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Old 24th August 2008, 11:22 PM   (permalink)
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Quote:
Originally Posted by Hero999 View Post
No you didn't, you've just edited your post.

Just admit you made a mistake as I have done.
Odd you would say that. I did not ever say bi-phase is 1/2 wave, and nope I did not edit any text in which I said bi-phase is 1/2 wave.

In fact I am well aware of the difference albeit many are not aware of the the subtleties of bi-phase versus bridge.

I suggest you look to my first posts in my thread in which I address both bi-phase and half-wave as separate entities and discuss these implications.
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Old 25th August 2008, 12:34 AM   (permalink)
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Chumly,

May I ask where you are located (which country), please?

Mike
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Old 25th August 2008, 11:59 AM   (permalink)
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Originally Posted by Leftyretro View Post
"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
Yes, EXACTLY the same - say you have a supply for a power amplifier, of +40V/-40V - these are measured from chassis, which connects to the centre tap.

If you now connect the -40V rail to chassis instead, and measure from chassis again, you now have +80V/+40V. Exactly the same circuit, except for the reference point you're measuring from.

Understanding the reference point for any measurements is a crucial simple part of electronics, and something you need to be VERY aware of.
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Old 25th August 2008, 04:39 PM   (permalink)
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Originally Posted by Nigel Goodwin View Post
Yes, EXACTLY the same - say you have a supply for a power amplifier, of +40V/-40V - these are measured from chassis, which connects to the centre tap.

If you now connect the -40V rail to chassis instead, and measure from chassis again, you now have +80V/+40V. Exactly the same circuit, except for the reference point you're measuring from.

Understanding the reference point for any measurements is a crucial simple part of electronics, and something you need to be VERY aware of.
Thanks, I think you finally made the light turn on for me

Lefty
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Old 26th August 2008, 06:47 PM   (permalink)
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Originally Posted by Hero999 View Post
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.
It is nonsense to you. because you don't understand. And you actualy agreed to half the power or 50% . Are you just plain disturbed or you like to contradict THE POSTING no matter what. the duty cycle has nothing to do with anything . unless you have found A WAY to changge the duty cycle of a 120v ac line by rectifier and or transformer setup. Read next time and understand not just look at the pictures. CT you get half the output voltage so how can you get more power by magic?
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Old 26th August 2008, 07:13 PM   (permalink)
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Originally Posted by neon View Post
It is nonsense to you. because you don't understand. And you actualy agreed to half the power or 50% . Are you just plain disturbed or you like to contradict THE POSTING no matter what. the duty cycle has nothing to do with anything . unless you have found A WAY to changge the duty cycle of a 120v ac line by rectifier and or transformer setup. Read next time and understand not just look at the pictures. CT you get half the output voltage so how can you get more power by magic?
hi,
I have reading these posts out of interest and with respect I think you are not reading what he has posted.

The way I read it is that when he talks of the duty cycle he means the half wave mains cycle during which the transformer/diode are supplying current to the load.

He didnt say it would double the 'power', I believe he is saying that as the transformer is only supplying current to the load for half the mains cycle it wouldnt be a problem for the winding to supply double the current if required to do so.

Its unfortunate that you feel its necessary to insult him, even if you disagree with his technical description.
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Old 26th August 2008, 07:29 PM   (permalink)
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That's exactly what I was saying.

Although I was incorrect because the power dissipation in the winding quadruples as the current doubles. This means that if you double the current for 50% of the duty cycle the copper loss doubles meaning it will over heat.

The truth is you that can safely draw √2 the rated current on a bi-phase rectifier. As you've correctly pointed out, the voltage is halved meaning that the total power rating is divided by √2.

Read the last pages of me discussing this with Chumly for more information and I'll be happy to ask any more questions you may have.
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Old 27th August 2008, 10:23 AM   (permalink)
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Originally Posted by Hero999 View Post
That's exactly what I was saying.

Although I was incorrect because the power dissipation in the winding quadruples as the current doubles. This means that if you double the current for 50% of the duty cycle the copper loss doubles meaning it will over heat.

The truth is you that can safely draw √2 the rated current on a bi-phase rectifier. As you've correctly pointed out, the voltage is halved meaning that the total power rating is divided by √2.

Read the last pages of me discussing this with Chumly for more information and I'll be happy to ask any more questions you may have.
THERE IS NO DUTY CYCLE ON MAINS 120 VAC. THERE IS NO PHASE EITHER. THERE IS NO BY PHASE RECTIFICATION ALSO. there is full wave and half wave and pulseting DC. YOU may not choose to use this terms and throw it into a discussion because that doesn't make any sense to me.But then again what do i know? you may mention phasing only if a triac or scr conduct part of a cycle to indicate cunduction during part of a cycle.

Last edited by neon; 27th August 2008 at 11:01 AM.
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Old 27th August 2008, 11:19 AM   (permalink)
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Quote:
THERE IS NO BY PHASE RECTIFICATION

In that case, whats this circuit.?
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Old 27th August 2008, 11:27 AM   (permalink)
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Quote:
Originally Posted by neon
YOU may not choose to use this terms and throw it into a discussion because that doesn't make any sense to me.
They obviously don't make any sense to you because you obviously don't understand them so I'll explain them for you.

You can use the term phase; you can have in phase and anti-phase (0° and 180°) which are produced by the centre tapped transformer.

You can also use the term duty cycle, which is relevent here because half wave rectification results in a pulsed DC waveform of 50% duty cycle.

Quote:
THERE IS NO BY PHASE RECTIFICATION ALSO.
I don't know what you mean by this; just looks like poor grammer to me.

I'll try explaining my previous post with some calculations.

Adding a bi-phase rectifer to a centre tapped transformer means you can safely draw √2 times the rated current and here's the mathematical proof.

Suppose the transformer has a rated current of 1A and a secondary resistance of 100mΩ from the centre tap to either phase giving 200mΩ in total.

With a bridge rectifier and a 1A load, the IČR losses will be 1Č*0.1 = 100mW for each side of the winding, giving a total secondary power dissipation of 200mW which is the maximum rated power dissipation of the secondary.

With a bi-phase rectifier and a 1A load, the IČR losses will be 1Č*0.1/2 = 50mW (remember we've divided it by half as each winding is conducting a pulsed DC current of 50% duty cycle) for each side of the winding, giving total secondary power dissipation of only 100mW which is half the maximum rated power dissipation of the secondary.

With a bi-phase rectifier and a 2A load, the IČR losses will be 2Č*0.1/2 = 200mW for each side of the winding, giving total secondary power dissipation of 400mW which is double the maximum rated power dissipation of the secondary so it will overheat if operated at this current for too long.

With a bi-phase rectifier and a 1√2 load = 1.414 the IČR losses will be 1.414Č*0.1/2 = 100mW for each side of the winding giving 200mW in total which is the maximum rated power dissipation of the secondary coil.
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Old 27th August 2008, 03:18 PM   (permalink)
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Originally Posted by ericgibbs View Post
Quote:
THERE IS NO BY PHASE RECTIFICATION
In that case, whats this circuit.?
That's a Full-Wave Rectifier Eric. Calling it a bi-phase rectifier is incorrect in North America. The entire circuit however is connected to a single phase AC signal.



With a 1000 vct / 1 amp transformer you could expect to drive a load with 500 v (rms) at 1 amp using the full wave rectifier.



With a 1000 vct / 1 amp transformer you could expect to drive a load with 1000 v (rms) at 1 amp using the full wave bridge rectifier.

Last edited by Mike, K8LH; 27th August 2008 at 03:57 PM.
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Old 27th August 2008, 03:59 PM   (permalink)
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[quote=Mike, K8LH;341329]That's a Full-Wave Rectifier Eric. Calling it a bi-phase rectifier is incorrect in North America.
[/QUOTE

hi Mike,
I can understand the difference in terminology and I am sure we both understand each other now that the difference has been explained.

IIRC in the UK its a hang over from the olden days when we used glass dual anode valve/tube rectifiers.

A typical example of the difference in electrospeak between us is the valve/tube.
When you refer to a tube I know exactly what you mean.


BTW I wouldnt nowadays refer to it as a bi-phase rectifier, its just that one poster said
because he hadnt heard off it, it didnt exist, not so.

EDIT: Googled for uses of Biphase rectification, the term is still in common use worldwide, which suprised me.

Bi-Phase Rules OK.!
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Last edited by ericgibbs; 27th August 2008 at 04:23 PM.
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Old 27th August 2008, 04:42 PM   (permalink)
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Quote:
Originally Posted by Mike, K8LH View Post
The entire circuit however is connected to a single phase AC signal.



With a 1000 vct / 1 amp transformer you could expect to drive a load with 500 v (rms) at 1.414 amp using the full wave rectifier.
Corrected.
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Last edited by Hero999; 27th August 2008 at 04:44 PM.
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