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Personally, I prefer to estimate the minimum voltage as Vp-2Vd-Vr(p-p) and add to it Vr(rms) as rms value of positive sawtooth.
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Introducing load causes capacitor discharging and ripple voltage calculations take care about it. Then the steady state dc voltage on capacitor is Vp-2Vd. Here, the ideal case is when current is zero. In your calculations you assumed Vo to be average voltage and it would indeed be a result of ideal filtering of a voltage signal. Nonideal effects you've included as lowering of this voltage. However, for the case at hand, it's not a valid approach and produces unrealistically low voltages. I wouldn't be surprised if much more complicated models gave results similar to what can be done on a small piece of paper, as the resulting ripple voltage estimate works amazingly well considering the assumptions and approximations. To tell the truth, I suspected that Vo and Vo1 were parts of different pictures but your presenting of the diagram again confused me. (I don't mean the math I've presented explaining my point to you, although it's not above highschool level). I can only return your curtesy as obviously you credit me with your own attributes.Īlthough this topic might seem at first sight not appropiate for a PIC´s specific forum, I think it could be of interest for people who might want to power up a PIC´s project using their own selected resources. Certainly not all designs run on batteries and I always thought that a little of simple practical math would not hurt anybody.
BRIDGE RECTIFIER CALCULATOR EIGHT DIODE SERIES
Of course it is the same equipotencial node as seen in the image It´s just that I dind´t want to break up the graphic in sections (before the cap is connected, and once it is paced in the circuit., so on).īTW, the ripple calculations give the simple result Vr(rms)=i/(4√3 fC) based on assumption that the ripple is an ideal sawtooth (indeed reverse one ), i.e.ĭon´t quite agree, floor function and Fourier Series are certainly diverse, but this is not to be discussed here, since it´s getting off the forum´s scope.Īnyway I´ll examine my basis and resources, in order to look for every feasible inconsistency. Your laborious urge to clear up things properly, in behalf of the benefit of the forum users is quite valuable.Īlthough this topic might seem at first sight not appropiate for a PIC´s specific forum, I think it could be of interest for people who might want to power up a PIC´s project using their own selected resources. Ooops, forgotten about factor 2 in rms corrected nowįirst of all, Thanks a lot Janni for your time and generosity. V(t)=Vp-Vr(p-p)*t/T (T being half of mains period, 1/(2f)) Reality is obviously different so there is slight overestimate of the real ripple voltage. Anyway, the filtered voltage value is important for regulator operation estimate and it seems that you underestimate that value in your calculations.īTW, the ripple calculations give the simple result Vr(rms)=i/(4√3 fC) based on assumption that the ripple is an ideal sawtooth (indeed reverse one ), i.e. Luis, frankly, I don't see a difference between Vo and Vo1 in your picture as both are drawn above an equipotential line.
BRIDGE RECTIFIER CALCULATOR EIGHT DIODE PLUS
This also means that the regulator will have to dissipate 7W of power.Ībout the diodes, the momentary current through each of them will reach 10A, but only for about 1/10 of the time (while the rectified voltage is higher then the capacitor voltage plus 2Vd) - don't worry, rectifier diodes (or bridges) are designed to work in such conditions. The diodes will have, on average, to dissipate 4*0.8V*1A=3.2W of power.
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If you use 24V voltage regulator after the filter, you'll have 6V regulation reserve so there is no need for further ripple reduction. The purely dc voltage you will get at the filter's output is not 24V, but 30V, so there is about 4% ripple. Vr(p-p) for 1A current and 50Hz mains will be circa 2V, hence Vr(rms)=Vr(p-p)/2√3 (it's sawtooth-like, not sinusoidal) which gives 0.6V.
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The practical capacitor value is 4700uF and lets start from there.