Modified Sine Wave Low Cost Inverter

Electricity is the present era’s OXYGEN. How restless we grow when there is a power cut in summer nights especially when we are about to go to bed, or having our dinner, watching TV or having exam the next day! Well these have become a matter of past as INVERTERS have protruded the whole market. Although last institutes, hospital and other large load rely on DG set, inverters have captured almost every home.

Won’t it be interesting to begin designing our own inverter? So let’s get started!!

Before actually designing an inverter, let me first clear what an inverter is and on the first place why do we require it. Why can’t we connect the battery set directly to our house equipment?

Inverter is an equipment which converts DC to AC signal. Almost all the household loads are designed for a specific voltage and frequency. In India it is 230V, 50Hz AC. Most of the lead batteries have an output of 24V or 12V. A large number of these batteries will be required to get the required voltage. And even though we get the required voltage, it will only be able to run linear load with ease. Loads like fan, tube-light will fail to run. These devices require AC signal to run.

Inverter gives us an easier way to convert the DC voltage from batteries to AC and they step it up to the required voltage.

Let us see how it does so! We will be designing a modified sine wave inverter. This is the most basic inverter and low cost but have lot of harmonics. Other option is a pure sine wave inverter. Although costlier, they are as good has AC supply that we get from supply.

However as a designer, modified sine wave inverters are better to begin with as it clears the concept behind.

Parts Required

1. Transformer- 12-0-12V
2. Mosfet(2) - IRZ540
3. 12v DC source
4. IC555
5. CD4047
6. Capacitor - 10uF (2), 1uF(2)
7. resistances - 1k(2), 300ohm (2)
8. Potentiometer - 10k(2)
9. Transistor (2) 2n2222

Circuit Diagram

How it Works?

The whole circuit is divided into 2 parts:

• Control Circuit
• Power Circuit

The CD4047 IC is tuned at 50Hz with 50% duty cycle. Formula to choose the required capacitance and resistance and relative connected can be found in its data sheet easily. I suggest you to use potentiometer as resistance so that you can fine tune it looking the waveform in DSO.

Next tune IC 555 to a frequency higher around 300- 400Hz. Take care that it does not cross the switching time of the transistors and mosfets used. Given range is a safe range.

The output of CD4047 – Q (pin 11) and Q bar (pin 10) is fed to the base of the transistors 2N2222. The emitter of the transistor are connected to the output of ic555 (pin 3).  The collector of the transistors are connected to gate of the mosfet through the required resistance.

This configuration gives us a modulated square wave of 50 Hz switching at 300Hz frequency for half or the cycle and remaining ideal for half. The other transistor output is its complement.

Important question that may arise is why we need high frequency at all!

Why not just on ic with 50hz cycle. Actually using 50Hz will saturate the transformer and this would lead to more loses as the current would increase. Switching it at higher frequency reduces this loss.

The mosfet and transistor are configured to work in switching mode.

Now come to the power circuit side. First precaution is to use thick wires as the circuit will draw heavy current (in amps) when loaded. Ground the source of both the mosfets and connect the drain terminals to the end terminals of the (12-0-12 V) centre tapped transformer.  Connect the centre tap to DC supply terminal.

The control circuit makes the mosfet switch thus providing the change in flux which is essential for the transformer action to take place. The wave form that we obtain is something like this:

without filter output---(no load current- 0.01A)

To smooth these wave we need a filter. We need to connect an appropriate capacitor so that the output is filters and an approximate sinusoidal waveform is generated. We can do this by giving 2 capacitors filter at the input or a single capacitor at the output. Do take care to check the voltage rating of the capacitor.

Wave forms below are obtained for various position of capacitor discussed earlier and the no load current drawn by it.

 100ohm input side b/w 12v -0v ---(no load current- 0.432A)

with 10ohm input side b/w 12v -0v ---(no load current- 0.08A)

with 100ohm output side b/w 230v -0v ---(no load current- 2A)

For me putting filter in the input side proved to be the best configuration.

The rating of the inverter would depend upon the rating of the transformer. I have used a 5A rating transformer so it can ideally drive load up to 60W. Also don’t forget to place proper heat sink for the protection of mosfets from excessive heating.

For the product to safety, fuse of appropriate rating should be connected.

Here’s the final product.

If you have any query, don’t hesitate to leave a comment or shoot me a mail.

*working on it to make it better!!