confirming my Power Calculation

dpavisic

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I know that this has probably been asked a million times, but I am anxious to see if my calculations are right for my first lot of pixel modules I am using. I am using the following modules from Ray Wu

20pcs/string waterproof led pixel module,4pcs SMD RGB 5050,1pcs WS2801,256 gray level,DC12V,0.96W

If I have 100 of these am I right in saying that they use

100 x 0.96W= 96 watts? Is it as simple as that?

So for a 350watt PSU, I can have 350watts x 80% = 280 watts usable or about 291 modules in total, less the resistance in the supply cable (see next section)

My lights are about 10m away from the start of the lights and I plan on using some low voltage cable I have which is heavty duty 3.3mm or 12 Guage. The resistance on the wire is 0.0057 watts per meter and there are 10 metres for the + and 10 metres for the N wires = 20 metres in total which gives a total of 0.114 watts which looks pretty low. Is this correct? Should I also take into account the fact that there is thin wire between each module which would also have some higher amounts of resistance?

Many thanks
 

fasteddy

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DannyP said:
I know that this has probably been asked a million times, but I am anxious to see if my calculations are right for my first lot of pixel modules I am using. I am using the following modules from Ray Wu

20pcs/string waterproof led pixel module,4pcs SMD RGB 5050,1pcs WS2801,256 gray level,DC12V,0.96W

If I have 100 of these am I right in saying that they use

100 x 0.96W= 96 watts? Is it as simple as that?

So for a 350watt PSU, I can have 350watts x 80% = 280 watts usable or about 291 modules in total, less the resistance in the supply cable (see next section)

My lights are about 10m away from the start of the lights and I plan on using some low voltage cable I have which is heavty duty 3.3mm or 12 Guage. The resistance on the wire is 0.0057 watts per meter and there are 10 metres for the + and 10 metres for the N wires = 20 metres in total which gives a total of 0.114 watts which looks pretty low. Is this correct? Should I also take into account the fact that there is thin wire between each module which would also have some higher amounts of resistance?

Many thanks
Your correct its that simple, for working out your total load in watts, if you have the watt rating of lights then you just multiply by how many you are using. Then you do what you did an use the 80% or 85% rule to get the total amount you can run from a power supply. But the cable resistance doesnt play a part in the total load you can run, instead it plays a part in voltage drop
Now the reistance of the cable will be in Ohms/m and not watts/m but besides that your assumption is correct, 12 guage cable will give you a very low resistance over that distance but remember that resistance is only one part and total current on the cable will also play a part in voltage drop over tha distance
Now when using the modules you must inject power a the beginning and then between every second string and then again at the end, so that is string 1, between string 2 and 3, between string 3 and 5 and so on as you will get voltage drop through the thin guage wires in the string if you dont do this.
 

AAH

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Hi Danny
Can you link to the modules so that we can confirm the 0.96W per module as if this is wrong then everything else will be based on something that is wrong.
Resistance is given in ohms and not Watts. Watts are power and are given by multiplying the volts by amps. 96W @ 12V=8A. If the number (but not the unit was correct) then I assume it is ohms per metre and therefore you would calculate the voltage drop as 20m x 0.0057ohm/metre which will give you a resistance of 0.114 ohms. Multiply ohms by amps and you get the voltage drop. 8 x 0.114 equals 0.912 V. The power dropped in the wire (which isn't usually calculated) will be P=IxIxR=VxI so there is 0.912V x 8A or 7.3W lost in the 20M of cable. All of these calculations are based on the assumption that you supplied the correct figure of 0.96W (plus Ray sometimes gives wrong numbers) and the assumption that the number of Watts/metre is really Ohms/metre.
 

dpavisic

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thanks guys. This is the link to the modules

http://www.aliexpress.com/snapshot/213979456.html

I have reread the 101 manual and yes it is ohms, so I get this now.

Eddy when you say inject between each string, if i can just confrim my understanding, it will be -

inject
string 1
string 2
Inject
string 3
string 4
inject etc

Also how long is a piece of string??? :) 20 Modules?
 

fasteddy

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DannyP said:
thanks guys. This is the link to the modules

http://www.aliexpress.com/snapshot/213979456.html

I have reread the 101 manual and yes it is ohms, so I get this now.

Eddy when you say inject between each string, if i can just confrim my understanding, it will be -

inject
string 1
string 2
Inject
string 3
string 4
inject etc

Also how long is a piece of string??? :) 20 Modules?
You stated in your description that they are a 20 string module and modules generally come in a string of 20

But you are correct because all you are doing is back feeding every even numbered string , eg 2, 4, 6. Just make sure that the cable is big enough to take the load of 2 strings because you are also forward feeding every even string and if using the 12 guage then it definetly will handle that load easily.
 

fasteddy

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I reakon because these are 4 x 5050 LEDs then they would have two series 5050 LEDs in parralel so the 5050 LED is rated at 20ma per colour so that would be 60ma per LED circuit for a total of 120mA per module . This at 12vdc means that each module should draw 1.44 watts, but then because they want to reduce the power requirements then i suspect they instead underdrive the LEDs with a current of 13.3mA per LED to get the rating of 0.96 watts per module this means they can still use the low guage wire for the modules. Thats just my assumption, I would actually take a reading and see what they draw to make sure.
 
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