I studied power supply circuits in the 60's and so I'm more familiar with the terms "half wave", "full wave", and "full wave bridge" as used in the ARRL Radio Amateurs Handbooks. This "bi-phase" term seems a bit cryptic in the context of rectification schemes (in my opinion).

Mike

 

hi Mike,
It originally started when the OP asked about the bi-phase rectifier.


I posted this link earlier.

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

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I'm sorry but you are wrong!

It's got a center tap secondary and only two diodes. Only 50% of the secondary of the power transformer is used at a given instant, thus if you have a 1000 VA power transformer you can only have a 500 VA (notwithstanding losses) DC power supply. You are forgetting the fact that you can't exceed the current rating of the secondary given VA in must equal VA out. Do the math and check a given transformer's ratings and see for yourself that this can easily be the case.

The current rating of the secondary is not simply a function of the duty cycle. Why? Because of the fact that you need to assess the current squared, magnetic saturation, impulse heat dissipation, the resistive thermal coefficient of the windings etc.

Remember double the current for 1/2 the time is twice the heat (at the very least).

Q: What is the current rating of a conductor based on?

A: Its ability to dissipate heat.

Q: It's ability to dissipate heat is based on?

A: The thermal resilience of the insulation.

Q: If you have a 1000 VA transformer rated at 240 / 120 and the secondary is center tapped what would you expect the maximum secondary current to be?

A: 1000 / 120 = 8.33 amps

Q: Can you then take out double the current for 50% of the time? Why not? You are taking out double the watts for 1/2 the time right? So it should not overheat right? And isn't the current carrying capacity of the conductor limited by the thermal resilience of the insulation?

A: No you can't because if we assume an internal impedance of the secondary of 1 ohm, then at the rated secondary current of 8.33 amps we would expect the secondary to dissipate 8.33 x 8.33 x 1 = 69.4 watts over time. However if we then say we are going to double the current for half the time we get 16.66 x 16.66 x 1 = 277.56 watts / 2 = 139.77 watts over time. Now you have twice the maximum rated wattage dissipation of the secondary of the transformer!

 

It's real world nomenclature, it's full wave but it's not a bridge. I also was alive and well and up on tube rectifiers in the 60's, if that's of consequence. I teach (some of) this stuff now and yep that exact term is part of the curricula (for better or worse).

 

Possibly OT but maybe not as this thread seems to be going circular. I was always fascinated by the attached rectifier circuit. I first ran across it in the ARRL handbook and they did have some kind of name for it. It basically allowed one to generate two different voltage (for both plate and screen voltages) levels from one transformer secondary winding.

So in reference to this thread what words would best describe this rectifier circuit? Possibly full wave, center tapped bridge? Or possible it fits the bi-phase term many of us are struggling with here

Lefty

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Like most people here, I've never heard of bi-phase either - the circuit in question is two fullwave rectifiers (or a single bridge if you like). It's the same circuit used in almost anything with a split supply (such as audio amplifiers), just with the chassis connection at a different point (actually on the negative rail).

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hi,
In the early days when using glass tubes/valves it was cheaper and easier to connect a centre tapped transformer to a dual glass tube rectifier in order to get fullwave rectification, it was called bi-phase rectification.

To get fullwave would have required at least two/four dual glass tube rectifiers.

Now we use a semiconductor FWB...

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"It's the same circuit used in almost anything with a split supply (such as audio amplifiers), "

Well not exactly the same circuit. The common split supply uses the center tap to establish the common (ground) reference and generates two equal but opposite polarities, where this supply uses one half of the bridge diode to establish the common reference and generates two different voltage levels but at the same polarity. Surly it deserves a unique name

Lefty

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"Bridge rectifer with chassis connected to a different point" - hardly a snappy title

Exactly the same circuit, just using a different reference point.

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Yes that's true, I forgot I²R so doubling the current quadruples the power which means double the current at 50% duty doubles the copper losses. This means that a 100VA transformer can only be used for 100/√2 = 70.7W.

However it will not saturate even if run at 100W (although it will overheat if run continiously) because the flux generated by the secondary will still be 100VAs worth.

However bi-phase, still is full wave rectification because power is being used from both the positive and negative cycles of the mains supply. If it were half wave rectification then the transformer core would saturate if run at 70.1% of the rated power and overheat.

 

Then why does one circuit generate two different voltage levels and the other two equal (but opposite) voltage levels, if they are the same circuit? I think there is more going on here with my circuit then the typical split rail bridge circuit. I'll settle for nothing less then a full circuit anaylsis of the two cases

Lefty

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I would hope you recognize that I never said bi-phase was half-wave, and that I always said it was full-wave! As to the balance of your post, I'll have to respond in detail in kind later, as it's time go shopping.

 

 

You have been reported to the moderator for personal attacks

An ad hominem argument consists of replying by attacking the poster. The process of proving or disproving the claim is thereby subverted, an ad hominem works to change the subject and attempt to discredit the poster.

 

No you didn't, you've just edited your post.

Just admit you made a mistake as I have done.

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