Introduction: Simplest 12V to 220V DC to AC Power Inverter DIY

Picture of Simplest 12V to 220V DC to AC Power Inverter DIY

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
  5. a load bulb

That's it!

Step 1: Transformer & Relay

Picture of 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:

Picture of 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:

Picture of 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

Comments

HectorJ19 (author)2017-11-16

Can you send the Circuit diagram

omars2 (author)HectorJ192017-11-18

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

jompon2547 (author)2017-11-04

i have other delay

can u recommend me some plz?

omars2 (author)jompon25472017-11-05

the relay i have used is from a ups inverter.

jompon2547 (author)omars22017-11-06

thx!

JimG50 (author)2017-09-26

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.

johnny3h (author)JimG502017-10-17

Car radios used a hefty, specifically designed and dedicated MECHANICAL VIBRATOR SWITCH similar to the model T Ford SPARK COIL. As I recall, upon disassembly it looked much like a heavy duty double pole reed switch.

DabeAltis (author)2017-09-27

That certainly is a simple inverter, and could be useful for many applications. My concern, as someone else already pointed out, would be in the longevity of the relay contacts. If you could make it work with a solid state relay it might hold up a lot longer, only problem is I think most solid state relays require the zero crossing of the ac sine wave to turn off. By the way, I'll not mention any names, but someone here likes to blow a lot of smoke, and hasn't a clue what he's talking about! LOL Nice ible, thanks for posting it.

JohnC430 (author)DabeAltis2017-10-09

put a cap across the contacts and it will absorb the spikes which usually arc across and burn the contacts

JohnM90 (author)2017-09-25

It is not a square wave AC output, just pulsing DC. There is no alternating voltage.

dcripe (author)JohnM902017-09-26

You cannot pass DC through a transformer. The voltage output is AC, but not a clean sine wave.

JohnC430 (author)dcripe2017-10-09

who cares about clean sine wave? chopping the DC is enough to activate the primary winding. flux changes in the core and transfers energy to the secondary. thats all thats needed. if he added a capacitor to resonate with the secondary it would give him rounded edges. but again who cares about sinewave. Most of the commercial inverters dont make "clean" sine wave. they make modified sine wave which are square waves with a gap in the middle.

youcantoo (author)dcripe2017-09-26

they are not passing DC through the transformer it is a pulsed DC that is being passed and then stepped up by the transformer, in this case 220V this is not AC but a pulsing DC IF you look at it on a O-scope you would clearly see it is not a AC signal but a pulsating DC square wave. Anyone with a very basic understanding of electronics would know that.

You are introducing a 12 V square way and the output is a stepped up pulsating DC Square wave. All transformers can indeed pass a pulsating DC voltage.

DabeAltis (author)youcantoo2017-09-27

It certainly is AC on the secondary side of the transformer! When the magnetic field is building up it induces a voltage in the secondary in one direction, when the field collapses it induces a voltage in the opposite direction. Consider the sawtooth wave used for horizontal deflection on a CRT. It is produced by a DC pulse applied to a transformer, but the result is AC on the secondary. If it were not so, then the electron beam would only deflect in one direction and would never pass center in the other direction. Yes, anyone with a very basic understanding of electronics would know this!

dcripe made it! (author)youcantoo2017-09-26

DC denotes direct current, having a non-zero average value. AC is alternating current, which may be pulsed or non-sinusoidal, but has a zero average value. A waveform may have AC superimposed upon a DC term, but you cannot transmit DC through a transformer, whether pulsed or not. You can simulate the circuit using LTSPICE, and see that the output waveform looks like this, which has _NO_ DC term:

DabeAltis (author)JohnM902017-09-27

With all due respect, it is indeed an AC output. Read my response to youcantoo for an explanation.

jwzumwalt (author)2017-09-25

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.

dcripe (author)jwzumwalt2017-09-26

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

jwzumwalt (author)dcripe2017-09-26

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.

dcripe (author)jwzumwalt2017-09-26

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.

jwzumwalt (author)dcripe2017-09-26

Your math is wrong. Its 6W

dcripe (author)jwzumwalt2017-09-26

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.

jwzumwalt (author)dcripe2017-09-26

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

morgan_flint (author)jwzumwalt2017-09-27

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

dcripe (author)jwzumwalt2017-09-27

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.

morgan_flint (author)dcripe2017-09-27

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...

ribeirovidal (author)dcripe2017-09-26

Whoever does this role is the transformer!

MillerI (author)dcripe2017-09-26

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.

frarugi87 (author)2017-09-27

Cool for demostration purposes, but with a relay rated even for 100k operations using it continuously for 140 operations per second will make it last for about 12 minutes...

EldarM1 (author)2017-09-27

Just some little tricks with capacitors and resistors.

adding a 230V 0.1microF on the transformer output, will provide you a more sinusoidal signal.

Adding a capacitor on the relay coil will reduce the switching frequency. Typically to get 50Hz or 20 Hz.

Adding a resistor (some Ohms) on the relay coil will increase the switching frequency. To its mechanical limit of course.

MichaelAtOz (author)2017-09-26

So how dangerous is this, with the size of battery shown?

Left hand + right hand = heart stopper?

burningsuntech (author)2017-09-20

This basic circuit was in use to raise the potential of 6 or 12 volt dc to several hundred volts to operate a vacuum tube style radio before transistors were all the rage. The output of the relay was fed to a transformer to "step up" the voltage which was then rectified into high voltage DC. You can still find this type of mechanical inverter at Amateur Radio shows.

vpu417 (author)burningsuntech2017-09-25

In fact, this is how all tube car radios worked.

charles543 (author)vpu4172017-09-26

Untill they came out with 12V tubes.

Yes. I remember those. They were called Vibrators. Old machine controls had them too.

Josey94 (author)2017-09-19

Is square wave not harmful to home gadgets?

jwzumwalt (author)Josey942017-09-26

It is VERY harmful for analog circuits not designed for it - which is pretty much most things. Many components such as triacs and certain diodes will only work with a stable AC wave form.

omars2 (author)Josey942017-09-19

At the output,you can place a high voltage capacitor to ensure that the square wave is some where near the sine wave.

youcantoo (author)omars22017-09-26

that would be called a modified square wave and really shouldn't be used on sensitive electronics IE: computers, or inductive loads like a refergrator

robtservice (author)Josey942017-09-19

This is a VERY good instructable, but you need to know its limitations:

This supply outputs a SQUARE WAVE. It is good for filament-type loads (tungsten and halogen lamps), MAY be good for other loads which will tolerate a square-wave power-supply, and will DEFINITELY HARM others which need a sine-wave supply.

Also, experimenters with 120 vac mains need to substitute a transformer with a 120 vac winding.

omars2 (author)robtservice2017-09-19

Nicely explained.

Quantumdust (author)2017-09-26

The physics and electrical connections are essentially correct. The project is useless. You obviously have an interest in electronic design. Combine that interest with the time it took to assemble the "circuit" upload the pictures and text to all the various places and you will have enough time to design something interesting and teachable to people actually wanting to make something new that they can be proud to show their friends. Skip the stuff that works for a few minutes and starts a fire.

JimT86 (author)2017-09-26

Where is a schematic?

schmitta (author)2017-09-26

Where is the SCHEMATIC

alexwhittemore (author)2017-09-26

Clever and simple, I like it!

But you definitely shouldn't use this in any practical application. As others mention, it's got some limitations. The most important one, as far as I'm concerned, is that the relay will only last about 30 minutes before falling apart, and when it does, it'll likely fail in such a way that the transformer sees only DC from a very high power source, likely resulting in wires catching fire. Relays are typically rated for 100,000 actuations, which get used up pretty quickly when you switch at 60Hz.

jimdooris (author)2017-09-26

Very Nice, it is a variation of the vibrator used in old car radios used when vacuum tubes were used. simple and reliable.

TexBoyd (author)jimdooris2017-09-26

I remember the old vibrator relays in car tube type radios. I remember they would burn out after many hours of play. My question is, what was special about them that made them last as long as they would? Maybe the contacts were in a vacuum?

runemadsen (author)2017-09-26

Putting square wave into an transformer will make peaks with very high voltage. This is potentially killing people!

axgnUW5VU8 (author)2017-09-26

The idea works mostly because of the inductance of the transformer. The transformer windings 'resist' the instantaneous flow of current when the relay first turns on -- the current flowing to the bulb is not a 'square wave'. When the relay turns off the field in the transformer collapses. The result is a quite large voltage in the reverse direction. If you were to look at the relay contacts you would see quite a bright spark from this. The wear on the relay is substantial and it will fail early on. Electrical engineers take great pains to avoid such conditions and this sort of design is no longer used. In an emergency this would be a Mcgyver-worthy approach. It is also a nice demonstration of some basic electrical principles but the analysis of how and why it works is, as we so often say, beyond the scope of this text.

MichaelS163 (author)2017-09-25

Generates shed load of RF noise and potentially very dangerous! Should be taken off Instructables immediately.

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