selecting an HLC series driver

BOBBY_G

Well-Known Member
based on several different discussions, im trying to spec the most universal driver for my CXB3590 system so i thought id think out loud and start a discussion on it

my parameters:
typical use = 49W 56.3% eff
'efficient' mode - would like to be able to turn down to 15W/cob efficiently
minimum load = 10W (at some driver efficiency penalty)
'overdrive' mode - would be like to drive up to 75W (~50% efficient)

tough to find a driver that can do all of the above so where do we make compromises, lets see....

HLG-240H-C2100 will soon be available to run 36V cobs at 2.1 A, but has the following disadvantages:
-at 115V, is less efficient across the board than a 185H series
-runs an odd number of cobs at that current - (3.3). running 4 would top them out at 1.75A

in this case, despite the driver cost, it might be best to double up on HLG-185H-C1050 and run 72V cobs instead of 36V

running at full current of 1.05A = 75W, 2 cobs per driver

looking at the datasheet for the HLG-185H-C series: (http://www.mouser.com/ds/2/260/HLG-185H-C-SPEC-806161.pdf)

efficiency at 93% for 115V operation is damn near linear down to 55%. so what is 55%? 0.58A

looks like it can go down to 39W @ 59% cob efficiency. with no penalty on driver efficiency. eggcelent.

so what happens when we try to turn it down to 15W?
current = 0.24A
% of full load current = 0.24/1.05 = 23%
driver efficiency at 23% = a respectable 89%
cob efficiency at 15W = 67%
overall driver+COB efficiency = 59.6%
ultimate eff. at same wattage with smaller driver at max 93% eff. at that current = 62.3%

for occasional low wattage use at that low load, ill take a little efficiency hit (id rather they be more efficient at high wattages and in the ranges im more likely to use

at minimum 10W/cob
current = 0.17A
% of full load current = 0.17/1.05 = 16%
driver efficiency at 16% = a fair 85%
cob efficiency at 10W = 69%
overall driver+COB efficiency = 58.7%
ultimate eff. at same wattage with smaller driver at max 93% eff. at that current = 64.2%

here were starting to see an efficiency hit as im really out of the range of my driver's efficient zone. but you know what? it really doesnt matter to me to lose 5% of efficiency when im turning the cobs down from 15W to 10W and saving almost 50% of dissipation.

but if i was going to run like this all the time id be better off with a 700 or 500 mA driver....

-----------------

but what about heat sinks, you say?

well how about a dual active/passive configuration?

a nice sized heatsink that can handle passive cooling of the cobs at 49W, with some fans on there that only get used when run at 75W. surface area TBD based on actual hat load as thread develops

thoughts? tweaks? flames?
 

nogod_

Well-Known Member
When calculating load on those drivers you use voltage (I believe).

So at 10 watts you'd still be at 34v or something on a 36v cob. On a 143v max driver, 50% load would be 72v. With 4 cobs on the string it's actually impossible to go below 50% load (72v).

So choose the max driving current you want and dim it down as needed and know that the closer you are to max, the more likely you are to suffer no efficiency penalty at all even when fully dimmed.

based on several different discussions, im trying to spec the most universal driver for my CXB3590 system so i thought id think out loud and start a discussion on it

my parameters:
typical use = 49W 56.3% eff
'efficient' mode - would like to be able to turn down to 15W/cob efficiently
minimum load = 10W (at some driver efficiency penalty)
'overdrive' mode - would be like to drive up to 75W (~50% efficient)

tough to find a driver that can do all of the above so where do we make compromises, lets see....

HLG-240H-C2100 will soon be available to run 36V cobs at 2.1 A, but has the following disadvantages:
-at 115V, is less efficient across the board than a 185H series
-runs an odd number of cobs at that current - (3.3). running 4 would top them out at 1.75A

in this case, despite the driver cost, it might be best to double up on HLG-185H-C1050 and run 72V cobs instead of 36V

running at full current of 1.05A = 75W, 2 cobs per driver

looking at the datasheet for the HLG-185H-C series: (http://www.mouser.com/ds/2/260/HLG-185H-C-SPEC-806161.pdf)

efficiency at 93% for 115V operation is damn near linear down to 55%. so what is 55%? 0.58A

looks like it can go down to 39W @ 59% cob efficiency. with no penalty on driver efficiency. eggcelent.

so what happens when we try to turn it down to 15W?
current = 0.24A
% of full load current = 0.24/1.05 = 23%
driver efficiency at 23% = a respectable 89%
cob efficiency at 15W = 67%
overall driver+COB efficiency = 59.6%
ultimate eff. at same wattage with smaller driver at max 93% eff. at that current = 62.3%

for occasional low wattage use at that low load, ill take a little efficiency hit (id rather they be more efficient at high wattages and in the ranges im more likely to use

at minimum 10W/cob
current = 0.17A
% of full load current = 0.17/1.05 = 16%
driver efficiency at 16% = a fair 85%
cob efficiency at 10W = 69%
overall driver+COB efficiency = 58.7%
ultimate eff. at same wattage with smaller driver at max 93% eff. at that current = 64.2%

here were starting to see an efficiency hit as im really out of the range of my driver's efficient zone. but you know what? it really doesnt matter to me to lose 5% of efficiency when im turning the cobs down from 15W to 10W and saving almost 50% of dissipation.

but if i was going to run like this all the time id be better off with a 700 or 500 mA driver....

-----------------

but what about heat sinks, you say?

well how about a dual active/passive configuration?

a nice sized heatsink that can handle passive cooling of the cobs at 49W, with some fans on there that only get used when run at 75W. surface area TBD based on actual hat load as thread develops

thoughts? tweaks? flames?
 

BOBBY_G

Well-Known Member
interesting. i assumed max load = 100% rated current and dimmed to 50% current = 50% load but i can see what youre saying, where 100% load would be maxing out the wattage of the driver

somebody here should know the answer fo rsure
 

SupraSPL

Well-Known Member
Yes efficiency of these drivers relates to the voltage. So as nogod said, if you have a high voltage and the driver is dimmed way down, you still get high driver efficiency.
 

BOBBY_G

Well-Known Member
ok so the rule you gave in the other thread, spec a driver close to the number of cobs you are running (3.9-4.1 etc) and then dim down to 10% and it will still be in the 90s in efficiency?

has anybody measured driver efficiency vs current? i notice the power factor curves are always given in addition to efficiency vs load curves. what use is that info?
 

SupraSPL

Well-Known Member
I tested an HLG-185H-C1050A (maxes at 190V) for efficiency at various voltages and various levels of dimming. The driver was powered by 120V AC.

@1130mA, 190.1Vf, 93.6%
@525mA, 177Vf, 93%

@1130mA 144Vf, 92.8%
@706mA, 137Vf, 92.3%
@479mA, 133Vf, 91.3%

@1130mA, 108.5Vf, 90.9%
@480mA, 100.5Vf, 88.5%

I would like to do a more thorough test with a B version driver now that I have some better equipment, but this gives you the general idea.
 

SupraSPL

Well-Known Member
I am no expert but this is how I understand it. The power factor in the case of AC-DC switching power supplies refers to harmonic distortion in the waveform
harmonic distortion.jpg

The HLG drivers are power factor corrected so they attempt to correct this distortion as much as possible. The result of an imperfect power factor is that the wires powering the device are subjected to an increased current. In residential situations the power company pays for that extra current, but it does result in more pollution to generate the extra power. So when powering large systems it makes a lot of sense to use power factor corrected power supplies IMO.
 
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