defaced Posted September 23, 2006 Report Posted September 23, 2006 I picked up a transformer with a 12VAC (I'm assuming that's RMS voltage) secondary, rectified and filtered it and I get 18VDC. Now I know why this happens, but I'd like to find the forumla that models this. Any help? The end result is that I need 12VDC, so a tranny with a secondary that when rectified and filtered can produce this. My guess is I need a tranny with a 8VAC secondary. Quote
ante Posted September 23, 2006 Report Posted September 23, 2006 Hi defaced,This should be helpful I hope! Quote
defaced Posted September 23, 2006 Author Report Posted September 23, 2006 I want to make sure I'm reading this correctly. For a few of the examples, there are two equations for E. Looking at what reading I'm taking from my circuit and compairing it to those forumlas, E = .9 Erms is the voltage after rectification and E = 1.4 ERms is the voltage after filtering. Is that correct? Quote
nickagian Posted September 24, 2006 Report Posted September 24, 2006 Hi defaced!Well this is what I have learned in my university:1)At a half-rectifier circuit (look at the first two circuits at the picture posted by Ante) the output voltage amplitude is given by the formula Vout(amp)=E(rms)xsqrt(2)-VD0[sqrt->square root, E the voltage at the secondary of the transformer and VD0-> the forward voltage of the diode, about 0.7~0.8V]Since sqrt(2) is approximately 1.4 you can say that Vout(amp)=1.4E(rms)-VD0 so roughly Vout(amp)=1.4E(rms).This is one of the formulas that you can see in the picture.Moreover the DC output voltage (practically the average voltage) can be calculated by the formulaVDC=1.4E(rms)/pi - VD0/2and since 1.4/pi is approximately 0.45, you can say that VDC=0.45E(rms)(That means probably the second formula at the picture)However if you use a simple capacitor for filtering, the output voltage roughly equals the output voltage amplitude, thus 1.4E(rms).2)For the full-rectifier circuit (both with the two diodes and the bridge-rectifier) the output voltage amplitude equalsVout(amp)=2*1.4E(rms)-VD0 thus approximately 2.8E(rms)and the DC levelVDC=2*1.4E(rms)/pi - VD0 so approximately 0.9E(rms)That`s all! 8) ;) Quote
MP Posted October 1, 2006 Report Posted October 1, 2006 defaced,Your unloaded output voltage measurement is pretty meaningless. You need to measure the voltage loaded to get an accurate measurement. Also, if you want a well regulated 12 VDC supply, you should use a regulator. Most regulators require an input voltage which is 2 to 3 volts (or more) above the output voltage. Just get a 7812 voltage regulator for this power supply and you will have exactly what you need. If you use a transformer with an 8 volt secondary, you will not have enough voltage after loading occurs on the output.MP Quote
aartak Posted October 16, 2006 Report Posted October 16, 2006 MP is rightalso check the resistance of secondary windingif it is more than 3-4 ohm you may have problems at high loadsbecause 7812 needs minimum 13.5-14V to work properlydont forget to put large cap for filtering Quote
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