What happens

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ptonsparky

ptonsparky

Tom
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EC - retired
I have a couple three phase rectifiers I took out of failed drives. I made a crude 24 volt power supplies out of one, just because. No, not used for anything.
I used stepped down from 120 to 24/12 via a .5kva transformer. Not smooth 120 hz.

What would happen if I attached the center tap to the third pole of the rectifier? I suspect a short, but not sure.
 
ptonsparky said:
I have a couple three phase rectifiers I took out of failed drives. I made a crude 24 volt power supplies out of one, just because. No, not used for anything.
I used stepped down from 120 to 24/12 via a .5kva transformer. Not smooth 120 hz.

What would happen if I attached the center tap to the third pole of the rectifier? I suspect a short, but not sure.
Click to expand...
Not a short, the back to back diode design prevents that.
But the AC voltage available from a connection using the center tap as one end will see a lower AC voltage and so will not contribute to a capacitor filtered output. But they may make the zero crossings in the unfiltered voltage/current output waveforms become just large dips instead.
 
ptonsparky said:
What would happen if I attached the center tap to the third pole of the rectifier? I suspect a short, but not sure.
Click to expand...

Of the three AC inputs to the bridge, the one that's the most positive (relative to the others) at any given time during the voltage waveforms will be the one that's connected to the positive output terminal of the bridge by a diode. And the input that's the most negative relative to the others will be connected by a diode to the negative output terminal of the bridge. But the voltage from the center tap to one end of the transformer secondary will never be more positive or negative than the voltage between both ends of the secondary. And in fact, the voltage will be half way in between. And so the center tap will always come up short of being the most postive or negative relative to the other secondary outputs. And so I don't think connecting the center tap will do much of anything.

The voltage will need to exceed the voltage drop across two diodes (approximately 1.4V) before conduction occurs. Of course, there also has to be some load current for the diodes to conduct. And as GoldDigger mentioned, the voltage on a shunt filter capacitor will need to be overcome before diode conduction occurs.
 
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My motorcycle's alternator has a fourth pair of diodes from the center of the wye, like this diagram:

1708218943485.jpeg
 
LarryFine said:
My motorcycle's alternator has a fourth pair of diodes from the center of the wye, like this diagram:

View attachment 2570128
Click to expand...

Those extra two diodes are apparently quite common in alternators because they can increase the maximum output current capability of an alternator by as much as 15%. Alternators in vehicles typically operate as current sources, and with the peaks of their open circuit voltage waveforms exceeding the battery voltage by a relatively large amount, especially at higher RPMs. And so even though the open circuit L-N AC voltage is a factor of 1/√3 ≈ 0.577 smaller than the open circuit L-L voltage (and shifted by 30°), its peaks can still substantially the exceed the battery DC voltage and therefore contribute a significant amount of extra charging current.
 
synchro said:
Of the three AC inputs to the bridge, the one that's the most positive (relative to the others) at any given time during the voltage waveforms will be the one that's connected to the positive output terminal of the bridge by a diode. And the input that's the most negative relative to the others will be connected by a diode to the negative output terminal of the bridge. But the voltage from the center tap to one end of the transformer secondary will never be more positive or negative than the voltage between both ends of the secondary. And in fact, the voltage will be half way in between. And so the center tap will always come up short of being the most postive or negative relative to the other secondary outputs. And so I don't think connecting the center tap will do much of anything.

The voltage will need to exceed the voltage drop across two diodes (approximately 1.4V) before conduction occurs. Of course, there also has to be some load current for the diodes to conduct. And as GoldDigger mentioned, the voltage on a shunt filter capacitor will need to be overcome before diode conduction occurs.
Click to expand...
That's the part that had me wondering what would happen.
 
ptonsparky said:
I have a couple three phase rectifiers I took out of failed drives. I made a crude 24 volt power supplies out of one, just because. No, not used for anything.
I used stepped down from 120 to 24/12 via a .5kva transformer. Not smooth 120 hz.

What would happen if I attached the center tap to the third pole of the rectifier? I suspect a short, but not sure.
Click to expand...
Have a diagram? Is this a Wheatstone bridge full wave rectifier. Bad capacitors yields not a smooth dc current and ripples
 
Remove the center point of Larry’s post #4 and you have my rectifier. I am powering it with 24 volt single phase. I get a very nice sign wave of 120 Hz as you would suspect. Adding about 600 mfd smooths It considerably. I have no intentions of powering anything other than simple experiments for my entertainment and a bit of education.
 
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