You are mistaken. A half-wave only uses half the power put in and this is a loss of half the power that is provided. The fact is two cycles AC in and only use one cycle is inefficient.

What for Pete's sake is a bi-phase rectifier? Never heard of it. Please explain. Do you have a diagram to enlighten me with?

Only thing I can think of is you are talking about a full-wave.

 

Look up rectifier on Wikipedia, and look for the second of the full-wave rectifier circuits (third of the nice green AC half cycle drawings). I believe this is what the OP is targeting...

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Ok, Don't we call that a Full-wave?

 

Yes, it's full wave, but there are two full-wave schemes, the bridge and the ?? (that's where we stumble into "bi-phase").

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Nope you are wrong again!

The 50% of the time the input sine-wave is not providing output power, said power is not (as you call it) "put in" at all.....it is in fact simply not used!

Again, there is no (as you call it) "loss of half the power", notwithstanding transformer losses!

VA in must equal VA out as discussed many times already, notwithstanding transformer losses. You do not understand how a transformer works vis-a-vis power transfer through mutual inductance.

 

------------Edit------------

I got a little hot under the collar, but cooler head prevailed and I retract my original input.

 

Sure:

A) "Requires a slightly beefier transformer build (Pt/Pd of 1.5 vs. 1.23)" means the ratio of input power to output power. This is specified for all five standard rectification schemes, from half-wave to 3-phase bridge. For your "bi-phase" circuit it's 1.5, meaning if you want 100VA out, your transformer input will be 150VA (or 123VA transformer input for the full-wave bridge). And since there are standardized transformer core sizes specified in VA, you go up a size or two for the "bi-phase." (A half-wave circuit requires 310VA in for 100VA output, the worst of the bunch).

B) "Requires higher voltage diodes (Vrrm/Vrms of 3.12 vs. 1.56)" means the ratio of diode peak inverse voltage rating (e.g. 50V for a 1N4001) to the secondary output voltage in volts RMS. For example, a 25.2VCT transformer secondary, Vrms = 12.6V, which means the rectifier must sport a Vrrm of 39V (or 20V for the full-wave bridge).

C) "Superior in one way: Irms/Iout of 0.78 vs. 1.11 for a bridge (or 1.11 vs. 1.57 with filter capacitor)" means the ratio of AC output current in amps RMS from the transformer secondary to the DC output current through the load (resistively and capacitively, respectively). For example, for 1A DC load current through a resistor, the "bi-phase" circuit pumps in 780mA RMS from the transformer (or 1.11A RMS using a bridge).

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Yes exactly, a DUAL BI-PHASE circuit.

I cant help but think that your argument is but a storm in a tea cup.
Just because a technique is old, it does not mean that it is totally obsolete and should be ignored.

And yes I am familiar with old valve circuits, I started with them in the early 1960s.

JimB

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CT means centertapped
28VCT means 28 volts centertapped or 14 volts each side of centertap.

If I replace the filter capacitor in my description with a resistor it is the exact circuit of the second full wave rectifier in your reference on wikipedia.
I still maintain it is a valid configuration in use today. I also believe it is a efficient type of power supply, and it can be more efficient than the full wave bridge circuit as there is one less diode voltage drop.

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hi,
I agree, I also use the CT, bi-phase connection in some projects.

Ref:
http://books.google.co.uk/books?id=w...um=9&ct=result

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Bill's: www.blueroomelectronics.com/

 

Some uses of the "bi-phase" rectifier:

>>> When the voltage drop across each diode is big (with respect to the output voltage): Inside a PC (personal computer) power supply, the "main" supply (5v or 3.3v) output stage has a small transformer (operating at a high freccuency) with a center tapped secondary, and tho diodes in a bi-phase configuration.

>>> When you need a DC supply and an AC supply that have a common "ground"

>>> Dual (+/-) supplies. They may be seen as two bi-phase rectifiers that use the same transformer or as a bridge with a grounded center tap.

>>> And, of course, vintage circuits and "antiquated holdovers" in new designs made by people that have learned in the "old days" and still use that configuration.

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A bi-phase rectifier is a form of fullwave rectifier becuase the power is being used from both sides of the AC cycle - it's silly arguing about it.

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Only 50% of the secondary of the power transformer is used at a given instant, thus it is not the same as other full-wave rectifier schemes.

 

first of all with a tube full half wave rectification you loose about 90 volts across the tube. 5u4 and so on. half wave you get the half the rectification as a pulsating DC needs big caps to smooth out the ripple full wave as in CT trasformers you will get half the voltage available. with a bridge rectifier and the same transformer you will get twice the voltage. current available will be the same as kva. You like to know more?

 

That's nonsense 100% of the secondary power is used and more importantly 100% of the primary power is used, i.e both positive and negative cycles are used which by definition full wave rectification.

You could use a 100VA transformer with a bi-phase resctifier and still happily take 100W from the output because although each side of the centre tap on secondary is passing double the current the duty cycle is halved so it's alright.

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