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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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| Chumly, May I ask where you are located (which country), please? Mike | |
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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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Lefty
__________________ Measurement changes behavior | ||
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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.
__________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/ | ||
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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.
__________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez | |
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Last edited by neon; 27th August 2008 at 11:01 AM. | ||
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In that case, whats this circuit.?
__________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/ | ||
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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:
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.
__________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez | |||
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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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| [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.!
__________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/ Last edited by ericgibbs; 27th August 2008 at 04:23 PM. | |
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| Corrected.
__________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez Last edited by Hero999; 27th August 2008 at 04:44 PM. | |
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