Author Topic: PSU Efficiency + LED Driver  (Read 3425 times)

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Offline made2hackTopic starter

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PSU Efficiency + LED Driver
« on: January 22, 2014, 03:02:00 pm »
Hello all,

I have a question regarding a luminaire that I am building. It will be an LED Luminaire. It was previously covered here https://www.eevblog.com/forum/projects/how-to-drive-400w-of-high-powered-led%27s-in-parallel-series/.

My question is as follows:

I am considering a Meanwell USP-500 PSU http://www.meanwell.com/search/USP-500/USP-500-spec.pdf with a rated power of either 408W or 504W (the 24V Model). The tech specs stipulate 89% efficiency IN. So, does that mean that if I want let's say 408W, the unit will in fact use 458 Watts of power? Such as 408W / 0.89 ?

And I noticed that using only convection, the unit produces 408W. With a fan, I can get it to 504W. Does it mean that if I draw more than 408W without cooling it with a fan, it will burn?

Finally, my LED Drivers run @ around 95% efficiency. I need 390 Watts to drive my LEDs, so, yet again, using this formula, I need 390 / .95 = 410.1 (say 411 Watts) to run the module?

Should I make sure that I get the meanwell with a FAN so that I can have max 504 Watts (to cover my 411 Watt draw) ?

What kind of overages can a PSU handle? Or they can't. Should I always make sure there is "extra" juice just in case my draw is larger? So if I have 504W total to power 411W, am I covered?

Thanks everyone

Offline made2hackTopic starter

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Re: PSU Efficiency + LED Driver
« Reply #1 on: January 23, 2014, 11:53:58 pm »
anyone?

Offline sleemanj

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Re: PSU Efficiency + LED Driver
« Reply #2 on: January 24, 2014, 12:10:37 am »
1: The tech specs stipulate 89% efficiency IN. So, does that mean that if I want let's say 408W, the unit will in fact use 458 Watts of power? Such as 408W / 0.89 ?

2:  Does it mean that if I draw more than 408W without cooling it with a fan, it will burn?

3: Finally, my LED Drivers run @ around 95% efficiency. I need 390 Watts to drive my LEDs, so, yet again, using this formula, I need 390 / .95 = 410.1 (say 411 Watts) to run the module?

4: Should I make sure that I get the meanwell with a FAN so that I can have max 504 Watts (to cover my 411 Watt draw) ?

5: So if I have 504W total to power 411W, am I covered?

1: Yes, in the best case you need 458W input power.
2: The PSU states it has over temperature shut down and auto recovery.  When it gets too hot it should shut down until it cools off then start again.
3: Yes, you need 411 watts to supply your drivers
4: Yes, sounds like you should have at least the 504W output model.
5: Probably.  You could always reduce the current to the LEDs (or number of leds) to reduce the power consumption.
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Offline Marco

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Re: PSU Efficiency + LED Driver
« Reply #3 on: January 24, 2014, 12:14:56 am »
The rated power is output power generally ... but you can also just look at the current range spec, multiplying the output voltage by the upper output current range gives the rated power.

It has temperature/overload protection so presumably it won't burn, it will turn off.

The efficiencies are typical and in the case of your boost converters probably wildly optimistic, your math is right but you need a fair bit of safety margin.
 

Offline mariush

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Re: PSU Efficiency + LED Driver
« Reply #4 on: January 24, 2014, 12:34:16 am »
I don't think the meanwell will do 89% efficiency. Maybe it reaches that at about 30-50% load and then slowly goes down towards 75-80%.

Yes, to get 408 watts out to devices, the power supply will take 458w from the mains socket if it really has 89% eff. at 400w, which i doubt. 

I encourage you to double check datasheets and do actual measurements because the real world is much different that what says on datasheet - if it says that the boost driver has 95% doesn't mean it will actually have that much efficiency in real world (it may achieve that with particular Vin, particular Vout, a specific inductor etc)

The PSU can probably do 500w with no problems, but I'm not sure how long it would last at continuous 500 watts. Look in the datasheet how cramped everything is, with capacitors very close to heatsinks. At 500w output, even with 90% efficiency you have maybe 30-50w wasted as heat.  The fan is a 23 cm one.. it's going to be noisy. Unless you hack the psu and put a 80-120mm on top of it to keep it cool (and noisy)

How much is this power supply? I see it's $240 at mouser. As long as you do the design using boost converters, maybe you can just go for a regular pc power supply and run everything from 12v... a very good 650w gold efficiency Seasonic power supply with semi-passive fan (only turns on at slow speed at around 200 watts) is only $99 : http://www.newegg.com/Product/Product.aspx?Item=N82E16817151118

Personally, I would use a 48w power supply and use buck drivers, especially since you have those 39v leds. 70.2 watts at around 90% efficiency is 80 watts ... divided by 24v is about 3.3 amps... a lot of current for boost drivers.

Plenty of drivers that support 48v and you also save money by buying 10-20 chips of the same kind, with same (or similar) parts required for each one.
 

Offline made2hackTopic starter

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Re: PSU Efficiency + LED Driver
« Reply #5 on: January 27, 2014, 01:02:19 pm »
Thank you all for the replies.

I will more than likely go for a 24V @ 400W and under drive the 39V LEDs by about 200mA. So, probably 1600mA @ 38V or so (judging by the I/V curve on the Nichia website). I cannot use 48V in with the drivers I have since they are max rated 10V - 32V DC IN.

Also, I am running each one with a separate driver, so each boost driver will only put through 1600mA (in my case). But the idea of using a 12V PSU from a pc (say a 650W or a 500W) to have some leeway might be a better option. Technically, I believe the boost will be lower efficiency (since it has to go from 12V to 38/39V), is this correct?

However, I am over compensating with a much more powerful PSU that will run quieter and cooler. And, a good brand PSU has loads of protection built in.


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