baghu
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No problem .
But you should tell the value of your out put current and out put voltage and the frequency of your square wave , and the number of harmonic that you want and your out put impedance ( your minimum load and maximum load .
Respect
Goldsmith
I am planning to buy a 220v to 12v 2Amp step down transformer and connect it in reverse windings.
Hi,Did you measure it's impedance with multimeter , really ?!
No , it isn't , the way !
I didn't say , resistance ! you measured the resistance !
Do you know about meaning of impedance ? If yes , no problem , but if no :
The impedance , Is the stand due to the AC signal ! You should measure it's inductance with LCR meter , or the catalog of your thing . and then : XL= 2*pi*F*L or omega*L .
So as you told , F=50HZ , and pi =3.14 ( approximately ).
And 2 is a constant . and you measured the L value instead of henry .
Try this way.
Good luck
Goldsmith
You can get by with a single coil. It will cause current to rise or fall gradually in response to voltage reversals. This produces a sine wave.
As to what value coil...
You want the coil to reach equilibrium just before each voltage reversal takes place. To reach equilibrium takes about 5 time constants. There is .01 second between voltage reversals. So you want a time constant of .002.
The TC formula for inductance is L / R.
Or, L = R * TC.
Thus if your loop impedance is 50 ohms, then your coil should be 100 mH.
If loop impedance is 100 ohms, then your coil should be 200 mH.
If loop impedance is 200 ohms, then your coil should be 400 mH.
Etc.
If the transformer were to be of the right henry value, then it may provide sufficient filtering by itself. However the extra coil will still be of use to reduce buzzing in the transformer.
---------- Post added at 20:39 ---------- Previous post was at 19:46 ----------
Edit:
I'm sorry, a single coil won't give you a sine wave. More like a lopsided triangle wave. I'll see what adding a capacitor can do.
I still don't believe you absolutely need two coils.
---------- Post added at 22:09 ---------- Previous post was at 20:39 ----------
Yes, adding a capacitor is sufficient to give an approximation of a sine wave.
This is the layout I come up with. Values are only tentative. The waveform can be improved by some experimentation.
I am really sorry Ali. I don't have a LC meter to measure the inductance. Is there any other way to do so? Thanks.
Dear BradtheRadThe two circuits will display and run simultaneously in the same window. Each has a different loop impedance. Values have been adjusted to suit.
$ 1 5.0E-6 8.63434833026695 50 5.0 43
r 128 48 256 48 0 50.0
c 400 48 400 208 0 6.8E-5 -2.5314959835075426
l 400 48 256 48 0 0.06 0.04938390261801429
v 128 48 128 208 0 2 50.0 5.0 0.0 0.0 0.5
w 400 48 544 48 0
w 400 208 544 208 0
w 128 208 400 208 0
w 128 416 400 416 0
w 400 416 544 416 0
w 400 256 544 256 0
r 544 256 544 416 0 500.0
l 400 256 256 256 0 0.6 -0.004938765565212646
c 400 256 400 416 0 6.799999999999999E-6 2.5337314745240382
r 128 256 256 256 0 500.0
r 544 48 544 208 0 50.0
v 128 256 128 416 0 2 50.0 5.0 0.0 0.0 0.5
o 14 64 0 34 5.0 0.05 0 -1
o 10 64 0 34 5.0 0.003125 1 -1
Why not use a sine wave oscillator instead and completely remove the hassle?I want to convert this square wave output of the oscillator to a nearly sinewave output without much hassle.
Yes, it's quite easy.... is there a very easy way to make a sinewave oscillator and amplifier, ..... Can I do it myself on a veroboard?
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