What is the advantage of a four diode bridge full-wave rectifier over a two diode full-wave rectifier. There must be some, else why accept the extra cost?
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"He who will not reason is a bigot;
he who cannot is a fool;
What I need is a bridge (the transformer I'm considering doesn't has a centre tapped secondary) that will handle 420 volts, 3.5 amps. Also, I'd like recommendations for capacitor and inductor values for smoothing.
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"He who will not reason is a bigot;
he who cannot is a fool;
Also you get better transformer utilization because the whole winding conducts on both half-cycles. That improves the RMS-to-average ratio and reduces transformer heating (other things being equal).
Very few amps use Pi filters these days (capacitor inductor capacitor) that's more common back when capacitors were these huge metal cans with aluminum foil paper and oil and had a rating of only ~1-10 microfarads. Today they use electrolytic caps with much higher capacity and avoid the expense of big heavy inductors.
I've wondered about that. Seems to me you have two windings conducting half the time with a CT so the average current and wire size is smaller (or could be) and you only have two diode-drops instead of four. So I'd expect the two diode version to be more efficient (in low voltage power supplies at least, because the diode drop represents a larger amount of the total V output).
They make those too. Or you could use half of a full wave bridge if heat sinking is a consideration. You're not designing this for mass production where every penny is going to be counted.
I thought you were working at ~400 V? The storage time is insignificant in linear mains supplied transformers - it becomes significant when you are rectifying square waves or high frequencies or both.
Not very for 50-60 cycles. A little extra head-room on the voltage is nice to accommodate power line surges without shorting the diodes.
Efficiency is a slightly different issue. Neglecting diode drops, a half-wave supply has much worse copper losses because (a) you have to use smaller wire to get twice the secondary voltage, and (b) you're drawing twice the current for half the time. Item (b) costs you because the copper loss goes as I**2, so twice the current for half the time doubles the losses.
At very low voltages the diodes become more important, but it's the transformer that costs the money. You can make a DC-DC converter for a dollar or so if that's an issue--a buck converter in both senses. ;)
There's copper losses, and diode-drop losses, and core size difference. For a center-tapped winding, you need longer wire for the coil, and that raises the core size (unless you accept higher resistive heating from thinner wire). With a full-wave bridge, four diodes, you have TWO diode drops on each conducting half-cycle, but with a center-tapped two diode fullwave rectified circuit, there's only ONE diode drop.
And, given a choice, a four-diode bridge and center-tapped coil gives you TWO power supplies. That's very convenient.
They got them, it is just a matter of learning how to use their selector pa ges. Problem is they changed them and while it all still works it seems you can't get back to "More Filters" and have to back out and reapproach.
They also have diodes in series in those packages.
You want to know about filtering ? Well you got 420 volts at 3.5 amps, that means pretty much a resistance of 120 ohms. Here comes the math, now if yo u are going brute force there is no inductor, just a ton of capacitance. Fi gure out how much ripple you can handle and get a cap big enough to not dis charge more than that in 1/60th of a second.
That amount of ripple going into a choke will result in a certain voltage l oss, which will be half the amplitude of the ripple. If you got 450 and 20 of ripple you got 440. that is what you'll get out of a choke. Into the nex t cap the whole thing is different because you have an impedance feeding th at cap. The selection of the value of that next cap depends on the load and the variations in the load.
Too much inductance you lose all semblance of regulation, less inductance y ou need bigger caps.
If you have to much trouble figuring this out just adapt an extant design.
I had a computer that would reset when I turned off the amplified speakers before turning off the computer. A small common mode filter on the amp cured the problem.
Ya think? Generally speaking, they are damn serious when they talk about "absolute maximum values.."
And when they specify a mosfet (for instance) at a current capacity that would cause its leads to melt, or unsolder itself from the board
You think? I happen to stumble across a MOSFET in a TO-227 that was spec'ed for 300A & 1500W. How is that useful? Shouldn't they also say that it requires a LN2-cooled heatsink? And if you do need, say, a 15A MOSFET, how do you find it, what with all the bogus spec's?
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