This Curcuit dont do diddley right :-) Single Li-ion

VidPro

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http://www.dealextreme.com/details.dx/sku.1886
this is the curcuit

http://www.micro-bridge.com/data/ADD/AMC7135.pdf
and this is the chip type it uses.

so suppose somebody wanted to use these for a SINGLE li-ion (as indicated), after all they wont work with 2x li-ion with max voltage at 6v.
my entire post is based on and intended for SINGLE li-ion , with 4.2-3.0V capability
Please speak in english , so i can understand, and correct any of my mis-conceptions, or help me to understand this stuff.

they are JUST Current control items right? and it would REQUIRE the VF of the input battery to be higher than the Vf of the led , cause it aint a boost right?

they get to 1400ma by parelleling a bunch of them together , even though each chip is only capable of about 350ma, wouldnt there be harmonic current distortions from parelelling multiple items like this, that could cause wide changing fluxuations in total output current. (note graphs seem to display this possibility, scope would be needed to see it occuring) A graph for a similar item is here https://www.kaidomain.com/WEBUI/ProductDetail.aspx?TranID=1806

the diode is going to drop a lot of voltage, but there are 2 diodes, so will the diodes drop even more in the configuration used here, or is it one diode drop total, not 2 total in series in the curcuit? THIS WAS WRONG, diode does not effect

so this puppy is going to be dropping into direct drive as soon as the Input voltage drops below the led voltage for the drive current being applied?

if you want to push some 1.4A through a Cree Q5 you gonna need some 3.9V about , so your in direct drive WITH the additional curcuit losses some 15 minutes into the run when the voltage drops.

removed wrong part

so potentially all this thing is going to accomplish is knocking down the high end of the li-ion voltage for a few minutes, IF that, while the rest of the time it wastes power for no usefull reason?

ok so they put the INput voltage at 2.7v in the specs. well so what, without a very low LED voltage, it was in direct drive long ago, mabey usefull for li-ion and a red led?
Because it doesnt Boost, reguardless of the fact that the chip will operate at 2.7v there aint gonna be 3.9v going to the led anyways right , unless the battery voltage is higher. the input voltage of 2.7v does not magically make it function as a boost item, its just the range of possible input voltages right?

so this type of curcuit is usefull for about ONLY some 6V worth of declining battery, chip away some voltage for the curcuit losses, and slam the led with 1.4 amps pushing up the Vf of the led?

BUT they say , hey this will work with a li-ion, BECAUSE the chip still operates at 2.7v, well so what , the led isnt going to be running at the 1.4a long before that time, and the extra diode should have your average white led not even triggering, before the curcuit ceases to function.

so it might "work" with a Single Li-ion, but it will be as regulated as the battery itself, but wasting power in some chip thingee?.

is my assessment of this accurate? me dont see no caps, or inductors, or switching curcuits, just mabey some current limiter, with the Usual Very short range of voltage possibilites.

begin Rant: ------------------------------------------------
i am getting aufully tired of seeing little out there that does what it says it does, jumping back to providing excelent graph of the battery discharging,and people calling that regulation.
Today i have observed in THIS FORUM more than 8 flashlight graphs that show that the lower voltage curcuits went into Direct drive with a fully charged Li-ion, the worst Possible time to be in DD, and the only major need to have current regulation. Lights that are sold as claiming to be good for li-ion bateries

There are graphs that show the higher voltage curcuits (opposite) , fell out of regulation the opposite direction and went into direct drive as soon as the voltage dropped a bit to around 3.6-3.8v. leaving the battery to do the regulation, just like a Resistered drive would have.

YET, i still have to hear people IN CPF, telling me that this wonderfull flashlight was Regulated :) oh it was WHEN ? Well it was regulated when it LEAST needed to be regulated. it was not regulated when the voltage was too high, went into direct drive at the worst possible moment. (then reverse that) The high voltage regulation falls out of regulation, when its MOST needed, to keep the voltage up across the led and maintain the current.

What kind of garbage are they selling us into, when the electronics geniouses dont call them on this flim flam sham of regulation, that does nothing more than waste power, and have a cute curcuit in there :). this is a great wonderfull curcuit, WHEN used with a very specific voltage of battery, and li-ion is not it?

single li-ion current controlled curcuits, are a difficult voltage area for silicoln gates, but why dont they admit, that, quit BS ing us, and quit wasting power and chips for little to no reason? there ARE curcuits out there that are fully regulated at each point of the li-ion voltage capability, but it seems to take more than your average boost or buck curcuit to do so, some sort of high lossy DC-DC device instead.

how am i supposed to find a efficient current control for a single li-ion, when the specs are all Muddled about, incorrect, or factually wrong statements are made about the capacity of the curcuits to do the job. and i aint just talking about cheap ones. sombody is just trying to fool us electronic ignorants.
the graphs on flashlights and curcuits are showing whats going on, but does anybody say "ummm err hey, that is out of regulation"
 
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Hi,

I just thought i would mention that the dropout voltage is not zero, it is
around 0.1v or a little higher. This means if you want to drive an LED
at 3.9v at 350ma you need a min of 4.0 volts, not 3.9 volts, and at
4.0 volts you will start to see a decrease in current below 350ma and
how low it goes depends a lot on the curve of the LED.
If you are using four of the basic devices in parallel, then the same applies
except the current is 4 times higher.

If you are also asking about the stability of the devices when 4 of them
are paralleled, that is difficult to answer as you already seem to have
guessed. It depends on how the circuit is designed. Some circuits work
just find when paralleled, and are even made to be paralleled. This one
sounds like it is, but sometimes the only way to tell is to buy 4 of them
and try it. You could also try getting the internal schematic if the
company makes it available as that could help determine the operation
when two or more are paralleled. Another idea is to contact the company
and ask them if they have ever paralleled their devices, and there may
even be something about it on the data sheet.
If there does happen to be instablity sometimes a small capacitor placed
at the right place in the circuit helps stop any oscillations. It's hard to say
if this would work with this circuit or if it even needs it.
Also, sometimes a circuit oscillates a little but as long as it doesnt get
hot as a result it might still function ok.
 
afaik, the diode is there to protect the "regulators" againts reverse-polarity. it is not connected "directly" anywhere in the path between the led and power source, hence, virtually 0 voltage drop to the LED.

as for being "regulated", well, the amc7135 *is* a current regulator so a light that uses them is a "regulated" light. in this case, the regulator is there to "limit" current since it is what the amc7135 is designed to do.
 
ok so the diode is on the battery side, like i mentioned, which further offsets the Battery Vf from the LEd Vf, and drops voltage at the input side.
IF this curcuit is only a current clipper, and not any sort of boost capability, then the VF of the battery and the Vf of the LEd are going to be rather important right?
and if i clip off .7v of 4.0v be it at the battery or at the led, i still have a direct resulting loss of .7/4 ? or some 12-17% loss right off the top.

the only problem i have, is will it Work correctally with a SINGLE li-ion, because that is what i got, and i cant dtetermine how in heck this accomplishes that.
 
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There are examples all over the Candlepower forum of the regulation that exists. you can also hear it in How people talk about how "bright" or long the runtime is, Depending on the battery used. or the state of the battery.

but the best clue to all of this, is the graphs.
i present 5 graphs that are generally displayed for Time Vrses Output of these lights

Graph1.gif

the hedgetrimmer, one of the few graphs that represents an actual full regulation, with proper declining output as the batteries are exhausted, so the battery is not ruined.
this can usually be found when the light is being run with a curcuit that works, at the voltages of battery that are being put into it.
it would be Rare to see a li-ion single battery putting out this kind of graph with the present curcuits being used.
generally this is seen when the curcuit is using primary 3V cells and is your low voltage curcuit.
or when using high voltage current controls and the input voltage is sufficiently higher than the LED voltage.
or when using DC-DC current controls that have proper cutoffs , and dont wipe out a battery.

Graph2.gif


The RamerJammer, the "wow" curcuit, wow that is bright, ya well its in DD then.
Some Clue that it was going into overdrive would have been the 255Lumens comming out the front of the flashlight :)
this battery curcuit led combination goes into severe overdrive, just like Direct drive would have, but only when the voltage of the battery is to high for the curcuit.
The silicon resistered Direct drive :) it still regulates, once it gets around to it, and tapers off properly so it doesnt reverse charge the cells.
other than the huge overdrive at the first, its regulated, just not when it would be MOST needed.


Greph3.gif

The half pipe, poor little current control curcuit, cant do anything right :)
it goes into overdrive when the voltage is to high, and when the voltage it to low for it, the poor thing tries to hard to keep it going, and fails again
nothing wrong with it struggling near the end there, as that wont effect anything , but it still is in overdrive when the voltage is to high.
the halfpipe doesnt usually destroy batteries because it does have trouble near the end, and usually becomes a dwindler

Graph4.gif


The dwindler, you see this graph often when a single li-ion is used in place of 2 Primary 3v cells, or 4 primary alkalines
this is almost direct drive, cept for the total waste of the curcuit exisiting in the path.
with this combination, a person would have been better off to toss the curcuit and get 20% more runtime or output.
this is what you often get with the higher voltage curcuits, when there is not higher voltages.

The Dwindler is a battery discharge curve :) not regulation, and generally tossing the curcuit and using DD would have worked better for longer.

Graph5.gif


and finnaly Killer, the Reverse charger of batteries, this curcuit will drain a battery to complete 0, will draw way more current as the battery voltage is declining, and has the potential to reverse charge SERIES cells.
this can often be found in boost drivers that know no end. or DC-DC current devices with no end.
its perfect regulation, but with the wrong battery combo, it kills batteries.

Graph6.gif


one more here, LIMPEY current control that fluxuates a bit with input voltage.
still usable current control it just isnt straightlined, is a bit nicer on the batteries when they are low, but still doesnt maintain some perfect regulation. its regulated and current controlled , just the curcuit doesnt keep it perfect, and fluxuates a bit with input.
 
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ok so the diode is on the battery side, like i mentioned, which further offsets the Battery Vf from the LEd Vf, and drops voltage at the input side.
IF this curcuit is only a current clipper, and not any sort of boost capability, then the VF of the battery and the Vf of the LEd are going to be rather important right?
and if i clip off .7v of 4.0v be it at the battery or at the led, i still have a direct resulting loss of .7/4 ? or some 12-17% loss right off the top.

the only problem i have, is will it Work correctally with a SINGLE li-ion, because that is what i got, and i cant dtetermine how in heck this accomplishes that.

the only thing that's "seeing" the voltage drop of the diode are the regulators. more info here.

you might be asking too much for such a "cheap" regulator that only does "passive regulation"(ie. limit current). its not and never was designed to be either a voltage regulator or a constant-current regulator. to me, it's basically just a "glorified" resistor. :grin2:
 
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the only thing that's "seeing" the voltage drop of the diode are the regulators. more info here.

you might be asking too much for such a "cheap" regulator that only does "passive regulation"(ie. limit current). its not and never was designed to be either a voltage regulator or a constant-current regulator. to me, it's basically just a "glorified" resistor. :grin2:


ok so it IS just a current limiter, that is probably the main thing i needed to know.
and if i put a single li-ion on, and the voltage of the battery drops below the voltage needed on the LED, then i am not going to have 1.4Amps of current?

well i wasnt asking to much of the curcuit :) it would work great with 4AAs in a 5mega adapter or something. i was asking for truth in advertising. or awareness in selling :)
that way i can buy the right item for the battery and led. this stuff says "Can use Li-ion" all over it, flashlights Too. but they go into direct drive, or are poorly regulated, i can do that with a 2c resister :)

if they are going to toast my led anyways, or cant keep up the power, well heck then i will use the stupid resister again.

ya see george doesnt have any MaxFlexes, (sniff sniff) and i am trying to assemble a light, and about to go with a 1ohm resister , which works wonderfully. i hear all about the miracle drivers, but for Single li-ion , i dont see a miracle.

i got one huge li-ion, and 3 leds i want to drive HARD, at least when I want to, not always. Know any curcuits that would do that with say less than 20% losses?
 
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its not and never was designed to be either a voltage regulator or a constant-current regulator. to me, it's basically just a "glorified" resistor. :grin2:
The AMC7135 works well as a glorified resistor. As long as the input conditions are met, it is a constant current regulator. Why do you say it's not designed to be that?
 
the only thing that's "seeing" the voltage drop of the diode are the regulators. more info here.

that is a great thread, in an unexpected location, shoulda been in lectronics or modding.

i see now what your saying about the diodes, its part of the chips external parts (as shown in the chips data sheet) , not reverse protection for the whole curcuit itself, so it doesnt drop the voltages. now i understand. (somebody woke up finnaly)

and its a glorified resister, because to regulate it takes the extra power into itself, heating up itself, and wasting the power, instead of "converting" the power to usefullness , it blows it out its backside like a resister would :)
 
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The AMC7135 works well as a glorified resistor. As long as the input conditions are met, it is a constant current regulator. Why do you say it's not designed to be that?

the run time graphs cleary shows that it is not. ;)

a constant-current regulator is supposed to maintain constant output during regulation, hence, a relatively flat run time graph. it "actively" tries to draw increasing current as input voltages goes down in order to maintain constant output.

the amc7135 just limits current. if there's too much voltage that's causing too much current, it is held back by the regulator to the rated current limit and turns it into heat. with the ic having an input voltage range of 2.7-6v, is there any other reason that its not getting the "input conditions" required to maintain constant output? the runtime graphs that i've seen shows that output starts dropping as soon as you turn the light on. :grin2:
 
so your saying that even if the voltage of the battery was totally correct (high enough) for the curcuit, that its still gonna be just party regulated, and not perfectly linear?
the slow lean downward, as the input voltage droops. i have seen that graph too, close enough current control, but still droops slightly as the battery does.

i thought that most of the graphs they showed , in that thread about the chip, they didnt have enough input voltage , and so there was not continual clipping, it had "fallen out of regulation".


i added in one more graph to my list, dont see it much in graphs, but some lights seem to operate that way.
 
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This is the perfect chip to use for a single lion 3.7 PROVIDED its a 18650 with at least 2Ah and is driving a led with 3.3v vf or less and not more than 1a. The chip will drop the input v by at least 0.1v and. The 18650 fully charged to 4.2v will drop to 3.7 within 5 min of 1a draw and therefore the chip is perfectly matching the vf and batt voltage to less than 0.4v for very low loss as heat. It will also not kill the 18650 by overdischarging it via sucking like a boost circuit will.
 
the run time graphs cleary shows that it is not. ;)
Show me your graph. And then explain why the graph in the datasheet is wrong. You know, the one that clearly shows a very tightly regulated output current between 0.2V and about 3.7V (just 3mA/V according to the specs). :shakehead

Are you sure you're not getting confused between a constant current linear regulator and a constant current buck/boost regulator? You do know that a linear regulator can still be a constant current supply, right?
 
Wait a mo... I bet you're confused about how the AMC7135 doesn't maintain regulation when the battery voltage falls beneath the operating voltage of the LED plus the dropout voltage. :ohgeez:

:ironic:

Edit - From the datasheet: The AMC7135 is a low dropout current regulator rated for 350mA constant sink current. Sorry, but it certainly sounds like it is designed for it.
 
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This is the perfect chip to use for a single lion 3.7 PROVIDED its a 18650 with at least 2Ah and is driving a led with 3.3v vf or less and not more than 1a. The chip will drop the input v by at least 0.1v and. The 18650 fully charged to 4.2v will drop to 3.7 within 5 min of 1a draw and therefore the chip is perfectly matching the vf and batt voltage to less than 0.4v for very low loss as heat. It will also not kill the 18650 by overdischarging it via sucking like a boost circuit will.

but the high powered leds dont work at such low voltages WHEN run at such high amperages. even if you get some low bin thing.

here is a most usefull chart that shows the voltages needed for Par leds, and the changes over time
http://myweb.cableone.net/evan9162/vfcurve4.png

sure there are different voltage bins, but the Gate in the led die is very similar, and the binning is based on 350ma not really driving the led hard.

there is also various tests all over the forum, that basically show that the gates of most of these leds is similar, and relative more to drive currents and time and heat.

my own experience with the different bins, shows that over time they basically all about the same, depending on how hard you drive them. 1400, is a fat overdrive, gonna be a high Vf on the led, unless its not a blue gate in the die, like red or orange or yellow leds.

mabey the math would work with 100+ 5mm leds and the single li-ion, or multiple low driven leds, just not hard driven single led items., because they need more votlage to be driven like that.

The 18650 fully charged to 4.2v will drop to 3.7 within 5 min of 1a draw
What? is this an ultrafire cell , cause it sure has some horrible resistance if its acting like that.

here is the chart for a half decent li-ion cell on a 1amp discharge curve.
http://img.photobucket.com/albums/v482/SilverFoxCPF/LG186502400atVariousRates.gif
it doesnt reach 3.7 volts till 1/4 of its capacity is gone from what i am seeing.
 
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