# Which driver to use?



## Ron99 (Aug 6, 2015)

Howdy folks. I have a fair bit of experience working with LEDs using AC powered drivers to build aquarium lights. But now I'm tackling a different project. I'm trying to build an underwater video/focus light using one of these 30W, 10 LED modules:

http://www.ebay.com/itm/351221956181?_trksid=p2057872.m2749.l2649&ssPageName=STRK:MEBIDX:IT

I want to power it with two or maybe three 26650 batteries. Would one of these drivers do the trick:

http://www.ebay.com/itm/DC-DC-30W-C...pply-Module-/321823197737?hash=item4aee284a29

Or could I go with one of these adjustable ones and turn it down:

http://www.ebay.com/itm/DC-DC-100W-...-LED-Driver-/281541980418?hash=item418d35b102

http://www.ebay.com/itm/DC-to-DC-Co...-LED-Driver-/261907091027?hash=item3cfae10e53

Thanks


----------



## DIWdiver (Aug 6, 2015)

I'd say any of those drivers ought to work. 

The first one says the input voltage is 12V, so you don't really have any idea what the actual working range is. It's pretty likely it would handle three cells in series, but they don't actually say that.

The second one is much beefier (which I like - most things last much longer if you don't push them to their limits) but the input and output ranges are barely (or possibly not quite) enough.

The last one is actually my favorite for several reasons:
- it's the cheapest, with shipping
- it's plenty beefy
- it's designed to attach to your heatsink (the wall of your enclosure?) which is probably the optimal thermal path if it's enclosed.
- you are going to run it so far below its rating that a heatsink may not even be needed.
- it has input and output ranges well past what you need.
- you can adjust it to suit your current or future needs/desires.
- it's more transferrable to another project.


----------



## m.pille.led (Aug 7, 2015)

In my personal opinion the way I would go is a 3 Xml board off Ebay for about 10-15$
http://www.ebay.com/itm/30W-White-L...m-PCB-Board-/361290812718?hash=item541e9c4d2e

For those you find good drivers, for exemple this one here. you can adjust current is small steps
http://www.ebay.com/itm/LED-driver-...784?pt=LH_DefaultDomain_0&hash=item1a02bd8a18

I`m not shure if this driver is a buck-linear, if yes the input voltage should be about 11.5v or higher for proper work

this Is how I would go, it`s about the same price and works well. (proven system on flashlights)
Regards


----------



## Anders Hoveland (Aug 7, 2015)

Although LEDs have very little resistance, the fact that they do act as a resistive load is still fundamental to how a constant current driver works. The current flowing through the LED is determined by the voltage from the power supply divided by the resistance of the LED (Ohm's law). The driver will automatically increase or decrease the output voltage if the current flowing through the circuit becomes less than or greater than the driver's rated current (mA), in a sort of feedback regulator.

The first thing you want to do is to make sure the current range of the LED emitter matches up with the constant current rating of the driver. If the LED does not have a current rating, you can usually calculate this by dividing the maximum wattage rating by the voltage. Usually it is okay if the LED emitter has a rated current up to 3 times greater than the constant current driver, that also helps give you a margin of safety to avoid the risk of the LED being overdriven.

Usually the cheap Chinese constant current drivers have a fairly narrow voltage range. One labeled "12 volts" will typically only be able to drop down to 4v minimum, while one labeled "48 volts" will often only be able to drop down to 36v minimum. Since the actual current running through the resistive load (in this case the LED emitter being powered) is determined by the voltage, if the voltage drop across the load is less than the minimum voltage the driver is able to drop down to, the current flowing through the load will be greater than the rated constant current indicated on the driver, and you could risk burning out your LEDs. If the LEDs burn out, it's usually instantaneous as soon as the LED is connected to the power.

So a "constant current" driver here is really only actually constant current if the voltage drop in the load stays within the proper voltage range.

As an example, if the LED is rated 12v 350-700mA, then you will want to find a driver that has a constant current rating of 700mA (or less, but if the current is less the LEDs will have less power), and the voltage range of the driver could be 6-24v. But if the driver does not list a voltage range, and only lists a single rated voltage, you should be cautious and assume the minimum voltage is only about 25% less than the rated voltage. So in this example, you would definitely not want to use a driver that was just labeled "48v".

2 LEDs wired in series that are each rated 3.2v will have a total voltage drop of 6.4v
Usually when two lights are wired in parallel they consume twice the current, but with LEDs this gets more complicated, and we usually want to use less than the theoretical current draw so all the current does not all flow through just one of the LEDs, burning it out.


----------



## DIWdiver (Aug 8, 2015)

Anders Hoveland said:


> Although LEDs have very little resistance, the fact that they do act as a resistive load is still fundamental to how a constant current driver works.


It's true that LEDs have resistance. However when discussing current regulators, it's not a useful concept. It's also NOT fundamental to how most current regulators work. A well-designed current regulator can work with a load that has positive, zero or even negative resistive component to its impedance. While no simple passive component can have zero or negative resistance, more complex elements can. For example, a buck regulator driving an LED is a load that has a negative resistance.



Anders Hoveland said:


> The current flowing through the LED is determined by the voltage from the power supply divided by the resistance of the LED (Ohm's law). The driver will automatically increase or decrease the output voltage if the current flowing through the circuit becomes less than or greater than the driver's rated current (mA), in a sort of feedback regulator.


If you had said 'impedance' instead of 'resistance', I would agree with this. The distinction between resistance and impedance is far more than semantic. When we say 'resistance' we tend to think of a fixed value that we can use in calculations. LEDs do not exhibit a fixed resistance that's useful in calculating anything. They exhibit a complex impedance that changes dramatically with operating point (current, temperature). Oh, and eliminate 'sort of'. It's exactly a feedback regulator. 



Anders Hoveland said:


> The first thing you want to do is to make sure the current range of the LED emitter matches up with the constant current rating of the driver. If the LED does not have a current rating, you can usually calculate this by dividing the maximum wattage rating by the voltage. Usually it is okay if the LED emitter has a rated current up to 3 times greater than the constant current driver, that also helps give you a margin of safety to avoid the risk of the LED being overdriven.


Actually what you want is to make sure that the regulator can be set to an output current that's no higher than the maximum rating of the LED. Very few LEDs have a minimum current rating. Luminus is the only manufacturer that I have ever seen specify one. The one I remember was 1/6th of the maximum rating. I could hardly believe this, as most LEDs work quite well below 1% of their max rating, though the color output is questionable.



Anders Hoveland said:


> Usually the cheap Chinese constant current drivers have a fairly narrow voltage range. One labeled "12 volts" will typically only be able to drop down to 4v minimum,


So 4V to something over 12V is a 'fairly narrow' range?



Anders Hoveland said:


> while one labeled "48 volts" will often only be able to drop down to 36v minimum.


Can you point to a specific device, or is this purely conjecture?



Anders Hoveland said:


> Since the actual current running through the resistive load (in this case the LED emitter being powered) is determined by the voltage, if the voltage drop across the load is less than the minimum voltage the driver is able to drop down to, the current flowing through the load will be greater than the rated constant current indicated on the driver, and you could risk burning out your LEDs. If the LEDs burn out, it's usually instantaneous as soon as the LED is connected to the power.


Do you have experience with drivers connected to loads below their minimum output voltage? I would have guessed that a 12-36V output driver connected to a 3.3V LED would under power it. You imply that the 12V minimum would overdrive the 3.3V LED. Is there some real-world experience here? That would be far more valuable than my (or your) conjetcure.



Anders Hoveland said:


> So a "constant current" driver here is really only actually constant current if the voltage drop in the load stays within the proper voltage range.


Yes.


Anders Hoveland said:


> As an example, if the LED is rated 12v 350-700mA, then you will want to find a driver that has a constant current rating of 700mA (or less, but if the current is less the LEDs will have less power), and the voltage range of the driver could be 6-24v. But if the driver does not list a voltage range, and only lists a single rated voltage, you should be cautious and assume the minimum voltage is only about 25% less than the rated voltage. So in this example, you would definitely not want to use a driver that was just labeled "48v".


True, but it's probably an answer looking for a question. Nobody in this thread has mentioned a driver that doesn't have a far better characterization of its output than just "48V".


Anders Hoveland said:


> 2 LEDs wired in series that are each rated 3.2v will have a total voltage drop of 6.4v
> Usually when two lights are wired in parallel they consume twice the current, but with LEDs this gets more complicated, and we usually want to use less than the theoretical current draw so all the current does not all flow through just one of the LEDs, burning it out.


Another answer looking for a question. The only LEDs mentioned here are modules with all the emitters in series.


----------



## Anders Hoveland (Aug 8, 2015)

DIWdiver said:


> It's true that LEDs have resistance. However when discussing current regulators, it's not a useful concept. [...] The distinction between resistance and impedance is far more than semantic. When we say 'resistance' we tend to think of a fixed value that we can use in calculations. LEDs do not exhibit a fixed resistance that's useful in calculating anything. They exhibit a complex impedance that changes dramatically with operating point (current, temperature).


An LED behaves similar to a resistor over a certain limited voltage range, although even over this interval the voltage drop slightly varies depending on the current.




DIWdiver said:


> Do you have experience with drivers connected to loads below their minimum output voltage?


Yes, I have burned out many LEDs that were connected in series when I accidentally used a constant current driver that had too high of a minimum voltage. Usually only one of the LEDs in the series circuit burns out, it acts like a fuse, but there is often damage to a second LED somewhere in the series as well. 
Usually this is not a worry when the LED is being greatly underdriven, for example powering an emitter rated to handle 1000mA with a 350mA constant current driver, because even if the voltage running through the circuit is higher than it should be, there is no way such a driver is going to supply anywhere near 1000mA.



DIWdiver said:


> I would have guessed that a 12-36V output driver connected to a 3.3V LED would under power it.


It all depends on the current flowing through the LED. I have a 12-24v 350mA driver and it will burn out a 3.2v or 2.7v LED rated for 1W max, even though the calculated current of the LED is quite similar to the rated constant current of the driver. Obviously the driver is putting out substantially more than its rated constant current when the voltage drop in the load is too low.
It takes at least 3 of these LEDs in series to avoid a burn out. 

I have burned out LEDs multiple times with two different "constant current" drivers, when the rated current on the driver should have matched up to what the LEDs could handle. In all these cases the output voltage of the driver was too high, or there were not enough LEDs in series when the power was supplied. It depends on the driver, because I have a driver that can power a single 3.2v LED just fine without any problems, but not all drivers can drop their voltage that low to maintain a constant current.


----------



## SemiMan (Aug 8, 2015)

Anders, I mean really. I know you are not an electronics expert and I know that DIWDiver is.

w.r.t. to the comment on "Cheap Chinese drivers" only being able to operate over a narrow range, say 4-12, or 36-48V, that is a comment out of ignorance of how drivers are designed and architected. The same comment could be applied to products from any driver company on any continent and of any quality. It's an architectural decision, pure and simple and is almost always related to how the voltage for the IC control and feedback circuits are generated. They are often a tap off the transformer that drives the LEDs. As the voltage is a ratio, AND the LEDs can often be dimmed, there is a limited range where you can reliably supply the control IC(s). As well, the range may be listed for other factors such as maintaining PF/THD to a market/regulatory requirement.

The "resistance" of the LED, plays 0 part in setting the output current of the LED driver. It does not even come into play. The output current to the LED is measured directly (voltage is not except for over/under voltage). Resistance does come into play, but not remotely what you listed it for. It can impact the stability of feedback loops, as well as the amount of ripple current (line rate flicker).

That you have burned out many LEDs at below the rated current does not mean that the driver was outputting higher current. More likely is you connected the LED while the driver was turned on. The driver would float to it's maximum output voltage and it's output caps would be fully charged. Hook up the LED, and those caps dump all their energy into the LED. When you have a bunch of LEDs, the power is spread out. When you only have one or two, it will take the full hit and turn the bond-wires into fuses. Most low cost drivers will actually detect really low output voltages as shorts and go into hiccup mode. I am not saying that no drivers will have issues that could lead to LED failures, but it would be the rare case and because of architecture, would not be the cheapest versions, as it will generally only happen with secondary feedback where the control circuits are not properly powered are not properly feeding back current. That you have one driver that will do it, does not mean it is normal. I am quite certain I have had many more drivers in my hands with far more sophisticated equipment for testing. ...... Contrary to what you wrote as well, that 350mA driver could quite possibly put out over 1A, especially in a wide input voltage supply with a good temp rating, as the only physical limit for a period of time is the saturation of the transformer if the feedback circuits are not working.

Semiman


----------



## DIWdiver (Aug 8, 2015)

Now, now, no need to be nasty.

Also, apologies to the OP for the hijack of this thread.

While we're at it Semiman, this thread is about battery powered drivers. I don't remember ever seeing one with a transformer. Line voltage drivers have them, and they are even sold as 'transformers'. But your comments that relate to line voltage drivers don't apply here, or at least ought to be distinguished as such. 

You are right about attaching LEDs to a powered driver. Very bad idea. I blew my expensive SST-90 (rated to operate at 9A) that way. And the driver doesn't even have a big output cap. It has a large control FET that was saturated. It was the beefy power supply that dumped through the saturated FET and fried the LED. Still, I think it's a bit rash to _assume_ that this was the problem Anders had, multiple times. It would be more diplomatic to suggest this as a possibility and see where the conversation goes.

I'm completely open to the possibility that my guess is wrong, and that some drivers _will_ output excessive current when presented with a load that has too low a voltage at design current. I absolutely agree with you that this is an architecture issue not a quality issue, and that the warning should be addressed at all drivers, not only those from certain sources, but I am not ready to dismiss Anders' experience as unimportant. I'd very much like to hear more of what he (and hopefully others) have to say about it before reaching any conclusion.

Also, Anders never suggested, as you implied, that the driver would measure voltage. What he said was "The driver will automatically increase or decrease the output voltage if the current flowing through the circuit becomes less than or greater than the driver's rated current (mA), in a sort of feedback regulator," which is actually accurate in most cases, even if it's mostly by accident. Of course you and I both know that the_ intention_ of the driver (and its designer) is to increase or decrease the output _current_, based on the measured current, but the voltage is usually inextricably linked and the actual driving forces can be obscured because of the physics and complexity of the devices involved. Most people don't need to know the fine points of that discussion, and those that do need to probably already know. So pointing this out might be construed as being a bit pedantic.


----------



## SemiMan (Aug 9, 2015)

DIWdiver said:


> If you had said 'impedance' instead of 'resistance', I would agree with this. The distinction between resistance and impedance is far more than semantic. When we say 'resistance' we tend to think of a fixed value that we can use in calculations. LEDs do not exhibit a fixed resistance that's useful in calculating anything. They exhibit a complex impedance that changes dramatically with operating point (current, temperature). Oh, and eliminate 'sort of'. It's exactly a feedback regulator.



Unless you are talking 10's or hundreds of megahertz, LEDs do not exhibit complex impedance. Complex impedance (impede current), requires a non-linear time relationship between voltage and current, which implies reactance (capacitance and inductance) of which LEDs have very little and only becomes an issue at very high frequencies. The relationship between voltage and current is non-linear, but not complex (in the mathematical sense).


----------



## Anders Hoveland (Aug 9, 2015)

That was very eloquently put, DIWdiver.

And of course, just my luck, SemiMan posted right after your post while I was still in the middle of typing this post. :hairpull:


----------



## SemiMan (Aug 9, 2015)

I have actually used transformers several times in battery powered drivers, In volume, it is quite a bit cheaper than using a multi-switch buck-boost IC and quite a bit more efficient than using a sepic configuration. Also regularly use it for 24/48V distribution lighting applications.

I am not being nasty DIW, I am just being more honest about being dismissive ;-)

I did not address Anders experience as unimportant, I just gave it context. I have tested probably close to 100 drivers. It's a rare driver that will blow a LED at below its rated voltage. 

My comment w.r.t. to voltage was to be completely clear that current output LED drivers purely regulate current (and not voltage) even if as you point out, when we get to the physics, its a more complex answer.

I am not sure why you pointed out the "saturated" FET. Can you explain that? In switching power supplies or on/off control circuits, the FETs would always be moving from saturated to saturated states. They can normally switch of out saturation (or/off) in the 10's or low 100's of nanoseconds. Was the issue that the transistor was in saturation or how fast it was able to regulate current? Power LEDs can take 10's of amps or more for very short durations. I expect (guessing) for Ander's supply it was either caps discharging and/or not enough voltage to power the regulation circuit. At low voltage, the transformer would be far from saturation and could put out quite a bit of current.




DIWdiver said:


> "LEDs do not exhibit a fixed resistance that's useful in calculating anything".



Perhaps not "fixed" but piecewise linear about the operating point, especially if that operating current is near the rated LED current. It is actually quite useful for calculating things .... namely control loop stability in DC or AC drivers and ripple current in AC single stage drivers.


----------



## DIWdiver (Aug 9, 2015)

The driver with a saturated FET was a linear driver, with an op-amp driving the FET. The op-amp had a (deliberately) reduced bandwidth to ensure stability in an unknown user application. As such, when the LED was connected, the op-amp could not turn off the FET fast enough to prevent destruction of the LED. I was just pointing out that there could be more than one reason to not connect an LED to a powered driver (even one with no large output caps).

I admit that 'complex impedance' was an unfortunate choice of words which has a specific technical meaning and did not express my intention to those that understand that meaning. I meant to say that the resistance/impedance is not simple to express in a single number, and that only when given an operating point can a resistance be expressed that will be useful (therefore it's not 'fixed'). I'll add here that the ability to measure or calculate, understand, and use such a value would require one to be at the upper end of the tech spectrum of DIYers. No affront meant to DIYers, its just that it's sort of specialized knowledge that most don't need, and so often don't have. A 'gut' understanding of the V-I curve of an LED is often sufficient without the ability to mathematically or numerically express and make use of the impedance or resistance values at an operating point.

I'm also curious, in what capacity have you tested so many drivers? Are the results published and/or available to CPF? Are they drivers that we would be interested in and able to get our hands on? Spill!


----------



## SemiMan (Aug 9, 2015)

DIWdiver said:


> The driver with a saturated FET was a linear driver, with an op-amp driving the FET. The op-amp had a (deliberately) reduced bandwidth to ensure stability in an unknown user application. As such, when the LED was connected, the op-amp could not turn off the FET fast enough to prevent destruction of the LED. I was just pointing out that there could be more than one reason to not connect an LED to a powered driver (even one with no large output caps).
> 
> I admit that 'complex impedance' was an unfortunate choice of words which has a specific technical meaning and did not express my intention to those that understand that meaning. I meant to say that the resistance/impedance is not simple to express in a single number, and that only when given an operating point can a resistance be expressed that will be useful (therefore it's not 'fixed'). I'll add here that the ability to measure or calculate, understand, and use such a value would require one to be at the upper end of the tech spectrum of DIYers. No affront meant to DIYers, its just that it's sort of specialized knowledge that most don't need, and so often don't have. A 'gut' understanding of the V-I curve of an LED is often sufficient without the ability to mathematically or numerically express and make use of the impedance or resistance values at an operating point.
> 
> I'm also curious, in what capacity have you tested so many drivers? Are the results published and/or available to CPF? Are they drivers that we would be interested in and able to get our hands on? Spill!




What capacity ... many years as a consultant in the lighting space on fixtures and electronics, and involvement (design, mktng, etc.) in a driver company, and consulting in the semi-industry in the lighting space. I don't get into the hard core design like I used to, but my lab is extensive. Unfortunately, all that work is paid for, and hence the results cannot be publicly shared. I can answer in generalities of course. Testing generally involved teardown, sometimes destructive. In terms of interest to the community, most were AC input, and from all the majors and some not so major vendors. As well 24/48 input stuff. Not too many DC\DC but only because most are pretty rudimentary.

Effective resistance at the operating point is something many DIYers could understand and is relatively useful as it is a good way to estimate how much current will change w.r.t. voltage. It is also useful for estimating string matching. No time for the next few days, but I could throw a tutorial together on it.


----------



## Anders Hoveland (Aug 10, 2015)

The only thing basic DIYers need to understand is that the driver needs to supply the right amount of current (not more than it can handle), and the voltage should not be _too much_ higher than what they need either. The amount of current the LED was designed for can easily be calculated by taking the maximum Wattage divided by the LED's rated voltage. It is very simple, and does not have to be more complicated than that.

If the power supply is rated for 12-36v, and your LED chip is rated for 12v, it should work just fine. Assuming the rated voltage of the LED is within the rated voltage range of the driver, the power flowing through the LED will be a product of the driver's output current multiplied by the LED's rated voltage.


----------



## peter78 (Aug 31, 2015)

What do you guys think of using this driver 

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=1510-1159-ND

with this led

I found this part with the #digikey mobile app.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=976-1271-ND

Is it a good combination? Can I use 2 cobs per driver?


----------



## SemiMan (Sep 1, 2015)

peter78 said:


> What do you guys think of using this driver
> 
> http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=1510-1159-ND
> 
> ...



1.05A is the test current for the Bridgelux LED. You can actually drive them harder for more light output (with better heat sinking of course). You may want to consider a driver that will do 1.4A.

You could likely use 2 COBs in parallel with one driver as they tend to be somewhat well matched for voltage, but one LED will be a bit brighter than the other. With this driver, you are not going to get twice the light .... you would be splitting the power to each COB.

What are you using this for?


----------



## peter78 (Sep 1, 2015)

SemiMan said:


> 1.05A is the test current for the Bridgelux LED. You can actually drive them harder for more light output (with better heat sinking of course). You may want to consider a driver that will do 1.4A.
> 
> You could likely use 2 COBs in parallel with one driver as they tend to be somewhat well matched for voltage, but one LED will be a bit brighter than the other. With this driver, you are not going to get twice the light .... you would be splitting the power to each COB.
> 
> What are you using this for?



Vegetables. It is in planning stage though. Trying to figure out a good combination. Still learning to read the spec sheets. Price matters. Trying to save a buck from the ones sold commercially. I plan to mount them directly on a 10x6 heat aluminum heat sink with a fan blowing on it. 

How do you calculate how many cobs you can run per driver?

Thanks!


----------



## SemiMan (Sep 1, 2015)

Well with plant growth we are usually into large power draw so efficiency can matter. Figure out what you need in equivalent lumens and lets take it from there.


----------



## DIWdiver (Sep 1, 2015)

Also, plant growers talk a lot about what colors are in your light. Any given plant will have colors it needs and colors it doesn't (or needs less). This can make a substantial difference in your power requirements. You might want to do some investigation.


----------

