Bi-Phase & Half-Wave Rectification

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

Chumly said:

Could you expand a bit more on the acronyms and other terms you are using (and the math too) so I can be sure of what you are talking about please?

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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).

Chumly said:

Nope, that's a four diode, dual-DC-supply using a center tap transformer, that's not a bi-phase circuit as described (a two rectifier configuration with the secondary of the power transformer having a center tap).

Maybe dual bi-phase, but that's not what I'm referencing, if you read my text carefully I spec two rectifiers and a center tap. I suggest you check out the history of tubes in this context for background on the original rationale of reducing the number of rectifier tubes. But in modern solid state such a configuration is iffy for the reasons given. You're welcome to show otherwise with actual modern popular solid state DC power supply examples.

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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

Chumly said:

Nope, I’m sorry to say. No filtration of any kind has been referenced and you have not described a bi-phase configuration. It's all there if you read my posts carefully, and do the needed research.

Would you please clarify the following acronyms with full word definitions:

CT you mean?
28VCT you mean?

Cheers!

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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.

k7elp60 said:

.
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:
https://books.google.co.uk/books?id...&hl=en&sa=X&oi=book_result&resnum=9&ct=result

Chumly said:

To further edify:

The main point of the thread however is bi-phase, it's in that context I posted, I just threw the other stuff in there as it was no more work to post, it has some commonalty and I must teach it as well.

You see I am a teacher, and bi-phase comprises a substantive amount of the mandatory course material as it relates to power electronics.......but it would appear bi-phase (as it relates to power electronics at the least) only has a place in tube rectifier circuits.

I am not teaching vintage audio, guitar amps, nor tubes at all nor are they referred to at all in the power electronics section of my course.

I am teaching apprentice electricians actually, so power electronics is applicable to the learning but I don't see how an emphasis on bi-phase is of any net benefit when time is short and there is much material to cover.

I suspect this ongoing emphasis on bi-phase in the power electronics course material (and on the exam) is an antiquated holdover from the halcyon days of big industrial tube DC power!

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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.

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.

Hero999 said:

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.

bridges

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.

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

Mike said:

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

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


I posted this link earlier.

https://books.google.co.uk/books?id...&hl=en&sa=X&oi=book_result&resnum=9&ct=result

Hero999 said:

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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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!

Mike said:

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

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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

Leftyretro said:

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

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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).

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...

Nigel Goodwin said:

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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"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

Leftyretro said:

"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

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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.

Chumly said:

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!

Click to expand...

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.