simple inverter 12v to 220v

Thread Starter

sendbad

Joined Jan 8, 2015
43
i want to make simple enverter 12v dc to 220v ac

i used
transformer 2a 12v without center point
Oscillator 50 hz
12v dc 500 ma Expermentlly

output of transformer was jast 8 v ac

how can i get 220v ac
 

MCU88

Joined Mar 12, 2015
359
Hello....

There is no such thing as an simple electronics circuit. It is THE hardest subject to learn. Please post an schematic of your work.
 

DickCappels

Joined Aug 21, 2008
6,192
About the transformer. It sounds as though you are driving the 12 volt winding of a 220V to 12V step-down transformer.

12 VAC is 33.8 V P-P but you need more than that.

Both windings of your transformer have resistance and the winding that is being driven will have to supply magnetizing current (this is current that results from driving an inductor with AC) and some core losses, and this will result in a voltage drop within the driven winding. Further, if you load the un-driven winding there will be an additional voltage drop in that winding.

If you go to a higher frequency the magnetizing current will get smaller quickly. If you go to too high of a frequency other losses start to take over, so for a given transformer and a given waveform there is an optimum frequency.

A product that I am designing generates 230 VAC (sine) waveforms that are loaded very lightly, but even at that I have to drive the 5 volt windings of a little 1.1 VA transformer with 19.7 V P-P.

The yellow trace in the picture below shows the magnetizing current waveform in a 1.5 VA transformer I tested. The vertical scales are 200 am/div for magnetizing current (yellow) on a 6 volt winding and 100 volts/division for the 220 VAC winging (blue). Notice that the magnetizing current is 1.4 amps P-P.

 

Thread Starter

sendbad

Joined Jan 8, 2015
43
About the transformer. It sounds as though you are driving the 12 volt winding of a 220V to 12V step-down transformer.

12 VAC is 33.8 V P-P but you need more than that.

Both windings of your transformer have resistance and the winding that is being driven will have to supply magnetizing current (this is current that results from driving an inductor with AC) and some core losses, and this will result in a voltage drop within the driven winding. Further, if you load the un-driven winding there will be an additional voltage drop in that winding.

If you go to a higher frequency the magnetizing current will get smaller quickly. If you go to too high of a frequency other losses start to take over, so for a given transformer and a given waveform there is an optimum frequency.

A product that I am designing generates 230 VAC (sine) waveforms that are loaded very lightly, but even at that I have to drive the 5 volt windings of a little 1.1 VA transformer with 19.7 V P-P.

The yellow trace in the picture below shows the magnetizing current waveform in a 1.5 VA transformer I tested. The vertical scales are 200 am/div for magnetizing current (yellow) on a 6 volt winding and 100 volts/division for the 220 VAC winging (blue). Notice that the magnetizing current is 1.4 amps P-P.

so you think current of 12v dc little
 

DickCappels

Joined Aug 21, 2008
6,192
If I understand your statement to mean that I think that driving the 12 volt winding with 12 V peak-to-peak is too little to obtain 220VAC on the output, yes, I agree, 12V is too little. Remember that P-P voltage = 2.82 x VAC (which refers to the RMS voltage of a sine wave).

I think that if you drive the 12V winding of a 220V to 12V transformer with 12 V P-P, the unloaded output of the 220V winding will be less than 220 V P-P. If the signal is a square wave with high quality, 220 V P-P would look like 120 VAC to a good quality True RMS voltmeter, but because of internal losses in the transformer, you would not get even that much.

It is not clear what the 8 volts you measured really represents unless you were using a True RMS meter or an oscilloscope, but 8 volts on the 220 volt winding (still not 100% clear to me if that is where you measured it) sounds very low. L1 is not known because I can't read the inductance and the resistance is not known, but if it has a high resistance that might account for the surprisingly low output voltage. The voltage losses when driving the low voltage winding can be huge.

While posting photos as MCU88 suggested, would you also please measure the resistance of L1?
 

Thread Starter

sendbad

Joined Jan 8, 2015
43
مUOTE="DickCappels, post: 833610, member: 27805"]If I understand your statement to mean that I think that driving the 12 volt winding with 12 V peak-to-peak is too little to obtain 220VAC on the output, yes, I agree, 12V is too little. Remember that P-P voltage = 2.82 x VAC (which refers to the RMS voltage of a sine wave).

I think that if you drive the 12V winding of a 220V to 12V transformer with 12 V P-P, the unloaded output of the 220V winding will be less than 220 V P-P. If the signal is a square wave with high quality, 220 V P-P would look like 120 VAC to a good quality True RMS voltmeter, but because of internal losses in the transformer, you would not get even that much.

It is not clear what the 8 volts you measured really represents unless you were using a True RMS meter or an oscilloscope, but 8 volts on the 220 volt winding (still not 100% clear to me if that is where you measured it) sounds very low. L1 is not known because I can't read the inductance and the resistance is not known, but if it has a high resistance that might account for the surprisingly low output voltage. The voltage losses when driving the low voltage winding can be huge.

While posting photos as MCU88 suggested, would you also please measure the resistance of L1?[/QUOTE]
l1 8 ohm

i tray to rples Oscillator with that and no thing chang
 

DickCappels

Joined Aug 21, 2008
6,192
18 ohms for L1? I don't know why it was put in the circuit, but it can be a large part of the problem -that is a lot of resistance. What happens to the output voltage of your circuit if you short out L1?

I think MCU88 wanted you to post a photograph of your circuit.
 
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