500 Watt Inverter Circuit with Battery Charger

In this post we will comprehensively discuss how to build a 500 watt inverter circuit with an integrated automatic battery charger stage.

Further in the article we will also learn how to upgrade the system for higher loads and how to enhance ot into a pure sine wave version.

This 500 watt power inverter will convert a 12 V DC or 24 V DC from a lead acid battery to 220 V or 120 V AC, which can be used for powering all types of loads, right from CFL lights, LED bulbs, fans, heaters, motors, pumps, mixers, computer, and so on.

Basic Design

An inverter can be designed in many different ways, simply by replacing the oscillator stage with another type of oscillator stage, as per user preference.

The oscillator stage is basically an astable multivibrator which could be using ICs or transistors.

Although an astable based oscillator can be designed in various ways, we will use the IC 4047 option here since it is a versatile, accurate and a specialized astable chip designed specifically for applications like inverers.

Using IC 4047

Making any inverter using the IC 4047 is probably the most recommended option due to high accuracy and readability of the IC. The device is a versatile oscillator IC which provides a dual push pull or flip flop output across its pin10 and pin11, and also a single square wave output at pin13.

A basic 500 watt inverter with a square wave output can be as simple as above to build. However, to upgrade it with a battery charger we may have to employ a charger transformer rated appropriately as per the battery specifications.

Before learning the charger configuration let's first get acquainted with the battery specification required for this project.

From one of our previous post we know that the more appropriate charging and discharging rate of a lead acid battery should be at 0.1C rate or at a supply current that's 10 time less than the battery Ah rating. This implies that to get a minimum of 7 hours back up at 500 watt load, the battery Ah could be calculated in the following manner

Operational current required for a 500 watt load from a 12V battery will be 500 / 12 = 41 Amps approximately

This 41 amps needs to last for 7 hours, implies that the battery Ah must be = 41 x 7 = 287 Ah. However, in real life this will will need to be at least 350 Ah.

For a 24 V battery this may come down to 50% less at 200 Ah. This is exactly why a higher operational voltage is always advised as the wattage rating of the inverter gets on the higher side.

Using 24 V Battery

In order to keep the battery and the transformer size smaller and cables thinner, you may want to use a 24 V battery for operainf the proposed 500 watt design.

The basic design would remain as is, except a 7812 IC added to the IC 4047 circuit, as shown below:

Schematic Diagram

Battery Charger

To keep the design simple yet effective, I have avoided the use an automatic cut off for the battery charger here, and have also ensured a single common transformer is used for the inverter and the charger operations.

The complete circuit diagram for the proposed 500 watt inverter with battery charger can be seen below:

The same concept has been already elaborately discussed in one of the other related posts, which you can refer to for additional information.

Basically, the inverter uses the same transformer for charging the battery and for converting the battery power to 220 V AC output. The operation is implemented through a relay changeover network, that alternately changes the transformer winding to charging mode and inverter mode.

How it Works

When grid mains AC is not available, the relay contacts are positioned at their respective N/C points (normally closed). This connects the drains of the MOSFETs with the transformer primary, and the appliances or the load connect with the secondary of the transformer.

The unit gets into inverter mode and begins generating the required 220V AC or 120 V AC from the battery.

The relay coils are powered from a simple crude transformerless (capacitive) power supply circuit using a 2uF / 400V dropping capacitor.

The supply is not required to be stabilized or well regulated because the load is in the form of the relay coils which are quite heavy duty and will easily withstand the switch ON surge from the 2uF capacitor.

The coil for RL1 relay which controls the mains AC side of the transformer can be seen connected before a blocking diode, while the coil of RL2 which controls the MOSFET side is positioned after the diode and in parallel to a large capacitor.

This is intentionally done to create a small delay effect for RL2, or to ensure RL1 switches ON and OFF prior to RL2. This is for safety concerns, and to ensure that the MOSFETs are never subjected to the reverse charging supply whenever the relay moves from inverter mode to charging mode.

Safety Suggestions

As we know, in any inverter circuit the transformer works like an heavy inductive load. When such a heavy inductive load is switched with a frequency, it's bound to generate a massive amount current spikes which may be potentially dangerous for the sensitive electronics and the involved ICs.

To ensure proper safety to the electronic stage, it may be important to modify the 7812 section in the following manner:

For a 12V application, you can reduce the above spike protection circuit to the following version:

Battery, MOSFET and Transformer Determine the Wattage

We have discussed this many times through different posts that it is the transformer, the battery, and the MOSFET ratings that actually decide how much power an inverter can produce.

We have already talked about the battery calculations in the previous paragraphs, now let's see how the transformer can be calculated for complementing the required power output.

It is actually very simple. Since the voltage is supposed to be 24 V, and power 500 watts, dividing 500 with 24 gives 20.83 amps. Meaning the transformer amp rating must be above 21 amps, preferably up to 25 amps.

However, since we are using the same transformer for both charging and inverter modes, we have to select the voltage in such a way that it suits both the operations optimally.

A 20-0-20 V for the primary side appears to be a good compromise, in fact it is the ideally suited rating for the overall working of the inverter across both the modes.

Since, only one half winding is used for charging the battery, the 20 V RMS rating of the transformer can be used for getting a 20 x 1.41 = 28.2 V peak Dc across the battery with the help of the associated filter capacitor connected across the battery terminals. This voltage will charge the battery at good rate and at the correct speed.

In the inverter mode, when the battery is at around 26 V, will allow the inverter output to be at 24/26 = 220 / Out

Out = 238 V

This looks a healthy output while th battery is optimally charged, and even when the battery drops to 23 V, the output can be expected to sustain a healthy 210V

Calculating MOSFET: MOSFETs basically work like switches that must not burn while switching rated amount of current, and also must not heat up due to increased resistance to switching currents.

To satisfy the above aspects, we have to make sure that the current handling capacity or the ID spec of the MOSFET is well over 25 amps for our 500 watt inverter. Also to prevent high dissipation and inefficient switching the MOSFET's RDSon spec must be as low as possible.

The device shown in the diagram is IRF3205, which has an ID of 110 amp and RDSon of 8 milliohms (0.008 Ohms), which actually looks quite impressive and perfectly suitable for this inverter project.

Parts List

To make the above 500 watt inverter with battery charger, you will need the following bill of materials:

• IC 4047 = 1
• Resistors
• 56K = 1
• 10 ohms = 2
• Capacitor 0.1uF = 1
• Capacitor 4700uF / 50 V = 1 (across the battery terminals)
• MOSFETs IRF3205 = 2
• Diode 20 amp = 1
• Heatsink for the MOSFETs = Large Finned Type
• Blocking Diode Across MOSFETs Drain/Source = 1N5402 (Please connect them across drain/source of each MOSFET for added protection against reverse EMF from the transformer primary. Cathode will go to the drain pin.
• Relay DPDT 40 amp = 2 nos

Upgrading to Modified Sinewave Inverter

The square wave version discussed above can be effectively converted into a modified sinewave 500 watt inverter circuit with much improved output waveform.

For this we use the age old IC 555 and IC 741 combination for manufacturing the intended sine waveform.

The complete circuit with battery charger is given below:

The idea is the same which has been applied in a few of the other sinewave inverter designs in this website. It is to chop the gate of the power MOSFETs with calculated SPWM so that a replicated high current SPWM is oscillated across the push pull winding of the transformer primary.

The IC 741 is used as a comparator which compares two triangle waves across its two inputs. The slow base triangle wave is acquired from the IC 4047 Ct pin, while the fast triangle wave is derived from an external IC 555 astable stage. The result is a calculated SPWM at pin6 of the IC 741. This SPWM is chopped at the gates of the power MOSFETs which is switching by the transformer at the same SPWM frequency.

This results in the secondary side with a pure sinewave output (after some filtration).

If you have questions or doubts regarding the above concept, the comment box below is all yours.