LED wiring Series or Parallel 110 or 12v (for Power Efficien

[Michael A. Terrell]

Chris Carlton wrote:

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

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]

On Wed, 4 Jan 2012 03:57:08 -0800 (PST), Chris Carlton
<personalgrowthnow@yahoo.com> wrote:

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.

---
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]

Chris Carlton wrote:

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

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]

Chris Carlton wrote:

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

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]

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!!

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 4 Jan 2012 05:00:26 -0800 (PST), fungus <tooby@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.

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.

 

[ehsjr]

Chris Carlton wrote:

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!!

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]

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]

"Chris Carlton" wrote in message
news:9edb98f9-1dc3-49d0-8fb4-c5e8a359130d@q17g2000yqh.googlegroups.com...

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

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 100ľ
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 47ľ
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]

Chris Carlton wrote:

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

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]

@ 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]

On Thu, 5 Jan 2012 11:07:10 -0800 (PST), Chris Carlton
<personalgrowthnow@yahoo.com> 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
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!

---
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
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WIRE -96 288 -96 144
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WIRE 128 304 128 272
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WIRE -160 400 -448 400
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WIRE -96 400 -160 400
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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 220ľ
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 20ľ
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]

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

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!

 

[P E Schoen]

"John Fields" wrote in message
news:6nfcg79g23qtp78asla3muru189dmpsase@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]

On 2012-01-04, ehsjr <ehsjr@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]

On 2012-01-06, John Fields <jfields@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 news@netfront.net ---

 

[ehsjr]

Jasen Betts wrote:

On 2012-01-04, ehsjr <ehsjr@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

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]

"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
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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]

On Sun, 08 Jan 2012 00:43:27 -0500, ehsjr <ehsjr@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]-+
| | | | |
| | | +------------+
| | [3.9ľF] |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]

"John Fields" wrote in message
news:spqpg7tkng4dba3vcaqrbntg46ljk9efee@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