**“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 convert** DC to AC** by the inverter. Measure of charge is Coulomb and each electron carries **1.602 e- 19 coulomb charge**. As a rule, when **1 amps **current passes through a conductor in **1 second**, it uses **1 coulomb charge**

So charge **Q = It**

If **1 amps **current flows through the conductor in **1 hour**, 3600 **coulomb charge** will be utilized.

The amount of charge in the battery is represented in **Amps hour (Ah)**. That is **Amps 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 is Watts 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 drawing **120 mA** current in 24 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 takes **120 mA current** in **1 hour**, maximum capacity of the battery should be **0.12A 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 for **20 hours** discharge 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 discharge of the battery at the rate of **1C **(10 Amps in 1 hour), the capacity reduces to 5 Ah in 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 at 20 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 a **25 Ah** battery can give **20 Amps **current for 1 hour to 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 a 50Ah battery to run the load at 20 Amps per Hour to keep 50% charge in the battery.

Suppose the load is not drawing **20 Amps** continuously 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 has 1470 watts hour instead 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.
**

**D.Mohankumar
**

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Inverter selection

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Thanks for the feedback.Expect more articles in the blog

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