--- 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:
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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.
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:
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.
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.
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!!
"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.
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
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.
"Chris Carlton" wrote in message=20 news: snipped-for-privacy@q17g2000yqh.googlegroups.com...=
I made a very simple LTSpice circuit that uses a 24 VAC "Wallwart", 2=20 capacitors, and two diodes, which provides about 28 mA into a string of =
15=20 white LEDs with a voltage drop of about 57 VDC. And the current is = limited=20 to less than 150 mA if the output is shorted (use 1 ohm for R1). Here is = the=20 error log (use Ctrl-L), which shows the efficiency to be better than =
94%:
.OP point found by inspection.
iled: AVG(i(d1))=3D0.028176 FROM 0.999989 TO 1.19998 p_in: AVG(i(v1)*v(ac1))=3D-1.70889 FROM 0.999989 TO 1.19998 p_out: AVG(i(d1)*v(v+))=3D1.60876 FROM 0.999989 TO 1.19998 efficiency: 100*p_out/p_in=3D-94.1404 vled: AVG(v(v+))=3D56.9075 FROM 0.999989 TO 1.19998
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D 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 100=B5 SYMATTR Description Capacitor SYMATTR Type cap SYMATTR SpiceLine V=3D63 Irms=3D2.51 Rser=3D0.025 Lser=3D0 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=3D.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 47=B5 SYMATTR Description Capacitor SYMATTR Type cap SYMATTR SpiceLine V=3D63 Irms=3D2.51 Rser=3D0.025 MTBF=3D5000 Lser=3D0 = ppPkg=3D1 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=3D15 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=3D0 = td=3D1 RISE=3D1=20 targ V(AC1) val=3D0 td=3D1.2 RISE=3D1 TEXT 704 296 Left 2 !.meas tran p_in avg I(V1)*V(AC1) trig V(AC1) = val=3D0 td=3D1=20 RISE=3D1 targ V(AC1) val=3D0 td=3D1.2 RISE=3D1 TEXT 704 320 Left 2 !.meas tran p_out avg I(D1)*V(V+) trig V(AC1) = val=3D0 td=3D1=20 RISE=3D1 targ V(AC1) val=3D0 td=3D1.2 RISE=3D1 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=3D0 = td=3D1 RISE=3D1=20 targ V(AC1) val=3D0 td=3D1.2 RISE=3D1=20
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
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!
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.
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.
Nice. :-)
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.
It's only 60% efficient. 4.26W in from V1, 1 watt in the dropping = resistor=20 R1 and 400 mW in the MOSFET, for 2.58 watts out into R7 (presumed LED = load).
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
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:
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still that 100K is going to get warm.
--
?? 100% natural
--- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net
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?
Here is a design that provides a very accurate 18 mA over a supply range = of=20
35 to 48 VDC.
Parts are less than a dollar, except for the power supply. The same=20 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 =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D
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=3D10 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
--- 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.
Maybe add a 35V zener across the LM317 and a capacitor across the LED = string=20 so spikes will be limited by the 100R series resistor and voltages = across=20 the regulator and the LEDs will stay within their limits. The 100R could = be=20 replaced by a polyswitch self-resetting fuse.
I have about 400 surplus pieces of LM317HVH (rated for 57V) in the metal = can=20 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=20 version is in stock at Mouser for $11.60. I'll sell 'em for a buck a = pop.
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
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