bit scary.
>
>Yep. You seem to be hinting at selling
>this to the public. I don't know if that
>requires certification where you live, but
>it's likely (and for good reasons!).

"Requires" certification? Very few jurisdictions require any sort of
certification. It may be a lawyer license to not certify but it's not
generally required by law for end-user products.

ehsjr

unread,

Jan 5, 2012, 1:26:12 AM1/5/12

to

Here's a design for you which yields ~20 mA through the LEDs,
and doesn't cost an arm and a leg. Use a regulated power supply
that provides 28 volts, such as Jameco part # 1707083 for $9.95
That supply can provide a total of .64 amps

String1:
6 - 30mA IF 3.8VF = 22.8V
3 - 30mA IF 1.7VF = 5.1V
Total Vf = 27.9V

String2:
3 - 30mA IF 2.2VF = 6.6V
6 - 30mA IF 3.3 = 19.8V
Total Vf = 26.5V

String3:

1 - 30mA IF 3.8VF
1 - 30mA IF 1.7VF
1 - 30mA IF 2.2VF
1 - 30mA IF 3.3

Total Vf = 11.0V

r1
+28---+---/\/\/---String1---+
| |
| r1 |
+---/\/\/---String1---+
| |
| r2 |
+---/\/\/---String2---+
| |
| r2 |
+---/\/\/---String2---+
| |
| r3 |
+---/\/\/---String3---+
|
Gnd-------------------------+

r1 = (28-27.9)/.02 = 5 ohms
r2 = (28-26.5)/.02 = 75 ohms
r3 = (28-11.0)/.02 = 850 ohms

The supply specified can support 6 of the setups above,
or 30 strings total. Note that if you use the minimum
load resistance described below, that reduces to 5 setups
comprising 5 strings per setup, or 25 strings total.

850 is not a standard value - you can use an 820 ohm resistor
in series with a 30 ohm resistor to get 850 ohms. r1 will
dissipate .002 watts, and r2 will dissipate .003 watts. r1 and r2
can be 1/4 watt or even 1/8 watt, but you need higher wattage
for r3. It will dissipate .34 watts, and it is a good idea
to use double that or more, so use a 1 watt resistor for it.
(The 820 ohm resistor should be rated 1 watt, but the 30 ohm
resistor in series with it can be way lower - 1/4 or 1/8 watt
will be fine for it.

A note about the specified power supply: it needs a minimum
load that draws at least 60 mA. The circuit above draws
100 mA, so that satisfies the minimum current spec. However,
if you anticipate turning some portion of the LEDs off, you
could drop below the minimum spec. Therefore, it's a good
idea to add a load from +28 to ground that will always draw
(at least) that .06 amp minimum current. That load will
dissipate at least 1.68 watts (.06 * 28).

An easy way to provide the load is to use 2 5 watt, 820 ohm
resistors in parallel. That will yield 410 ohms resistance,
which will draw about 68 mA, and will dissipate about 1 watt
in each resistor (about 2 watts total). 5 watt resistors are
not required, but I don't think Jameco carries 2 watt resistors.
They do carry the others.

Ed

Chris Carlton

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Jan 5, 2012, 2:22:52 AM1/5/12

to

sorry...

make that...

and 1 other piece @

3 - 30mA IF 3.8VF

3 - 30mA IF 1.7VF
3 - 30mA IF 2.2VF

3 - 30mA IF 3.3

P E Schoen

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Jan 5, 2012, 4:59:13 AM1/5/12

to

"Chris Carlton" wrote in message
news:9edb98f9-1dc3-49d0...@q17g2000yqh.googlegroups.com...

I made a very simple LTSpice circuit that uses a 24 VAC "Wallwart", 2
capacitors, and two diodes, which provides about 28 mA into a string of 15
white LEDs with a voltage drop of about 57 VDC. And the current is limited
to less than 150 mA if the output is shorted (use 1 ohm for R1). Here is the
error log (use Ctrl-L), which shows the efficiency to be better than 94%:

.OP point found by inspection.

iled: AVG(i(d1))=0.028176 FROM 0.999989 TO 1.19998
p_in: AVG(i(v1)*v(ac1))=-1.70889 FROM 0.999989 TO 1.19998
p_out: AVG(i(d1)*v(v+))=1.60876 FROM 0.999989 TO 1.19998
efficiency: 100*p_out/p_in=-94.1404
vled: AVG(v(v+))=56.9075 FROM 0.999989 TO 1.19998

================================================================
Version 4
SHEET 1 1936 680
WIRE -16 144 -48 144
WIRE 64 144 -16 144
WIRE 240 144 128 144
WIRE 288 144 240 144
WIRE 416 144 352 144
WIRE 512 144 416 144
WIRE 528 144 512 144
WIRE 640 144 528 144
WIRE 240 192 240 144
WIRE 416 192 416 144
WIRE -48 224 -48 144
WIRE 528 224 528 144
WIRE 640 224 640 144
WIRE -48 384 -48 304
WIRE -16 384 -48 384
WIRE 240 384 240 256
WIRE 240 384 -16 384
WIRE 416 384 416 256
WIRE 416 384 240 384
WIRE 528 384 528 304
WIRE 528 384 416 384
WIRE 640 384 640 288
WIRE 640 384 528 384
WIRE 640 432 640 384
FLAG 640 432 0
FLAG 512 144 V+
FLAG -16 144 AC1
FLAG -16 384 AC2
SYMBOL diode 288 160 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL polcap 400 192 R0
SYMATTR InstName C2
SYMATTR Value 100µ
SYMATTR Description Capacitor
SYMATTR Type cap
SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 Lser=0
SYMBOL voltage -48 208 R0
WINDOW 3 31 200 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 24 44 Left 2
SYMATTR Value SINE(0 35 60 0 0 0 120)
SYMATTR SpiceLine Rser=.1
SYMATTR InstName V1
SYMBOL polcap 128 128 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C1
SYMATTR Value 47µ
SYMATTR Description Capacitor
SYMATTR Type cap
SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 MTBF=5000 Lser=0 ppPkg=1
SYMBOL diode 224 256 M180
WINDOW 0 24 72 Left 2
WINDOW 3 24 0 Left 2
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL LED 656 224 M0
WINDOW 3 -16 87 VLeft 2
SYMATTR InstName D1
SYMATTR Value NSCW100
SYMATTR Description Diode
SYMATTR Type diode
SYMATTR Value2 N=15
SYMBOL res 512 208 R0
SYMATTR InstName R1
SYMATTR Value 100meg
TEXT 296 408 Left 2 !.tran 2
TEXT 704 272 Left 2 !.meas tran Iled avg I(D1) trig V(AC1) val=0 td=1 RISE=1
targ V(AC1) val=0 td=1.2 RISE=1
TEXT 704 296 Left 2 !.meas tran p_in avg I(V1)*V(AC1) trig V(AC1) val=0 td=1
RISE=1 targ V(AC1) val=0 td=1.2 RISE=1
TEXT 704 320 Left 2 !.meas tran p_out avg I(D1)*V(V+) trig V(AC1) val=0 td=1
RISE=1 targ V(AC1) val=0 td=1.2 RISE=1
TEXT 704 344 Left 2 !.meas efficiency PARAM 100*p_out/p_in
TEXT 704 248 Left 2 !.meas tran vled avg V(V+) trig V(AC1) val=0 td=1 RISE=1
targ V(AC1) val=0 td=1.2 RISE=1

ehsjr

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Jan 5, 2012, 12:58:47 PM1/5/12

to

Then you'll need a different power supply and circuit than
what I posted. The above adds up to 33 volts, not 11 volts
as in your prior post.

Review what I posted - the method and math examples are
there so you can figure it out for yourself. You will
need a different power supply, one that is capable of
at _least_ 33 volts output. One possibility is part
number 2101315 from Jameco for $22.95

Ed

Chris Carlton

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Jan 5, 2012, 2:07:10 PM1/5/12

to

@ Ed

I could make the...

1 other piece @

> 3 - 30mA IF 3.8VF
> 3 - 30mA IF 1.7VF
> 3 - 30mA IF 2.2VF

> 3 - 30mA IF 3.3VF

into 2 strings around 16VF and use a couple 500 ohms R
I've noticed that if my strings are tight like your string with 5 ohms
and since I have to round up to the next available ohms R each time, I
get the same R answer no matter if I use 20 or 30mA in my math.

so this set up needs no LM317 type stuff, just the old school. I like
this for the next prototype, mainly cause, I want to finish it in the
next 2 days and I've got all those pieces or they'll be here tomorrow.
I've got a wall wart that puts out a steady 16.4V and makes 2amps. It
will run quite a few of my fixtures, good enough for the upcoming demo
with the $ boys. :)

I need to study JF's bridge rectifier route, and some of the other
LM317 type suggestions too since one of my main selling points is the
massive energy cost diff between my design and other products of it's
type on the market. It's already more than good enough to go to market
now, but this will make for more efficient models in the future. I
don't care really, I'll probably leave it up to the $ guys if they
want to go to market old school or new school. The new school shit is
a turn on and I'll have fun learning it more if I don't have a
deadline involved. If they want new school right away I'll just have
to post a help wanted on this thread and buy a set of registered plans
from one of you guys!

Thanks Again,
What a great board!

John Fields

unread,

Jan 5, 2012, 7:35:45 PM1/5/12

to

---
This isn't a board, it's USENET; the last unfettered in-your-face
access to free speech in the world.

Here's one just for grins; runs off 120 VAC mains and limits the
current into a 4400 ohm load to about 30mA.

No spike suppression and needs a little more cleaning up, but just
presented as a concept.

Version 4
SHEET 1 1580 992
WIRE -448 -192 -624 -192
WIRE -416 -192 -448 -192
WIRE -320 -192 -352 -192
WIRE -160 -192 -320 -192
WIRE 400 -192 -160 -192
WIRE -160 -144 -160 -192
WIRE -448 -128 -448 -192
WIRE 400 -128 400 -192
WIRE -160 -16 -160 -64
WIRE 400 -16 400 -48
WIRE 224 32 224 0
WIRE 192 48 128 48
WIRE -624 64 -624 -192
WIRE 352 64 256 64
WIRE 192 80 160 80
WIRE -160 144 -160 48
WIRE -96 144 -160 144
WIRE 16 144 -96 144
WIRE 224 144 224 96
WIRE 224 144 16 144
WIRE -160 176 -160 144
WIRE 16 176 16 144
WIRE 160 208 160 80
WIRE 400 208 400 80
WIRE 400 208 160 208
WIRE -624 272 -624 144
WIRE 16 272 16 256
WIRE 128 272 128 48
WIRE 128 272 16 272
WIRE 400 272 400 208
WIRE -160 288 -160 240
WIRE -96 288 -96 144
WIRE 16 304 16 272
WIRE 128 304 128 272
WIRE -448 400 -448 -64
WIRE -160 400 -160 352
WIRE -160 400 -448 400
WIRE -96 400 -96 352
WIRE -96 400 -160 400
WIRE 16 400 16 384
WIRE 16 400 -96 400
WIRE 128 400 128 368
WIRE 128 400 16 400
WIRE 352 400 128 400
WIRE 400 400 400 352
WIRE 400 400 352 400
WIRE -448 432 -448 400
WIRE 400 464 400 400
WIRE -624 528 -624 352
WIRE -448 528 -448 496
WIRE -448 528 -624 528
WIRE -416 528 -448 528
WIRE -320 528 -320 -192
WIRE -320 528 -352 528
FLAG 400 464 0
FLAG 224 0 0V
FLAG 352 400 0V
FLAG 224 144 21V
SYMBOL diode -432 -64 R180
WINDOW 0 71 7 Left 2
WINDOW 3 39 33 Left 2

SYMATTR InstName D3
SYMATTR Value MUR460

SYMBOL diode -464 432 R0
WINDOW 0 -57 3 Left 2
WINDOW 3 -103 32 Left 2

SYMATTR InstName D2
SYMATTR Value MUR460

SYMBOL diode -416 -176 R270

WINDOW 3 0 32 VBottom 2
WINDOW 0 32 32 VTop 2

SYMATTR Value MUR460
SYMATTR InstName D4
SYMBOL voltage -624 256 R0
WINDOW 3 24 104 Invisible 2

WINDOW 123 0 0 Left 2

WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 187 60)
SYMATTR InstName V1
SYMBOL res 384 256 R0
WINDOW 3 36 68 Left 2
SYMATTR Value 100
SYMATTR InstName R4
SYMBOL res 384 -144 R0
SYMATTR InstName R7
SYMATTR Value 4700
SYMBOL res -176 -160 R0
SYMATTR InstName R1
SYMATTR Value 12k
SYMBOL zener -144 352 R180
WINDOW 0 24 64 Left 2

WINDOW 3 24 0 Left 2

SYMATTR InstName D5
SYMATTR Value BZX84C15L

SYMATTR Description Diode
SYMATTR Type diode

SYMBOL cap -112 288 R0
SYMATTR InstName C1
SYMATTR Value 220µ
SYMBOL diode -416 544 R270

WINDOW 3 0 32 VBottom 2
WINDOW 0 32 32 VTop 2

SYMATTR Value MUR460
SYMATTR InstName D1
SYMBOL res 0 160 R0
SYMATTR InstName R5
SYMATTR Value 10k
SYMBOL res 0 288 R0
SYMATTR InstName R6
SYMATTR Value 1600
SYMBOL cap 112 304 R0
SYMATTR InstName C2
SYMATTR Value 20µ
SYMBOL diode -176 -16 R0
SYMATTR InstName D6
SYMATTR Value 1N4148
SYMBOL nmos 352 -16 R0
SYMATTR InstName M1
SYMATTR Value IRFP90N20D
SYMBOL Opamps\\LT1007 224 128 M180
SYMATTR InstName U2
SYMBOL zener -144 240 R180
WINDOW 0 24 64 Left 2

WINDOW 3 24 0 Left 2

SYMATTR InstName D7
SYMATTR Value BZX84C6V2L
SYMBOL voltage -624 48 R0
WINDOW 3 24 104 Invisible 2

WINDOW 123 0 0 Left 2

WINDOW 39 0 0 Left 2
SYMATTR Value SINE(0 1000 1000 1 0 0 2)
SYMATTR InstName V2
TEXT 288 424 Left 2 !.tran 2 uic
--
JF

ehsjr

unread,

Jan 6, 2012, 1:05:15 AM1/6/12

to

Chris Carlton wrote:
> @ Ed
>
> I could make the...
>
> 1 other piece @
>
>
>>3 - 30mA IF 3.8VF
>>3 - 30mA IF 1.7VF
>>3 - 30mA IF 2.2VF
>>3 - 30mA IF 3.3VF
>
>
> into 2 strings around 16VF and use a couple 500 ohms R

Excellent choice. :-) 3 at 3.8 plus 3 at 1.7 = 16.5 and
3 at 2.2 plus 3 at 3.3 = 16.5 so r = (28-16.5)/.02 = 575

You can use a 470 ohm in series with a 100 ohm to get 570 ohms,
which is close enough, or you can go with a standard 620 ohms
and get ~18 mA which is also close enough.

The dissipation in each resistor will be .02 * (28-16.5) or
.23 watts so the resistors used should be rated at 1/2 watt
or higher. It's almost always better to use resistors rated
well above the computed dissipation.

> I've noticed that if my strings are tight like your string with 5 ohms
> and since I have to round up to the next available ohms R each time, I
> get the same R answer no matter if I use 20 or 30mA in my math.

You should get different answers with 30 mA:
r = (28-16.5)/.03 = ~383 which would round up to a 390 ohm
standard value resistor, versus r = (28-16.5)/.02 = 575 which
would round up to a 620 ohm standard value resistor. By the
way, you can get closer standard values of resistors if you use
1% resistors but they're not available at Jameco so you'd have
to use an additional supplier with additional shipping cost.

>
> so this set up needs no LM317 type stuff, just the old school. I like
> this for the next prototype, mainly cause, I want to finish it in the
> next 2 days and I've got all those pieces or they'll be here tomorrow.
> I've got a wall wart that puts out a steady 16.4V and makes 2amps. It
> will run quite a few of my fixtures, good enough for the upcoming demo
> with the $ boys. :)

Nice. :-)

>
> I need to study JF's bridge rectifier route, and some of the other
> LM317 type suggestions too since one of my main selling points is the
> massive energy cost diff between my design and other products of it's
> type on the market.

When you use a regulated voltage supply in your project, it's best to
use the single resistor instead of the LM317. Constant current via the
LM317 route will never be more efficient than current limiting via the
single resistor route when using a voltage regulated supply, because
the LM317 needs some headroom. Besides the single resistor is cheaper.

JF's bridge rectifier approach is likely the most efficient of all.

Ed

P E Schoen

unread,

Jan 6, 2012, 6:17:57 AM1/6/12

to

"John Fields" wrote in message
news:6nfcg79g23qtp78as...@4ax.com...

> This isn't a board, it's USENET; the last unfettered in-your-face
> access to free speech in the world.

> Here's one just for grins; runs off 120 VAC mains and limits the
> current into a 4400 ohm load to about 30mA.

> No spike suppression and needs a little more cleaning up, but just
> presented as a concept.

It's only 60% efficient. 4.26W in from V1, 1 watt in the dropping resistor
R1 and 400 mW in the MOSFET, for 2.58 watts out into R7 (presumed LED load).

Paul

Jasen Betts

unread,

Jan 6, 2012, 11:31:54 PM1/6/12

to

On 2012-01-04, ehsjr <eh...@nospamverizon.net> wrote:
> Chris Carlton wrote:
>> sorry Ed, I must be too thick in the head like Dad used to say...

> You make 3 series strings:
> String 1: 12 3.8V LEDs; total Vf = 45.6
> String 2: 3 3.8V LEDs + 9 3.3V LEDs; total Vf = 41.1
> String 3: 15 1.7V LEDs + 9 2.2V LEDs; total Vf = 45.3
>
> Then:
> ------------- r1
> | 48V +|----+---/\/\/---String1---+
> | LED Supply | | |
> | | | r2 |
> | Jameco | +---/\/\/---String2---+
> | Part # | | |
> | 2101121 | | r3 |
> | | +---/\/\/---String3---+
> | | |
> | -|--------------------------+
> -------------
>
> If you choose 20 mA for the current:
> r1 = (48-45.6)/.02 or 120 ohms
> r2 = (48-41.1)/.02 or 345 ohms
> r3 = (48-45.3)/.02 or 135 ohms

Those resistors are way too small.

Vf on leds can vary by 15%

(here'a datasheet chosen at random)
http://www.hebeiltd.com.cn/led.datasheet/530AG7C.pdf

so you need to drop atleast 20% of the supply voltage to get
even half-way predictable behaviour.

If efficiency is important go the LM317 route because that requires
less head-room (3V (IIRC) instead of 20%) although this is less significant
with a 19V supply.

OTOH the numbers of each voltage of led are divisible by three,
so if you divide each type evenly among the strings
(producing three identical strings) and use 150 ohm resistors,
and the leds (of each type) all came from the same bin they'll
probably work OK

each string:
(5x3.8V) + (3x3.3v) + (3x2.2v) + (5 x 1.7V) + 150 Ohms

--
⚂⚃ 100% natural

Jasen Betts

unread,

Jan 7, 2012, 4:36:20 AM1/7/12

to

On 2012-01-06, John Fields <jfi...@austininstruments.com> wrote:
>
> Version 4
> SHEET 1 1580 992
> WIRE -448 -192 -624 -192
> WIRE -416 -192 -448 -192
> WIRE -320 -192 -352 -192
> WIRE -160 -192 -320 -192

yow! dissipathion in R1

why not instead of the mosfet use a darlington with (eg:) 100K pull-up on
the base and a TL431 watching the bottom resistor and stealing the base
current,

this does away with the op-amp and related low voltage DC supply.

82 ohms looks about right for the bottom resistor.

basically figure 14 here:
http://www.datasheetcatalog.org/datasheets/90/321931_DS.pdf

still that 100K is going to get warm.

--
⚂⚃ 100% natural

--- Posted via news://freenews.netfront.net/ - Complaints to ne...@netfront.net ---

ehsjr

unread,

Jan 8, 2012, 12:43:27 AM1/8/12

to

Something is out of whack. Your total Vf is 44 volts:
5x3.8 = 19V; 3x3.3 = 9.9V; 3x2.2 = 6.6V; 5x1.7 = 8.5V
19V + 9.9V + 6.6V + 8.5V = 44V
That, plus your 150 ohm resistor, yields 26.6 ma
through the LEDs. (48-44)/150 = 26.6 mA

So you're going to run the string at 26.6 mA instead of 20 mA,
and you haven't used the 15% variation figure in Vf that you
mentioned. How is running at 88.6% of maximum current better
than running at 66.6% of maximum current as in the previous
strings (below) run at 20 mA?

r1 = (48-45.6)/.02 or 120 ohms
r2 = (48-41.1)/.02 or 345 ohms
r3 = (48-45.3)/.02 or 135 ohms

Your string wastes 80 more milliwatts, exposes the LEDs
to 30% more current, and doesn't account for the variation
in LED Vf you assume. So - I don't get it.

The engineering decision between using an LM317 plus resistor
in constant currant versus a single resistor to accommodate
worst case Vf variation _requires_ a higher supply voltage.
If Vf varies by 15%, then you need at least 47.146 + headroom
for the LM317 for each string, figuring 7.5% higher. So that
means a design change to a 50 volt supply. Then, if the LED Vf
worst case variation is lower rather than higher, your 44 volts
becomes 35.2V. Each string will need to drop 14.8 volts at
26.7 mA in the 317, or about .395 watts, and with 3 strings
that's close to 1.2 watts wasted. That solution is bulletproof
to Vf variation, and costs a lot more for the supply, if you can
find one.

The proposed solution you object to dissipates .048 watts in
r1 (string1), .144 watts in r2 (string2), and ~.06 watts in
r3 (string3), or about .252 watts total wasted power. It uses
a cheap and available 48 volt supply designed for LEDs. It
is not bulletproof to Vf variation.

So, if "bulletproof" at higher cost trumps efficiency at lower
cost, choose the LM317 approach. If efficiency at lower cost
trumps "bulletproof", choose the single resistor. And if both
"bulletproof" and low cost have to be part of the design,
sacrifice efficiency and brightness and use higher value
resistors. The OP's primary requirement was efficiency. Next
was low cost - implied or stated, I don't remember which.
Then there was a question about whether connecting directly
to the mains was a safe approach. Mixing all that together
yielded the posted design. Do you have a design that does
all of that?

Ed

P E Schoen

unread,

Jan 8, 2012, 2:52:24 AM1/8/12

to

"ehsjr" wrote in message news:jebae3$sqc$1...@news.eternal-september.org...

[snip]

> So, if "bulletproof" at higher cost trumps efficiency at lower
> cost, choose the LM317 approach. If efficiency at lower cost
> trumps "bulletproof", choose the single resistor. And if both
> "bulletproof" and low cost have to be part of the design,
> sacrifice efficiency and brightness and use higher value
> resistors. The OP's primary requirement was efficiency.
> Next was low cost - implied or stated, I don't remember
> which. Then there was a question about whether connecting
> directly to the mains was a safe approach. Mixing all that
> together yielded the posted design. Do you have a design
> that does all of that?

Here is a design that provides a very accurate 18 mA over a supply range of
35 to 48 VDC.

Parts are less than a dollar, except for the power supply. The same
principle could be used for a direct line supply with a FWB and a higher
voltage MOSFET.

But still, the buck regulator ICs are most efficient and still very cheap.

Paul
===================================================

Version 4
SHEET 1 880 680
WIRE 208 -32 -64 -32
WIRE 336 -32 208 -32
WIRE 336 -16 336 -32
WIRE 208 64 208 -32
WIRE 336 96 336 48
WIRE -64 144 -64 -32
WIRE 208 176 208 144
WIRE 288 176 208 176
WIRE 336 240 336 192
WIRE -64 416 -64 224
WIRE 208 416 208 368
WIRE 208 416 -64 416
WIRE 336 416 336 320
WIRE 336 416 208 416
FLAG 336 416 0
SYMBOL res 320 224 R0
SYMATTR InstName R1
SYMATTR Value 50
SYMBOL diode 192 176 R0
SYMATTR InstName D1
SYMATTR Value 1N4148
SYMBOL diode 192 240 R0
SYMATTR InstName D2
SYMATTR Value 1N4148
SYMBOL res 192 48 R0
SYMATTR InstName R2
SYMATTR Value 100k
SYMBOL LED 320 -16 R0
SYMATTR InstName D3
SYMATTR Value NSCW100

SYMATTR Description Diode
SYMATTR Type diode

SYMATTR Value2 n=10
SYMBOL voltage -64 128 R0

WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2

SYMATTR InstName V1
SYMATTR Value PWL(0 0 .1 40 1 48)
SYMBOL nmos 288 96 R0
SYMATTR InstName M1
SYMATTR Value Si5515_N
SYMBOL diode 192 304 R0
SYMATTR InstName D4
SYMATTR Value 1N4148
TEXT -16 376 Left 2 !.tran 1 startup

John Fields

unread,

Jan 10, 2012, 9:11:57 PM1/10/12

to

On Sun, 08 Jan 2012 00:43:27 -0500, ehsjr <eh...@nospamverizon.net>
wrote:

>So, if "bulletproof" at higher cost trumps efficiency at lower
>cost, choose the LM317 approach. If efficiency at lower cost
>trumps "bulletproof", choose the single resistor. And if both
>"bulletproof" and low cost have to be part of the design,
>sacrifice efficiency and brightness and use higher value
>resistors. The OP's primary requirement was efficiency. Next
>was low cost - implied or stated, I don't remember which.
>Then there was a question about whether connecting directly
>to the mains was a safe approach. Mixing all that together
>yielded the posted design. Do you have a design that does
>all of that?
>
>Ed

---
Just for fun, I went back to your constant current circuit to see what
would be required to drive the OP's 48 LED, 30mA string from the mains
and built this, in the real world.

+-----+ R1 U1 R2
120AC>---|~ +|---[100R]-+-[LM317]---[39R]-+
| | | | |
| | | +------------+
| | [3.9�F] |R3
| | |C1 [LED STRING]
| | | |4400R
120AC>---|~ -|----------+-----------------+
+-----+

It works pretty well,

LINE ACIN I(R3)
VRMS mADC
-----+------+-------
LOW 108 28.24
NOM 120 30.37
HIGH 132 31.9

is dirt cheap and survives line surges, but I haven't done any spike
testing yet.

Maybe tomorrow.

Anyway, I just thought I'd run it up the flagpole and see if anyone
salutes. ;)

--
JF

P E Schoen

unread,

Jan 10, 2012, 9:39:25 PM1/10/12

to

"John Fields" wrote in message

news:spqpg7tkng4dba3vc...@4ax.com...

> is dirt cheap and survives line surges, but I haven't done any
> spike testing yet.

> Maybe tomorrow.

> Anyway, I just thought I'd run it up the flagpole and see if anyone
> salutes. ;)

Maybe add a 35V zener across the LM317 and a capacitor across the LED string
so spikes will be limited by the 100R series resistor and voltages across
the regulator and the LEDs will stay within their limits. The 100R could be
replaced by a polyswitch self-resetting fuse.

I have about 400 surplus pieces of LM317HVH (rated for 57V) in the metal can
TO-39 package if you let the magic smoke out of yours...

They are special order and $7 each (500pc) from DigiKey and Newark. The NOPB
version is in stock at Mouser for $11.60. I'll sell 'em for a buck a pop.

Paul

ehsjr

unread,

Jan 10, 2012, 10:40:13 PM1/10/12

to

Nice. R1 needs to be higher - say 470 - to accommodate 132V ACIN,
and that reduces the deltaV across the 317 for all line voltages.
Also, I'm a wimp where it comes to running LEDs near their ratings,
so I'd use 51 ohms or more for R2 to keep the current lower than
30 mA.

Ed

John Fields

unread,

Jan 12, 2012, 12:05:57 PM1/12/12

to

On Tue, 10 Jan 2012 22:40:13 -0500, ehsjr <eh...@nospamverizon.net>
wrote:

>John Fields wrote:
>> On Sun, 08 Jan 2012 00:43:27 -0500, ehsjr <eh...@nospamverizon.net>
>> wrote:
>>
>>
>>>So, if "bulletproof" at higher cost trumps efficiency at lower
>>>cost, choose the LM317 approach. If efficiency at lower cost
>>>trumps "bulletproof", choose the single resistor. And if both
>>>"bulletproof" and low cost have to be part of the design,
>>>sacrifice efficiency and brightness and use higher value
>>>resistors. The OP's primary requirement was efficiency. Next
>>>was low cost - implied or stated, I don't remember which.
>>>Then there was a question about whether connecting directly
>>>to the mains was a safe approach. Mixing all that together
>>>yielded the posted design. Do you have a design that does
>>>all of that?
>>>
>>>Ed
>>
>>
>> ---
>> Just for fun, I went back to your constant current circuit to see what
>> would be required to drive the OP's 48 LED, 30mA string from the mains
>> and built this, in the real world.
>>
>>
>> +-----+ R1 U1 R2
>> 120AC>---|~ +|---[100R]-+-[LM317]---[39R]-+
>> | | | | |
>> | | | +------------+

>> | | [3.9湩] |R3

>> | | |C1 [LED STRING]
>> | | | |4400R
>> 120AC>---|~ -|----------+-----------------+
>> +-----+
>>
>> It works pretty well,
>>
>> LINE ACIN I(R3)
>> VRMS mADC
>> -----+------+-------
>> LOW 108 28.24
>> NOM 120 30.37
>> HIGH 132 31.9
>>
>> is dirt cheap and survives line surges, but I haven't done any spike
>> testing yet.
>>
>> Maybe tomorrow.
>>
>> Anyway, I just thought I'd run it up the flagpole and see if anyone
>> salutes. ;)
>>
>
>Nice. R1 needs to be higher - say 470 - to accommodate 132V ACIN,
>and that reduces the deltaV across the 317 for all line voltages.

---
If I used 500 ohms it'd drop 15V at 30mA - which wouldn't let me get
30mA into the LED string with low line in - and it'd dissipate 450
milliwatts when there was 30mA through it.

The trick to why it works with 100 ohms in there is that the LM317
doesn't start waking up until the voltage differential between the
input and output is a volt or so, so until then it acts pretty much
like a saturated NPN series pass transistor and the voltage across the
load will follow the input voltage minus a volt or so.

Then, as the voltage across the load rises, the current through it
will increase until the voltage dropped across R2 increases to 1.25
volts, when the LM317 will start limiting the current into the load to
30mA until the voltage across R2 falls to less than 1.25 volts later
on in that mains half-cycle.

Doing it that way only drops 3 volts across R1 and causes it to
dissipate 90mW with 30mA through it.

Also, the value of C1 was chosen to allow it to charge only to the
voltage required to get 30 mA into the string at low line, with the
LM317 doing the required limiting to keep the current in the string
relatively constant from low line to high line.

I've posted some photos to abse:

news:ng0ug7hf2jrhh49sn...@4ax.com

I didn't post any of the output voltages into the LED string, and I
want to get this on its way, so I'll do it later on this afternoon
when I have more time.

>Also, I'm a wimp where it comes to running LEDs near their ratings,
>so I'd use 51 ohms or more for R2 to keep the current lower than
>30 mA.

---
Tsk, tsk, tsk. ;)

--
JF

ehsjr

unread,

Jan 13, 2012, 2:49:10 AM1/13/12

to

Thanks!

>
> I've posted some photos to abse:
>
> news:ng0ug7hf2jrhh49sn...@4ax.com

You're correct, and it's much appreciated. I have no idea where
I erred on the higher value for R1, except that I may have typoed
the calculator. Typo or not, it's a glaring error. :-(

I know your post made it to ABSE, because I see the post where
JT asked about it, and your response to him. The bad thing is
that your post didn't show up on news.eternal-september.org
Makes me think I'd better get a different newsgroups provider.

>
> I didn't post any of the output voltages into the LED string, and I
> want to get this on its way, so I'll do it later on this afternoon
> when I have more time.
>
>
>
>>Also, I'm a wimp where it comes to running LEDs near their ratings,
>>so I'd use 51 ohms or more for R2 to keep the current lower than
>>30 mA.
>
>
> ---
> Tsk, tsk, tsk. ;)
>

Hey, that's my story and I'm sticking to it. :-)

Ed

ehsjr

unread,

Jan 13, 2012, 2:04:35 PM1/13/12

to

In my prior reply I said: "I have no idea where

I erred on the higher value for R1, except that I may have typoed
the calculator. Typo or not, it's a glaring error."

Now I realize what I did - and I was correct, 100 ohms for
R1 is too low using the LM317 and figuring worst case Vin
high, but I didn't think of the high voltage version of the
LM317. Your circuit using the high voltage version is better. :-)

The wasted power will always be the dissipation of R1 plus the
dissipation of the LM317, so it really doesn't matter how much
is wasted in R1 in terms of circuit efficiency.

Ed

John Fields

unread,

Jan 14, 2012, 5:26:52 PM1/14/12

to

On Fri, 13 Jan 2012 14:04:35 -0500, ehsjr <eh...@nospamverizon.net>
wrote:>>

>In my prior reply I said: "I have no idea where
>I erred on the higher value for R1, except that I may have typoed
>the calculator. Typo or not, it's a glaring error."
>
>Now I realize what I did - and I was correct, 100 ohms for
>R1 is too low using the LM317 and figuring worst case Vin
>high, but I didn't think of the high voltage version of the
>LM317. Your circuit using the high voltage version is better. :-)
>
>The wasted power will always be the dissipation of R1 plus the
>dissipation of the LM317, so it really doesn't matter how much
>is wasted in R1 in terms of circuit efficiency.
>
>Ed

---
True, but with the mains varying +/- 10% about the 120V nominal, R1's
being equal to 470 ohms won't allow the current - over the range of
mains variations - into the LED string to be the 30mA the OP asked
for.

With R1 equal to 100 ohms, and a 4400 ohm load resistance, (the
equivalent resistance of the LED string) the voltage across the LM317
never exceeds the absolute maximum rating at high line and, with no
heat sink, the temp of the LM327's tab never exceeds about 65C with a
25C ambient, so the circuit's pretty safe with a standard LM317.

Transient protection is something else, again, and the HV may have an
advantage there due to its inherent standoff.

Any thoughts?

--
JF

ehsjr

unread,

Jan 15, 2012, 1:39:57 PM1/15/12

to

John Fields wrote:
> On Fri, 13 Jan 2012 14:04:35 -0500, ehsjr <eh...@nospamverizon.net>
> wrote:>>
>
>
>>In my prior reply I said: "I have no idea where
>>I erred on the higher value for R1, except that I may have typoed
>>the calculator. Typo or not, it's a glaring error."
>>
>>Now I realize what I did - and I was correct, 100 ohms for
>>R1 is too low using the LM317 and figuring worst case Vin
>>high, but I didn't think of the high voltage version of the
>>LM317. Your circuit using the high voltage version is better. :-)
>>
>>The wasted power will always be the dissipation of R1 plus the
>>dissipation of the LM317, so it really doesn't matter how much
>>is wasted in R1 in terms of circuit efficiency.
>>
>>Ed
>
>
> ---
> True, but with the mains varying +/- 10% about the 120V nominal, R1's
> being equal to 470 ohms won't allow the current - over the range of
> mains variations - into the LED string to be the 30mA the OP asked
> for.

Yup. That's why your 100 ohms is better than the 470 (which is
too small, anyway, with the regular LM317) if using the HV
version LM317.

>
> With R1 equal to 100 ohms, and a 4400 ohm load resistance, (the
> equivalent resistance of the LED string) the voltage across the LM317
> never exceeds the absolute maximum rating at high line and, with no
> heat sink, the temp of the LM327's tab never exceeds about 65C with a
> 25C ambient, so the circuit's pretty safe with a standard LM317.
>

Here's where I'm missing something: with AC in at 132, the peak
after the bridge is around 185. With 3V across R1, that's 182
on the input side of the LM317, and 132 on the output side,
exceeding the 37 V max Vin-Vout. Even the 470 is too small,
dropping only ~14.1 at 30 mA and yielding Vin-Vout ~39.

> Transient protection is something else, again, and the HV may have an
> advantage there due to its inherent standoff.
>
> Any thoughts?
>

Yes, but a different topology at a lot higher cost. I'd
use a low voltage LED supply & several LED strings. The
supply takes care of the transients and safety issues, and
current limiting can be the 317 circuit or a resistor.

If using the mains-bridge-limiter design, a TVS across the
'lytic seems a good way to go, and a .1 uF should be there too.

Paul proposed a zener across the 317 and a cap across the
LEDs which should help, and you can work on the AC in
side prior to the bridge. I'd also worry about a switch
placed between the cap and the 317.

Ed