# Help calculating power usage

#### Kingfisher

##### New elf
Hi there, I am hoping someone can help me here.
I'm having a bit of an disagreement (read: fight) with my dad over the power consumption of my lights. His concern (and mine to an extent) is the consumption of 240v power.

I have estimated that all of my pixels will use at the extreme 310A @ 5V.
I am hoping someone can give me a formula to calculate the Amps that would be used from mains.
I am aware that this won't be perfect but it's better then not having a real idea.

Thanks in advance for you help.
David.

#### BAZMick

##### Full time elf
Generous elf
Watts = Volts X Amps.
5V X 310A = 1550W
1550W ÷ 240V = ~6.5Amps. A bit more than a set of bathroom heat lamps with 4x bulbs in them.
That's what Dr Google tells me.
Cheers Bazzy

#### thewanderingpine

##### New elf
It’s worth noting that power supplies are not 100% efficient. So we can take that 1550W figure and bank on a power supply being 70% efficient. (Possibly low - but it works for the calculation)

that gives us 1550 divided by 70% = ~2200W. So we’re creeping up to the maximum draw from a single 10A circuit. Or about the level of a plug-in electric heater.

#### AAH

##### I love blinky lights :)
Community project designer
1550W at 240V will be about 2000W (depending on the brand and quality of power supplies). With that you can run off 1 power point if it's heavy duty 1mm-1.5mm2 extension leads. In reality that's the peak load. You are probably averaging under 10% for the duration of the show so it's probably on par with running a decent sized tv for the duration.

#### Kingfisher

##### New elf
Thanks everyone.
Just to make sure I understand this right;
watts are the same regardless of the voltage, where Amps are voltage dependent.
if that's right than that makes things easier.
Edit: add a bit of inefficiency for the power supply, ie; the power supply is only about %70 efficient.

Edit2: the calculation would be something like;
(Watts / 70%) / 240 = Amps

thanks again for all your help.
David

Last edited:

##### Apprentice elf
Wow thanks for the question ! I ran 60% 12V and 40% 5V last year and based on these calculations
it gives a good reason to consider the 5V route. YMMV

AAH

#### AAH

##### I love blinky lights :)
Community project designer
If you use P=VI (Power=Volts times Amps) you can scale from the 5V power usage up to the 240V usage.

80% efficiency is fairly low for a big power supply. In reality they should be getting closer to 90% but 5V will actually be a little lower due to constant losses within the power supplies. A 400W 12V supply could be 85% efficient whereas a 400W 5V one could be 80%. The 5V one however can power max 1400 pixels compared to the 12V running 600 max.

#### thewanderingpine

##### New elf
Please take AAH's numbers over mine. I was throwing out 70% as a worst case scenario. He actually knows what he is talking about.

#### i13

##### Dedicated elf
If you're concerned about tripping the circuit breaker, keep the power factor in mind. If the power factor is 0.5 then the circuit breaker will trip at 50% of its rated load. If the power factor is 0.95 then it'll trip at 95%.

I'm not sure how you calculated 310A but the maximum current that a WS2811 pixel can possibly draw is 0.0555A. Multiply your total number of pixels by 0.0555A for a worst case scenario estimation. In reality, pixels often draw less than this, even on full brightness white.

#### SAALTFAM

##### Sparky with Blinky Lights ;-)
Watts are Watts 5,12,110, 230 volts, watts is still the same. (Energy done) as AAH said.
Your 310A @ 5V = 1550VA (W) if you have 0.5PF = approx. 438A = 2190W @ 5V. Still under 10A @ 240VAC. But it can get a lot deeper than that.
The total amount of your power supplies in Watts should be the max as they should switch them selves off when you go over, unless they are really cheap ones.
I have 5 power supplies of 3 x 5V-350W, 2 x 12V-120W running approx 4000pixel and calculated cost for a show period is \$10 of electricity. ( the actual amount is less as it is harder to calculate the cost with the different colours as they use less than white. The Calc is 30% White the whole 3 hrs for 40 days.)
This is all of 1 power point.

But fused correctly is your best friend.
Just enjoy the lights and the thrill and happiness you give to others.

#### TerryK

##### Retired Elf
... I'm having a bit of an disagreement (read: fight) with my dad over the power consumption of my lights. His concern (and mine to an extent) is the consumption of 240v power. ...
... I am hoping someone can give me a formula to calculate the Amps that would be used from mains. ...
David, for amperage on the primary side, try this: InputCurrent = OutputPower / (InputVoltage * PowerFactor * Efficiency)

I've been watching this thread waiting to see if you would expand on the finer details of the father/son power debate. I planned on only watching until PF (Power Factor) entered the thread. The dilemma then became how to add to the thread without causing (more) confusion.

The formulae above is from one of Meanwell's technical manuals. PF by definition can only have a value from -1 to 1. Values from 0 to 1 are quite common (and the rule generally) and are caused by inductive loads such as motors and transformers. Values from 0 to -1 are caused by capacitive loads and are rather rare to encounter. Distributed capacitance in cables can cause a negative PF as odd as it may sound; capacitor banks are more common.
Efficiency is fairly understood so I'll leave it aside. PF though; has meaning only in AC circuits, not DC circuits. That is, PF is a ratio of sinusoidal waveforms. A PF = 1 is a situation where the load is entirely resistive and the voltage waveform is in phase or aligned with the current waveform. For motors or transformers the PF calculations are straightforward and where the PF goes < (less than) 1, one encounters Real Power, Apparent Power, and Reactive Power. Still rather straightforward but the rules change a bit when calculating the PF for non-linear devices. An AC/DC supply in this case where the Voltage waveform is sinusoidal but the current waveform is anything but.

Meanwell indicates that for AC/DC supplies without PFC (Power Factor Correction), the PF will be in the 0.4 to 0.6 range. Passive PFC can increase that to the 0.8 to 0.9 range and Active PFC can bump that further up into the 0.98 plus or minus a little range. But, (there's always a but) while PF has no meaning on DC, the DC load does reflect back to the amperage requirements on the AC side where PF has meaning. Put another way, DC load affects the PF seen on the primary side.

The graphic is one of Meanwell's HLG supplies. I choose this one because their datasheet has both PF/Load and Efficiency/Load graphs. The HLG series also has APFC as does their RSP series. I would be a bit surprised if a RSP supply PF/Load graph would greatly differ. Unfortunately, their LRS series does not have PFC and Meanwell does not indicate (that I have found so far) what the PF is in that series. PF = 0.4 to 0.6 perhaps?

I have left a lot out as the rabbit hole goes pretty deep regarding non-linear load PF calculations. There are formulaes for calculating PF with 'True RMS' meters but they need to correctly determine the True RMS with non-linear waveforms. Amperage in most situations as the voltage waveform is just about always sinusoidal. If I was to sum all of this it would be to add a bit of cost to one's display and purchase supplies with Active Power Factor Correction. I'm not sure about server supplies but I suspect the better HP/Dell models do have APFC designed in them. I would also suggest trying to keep your supplies loaded as much as possible for best PF and Efficiency as can be seen in Meanwell's graphs. Myself, I design or try to design 70% pixel drive to 100% supply capability. I use RSP supplies because they enter current limit on overload rather then Meanwell's 'Hiccup' mode.

As always, hoped I helped and apologies for the length. Cheers.

#### Kingfisher

##### New elf
Terry, thanks so much for your knowledge.

In regards to the father/son debate, well we where both wrong, dad was just a little more right then me.

I'm quite confident that my display will not go over my 10amp max, but out of caution and to make dad happy, I'm splitting the display across 2 separate circuits, a brand new dedicated one (that I am yet to get installed) and one that I'm confident is only just barely used.

I am not using Meanwell supplies and instead have gone with some cheap Chinese copies.

Thanks again.

#### stevelee

##### Apprentice elf
I use RSP supplies because they enter current limit on overload rather then Meanwell's 'Hiccup' mode.
Are the meanwell RSP supplies noisy ? I’m looking at getting a couple of RSP-1500’s

#### TerryK

##### Retired Elf
Difficult to answer specifically as it would depend upon what supplies one is comparing. My opinion is Meanwell does a good effort at producing a quality supply at a reasonable cost. From that perspective I do not consider the RSP-1500-5 (or -12) supplies exceedingly noisy. Meanwell specs both of these at a maximum of 150 mVp-p (milli Volt Peak To Peak). Comparing is made more difficult as some manufacturers spec ripple/noise in RMS which makes the measurement a bit 'wonky' as ripple and noise are not exactly sinusoidal; the measuring concern I mentioned above.

This is I think a situation where one 'gets what they pay for'. I have looked at the higher wattage Meanwell RSP supplies but have not yet purchased any of them and see no reason to avoid their purchase once I have the need. I would like to point out that I think one should try to keep them fairly well loaded as their efficiency drops under light loads. This can be seen in their Load/Efficiency graph that Meanwell commonly includes in a supply's datasheet.