Question

LED display power consumption and cable calculation

Answer 1

Regarding the power consumption of display screens and the cross-sectional area of ​​cables, this article specifically organizes the knowledge in this regard. We calculate the power consumption of electrical products by their electrical power, that is, how much "work" this appliance does per hour, and this work is obtained by multiplying the current by the voltage.

However, in actual use, the power calculation generally divides the load into two types, one is a resistive load and the other is an inductive load.

The calculation formula for resistive loads: P=UI, that is, power = voltage × current

The calculation formula for inductive loads: P=UIcosф, that is, power = voltage × current × power factor

And our LED display screen is used through a switching power supply transformer, which is an inductive load. The power factor of its switching power supply load is cosф=0.75. Different inductive loads have different power factors. When we uniformly calculate the power factor cosф of the LED display screen power supply switching power supply, we can take 0.8.

The driving voltage of the LED display screen is 5V, and the driving current is less than 20mA. The maximum power calculation method of the screen is as follows (based on 100 square meters):

1. The specifications of the outdoor LED display screen are: the dot pitch is 10mm, the dot density is 10,000 dots/square meter, each dot (pixel) consists of 1 red, 1 green, and 1 blue LED lights, and the scanning method is 1/4 scanning.

2. Calculation of power per square meter of display screen:   P= 10000 (dots) X 3 (pieces) X 5 (volts) X 0.02 (amperes) X 1/4 (scanning) = 750 (watts)

3. Power consumption of the entire display screen:   P= 750 (watts) X100 (square meters) = 75KW

4. Power consumption of the entire screen:

Because of the product loss of electric energy conversion, the general power is 10% higher than the theoretical calculated value. Auxiliary equipment between screens (air conditioners, fans, lighting, amplifiers, etc.) also generates a certain amount of power consumption. The power consumption of this screen auxiliary equipment is calculated according to the actual amount (we calculate the following formula temporarily based on 5KW).

P= 75KW+7.5 KW +5KW =87.5KW

Therefore, the total power of the display is 87500 watts, and the maximum current is:

I=P/Ucosф=87500 (watts) ÷ 220*0.8 (cosф) =318(A)

However, under normal working conditions, the display light-emitting diodes cannot be lit for a long time at the same time, so when the user's power capacity is small, a common coefficient formula can be used, and the common coefficient is generally 0.5. Therefore, the above calculation should be rewritten as:

I=P* common coefficient /Ucosф= ( 82500 (watts) *0.5+5000 (watts)) ÷ 220*0.8 (cosф) =168(A)

In other words, the total current value of this screen is 168A. Because the instantaneous current when the LED display is started is relatively large, the main air switch cannot use a three-phase 168A switch, but a 318A air switch, and 400A can be selected. Many of our peer engineers are accustomed to directly using the number of switch power supplies on the entire screen to calculate the total power of the large screen. This method is not standard. Some manufacturers reduce the number of switch power supplies in order to reduce costs, which eventually leads to incorrect power calculations.

5. Average power consumption:     The power consumption varies due to different playback content. For example, when playing advertising promotional videos, the colors are rich, mostly white, and the average power consumption is 50% of the maximum power consumption; when playing simple graphics such as text, most of the screen is black, and the average power consumption is 10% of the maximum power consumption. According to normal advertising statistics, the average power consumption is generally 40% of the maximum power consumption.

The general copper wire safety calculation method is:

The safe current carrying capacity of 2.5 square millimeters copper power line is 28A.

The safe current carrying capacity of 4 square millimeters copper power line is 35A.

The safe current carrying capacity of 6 square millimeter copper power line is 48A.

The safe current carrying capacity of 10 square millimeter copper power line is 65A.

The safe current carrying capacity of 16 square millimeter copper power line is 91A.

The safe current carrying capacity of 25 square millimeter copper power line is 120A.

If it is aluminum wire, the wire diameter should be 1.5-2 times that of copper wire.

If the current of copper wire is less than 28A, it is definitely safe to take 10A per square millimeter.

If the current of copper wire is greater than 120A, take 5A per square millimeter.

The current that can normally pass through the cross-sectional area of ​​the wire can be selected according to the total current that it needs to conduct. Generally, it can be determined according to the following jingle:

Ten is five, one hundred is two, two-five is three-five, four-three is three, seventy-nine-five is two and a half, and copper wire is upgraded.

Let me explain it to you. For aluminum wires below 10 square meters, the square millimeters can be multiplied by 5. If it is copper wire, it will be upgraded by one level. For example, 2.5 square copper wires are calculated as 4 squares. For wires above 100, the cross-sectional area is multiplied by 2, for wires below 25 square meters, it is multiplied by 4, for wires above 35 square meters, it is multiplied by 3, and for wires both 70 and 95 square meters are multiplied by 2.5. These few formulas should be easy to remember.

Explanation: It can only be used as an estimate, not very accurate. In addition, if you remember the indoor wire, copper wire below 6 square millimeters is safe if the current per square does not exceed 10A. From this perspective, you can choose 1.5 square copper wire or 2.5 square aluminum wire.

Within 10 meters, the current density of the wire is 6A/square millimeter, 10-50 meters, 3A/square millimeter, 50-200 meters, 2A/square millimeter, and less than 1A/square millimeter above 500 meters. From this perspective, if it is not very far, you can choose 4 square copper wire or 6 square aluminum wire. If the power supply is really 150 meters away (not to mention whether it is a high-rise building), 4 square copper wire must be used.

The impedance of the wire is proportional to its length and inversely proportional to its wire diameter. When using the power supply, please pay special attention to the wire material and wire diameter of the input and output wires. To prevent accidents caused by excessive current and overheating of the wire.

Answer 2

The wire diameter is generally calculated according to the following formula: Copper wire: S = IL / 54.4*U`

Aluminum wire: S = IL / 34*U`

Where: I——The maximum current passing through the wire (A)

L——The length of the wire (M)

U`——The allowable power drop (V)

S——The cross-sectional area of ​​the wire (MM2) Description:

1. The voltage drop of U` can be selected by considering the rated value of the power supply voltage used for the system power supply by the range of equipment (such as detectors) used in the entire system.

2. The calculated cross-sectional area is closer to the upper side.

Estimation of the current carrying capacity of insulated wires The relationship between the current carrying capacity of aluminum core insulated wires and the multiple of the cross-section Wire cross-section (mm2) 11.  52.  54610162535507095120

Current carrying is a multiple of the cross-section 9 8 7 6 5 4 3.  5 3 2. 5

Current carrying capacity (A) 9 14 23 32 48 60 90 100 123 150 210 238 300

Estimation formula: 2.5 times 9, minus one sign upward. 35 times 3.5, minus one sign upward for pairs. Add discounts when conditions change, 90% discount for high temperature copper upgrade. 234 for the number of pipes, 876% discount for full load current. Note: (1) The formula in this section does not directly indicate the current carrying capacity (safety current) of various insulated wires (rubber and plastic insulated wires), but rather expresses it as “cross-section multiplied by a certain multiple” and is obtained through mental calculation. It can be seen from Table 53 that the multiple decreases as the cross-section increases. “2.5 times 9, minus one sign upward” means 2. The current carrying capacity of aluminum core insulated wires with various cross-sections of 5mm’ and below is about 9 times the number of cross-sections. For example, the current carrying capacity of a 2.5mm’ conductor is 2.5×9=22.5(A). The relationship between the current carrying capacity and the multiple of the cross-section number of conductors from 4mm’ and above is that the multiple decreases by 1 as the wire number goes up, i.e. 4×8, 6×7, 10×6, 16×5, 25×4. “ Thirty-five times three point five, minus one point five for pairs”, means that the current carrying capacity of 35mm” conductor is 3.5 times of the number of cross-sections, that is, 35×3.5=122.5(A). For conductors of 50mm’ and above, the multiple relationship between the current carrying capacity and the number of cross-sections changes to two wire numbers in a group, and the multiple is reduced by 0.5. That is, the current carrying capacity of 50 and 70mm’ conductors is 3 times of the number of cross-sections; the current carrying capacity of 95 and 120mm” conductors is 2.5 times of their cross-sectional areas, and so on.  “ Add discounts when conditions change, and upgrade to copper at a 90% discount at high temperatures”. The above formula is based on aluminum core insulated wires and open laying at an ambient temperature of 25℃. If the aluminum core insulated wire is laid in an area where the ambient temperature is higher than 25℃ for a long time, the current carrying capacity of the wire can be calculated according to the above formula, and then a 10% discount can be applied; when the copper core insulated wire is used instead of aluminum wire, its current carrying capacity is slightly larger than that of the aluminum wire of the same specification, and the current carrying capacity of the wire with one line number larger than that of the aluminum wire can be calculated according to the above formula. For example, the current carrying capacity of 16mm’ copper wire can be calculated according to 25mm2 aluminum wire.

According to the previous calculation of the total current of 318A for a 100 square meter display screen, the maximum safe current that can be carried by three-phase electricity is 3 times the above, so the national standard 25 square three-phase five-wire (4+1) cable can be used normally. If the cable wiring exceeds a certain length, or is in a high temperature environment, or is concealed, it needs to be cumulatively discounted, otherwise it will affect normal use and cause the cable to heat up. In addition, it is recommended that the circuit breaker that is not installed with graded power supply should be replaced every five years, because frequent switching causes electric shock aging and poor contact of the circuit breaker. If the above two items are not taken seriously, serious electrical fires may occur.