Just Bought Some Rspec's... How to Decipher a Mean Well Spec Sheet?

Just pulled the trigger on some QB Rspec's. Getting several of the combos that have two boards and the double slate heat sink. Never built a light and still just not getting it on the drivers. If I go to the site HLG links to, this one for example: https://www.findchips.com/search/HLG-120H-C1400 I can understand that or think I do. It is 108 volts, 1400 mA and will draw about 150 watts. Looking at HLG's guide, that means each pair of boards on that driver would each be running at about 67 watts. So I'm adding the voltage from HLG, 47.7 x 2 and then seeing 95.4 which is both boards and is less than the 108v total of the driver so I'm fine. Is all of that correct?

Until this very moment, I have been unable to "get" the mean well site but just realized I've been looking at the wrong data sheets. I've been looking at the constant voltage + constant current drivers. OK so for the example above it is this one http://www.meanwellusa.com/webapp/product/search.aspx?prod=HLG-120H-C not http://www.meanwellusa.com/webapp/product/search.aspx?prod=HLG-120H

So the idea is to "use up" the volts basically right? So sticking with the same 1400 mA...If I used this one: http://www.meanwellusa.com/webapp/product/search.aspx?prod=HLG-240H-C I would have 179v and could run three boards for a total of 143.1? I could not run 4 as that would be 190.8. So what happens to the rest of the potential in the 240 in the case where I would be using 143v? Am I still drawing the same amount of watts from the wall as I would if I was using the full 179 it has available? I don't think so but just making sure I understand. My guess is I save some watts and heat and gain a fraction of efficiency but have essentially wasted money by buying the incorrectly sized driver.

Which leads me to the last question for this post at least . So back to the 240H, http://www.meanwellusa.com/webapp/product/search.aspx?prod=HLG-240H-C the 1750 mA driver has 143v. There is nothing on HLG's site showing the board running at 1750 mA. Looking at the 1400 and 2100 numbers, 3 @ 1750 would be just a hair over the 143v the 240H has. I know I can't go over but are the published numbers that exact? Could I do three boards at 1750 mA?

No need to decipher driver spec sheet.

Just buy an HLG-240H-C2100 for each pair of boards, wired in series

https://horticulturelightinggroup.com/pages/qb288-board-guide

Thanks but I don't want to run that much wattage or that intensity and I want to spread out the light a little more. I've gone for 4 pairs and I'll run either 3 or 4 pairs in my space depending on what I land on driver wise.

As an example on that page, it suggests a 240H for 4 boards @ 1050 mA and the 185H for 3 boards @ 1050 mA. Four boards @ 1050 should be ~ 188v. The 185H has 190v so couldn't I run 4 boards with it instead of 3 as HLG's site says? If I can, what is the difference between using a 185H or a 240H to run the same number of boards at the same mA? I realize there is a cost difference in the purchase but what is the difference in usage? Less heat with the 240h? Is the same current pulled from the wall either way?

My take on driver selection: use a driver that allows you to run the light engines at design current and just use the dimming function if you don't need that much light.

BodhiKarma said:

The 185H has 190v so couldn't I run 4 boards with it instead of 3 as HLG's site says?

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Yes. You could also run an ELG-200-C1050, which is a bit cheaper.

BodhiKarma said:

"If I can, what is the difference between using a 185H or a 240H to run the same number of boards at the same mA? I realize there is a cost difference in the purchase but what is the difference in usage? Less heat with the 240h? Is the same current pulled from the wall either way?

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A slight drop in efficiency. The HLG-185 is already 94% vs 93.5% for the 240 (less if you have 115V mains vs 230V), but you will also lose a couple of % driver efficiency running 70-80% load instead of 100%.

By supplying the same amount of current to the same number of boards, you will pull slightly more power from the wall with a 240 vs a 185. But it won't be much - no more than 4-5W.

Spec sheets are here

240C: https://www.meanwell.com/webapp/product/search.aspx?prod=HLG-240H-C

185C: https://www.meanwell.com/webapp/product/search.aspx?prod=HLG-185H-C

OK, I'm going to backtrack a bit. My post was based on the voltages for the QB288 V2 here: https://horticulturelightinggroup.com/pages/qb288-board-guide

The R Spec appears to be a slightly higher voltage board - 54V at 2800mA, according to HLG.

The way they have wired the board means the Osrams get twice the current as the Samsung LEDs, and that adds an extra 1-2V to each string from what I can see (looking at the Osram spec sheet at 2100mA) . . .

But don't take my word for it: ask HLG how many volts the boards drop at 1050mA to confirm if you can run the 185 driver. I suspect you may still be able to do it but don't going buying any drivers until it is confirmed.

Thanks Prawn. I think I have a better grasp since your answers. So in a nutshell, selecting a driver is kind of like the price is right. The closest without going over.... Well that is the ideal scenario to get the best bang for the buck and efficiency. On the flip side I guess a driver sized as close as possible might limit future use for new tech. Picking nits though as I am not too concerned about that.

So from what I gather I am supposed to be aiming for 30-35 watts per square foot? My space is 12.5 ( 2.5' x 5' ) so I should be aiming for around 375 to 440 watts. Assuming you are right on the voltage I'll just add 2.5 ( since site says give 1 anyway ) to all. Everything below except the 185 has headroom. Also I picture the length of the board running across the 2.5 side of my tent for about 3 inches of clearance on either side.

I could do 2 pairs @ 2100 and be right in there with 411.2 watts HLG 480 229v

I like the idea of at least 3 pairs though for better coverage and avoiding hot spots. I actually bought 4 pairs but don't think I need the 4th.

3 pairs @ 1400 would be 400.68 and be right in there. HLG 480 343v

Another other option is 3 pairs @ 1750. I don't see any numbers on it but guess it would be just shy of 85 watts for an very beefy wattage of ~ 508. I guess I would need 3 x ELG 200. Or I could do 2 x HLG 320's and run three boards per driver.

If the 185 panned out, I could run 4 pairs @ 1050 via two 185's for 395 watts.

Step up from that would be 4 pairs @ 1400 with 2 x HLG 320's 229v. That would be 534.2 watts and seems like over kill.

As I see it, those are my options. I think two pair in a longer space as I have will be the least attractive but I really don't know. Haven't seen a modern light so that may be completely wrong. I like the idea of 3 pairs for the most even coverage. Blowing all 4 pairs in the space feels like a waste of one of them. Am I correct on my options and thinking?

The best results will be around the 40W per square foot mark. It comes down to light coverage, efficiency, how high you hang your lights etc.

If you have four pairs of QBs and a 5'x2.5' room, I'd run them all like this:

I I I I

Four pairs gives you better coverage, better efficiency. And you've already got them, so why not? Unless you plan to do something else with them.

I run my boards in parallel with 48V or 54V CV/CC drivers. All Mean Wells are CC drivers. But the "H" series are also voltage adjustable (B series use pots, A series have built-in adjustment for voltage and current).

I would prefer having 48-54V to each board than 300+V across six etc. That's just me.

OK, bang for buck, look at getting 2x ELG-240-C1050A drivers - one driver for each 4 boards. This will give you up to 216V (you won't see that at lower currents) at up to 1.2A (according to the spec sheet, they will go up to 1.2+A). That's close to 500W, which is about what you want for 12.5sq/ft.

Even better bang for buck would be 2x ELG-240-48A drivers. The spec sheet says these will go up to 53V and deliver up to 5.76A. You only need about 5A per driver (10A total) to get around 500W, but these drivers would give you a bit of head room: run them hotter and harder in winter when you need a bit more warmth in the grow room, and run them softer in summer etc.

Now the 240-48A is a CV driver, but can also be used as a CC driver. Nevermind that for now . . .

You would need to wire each pair of boards in parallel (easy), then use a Wago connector or splitter from the main power from each driver to each pair of boards.

Makes sense? Power splits in two (one to each heatsink), and each pair of boards is in parallel for a total of 50+V to each board, which would draw 5A divided by four boards = 1.25A to each board (same as a 1250mA CC driver).

My preferred option would be the 48V drivers. If anyone has a better suggestion, let's hear it!

Also, do ask HLG if they have proper voltage specs for these new boards and then we can all be certain about which drivers will work and which ones won't.

Prawn Connery said:

The best results will be around the 40W per square foot mark. It comes down to light coverage, efficiency, how high you hang your lights etc.

If you have four pairs of QBs and a 5'x2.5' room, I'd run them all like this:

I I I I

Four pairs gives you better coverage, better efficiency. And you've already got them, so why not? Unless you plan to do something else with them.

Click to expand...

That is exactly the way I intend to orient them. I'm basically using 30" x 60" wire shelving and putting them "long way" would seem to give great coverage. I will eventually have other enclosures like this one. This is a test run on a home made RDWC system I'm working on. Never done DWC so we'll see how that goes. If it goes well, I'll replicate the set up. So If I ran three pairs, I would use the other as a temp spare and eventually in the future tent.

When I bought I was intending to use all 4. I've laid some boxes out about the same size on the 30 x 60 shelf and 4 seems crowded. But the goal is to get the most out of every inch of that space so if 4 is the way to go....

Prawn Connery said:

I run my boards in parallel with 48V or 54V CV/CC drivers. All Mean Wells are CC drivers. But the "H" series are also voltage adjustable (B series use pots, A series have built-in adjustment for voltage and current).

I would prefer having 48-54V to each board than 300+V across six etc. That's just me.

OK, bang for buck, look at getting 2x ELG-240-C1050A drivers - one driver for each 4 boards. This will give you up to 216V (you won't see that at lower currents) at up to 1.2A (according to the spec sheet, they will go up to 1.2+A). That's close to 500W, which is about what you want for 12.5sq/ft.

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I do like that option. One thing I neglected to mention. I have 86" in height but using about 18" off the bottom for a reservoir. So really I only have about 68" in height. One of the things I liked about more boards running with less juice is I'll conserve space by not having to position the lights so high,

Prawn Connery said:

Even better bang for buck would be 2x ELG-240-48A drivers. The spec sheet says these will go up to 53V and deliver up to 5.76A. You only need about 5A per driver (10A total) to get around 500W, but these drivers would give you a bit of head room: run them hotter and harder in winter when you need a bit more warmth in the grow room, and run them softer in summer etc.

Now the 240-48A is a CV driver, but can also be used as a CC driver. Nevermind that for now . . .

You would need to wire each pair of boards in parallel (easy), then use a Wago connector or splitter from the main power from each driver to each pair of boards.

Makes sense? Power splits in two (one to each heatsink), and each pair of boards is in parallel for a total of 50+V to each board, which would draw 5A divided by four boards = 1.25A to each board (same as a 1250mA CC driver).

Click to expand...

Sorry no I don't get it. If you note my first post, I was complaining I can't decipher a Mean Well Spec sheet. The HLG CC ones I get. I can look at the sheet and understand its voltage capacity. I don't get the others at all. So the spec sheet for the one you reference is: https://www.meanwell.com/webapp/product/search.aspx?prod=ELG-240 I don't get it. It just shows it is 48 volts. I don't get parallel and how to decipher the drivers. Sorry... I'm a complete newb and literally just grasp series and the other drivers yesterday.

Prawn Connery said:

My preferred option would be the 48V drivers. If anyone has a better suggestion, let's hear it!

Also, do ask HLG if they have proper voltage specs for these new boards and then we can all be certain about which drivers will work and which ones won't.

Click to expand...

Emailing them now. Hopefully will get a response.

BodhiKarma said:

I do like that option. One thing I neglected to mention. I have 86" in height but using about 18" off the bottom for a reservoir. So really I only have about 68" in height. One of the things I liked about more boards running with less juice is I'll conserve space by not having to position the lights so high,

Click to expand...

That's correct - the more LEDs you run, the softer you can run them and the closer to the plants you can hang them.

BodhiKarma said:

Sorry no I don't get it. If you note my first post, I was complaining I can't decipher a Mean Well Spec sheet. The HLG CC ones I get. I can look at the sheet and understand its voltage capacity. I don't get the others at all. So the spec sheet for the one you reference is: https://www.meanwell.com/webapp/product/search.aspx?prod=ELG-240 I don't get it. It just shows it is 48 volts. I don't get parallel and how to decipher the drivers. Sorry... I'm a complete newb and literally just grasp series and the other drivers yesterday.

Click to expand...

The simple way to remember it is, if you wire in series, you add up all the voltages and supply the same current to each board.

If you wire in parallel, you add up all the currents and supply the same voltage to each board.

Let's take two boards that are 54V each and 2.8A maximum current as an example.

In series, you need a 108V driver that puts out 2.8A (54V + 54V = 108V, but 2.8A remains the same)

In parallel, you need a 54V driver that puts out 5.6A (2.8A + 2.8A = 5.6A, but 54V remains the same).

In both examples the total output will be 302.4W (108V x 2.8A, or 54V x 5.6A) and you would need a 320W driver (or thereabouts). You could use either a HLG-320H-54A (54V, 5.95A driver) in parallel, or a HLG-320-C2800A (114V, 2.8A driver) in series.

Do you see now the relationship between voltage and current?

In series, the current passes through each board before it reaches the next. Voltage drops as it passes through each board - which is why you have to add all the voltages together (you have to supply enough voltage so that there is enough left over for the next board after it passes through the first).

In parallel, you provide the same current to each board - the current is split into two - so there is not the same voltage drop (because the current is not passing through one board before it reaches the next). However, because you need to split the current up into two - one wire for each board - then you need to double the current (amps) to supply them (as you are going to halve that current when you divide it in two).

4 boards in series = quadruple the voltage. 4 boards in parallel = quadruple the amperage. And so on.

So why would you choose one over the other? There are some considerations.

You cannot exceed the board connector rating of 9A - so you cannot wire lots of boards together in parallel if their combined current exceeds 9A (that is, if you connect all the boards through their on-board connectors).

In the above example, you could not wire 4 boards in parallel (2.8A x 4 = 11.2A), but you could wire them in series. The trade-off? Higher voltage to the boards (54V x 4 = 216V), so you need connecting wire that can handle at least 216V. And if you have a short on one of the boards, you now have 216V live that can zap you (instead of 54V).

There are also consequences if one of the parts fails or starts to hog the current that could lead to thermal runaway. In reality, you won't have to worry about this with matched components, plus you can prevent the driver from supplying too much voltage or current in the event it does. So don't worry about that at this stage.

Finally, CV drivers tend to offer more power (better value) because there are no wasted volts. The 320H-54A supplies the correct voltage to all boards and has lots of excess current that can be used up. The 320-C2800 can supply 114V - so 6V don't get used (114V-108V = 6V wasted), and the current output is limited to 2.8A, so cannot go any higher.

Again, in reality these drivers will put out a little bit more than their specs suggest, so the above is just an example. You will get more than 2.8A out of a 320-C2800 driver. And you will get more than 5.95A out of a 320H-54V driver. But overall, the 320H-54V will drive your boards higher than the 320-C2800.

Hopefully, this all makes better sense now.

Thank you. I really appreciate your response. I'm sure people come in here all the time asking basic level questions like that so thank you for taking the time to explain it. I think I get it but still have a couple of questions. So using this as an example : https://www.meanwell.com/webapp/product/search.aspx?prod=ELG-240 - The "54" has 54 volts and a "pool" of 4.45 A. The "48" supplies 48 volts and has a pool of 5A. Is that correct? If so I get how to read the sheets now.

So.... You mentioned the QB Rspec boards are maybe a v or two higher than the QB guide page. So lets use the 320 you had mentioned. My questions are using the 48 and 54 as examples in parallel.

So if I had the 48, I could not run the boards at 2400 mA. That is 49.5v. Even 2100 mA is a hair over 48. So is it possible to be running then at like 2043 mA ( assuming I'm just a hair short on the 48 if it were very strict per the QB Guide Page )?

If I were to use just two boards with the 54, I have 5.95A available. Does that mean in the case of two boards they would be using all that amperage and running over 2800 mA? Could I run just a single board with the 54 or would it pop?

If I can run just one board, how am I or am I able to control the amperage going to the board?

You mention not using the 6v in a parallel but I'm still missing something on the series about it being wasted. If I had the 54 your mention and wanted to run the boards at 2100 mA, lets say for argument sake that is an even 48v. Am I not still leaving 6v hanging in the breeze or am I actually utilizing all the power by being able to use up all the amperage ( back to the 2043 mA question above ).

I appreciate your patience. I feel like I'm on the verge of getting this. I understand the voltage not being utilized in series but I don't understand how the amperage the board sees is controlled in parallel.

Have a look at the driver specs here: https://www.meanwell.com/webapp/product/search.aspx?prod=HLG-320H

Now, click on "REPORT" at the top of the page to see how each of the different drivers tested. You will see that they each go beyond their respective ratings.

Look at the 320H-48, for example here: https://www.meanwell.com/webapp/product/search.aspx?prod=HLG-320H&pdf=SExHLTMyMEgtNDgtUlBULlBERg==&a=2

Scroll down to "OUTPUT FUNCTION TEST" then "No.3 Output Voltage Adjustment Range" then right to "Result" and see "53.48" - that is the actual maximum voltage of the driver.

Now scroll down to "No.4 Output Current Adjustment Range" . . . "Result" . . . "7.675" - that is the maximum current.

So you can see the 48A driver will supply up to 53.48V or up to 7.675A - whichever limit it reaches first.

The 54A driver will supply up to 59.08V and 6.620A (if you are on 115V mains - it will be 59.07V and 6.621A if you are on 230V mains).

Remember, not all drivers will test the same - this is just a random driver off the production line used as an example by Mean Well. But they will all exceed their rated specs and may be slightly better or worse than the test results.

That's how a 48V driver can supply more than 48V. But this only applies to the "A" type drivers (and "AB" type - but you need to wire your own potentiometer to the "AB").

BodhiKarma said:

So if I had the 48, I could not run the boards at 2400 mA. That is 49.5v. Even 2100 mA is a hair over 48. So is it possible to be running then at like 2043 mA ( assuming I'm just a hair short on the 48 if it were very strict per the QB Guide Page )?

Click to expand...

See above. The 48 will go over 48V, so you could supply 2400mA each to two boards (49.5V x 2.4A = 118.8W x 2 boards = 237.6W total).

BodhiKarma said:

If I were to use just two boards with the 54, I have 5.95A available. Does that mean in the case of two boards they would be using all that amperage and running over 2800 mA? Could I run just a single board with the 54 or would it pop

Click to expand...

You can adjust the current with the "A" type drivers. You set the voltage at whatever you want (50-54V) and then wind up the amperage until you have 2.8A per board. If one board is connected, you will supply 2.8A at 49.86V (according to the HLG spec sheet), If two boards are connected, you will supply 5.6A at 49.86V.

You control the voltage and amperage using a small screwdriver which fits inside a recessed hole under a rubber plug that keep moisture and dust out. You will see what I mean when you buy the drivers.

BodhiKarma said:

If I had the 54 your mention and wanted to run the boards at 2100 mA, lets say for argument sake that is an even 48v. Am I not still leaving 6v hanging in the breeze

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Yes. So you would select the 48 driver. It supplies a greater current at a lower voltage, but the maximum voltage it can supply will still exceed the maximum voltage of the boards so that it can supply its maximum current.

It is a balancing act to find the driver which best meets your requirements. That is why you need to have an idea of how many boards you want, how hard you want to drive them, and whether it is best to run them in parallel or series.

Prawn Connery said:

See above. The 48 will go over 48V, so you could supply 2400mA each to two boards (49.5V x 2.4A = 118.8W x 2 boards = 237.6W total).


You can adjust the current with the "A" type drivers. You set the voltage at whatever you want (50-54V) and then wind up the amperage until you have 2.8A per board. If one board is connected, you will supply 2.8A at 49.86V (according to the HLG spec sheet), If two boards are connected, you will supply 5.6A at 49.86V.

You control the voltage and amperage using a small screwdriver which fits inside a recessed hole under a rubber plug that keep moisture and dust out. You will see what I mean when you buy the drivers.

Click to expand...

I finally understand this! I get both types of drivers and can read the specs now. So the last question is in regards to controlling the amperage. Is it just a potentiometer that I'm manipulating beneath the dust cover? If so, it is slotted in some way or marked for different amperage settings? What I'm getting it is how will I know the amperage I am supplying? It is just like 1 click for 1400, next click turns to 1750, then to 2100 etc.For the dimmable ones, I'd have no way of knowing my wattage without another device to measure it. Is the setting to manipulate the amperage the same in that I would need something else to measure it or does the driver have set markings or "clicks" for the various amperages?

After you have explained this I like the versatility of parallell. If I bought 1050 mA CC drivers, I can never run anything at more than 1050 mA. I like the idea of being able to manipulate the amperage and have a wider application for the driver on future builds or even this one.

Ok so if I should be aiming for 500 watts, I think I have two options. The first option is 8 boards @ 1400 for a total of 534 watts. There are no numbers for 1750 but I bet it is near 85 watts. I'm going with 84.7. The second option would be 6 boards @ 1750 for about 508 watts. As I see it, those are my best two options

Wow... had a box sitting outside my door today. Thought it was an Amazon order and much to my surprise it is my HLG stuff. I'm impressed. I ordered a little after 7 PM Eastern on Tuesday and two days later I had all my stuff.

Hopefully I can add two photos.... I have laid out the heat sinks over a shelf that is the foot print of my area.


The above photo has each 3 inches from the top long ways. Each end heat sink is 6 inches from the sides and then there is 12 inches between heat sinks.



In the 4 configuration each end is about 2 inches from the side and then 8 inches between each of the heat sinks.

BodhiKarma said:

I finally understand this! I get both types of drivers and can read the specs now. So the last question is in regards to controlling the amperage. Is it just a potentiometer that I'm manipulating beneath the dust cover? If so, it is slotted in some way or marked for different amperage settings? What I'm getting it is how will I know the amperage I am supplying? It is just like 1 click for 1400, next click turns to 1750, then to 2100 etc.For the dimmable ones, I'd have no way of knowing my wattage without another device to measure it. Is the setting to manipulate the amperage the same in that I would need something else to measure it or does the driver have set markings or "clicks" for the various amperages?

After you have explained this I like the versatility of parallell. If I bought 1050 mA CC drivers, I can never run anything at more than 1050 mA. I like the idea of being able to manipulate the amperage and have a wider application for the driver on future builds or even this one.

Click to expand...

It is effectively a built-in potentiometer. You will need an ammeter or a watt meter (wall meter) to measure total output.

Obviously the ammeter will tell you how many amps and volts are going to the boards, and you can adjust accordingly.

Or you can use a wattmeter and factor in a 5-10% loss for the driver. So if the meter shows 320-330W being pulled from the wall/plug, you can assume there are about 300W going to the boards.

Personally, if I have to set up multiple drivers - like I did the other day (6x 480 drivers connected to 12 High Light boards) - I will adjust the current and voltage on one driver to the desired level using a couple of ameters, then read what it is pulling at the wall using the wattmeter. Then I adjust all the other drivers to whatever the wattmeter read on the first one so I don't have to keep wiring the ammeter up.

BodhiKarma said:

Ok so if I should be aiming for 500 watts, I think I have two options. The first option is 8 boards @ 1400 for a total of 534 watts. There are no numbers for 1750 but I bet it is near 85 watts. I'm going with 84.7. The second option would be 6 boards @ 1750 for about 508 watts. As I see it, those are my best two options

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Again, I would use all eight boards if you have them and don't need them for anything else.

It comes down to how much you want to spend on drivers and what you will use them for now and in the future.

The first thing you need to know are the voltage specs for the boards.

Then the choice comes down to one driver per pair of boards (ELG-150) or one driver for each two pairs of boards (ELG-240).

If you run eight boards, a couple of ELG-240-48A drivers will get you there as long as the boards don't exceed 52+V at around 1300mA.

If you run six boards, you'll want something like an HLG-185 or ELG-200 for each pair of boards (three drivers).

According to the email response I got the Rspec is .5v higher than the V2.

OK I'm sold on doing all 4 at 1400. The ELG-240-48A makes sense. I don't have a way to measure my output so considering they will *just* run 4 boards at 1400 mA I like that I'll pretty much know what they are at. If I decide to do three pairs, the same drivers could run three boards each at roughly 1750 mA cranked up. So those same drivers give me both options in parallel.

Need to go buy connectors, wire, etc now as well.

So the ELG's only run at 75% power on a 110v connection. So no the ELG-240 WILL NOT work.