Square wave inverter output too low.

[nonny_sgr]

Hello everyone,

I have built a square wave inverter using the TL494 IC on a breadboard, I want to get 230V RMS, 50Hz at the output of the center-tap transformer, but I am getting 170V RMS and 60Hz (I have connected a 6W lamp as my load, does this have any effect? )
I calculated my oscillator frequency to be 50Hz using the formula 1.1/2*Rt*Ct (Rt=11k, Ct=1uF), but when I measure it I get 60Hz. I am using an oscilloscope to take all measurements because it is more accurate. I want to get my output voltage to be higher but I don't know how. I also anted to ask if there is a way for me to modify this circuit to give me a modified square wave. Please help me, I have attached a multisim printed picture of my circuit.

 

[BradtheRad]

1.
Your topology sends square waves to the secondary. If you want 6W then you must draw average 0.5 Ampere from the 12V supply. (12 x 0.5 = 6W).

Volt waveforms at each end of your primary are square. However current waveforms through Q3 Q4 are sawtooth with peaks of 5 Amps. Can you verify that amount of current really goes through your primary section?

2.
Your transformer is labelled 10.2. That is the correct step-up ratio for the entire primary. It has two windings. So each winding, when energized with 12 V, in effect steps it up 20x to amplitude 240 V square wave. It is clearer to call it that rather than RMS volts.

RMS implies a sinewave. 240 V RMS has peak amplitude of 340V. Did you expect to obtain 340V amplitude peaks from your project?

 

[KlausST]

Hi,

There may be several problems
* a breadboard is not suitable for switching power applications
* a nominal battery will almost never output exactly 12V. A lead acid battery may have a range from 10.5V to 14.4V
* part tolerances may shift your frequency
* I recommend a decoupled supply for the IC
* I recommend fast bulk capacitors at 12V node

Klaus