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Transformer secondary values

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Johnson777717

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Good day all!

I'm having a brain cloud, and I can't seem to be able to figure this out for sure. Maybe I just need someone to confirm what I am thinking:

I'm ordering a power transformer. The datasheet states the secondary RMS Rating as such:

Parallel VAC 8.0 @1.600 Amps
Series VCT 16 @ .800 Amps

So here is where I get fouled up. I'm working on a dual polarity power supply, regulated down to 5VDC positive and negative. Using a center tap would give me 8Volts for each regulator (7805 & 7905) correct? Because the potential is measured between the center tap and either side of the secondary?

So if I were to run a transformer using parallel secondary (No center tap) I would be getting 4 volts for each regulator, because the potential is then measured between the secondary leads, without a center tap?

So if I use a hot ground, this would then give me 8 volts for each regulator?

So for this transformer, in order to obtain 8volts for each regulator, I either need a center tap, or employ the use of a hot ground. Correct?

Apologies If this sounds confusing. Maybe I can state it in a different way to make more sense?
 
Use center-tap config.
 

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If I were you, I would series the secondaries using the centertap as ground. This will give about 11.2 volts + and _ if you use a bridge rectifier and capacitor input filter.

So the 11.2 voltage is voltage with a pretty good load on it, correct? Or is this 11.2 a no load voltage. I suppose it doesn't matter, since I'll be regulating down to 5 Volts, which leaves 5 or more volts head-room (so to speak).

Another question if I may?
What are the disadvantages and advantages to a hot ground vs. a center tap? I've never used a hot ground, but I notice some circuits employ this. I'm assuming so they can get more amperage out of a transformer?

Thank you all for your advice, and your time! It is appreciated :D
 
Hi Johnson,
When your regulated supply is fully loaded then the input voltage to the regulators will be 10.3V or less since the conducting rectifier will have a 1V drop or more, and the transformer will not reach its calculated peak voltage since a rectifier feeding a capacitor will draw a much higher current at the peak than the calculated value would be with a resistive load.
Your fully loaded output current is limited by the power rating (2 X 8V X 0.8A = 12.8W) of the transformer, not simply its current rating, so with a peak voltage of about 11V from the transformer then the current can be only 0.58A (12.8W % 11V % 2) for each polarity so that it doesn't heat-up too much.
Speaking of heat, at full load each regulator will dissipate 2.9W (10.0 - 5.0V X 0.58A) so use heatsinks, and the rectifier bridge will dissipate 1.16W (1V X 0.58A X 2) and will not need a heatsink if it is big enough.
 
AUDIOGURU!!! Thank you for your detailed explanations and calculations! I appreciate your time.

I have some pretty good sized heat sinks for the regulators, which will hopefully be suitable for extended use. I just need to calculate the heat dissipation of the sink, and see where I stand. Thanks for making sure I have this sinked.

As far as the rectifier, I'll need to check the datasheets for the bridge rectifier to see what kind of heat it will operate under. I'm definately using a larger sized bridge, but I don't have the part number with me right now (I'm at work. :D )
 
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