Transformers, # of secondaries and VA

Hi all,
I need to buy some transformers for a PIC project I'm making for
a train station model, but there's something that puzzles me.

Here are the data of two of them:

Secondary Current VA
12V 0.167A 2VA
12+12V 0.167+0.167A 8VA

Why in transformers with just one secondary the VA is V*A, while in
those with two secondaries it's just the half?

I mean, 12V*0.167A is 2W, same as 2VA.. but if I use the two 12V
secondaries in serie, to get a 24V secondary, multiplied for the
current of 0.167A I get 4W, i.e. half of the 8VA.

Where is my mistake?

When I don't need two secondaries, should I buy a transformer with
just one secondary for maximum power/money ratio?

Moreover, if the secondary gives 9V AC, after the diodes bridge and
capacitor, I should expect 11.3V, right? ( i.e. 9*1.414-(0.7*2) )
The capacitor is what accounts for the *1.414, and the diodes for the
the 2*0.7 voltage drop. But what should I expect under load, even
assuming a low impedance secondary? My fear is that the *1.414 will
be significantly lower.

I have The Art Of Electronics by Horowitz and Hill, but I didn't
really understand this issue well (and it's a simple, basic one!).

One last question.. is there any way to get, from one single AC source,
the effect of two secondaries? I know I could use two resistors, but
that would mean a lot of power loss. Any better ways?

Thanks,
Mike

 

[Tim Wescott]

See answers below. Be careful about using "Watts" instead of "VA". The
difference is that a transformer has seperate limits on voltage and
current -- the transformer core construction and coil insulation limits the
voltage that can be applied without overheating the core or sparking over,
the transformer winding generates heat as a function of RMS current and coil
resistance and dissipates it as a function of it's thermal construction. If
you hook a transformer up to a great big inductor or capacitor you'll
deliver no power to the load, but you could still burn up the transformer.

<Mike@invalid.com> wrote in message news:gMDDb.797$0w.31020@news2.tin.it...

Hi all,
I need to buy some transformers for a PIC project I'm making for
a train station model, but there's something that puzzles me.

Here are the data of two of them:

Secondary Current VA
12V 0.167A 2VA
12+12V 0.167+0.167A 8VA

Why in transformers with just one secondary the VA is V*A, while in
those with two secondaries it's just the half?

I mean, 12V*0.167A is 2W, same as 2VA.. but if I use the two 12V
secondaries in serie, to get a 24V secondary, multiplied for the
current of 0.167A I get 4W, i.e. half of the 8VA.

Where is my mistake?

You can either add the voltages or the current, not both (think series or
parallel connection).

When I don't need two secondaries, should I buy a transformer with
just one secondary for maximum power/money ratio?

Probably.

Moreover, if the secondary gives 9V AC, after the diodes bridge and
capacitor, I should expect 11.3V, right? ( i.e. 9*1.414-(0.7*2) )
The capacitor is what accounts for the *1.414, and the diodes for the
the 2*0.7 voltage drop. But what should I expect under load, even
assuming a low impedance secondary? My fear is that the *1.414 will
be significantly lower.

It will be lower, and it will have ripple. The higher the value of the
capacitor the higher your average voltage will be, but the current will come
in tall, narrow "spikes" which will increase the RMS current. It's a big
tradeoff between how much power you burn in your regulator, how big your
transformer is, etc. If I remember correctly you should design your DC
output voltage to be lower than the AC RMS voltage by 10 or 20% at the
design current.

You may want to check the ARRL handbooks for charts on sizing
transformers -- I know that my older copies have this information, and I
_think_ the new ones do as well. These handbooks used to be _the_ way to
learn radio electronics, they're still pretty good.

I have The Art Of Electronics by Horowitz and Hill, but I didn't
really understand this issue well (and it's a simple, basic one!).

One last question.. is there any way to get, from one single AC source,
the effect of two secondaries? I know I could use two resistors, but
that would mean a lot of power loss. Any better ways?

Yes, but at the cost of more ripple: Ground one side of the secondary, then
use one diode for a half-wave + supply and another diode for a half-wave -
supply.

Thanks,
Mike

 

Mike@invalid.com wrote:

Hi all,
I need to buy some transformers for a PIC project I'm making for
a train station model, but there's something that puzzles me.

Here are the data of two of them:

Secondary Current VA
12V 0.167A 2VA
12+12V 0.167+0.167A 8VA

Why in transformers with just one secondary the VA is V*A, while in
those with two secondaries it's just the half?

I mean, 12V*0.167A is 2W, same as 2VA.. but if I use the two 12V
secondaries in serie, to get a 24V secondary, multiplied for the
current of 0.167A I get 4W, i.e. half of the 8VA.

Where is my mistake?

When I don't need two secondaries, should I buy a transformer with
just one secondary for maximum power/money ratio?

Moreover, if the secondary gives 9V AC, after the diodes bridge and
capacitor, I should expect 11.3V, right? ( i.e. 9*1.414-(0.7*2) )
The capacitor is what accounts for the *1.414, and the diodes for the
the 2*0.7 voltage drop. But what should I expect under load, even
assuming a low impedance secondary? My fear is that the *1.414 will
be significantly lower.

I have The Art Of Electronics by Horowitz and Hill, but I didn't
really understand this issue well (and it's a simple, basic one!).

One last question.. is there any way to get, from one single AC source,
the effect of two secondaries?

Sort of, but not really the effect of 2 secondaries. With
2 secondaries you get isolation - there is no electrical
connection between them. With a single secondary, you could
produce numerous DC supplies by rectifying and regulating -
but they would not be isolated. There would be an electrical
connection between them.

I know I could use two resistors, but
that would mean a lot of power loss. Any better ways?

The way Tim mentioned is probably best.

Thanks,
Mike