# Simplest 12V to 220V DC to AC Power Inverter DIY

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Hi!

In this instructable, you will learn to make a simple but powerful inverter at home.This inverter does not requires multiple electronic components but a single component which is a relay.The relay alone is responsible for performing the switching action which in terns, converts the DC from a battery into an AC voltage.

This type of inverter is a square wave inverter and is good for school or collage projects.

List of components required for the project:-

1. 12 volt battery
2. Some wires
3. A 5 terminal relay
4. a transformer single phase

That's it!

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## Step 1: Transformer & Relay

--> You will need a 12 volts to 220 volts transformer.

--> A 12 volt relay and the wiring of the relay should be exactly as shown in the picture.

One of the terminals of the relay will remain disconnected from the entire circuit since its not required in the project which means that only 4 terminals of the relay will be responsible for connecting the Transformer and the battery together.

Watch Full Video

## Step 2: Circuit Details:

The tripping coil terminals of the relay will be connected to the low voltage 12V winding of the transformer as shown in the picture.The load will be connected to the primary high voltage winding of the transformer 220V.

After doing that the "mains disconnect terminal" of the relay will be connected in series with the battery and the other terminal of the battery will be connected to one of the secondary terminal of the transformer.

"Project Collection"

## Step 3: Testing the Circuit:

After doing all the connections as instructed, the bulb should start glowing brightly.

The maximum power of this inverter depends on the size of the transformer and the input power supply.

The frequency of this circuit is around 60 to 70Hz and the efficiency of this circuit is around 63%

So guys that is all for this project.

Thankyou!

Watch Full Step by Step video --> Full Video

Check out our Youtube Channel -->creativElectron7M

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## 54 Discussions

I hate to spoil your fun ( I do understand the adventure of experimenting), but you can run incandescent light bulbs directly off the 12v battery. Buy choosing different wattage bulbs (different resistance) the bulbs will burn at different brightness.

12 replies

Hate to spoil YOUR fun. Can you explain how one would run a modern LED or CFL 120VAC light bulb from a 12V battery without an inverter?
K, thnx, bye.
Please don't try suggesting an incandesant again, that is a horrible waste of energy.

Can you explain how to run a 120v bulb from a 12v battery?

You can connect a 40 or 60W bulb directly to the 12v. The filament in the bulb is nothing more than a resistive wire (like a toaster). This technique is commonly used in remote places in an emergency (I lived in Alaska for 40yrs). For the wise crackers that commented already, resistive components could care less whether it's AC or DC. They run on current not volts. This is also the same principle hobbiest use for hot wire foam cutters.

Yes, you can hook a 40W or 60W bulb to 12v. However, in the case of a 60W, 120V bulb, you will only get 0.6W of light out, which is nearly useless.

A 60W, 120v light bulb has a filament resistance of 240 ohms. Power = voltage squared divided by resistance. 120v^2 / 240ohms = 60W. Now apply 12v. 12v^2 / 240ohms = 0.6W.

I have one setting on my desk and it is 25ohms

Colud be 25 ohm cold (see my answer to dcripe), but dcripe's calculations are right at rated voltage, once the filament gets hot.

Morgan is correct. That is the cold filament resistance. Tungsten has a positive temperature coefficient, increasing in resistance as it warms up.

This morning, I took a 60W, 120v bulb, and measured its cold resistance, reading 26 ohms (not far from what you measured). Then I hooked it to a 12v battery, and measured 130 mA current, about 1.5W, for about 92 ohms. Not even close to 6W.

Most significantly, the bulb barely glowed, very deep red in color, which could only be seen when all other lights in the room were extinguished, and is completely useless for any illumination.

That would be with an ideal resistor, but filament resistance is very dependent on temperature, which in turn depends on dissipated power, so the behavior of a light bulb is highly non linear, much more than a conventional resistor. Resistance increases as current increases. This is true for most resistors (except for some designed with very low temperature coefficient materials, as current measurement shunts), the effect is not too important if the working temperature range is not high. In light bulbs this is clearly not the case, as it goes from room temperature to about 2600ºC.

For example, I've just measured the resistance of a cold 230V 100W lamp and it is 40 ohms, while it's "nominal" resistance would be 230^2/100=529 ohm. If you connect this lamp to 12V DC it would dissipate 3.6 W (0.3A) in the first instant, after that the resistance would start to increase as the filament heats and the current and power would stabilize al a lower level (0.1 A measured with the same lamp, which means 1.2 W and 120 ohms) It's very little power to be useful (in fact, it doesn't even glow, just warms), but much higher than the theoretical 0.27 W that you would get with the nominal 529 ohms. This is with a 230V lamp, I imagine that a 120V one could get to glow at 12V (of course, much dimmer), but I don't have one to play with...

No, he can't and further couldn't explain how to run a 220v (mains) voltage lamp as used in the example from it either. Methinks he kinda missed the point of the whole 'ible.

I have connected the 12v battery the transformer relay and the Bulb but it Is just blinking what can be the problem

all the necessary details have been provided in the video.Please don't forget to watch it.

the relay i have used is from a ups inverter.

Perhaps a voltage clamp and R-C combo on the primary would assist in reducing arcs on the contacts and high voltage spikes. Otherwise used in 1920s to 50's on car radios.