Bi-Phase & Half-Wave Rectification

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Nigel Goodwin said:
"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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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
 
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.
 
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Abusive post deleted, such posts are NOT permitted here.

Moderator.
 
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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.
 
Hero999 said:
No you didn't, you've just edited your post.

Just admit you made a mistake as I have done.
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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.
 

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
 

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.
 
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.
 
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neon said:
YOU may not choose to use this terms and throw it into a discussion because that doesn't make any sense to me.
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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.

THERE IS NO BY PHASE RECTIFICATION ALSO.
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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 I²R losses will be 1²*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 I²R losses will be 1²*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 I²R losses will be 2²*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 I²R losses will be 1.414²*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.
 
ericgibbs said:
THERE IS NO BY PHASE RECTIFICATION
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In that case, whats this circuit.?
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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.

**broken link removed**

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.

**broken link removed**

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.
 
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Mike said:
The entire circuit however is connected to a single phase AC signal.

**broken link removed**

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.
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Corrected.
 
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Ever seen a Mercury Arc Rectifier? Typically they are six phase half wave and the ones used with radio transmitters can handle about a megawatt. Later on the Excitron was invented and these are basically a single mercury arc half wave rectifier. Twelve phase half wave was a popular arrangement and the absence of arc-backs was better for peoples nerves!

Car battery chargers still use bi-phase half wave and some of them bang out a lot more than four amps. Take the RAC HP027 battery charger/engine starter which is a fairly recent design. A boost rate of 100 amps is claimed and amazingly its bi-phase half wave! As to transformer losses one can surmise that the iron losses will be extremely similar to those of a bridge circuit so its only the copper-losses in the secondary that will be different. As large power diodes may drop almost a volt when fifty amps are flowing (often its 0.7 t0 0.8 volts) the diode loss at low voltages can be quite considerable. Even with a 0.6 volt diode drop 30 watts will be dissipated when 50 amps are flowing. Germanium diodes used to be more efficient than silicon with a mere 0.3 volt drop but leakage currents and rather iffy thermal stability put Germanium out of fashion decades ago.

Neither circuit can be said to be better than the other if the voltage and current are not specified.
 
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