Dumb Q About Center Tapped Power XFMR

[Paul Marossy]

So if you have a 6V-0-6V center tapped transformer and want to get 12V out of it, what is the correct way to do that? Is it as simple as connecting both 6V leads to a diode full wave rectifier and leaving the center tap unconnected or is there more to it than that?

I'm rusty... haven't been messing with this sort of thing for over ten years. 

[EBK]

So if you have a 6V-0-6V center tapped transformer and want to get 12V out of it, what is the correct way to do that? Is it as simple as connecting both 6V leads to a diode full wave rectifier and leaving the center tap unconnected or is there more to it than that?

I'm rusty... haven't been messing with this sort of thing for over ten years. 

Yep, it's just that easy (well, almost).  You wont be able to end up with a full 12 volts DC once you rectify and regulate. 

[Paul Marossy]

So if you have a 6V-0-6V center tapped transformer and want to get 12V out of it, what is the correct way to do that? Is it as simple as connecting both 6V leads to a diode full wave rectifier and leaving the center tap unconnected or is there more to it than that?

I'm rusty... haven't been messing with this sort of thing for over ten years. 

Yep, it's just that easy (well, almost).  You wont be able to end up with a full 12 volts DC once you rectify and regulate.

That's OK in this case. The other thing I was wondering about is what happens with the current rating. If it's say a 500mA transformer, is still capable of that when it's configured that way?

[EBK]

I do owe you a slightly better answer to your first question.  I'll write it up in a few minutes....

May take quite a bit longer.  Pediatrician appointment for daughter is becoming a struggle... 

[PRR]

> Pediatrician appointment for daughter is becoming a struggle...

I, OTOH, breezed in and out of my Physiatrist's (not Psychiatrist) appointment in 110 minutes. (After 2 months of delay....)

6V-0-6V. Ignore the "0". Take both 6V leads to the "~" pins of a full wave bridge rectifier. Big (1000uFd) cap across "+" and "-". This will run near 18V of DC (1.414 plus un-sag at no-load). You won't be at 12.0VDC without a regulator.

The current: since you got "more voltage" DC than AC, and there is NO free lunch, you will not get the 500mA rated AC current. Depending whether heat or sag is the problem, numbers of 1.6 to 1.8 apply. As 6VA is very-very small, use "2". You have 250mA of DC available. Unless the transformer is real saggy and won't stay near 16V DC at non-zero current.

[Paul Marossy]

> Pediatrician appointment for daughter is becoming a struggle...

I, OTOH, breezed in and out of my Physiatrist's (not Psychiatrist) appointment in 110 minutes.

6V-0-6V. Ignore the "0". Take both 6V leads to the "~" pins of a full wave bridge rectifier. Big (1000uFd) cap across "+" and "-". This will run near 18V of DC (1.414 plus un-sag at no-load). You won't be at 12.0VDC without a regulator.

The current: since you got "more voltage" DC than AC, and there is NO free lunch, you will not get the 500mA rated AC current. Depending whether heat or sag is the problem, numbers of 1.6 to 1.8 apply. As 6VA is very-very small, use "2". You have 250mA of DC available. Unless the transformer is real saggy and won't stay near 16V DC at non-zero current.

Makes sense on how to hook it up, I remembered that part correctly. But why does the current drop so much? Is it because of inefficiency of the transformer or is there something else going on?

[EBK]

> Pediatrician appointment for daughter is becoming a struggle...

I, OTOH, breezed in and out of my Physiatrist's (not Psychiatrist) appointment in 110 minutes. (After 2 months of delay....)

6V-0-6V. Ignore the "0". Take both 6V leads to the "~" pins of a full wave bridge rectifier. Big (1000uFd) cap across "+" and "-". This will run near 18V of DC (1.414 plus un-sag at no-load). You won't be at 12.0VDC without a regulator.

The current: since you got "more voltage" DC than AC, and there is NO free lunch, you will not get the 500mA rated AC current. Depending whether heat or sag is the problem, numbers of 1.6 to 1.8 apply. As 6VA is very-very small, use "2". You have 250mA of DC available. Unless the transformer is real saggy and won't stay near 16V DC at non-zero current.

Paul R., I didn't understand more than a quarter of what you wrote.  I revere you too much to refute you, but that sounded so unfamiliar to me that I will need to brush up on transformers again before I can be of more help to Paul M.    I will at least stick to the simple answer I gave originally: you can ignore the center tap and treat the secondary as an untapped 12-volt secondary, connect a full wave bridge rectifier, and, with suitable regulation, be well on your way to making a fine DC power supply.

(My daughter has strep throat, by the way.  I could say more, but if you have kids, you understand that I could describe any result from minor inconvenience to World War III and it would reasonably fit the brief statement of facts I've given you.  )

[PRR]

> ignore the center tap and treat the secondary as an untapped 12-volt secondary, connect a full wave bridge rectifier, and, with suitable regulation, be well on your way

Throw in a large cap. Otherwise this is *exactly* what I thought I said?
__________________________
> strep throat

Sorry to hear that. Been a long time since I was a kid. I hope it is mild and quickly over.

A dear friend long-ago progressed to rheumatic fever which is a real problem.

[EBK]

Throw in a large cap. Otherwise this is *exactly* what I thought I said?

Yes. It was everything else you said after that that threw me.

Sorry to hear that. Been a long time since I was a kid. I hope it is mild and quickly over.

Thank you.

[GibsonGM]

The 2nd part just looked like how to 'derate' the transformer's stated ampacity (500mA, ending up with 250mA at 12VDC).   I was glad to see that broken down a bit; never saw it done like that.  Transformers are many ppl's 'black box of unknown', I know they are to me.  I just pop a regulator on and load test them, often...

Can you talk a bit more about that, Paul?  How you arrived at your 1.6, 1.8 figure, ending up at 250mA?  I THINK I know, but then I would be assuming and you know how that ends up!    

[EBK]

The 2nd part just looked like how to 'derate' the transformer's stated ampacity (500mA, ending up with 250mA at 12VDC).   I was glad to see that broken down a bit; never saw it done like that.  Transformers are many ppl's 'black box of unknown', I know they are to me.  I just pop a regulator on and load test them, often...

Can you talk a bit more about that, Paul?  How you arrived at your 1.6, 1.8 figure, ending up at 250mA?  I THINK I know, but then I would be assuming and you know how that ends up!   

Agree.  I'm happy to learn the stuff I should have learned in engineering school (or relearn it in the event I've simply forgotten it from disuse).
I'm more used to looking up a rule of thumb formula, like this one:
For a full wave bridge capacitor input load, I_DC = 0.62 × Secondary_I_AC.  That would yield a max of 310mA DC. 
I just quickly grabbed that formula from this: www.hammondmfg.com/pdf/5c007.pdf, a document that helpfully adds, "Don't forget to take into account losses (not included in this guide)...." 

[GibsonGM]

...and what if the current rating of the transformer is unknown?  I understand you need to load the secondary, and when it drops SOME percent, you then 'label' it as "blah mA, blah blah volts"?    I've used a 10% sag on loading for this in the past but no idea if it's right...

[EBK]

...and what if the current rating of the transformer is unknown?  I understand you need to load the secondary, and when it drops SOME percent, you then 'label' it as "blah mA, blah blah volts"?    I've used a 10% sag on loading for this in the past but no idea if it's right...

I'd probably resort to comparing physical dimensions of my unknown transformer with a table (for example, Mouser catalog) of known transformers.  After doing that for a while, I'd weigh the cost of buying a new transformer with known current rating vs. my lack of confidence in determining the unknown.  Luckily, I'd already have the catalog in front of me.   

[Paul Marossy]

I'm more used to looking up a rule of thumb formula, like this one:
For a full wave bridge capacitor input load, I_DC = 0.62 × Secondary_I_AC.  That would yield a max of 310mA DC. 
I just quickly grabbed that formula from this: www.hammondmfg.com/pdf/5c007.pdf, a document that helpfully adds, "Don't forget to take into account losses (not included in this guide)...." 

I didn't know there was that kind of loss when full wave rectifying... I mean if you have a 500mA transformer and after rectification you only have 310mA, that's only like 62% of what you started with. That explains a few things on some previous builds that puzzled me. Interesting and good to know.

[EBK]

It's mostly an RMS thing (I hope I'm not wrong here...).  500mA AC = 354mA RMS. 310mA DC is more like 88% of what you started with.  I feel like I'm ignoring something very important about loading though....

I defer entirely to anything Mike and Paul R. have to say.  I sort of feel like I'm relearning the fact that I really know nothing.   

[Paul Marossy]

It's mostly an RMS thing (I hope I'm not wrong here...).  500mA AC = 354mA RMS. 310mA DC is more like 88% of what you started with.  I feel like I'm ignoring something very important about loading though....

I defer entirely to anything Mike and Paul R. have to say.  I sort of feel like I'm relearning the fact that I really know nothing.   

OH! *Slaps forehead*. Now I get it. I always thought of that applying to voltage, didn't think of it in the context of current. And then on top of that we add a little loss for a less than 100% efficient transformer. It makes sense now.

[GibsonGM]

I didn't know there was that kind of loss when full wave rectifying... I mean if you have a 500mA transformer and after rectification you only have 310mA, that's only like 62% of what you started with. That explains a few things on some previous builds that puzzled me. Interesting and good to know.

Oh, ya!  Transformers are efficient, but they're not THAT efficient.  They suffer I^R losses, hysteresis losses...theoretically I do think they approach 80% efficiency, and like Eric was saying, there is an RMS thing going on too.    I'd like to see what Paul says about this     

>>>My experience has been to simply measure the trafo unloaded, load it up til it sags oh 10%, 20%, check it.....not getting hot....OK, that will be where i run it at.  Probably could get MORE out of them if I did it right!   But it seems to work without a lot of bother for the ones you rip out of a CD player or something.  The 'square area' thing is a good one, too.  The old radio books talk about doing that. 

[EBK]

It's mostly an RMS thing (I hope I'm not wrong here...).  500mA AC = 354mA RMS. 310mA DC is more like 88% of what you started with.  I feel like I'm ignoring something very important about loading though....

I defer entirely to anything Mike and Paul R. have to say.  I sort of feel like I'm relearning the fact that I really know nothing.   

OH! *Slaps forehead*. Now I get it. I always thought of that applying to voltage, didn't think of it in the context of current.

Keep in mind that I've basically just admitted to being in over my head.  Stick around for the right answer from someone else.

[PRR]

> For a full wave bridge capacitor input load, I_DC = 0.62 × Secondary_I_AC.

If you know that, then Idc is Iac times (1/0.62) which is 1.6.

Ruben Lee's transformer book goes into this. 1.6 is actually the factor when heat is the limit. Sag is often the limit on small iron. If truly optimized, the factor is then 1.8. However in real life small transformers are "optimized" for COST. It is wise to take a factor of "2". 1 Amp of DC needs 2 Amps of AC.

> I just quickly grabbed that formula from this: www.hammondmfg.com/pdf...

Which is in fact *wrong* in several ways. When challenged, a Hammond tech rep said they "got it from the web" and said users should use (or hire) their own engineering expertise.

> what if the current rating of the transformer is unknown?

Load it up until it smokes. Get another and use less load.

Compare weight against comparable products in catalogs. There are no magic metals or varnishes. A basic simple transformer has to be "so big" for a given capacity. Some tradeoffs are different for EI cores or for toroids so stay with a same-type.

> I've used a 10% sag on loading for this in the past but no idea if it's right...

Very small iron can be rated for 20% even 30% sag. So a "10% rule" may be quite conservative. (FWIW, OTOH, *BIG* 25KVA iron is rated for 2% sag, and sucking it down 3% for a whole day takes a month off its life.) 

> that kind of loss

Cap-input rectifiers make ugly waves which heat/sag much worse than a dumb resistor AC load (how PTs are specified).

Not a huge loss. Remember your cap-input voltage went UP, 12VAC to 17VDC. TANSTAAFL. Current MUST be less or a law of energy would be broken. This fact alone means current must be reduced by 1/1.414 or to 0.707. The ugly wave-form in cap-input shaves that to 0.62 (1/1.6). In practical small-work, "half" is the safer bet.

[EBK]

In practical small-work, "half" is the safer bet.

Thank you again, Paul R., for coming to the rescue. 
So, to sum up:
If you are using relatively cheap, small transformers for low voltage power supplies, and you hook one up to a full wave bridge rectifier and large filter cap, you can expect you max DC current to be roughly half the spec'd AC current value.  Sound right?

Can we touch on the effects of adding a voltage regulator?  Say, for example, we stick an LM7812 on the end of this....