Position: Index > Inverter Circuit >

Inverter Backup. Some facts

2014-12-12 21:13  
Declaration:We aim to transmit more information by carrying articles . We will delete it soon, if we are involved in the problems of article content ,copyright or other problems.

“My Inverter is not giving sufficient backup”. This is the most common complaint raised against the home inverter system. We expect maximum backup time from the inverter system with a tubular battery, but unfortunately we get half or less backup time than we expect. What is the reason for that? Before blaming the inverter, let us see what is happening.

The back bone of the Inverter system is the battery. We use a high current tubular battery to give a maintenance free performance. The charge in the battery is used to convertDC to ACby the inverter. Measure of charge isCoulomband each electron carries1.602 e- 19 coulombcharge. As a rule, when 1 amps current passes through a conductor in 1 second, it uses 1 coulomb charge

So charge Q = It

If1 amps currentflows through the conductor in1 hour, 3600coulomb chargewill be utilized.
The amount of charge in the battery is represented inAmps hour (Ah). That isAmps times Hour.It is the amount of charge present in the battery. But Amps hours cannot be used to measure the charge level, since the voltage changes during discharge. So the measure of charge isWatts Hours. Watt hour can be calculated by multiplying the nominal voltage with the battery capacity in Amps hours.

E = C x Vg. where C is the capacity of the battery in Ah and Vg is the discharge rate.

Charge requirement of the load

The following equation tells you, how much battery charge is required for your load.
First find out the capacity of the battery. If‘X amps’current is drawn by the load in‘t hours’, then the capacity of the battery

C = Xt

Suppose the load is drawing120 mAcurrent in24 hours, then the capacity of the battery should be

C = 0.12A x 24 = 2.88 Ah

It is not a good practice to discharge the battery completely till the load shut off. Stop running the load, if the battery charge reduces to 20%.

Battery capacity

If the load takes120 mAcurrent in1 hour, maximum capacity of the battery should be0.12A x 24 = 2.88 Ah.Best method to keep the charge / discharge cycles perfect is to stop discharging the battery till it maintains 20% charge. Hence to retain 20% charge, the capacity of the battery should be

C / 0.8

If the battery requires 2.88 Ah in one hour, then to keep 20% charge in it, the capacity of the battery should be

2.88Ah / 0.8 = 3.6Ah.

Discharge Rate

Lead acid batteries have few Amps hour if the discharge rate is fast. Generally, the lead acid battery is rated for20 hoursdischarge rate provided the discharge rate is slow. At high discharge rate, the capacity of the battery drops steeply. Suppose the battery is 10 Ah and its discharge rate is 1C.One hour dischargeof the battery at the rate of1C(10 Amps in 1 hour), the capacity reduces to5 Ahin one hour. So the following tips will guide you to keep a steady discharge rate of the battery.
Suppose you want to run a load at20 Amps for 1hour.Then the capacity of the battery should be

C = It = 20A x 1 Hour = 20Ah.

Keep discharge rate to maximum 80%. Then the battery capacity should be

20Ah / 0.8 = 25 Ah

So a25 Ahbattery can give20 Ampscurrent for1 hourto run the load

But it is better to drain the battery to 50% only then,

25 Ah / 0.5 = 50 Ah.

Hence as a rule of thumb, it is better to use a50Ah batteryto run the load at20 Ampsper Hour to keep 50% charge in the battery.
Suppose the load is not drawing20 Ampscontinuously in 1 hour. During the first second, it draws 20 Amps and then 100 mA during the remaining period. So the average current drawn by the load is

20A x 1/ 3600 0.1 / 3600 = 0.1044 A

3600 is the total seconds in 1 hour.

In short, if the load is not drawing the charge in a steady manner, the capacity of the battery will be increased.

How to calculate the charge?

It is difficult to measure the current drawing by the load at different times. The easy way is to consider the power rating (Watts) of the load. Suppose the power rating of the load is 250 watts and is drawing current from the inverter system for 5 hours.

Then, its Watts hour is Watts x Hour = 250 x 5 =1250 Watts hour.

Consider the efficiency of the inverter as 85 %( no inverter is 100% efficient).

So 1250 / 0.85 =1470 watts hour

So the load has1470 watts hourinstead of 1250 watts hour.

Then calculate the capacity of the battery

As you know,Watt is Amps x Volt

So if the Watt hour is divided by Voltage of the battery, you will get the Amps hour

Watt hour / Volt = Amps hour

1470 / 12 = 122.5 Ah

Thus to run 1470 watts load, minimum capacity of the battery should be 125Ah to run the load for 5 hours. As already stated, it is better to use a 150 Ah battery to keep discharge cycle 50%.

How to select an Inverter

Total load to be connected =500 watts

Power factor =0.8(all inverters have a power factor between 0.6 to 0.8)

Inverter VA = 500 / 0.8 =625 VA

So select 800 VA inverter to run 500 Watts load

How to select the battery

Backup time = Watt / Battery voltage x Hours

500 / 12 x 3Hr = 125 Ah

12 volt tubular battery has a terminal voltage of 14.8 volts in fully charged condition. Usually the inverter has cutoff facility to protect battery from deep discharge. Most of the inverters are set for 80 % (Retaining charge) cutoff voltage. That is after 20% discharge, inverter will shut off.

Inverter cutoff voltage = 14.8 x 0.8 =11.84 volts
Watt hour = watt x hour = 500 watts x 3 hours =1500 watt hours
Ah of battery = watt hour / volt = 1500 / 11 =126 Ah.

In short, an 800 VA inverter with 126 Ah battery can power 500 watts load for 3 hours.


Reprinted Url Of This Article:
http://electroschematics.com/5730/inverter-backup-some-facts/