1)So it is ok to use this 7-0-7 as the main trafo that will be switched at high frequency in my push-pull design?
2)pls kindly Explain further what you mean by a lot of filtering and low current load especially the lot of filtering part.
Thanks.
Hello again,
The peak voltage with a 7v rms source is:
7*1/4142=9.8994 volts, which is approximately 9.9 volts.
Subtract 2 volts from that for the diodes, we now have:
9.9-2=7.9 volts.
Now the peak of a full wave rectified 60 Hertz sine occurs once every 8.333ms, so the voltage deviation estimate is:
dv=I*0.008333/C
where I is the DC current and C is the capacitance used as the filter.
With a 100ma current and 1000uf we get a voltage deviation of 0.833 volts, so now the DC can fall to as low as 7 volts.
With a 10ma current and 1000uf we get a voltage deviation of 0.0833 volts, so now the DC can fall to as low as about 7.8 volts.
Using 3300uf would work ever better, but there will always be some ripple without a regulator. If the ripple is objectionable in the application then you need to use a regulator, or two regulators if you have plus and minus voltages.
If the above voltage is not high enough then you need a transformer with a higher output voltage or else build a voltage doubler if the current is low enough.
So by "more filtering" i mean add more capacitance on the output of the rectifier.