90 watts on meanwell 60-48?

[nanojg]

Is this common. I just put my meanwell on a kill o watt and it pulled 90 watts on full power.


Edit: subtract 5 watts for the fan that was connected.

[jeffm59]

http://www.meanwell.com/search/eln-60/default.htm

Not a wiz at this but the cut sheet shows it typically pulls 1.2 Amps at 115 V AC so that calculates out to 138 Watts. It is rated at 62.5 watts of DC output.

Guess it depends on the number of LEDS that you have on the driver and their actual specs since they vary from one LED to the next

[bamf25]

Is this common. I just put my meanwell on a kill o watt and it pulled 90 watts on full power.


Edit: subtract 5 watts for the fan that was connected.

That sounds high. If you are running 12 led at say 700 mA with a fV of 3.3 that is 3.3 x 12 x 0.7 = 27.72 w being used on the dc side. Now if you account for the 88% efficiency 27.72/.88 = 31.5 w. I may be wrong but that is my understanding.

It is a 60 watts driver which means it should only be able to pull a maximum of 60 watts.

[reefmusic]

Seems like we are discovering these meanwell drivers are VERY inefficient. I'd be looking for something that performed much better than this.

[reefmusic]

It is a 60 watts driver which means it should only be able to pull a maximum of 60 watts.

It appears the 60 watts specified refers to the output.

[nanojg]

This is from 12 xpgs on the full 1.3 amps

[widmer]

Yea I'm really not an expert on the mechanics of electricity like this, but when it says that it runs 88% efficient and it can feed 60 watts, than I would think it should consume (1/.88)*60= 68 watts when running full steam ahead.

I'm guessing it has something to do with what the people who know more about this than me are calling "harmonics". But my question is, even if it isn't really pulling like 120 watts, if the Kill-a-watt can't tell the difference, can the electric meter on the outside of the house?

Kcress your needed here. :lol:

[bamf25]

^^^^

Still say 12 xp-g's with a fV of 3.5 is 3.5 x 1.3 x 12 = 54.6w. If the efficiency is 88% and we are going to say the output is 60w, then 54.6/0.88 = 62 w from the AC side. I am not sure where that 90 (or 85 if you take the fan out) is coming from.

Granted it is much closer, but still not there. Did you touch the voltage ajustment. I have read the you can put the dc voltage up to 52.8 on those drivers. So if you are doing the full voltage on the dc side 52.8 x 1.3 = 68.64. Then 68.64/0.88 = 78 w. That is getting close and maybe with harmonics and varience in signal can get you to 85?

[nanojg]

I didnt touch the SVR1 (think thats what its called), just the 2. Something else I should mention is that the svr2 did not work, I could not adjust the current on this driver. I originally had the xres on this driver but after I found out I could not adjust the current I put the XPGs to be safe.

Maybe I should hook it up to a multimeter and try to adjust the svr1?

[nanojg]

Overall, I am happy with the electricity consumption, but this reading (with whites on full) does not seem right. It read 135, if I remember correctly, with both drivers on full power (at 1.3 amps for the white and .7 amps for the blues).

The 2 drivers are only pulling 55 watts (with the fan) on the output I like for the tank (about 50% white and 55% blue).

[LargeAngels]

I'd test it with another meter before I assumed the "kill o watt" is correct.

[kcress]

Not a wiz at this but the cut sheet shows it typically pulls 1.2 Amps at 115 V AC so that calculates out to 138 Watts. It is rated at 62.5 watts of DC output.

You cannot do the math that way. It is not 1.2A x 115V. That is only true for a purely resistive load which an ELN is not even close to. The 1.2A figure is only for calculating wire ampacity.

It is a 60 watts driver which means it should only be able to pull a maximum of 60 watts.

The "60" refers to the drivers output. Not the input.

Technically the input will be 60/0.88 as was suggested.


If there is a large difference it could be several things.

1) Meanwell's data sheet could be BS.
2) The OPs KillOWatt may be out of calibration.
3) The K-O-W may not understand the horrid harmonics coming from the MWs.
4) The OP could be misinterpreting the K-O-Ws output?

I would suggest (3) is the issue.

[dwl]

Kcress,

If it is number 3 above, then what would the power company be seeing and charging? Would it be the 60/0.88 watts or would it be the 90 that the OP stated?

Thanks.

[reefmusic]

You cannot do the math that way. It is not 1.2A x 115V. That is only true for a purely resistive load which an ELN is not even close to.

Yes, and someone else suggested power factor was pretty bad i.e. non-resistive. So assuming something like 0.7 at least, and 60 watts load we are near the 90 figure measured (VA not watts). Maybe the KOW is dumb with things like this.

[laverda]

I thyink part of it is we are not running them the way they are desinged to be used. They are less efficient at less then max output.

[nanojg]

I thyink part of it is we are not running them the way they are desinged to be used. They are less efficient at less then max output.

Do 90 watts at max power is more efficient? No

[Chris27]

There is always loss when you're talking transformers and rectifiers, remember that the AC-DC circuit itself will suck up some juice.....which is part of the reason why you get the 88% efficiency number vs. 100%. I'm figuring the 90W to be about right.

[patmack]

I am in the planning stages of doing a DIY build for my 210 that will require 12 meanwell 48D drivers. If this is the case (90 watts per driver) my current t5 lights driven by 2 IC 660s will be much more efficient than the "super efficient" LEDs... bummer I was looking forward to LEDs

[widmer]

There is always loss when you're talking transformers and rectifiers, remember that the AC-DC circuit itself will suck up some juice.....which is part of the reason why you get the 88% efficiency number vs. 100%. I'm figuring the 90W to be about right.

90W is nowhere near correct if the data sheet specifies 88% efficient. As stated above, 88% efficient would mean that the driver which supplies 60 watts should consume 60*(1/.88) = 68 watts. If it consumed 90 watts, it would be ~66% efficient.

I am in the planning stages of doing a DIY build for my 210 that will require 12 meanwell 48D drivers. If this is the case (90 watts per driver) my current t5 lights driven by 2 IC 660s will be much more efficient than the "super efficient" LEDs... bummer I was looking forward to LEDs

:lol: Don't let this single thread that is discussing the inefficiency of one series of drivers deter you from doing an LED build. You have lots of options for drivers. Even if you feel like you must stay within the Meanwell brand, you can get the PFC series, which is up to 92% efficient, and by definition in name, is able to correct for the the power factor that is throwing off the kill-a-watt...

...Not to mention that there's certainly no proof at this point that the household power meter is thrown off the way that the as-seen-on-TV kill-a-watt is...

[patmack]

I just got done with a test fixture last night so tonight I am going to check the power consumption myself.

This is probably not the correct place for this but...
If you run 2 strings of 12 in parallel they will still use the same amount of power correct?
What are other good dimmable drivers?

[reefmusic]

I just got done with a test fixture last night so tonight I am going to check the power consumption myself.

This is probably not the correct place for this but...
If you run 2 strings of 12 in parallel they will still use the same amount of power correct?


No they will see half the current (approx) each string.

Edit: oh maybe you meant total power - then yes, but the each LED is seeing half.

[patmack]

OK I measured some currents with my kill-a-watt meter and got:
44 watts on the XPG set to 1000mA
34 watts on the royal blue Xpe at 750 mA

To answer the OP question I turned up the current to 1.5 amps on the xpgs and it was over 90 watts.

Looks like I was right in the first place, LEDs are more efficient than my current 6x80watt t5s.

[nanojg]

Thanks, the svr2 does not work on that driver. I am happy with the watts used at normal operating current (55 watts total for both drivers and fan). This is 24 leds on my 24 nano cube so I have them running at about 50% of max current and its perfect for me.

[nanojg]

OK I measured some currents with my kill-a-watt meter and got:
44 watts on the XPG set to 1000mA
34 watts on the royal blue Xpe at 750 mA

To answer the OP question I turned up the current to 1.5 amps on the xpgs and it was over 90 watts.

Looks like I was right in the first place, LEDs are more efficient than my current 6x80watt t5s.

I thought the max was supposed to be 1.3? Because the svr2 did not work I guess I didnt test the max amps when i first set it up (just started it dimmed down and tried to adjust the svr2 from there). But I have tested with a clamp meter the amps jumped up over 1.5 as I turned the dimmer up near 100%, which kinda scared me so I backed it down.

[patmack]

The 1.3 amps is the input current - Different than the output current. The clamp on amp meters are only used for AC voltage. The meter you need would be put in series between the driver and the LEDs. I went to harbor freight and got one for $7.

[nanojg]

I have a AC/DC clamp on amp meter

http://www.service.kleintools.com/cgi-bin/lansaweb?procfun+KD01+KD01025+PRD+ENG+FUNCPARMS+KD01YYCTN(P0020):1+KD01YYPRD(A0150):CL2000

[nanojg]

The Specs list 1.3 as the max output.

[patmack]

Both are verry true!! well you got me. sorry for the incorrect info..