Calculating DC Output Current From Unregulated AC Transforme

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.

How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?

Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.

Thanks,
JC

 

On Jul 9, 9:26 pm, emailaddr...@insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.

Perhaps a little more info about the project is in order.  Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries.  The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply.  

Another thing I am wondering is if a basic, typical small brick AC-DC
switching supply is stable powering this kind of load. Stripping down
the circuit to the basic topology, what if it had for example:

AC-DC SMPS -> LM317 -> 1.25 Ohm resistor (LM317 in current regulating
config) -> 1A into Battery Pack + LM317 feedback

Since the SMPS is trying to regulate to it's spec'd voltage, let's say
that is 13V even if an uncommon value, will it run stable doing so and
if so, is that at the same constant output wattage from the SMPS the
whole time, with the linear regulator simply dropping more voltage,
creating more heat at the beginning of a battery recharge cycle since
it's suppling constant current to batteries that are at a lower
initial, low state-of-charge voltage?

 

On Jul 9, 9:47 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

emailaddr...@insightbb.com

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC.  I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

** This is your error.

This what? I was looking for details.

The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.

Ok, but I'm not asking what it's not, I'm asking what it is. what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating. Is 20VA really the answer, since we're
talking about winding ratios? It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of air) 40V @ 0.5A for 20VA.

It is also a " free air " rating -  meaning
there must be plenty of ventilation and no heat sources next to the
transformer.

Yes, I am looking only for a free air rating. It all starts there.

Perhaps a little more info about the project is in order.  Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries.  The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply.  I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.

How much rectified DC voltage can that transformer maintain at 1.67A?

**   It would be overloaded delivering that amount of DC current and
overheat -  unless perhaps you blew cold air over it with a fan.

I am not interested in what overloads it, I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.

Only after I know that *overloaded* maximum can I begin to derate to
account for heat.

Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core )  but is in the range of  0.5 to 0.7.

So, if you really do need 1.67 amps DC at 16 volts  -  then pick a 40 VA
transformer.

As mentioned I need to know the transformer's capability, not 1.67A at
16V. Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.

Beyond that, I will use the actual equations to achieve at least
12.3VDC and a current depending on cost, transformer volume, etc.

I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

 

[ehsjr]

emailaddress@insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.

How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?

Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.

There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.

I'll mention some general things.

First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.

Next, figure the ripple voltage, Vr. To figure Vr, see
http://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-101Introductory-Analog-Electronics-LaboratoryFall2002/8DCA1C5C-EC80-48C8-A502-61CDC632D5B7/0/ripplevolts.pdf

(watch the line wrap)

Next, find out how much headroom your regulator needs, usually
available from the datasheet. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the datasheet. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.

By the way, the 12.3 volt figure is odd - how did you arrive
at it?

Ed

Thanks,
JC

 

On Jul 9, 10:57 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

emailaddr...@insightbb.com
"Phil Allison"
** This is your error.

This what?  I was looking for details.

** Read the whole post before making silly replies.

** Phil, isn't this a bit of irony since I clearly asked for something
specific and you replied throughout your entire prior post with
anything except what was asked?

The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.

Ok, but I'm not asking what it's not, I'm asking what it is.

**  Read the whole post before making silly replies.

See "** Phil" above.
I recall you're always a bit cranky but that's why I wrote what I did
two lines up, because I'm "asking what it is", for an equation, which
makes it seem as though you either choose not to provide one, or you
don't know. If you don't know it's not a problem, but isn't it
important to clarify what I am looking for?

what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating.

**  Read the whole post before making silly replies.

Because if I do, I'll see the equation(s) I was asking about?

Is 20VA really the answer, since we're
talking about winding ratios?  It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of  air) 40V @ 0.5A for 20VA.

**  Read the whole post before making silly replies.

See above.

It is also a " free air " rating - meaning
there must be plenty of ventilation and no heat sources next to the
transformer.

Yes, I am looking only for a free air rating.  It all starts there.

**  Don't complain when you get free, good advice -  fuckhead.

Phil, how do I put it so you understand? If you really want to build
this entire project for me then go right ahead, I'll email you the
parameters. Until then, all I needed was what was asked, not for you
to second-guess something that was even a first-guess since you have
no idea what current I plan to run from any particular transformer,
you only know that I asked how to convert a spec mathematically and
used a few numbers as random examples of a starting point.

I am not interested in what overloads it,

** You need to be  -  for safety and functional reasons.

This is not a fixed plan for a one-off design Phil, this is a generic
question about transformers. The numbers were just random, forget the
numbers because they have nothing to do with the topic. I am not
interested because I have no trouble building within margins once I
know what those are - which requires having the equations.

Because I am not dead set on using any particular transformer for any
particular thing, it is valid and correct to say I am not interested
in what overloads any one particular transformer, only in how to
mathematically determine the free air DC ratings from the free air AC
ratings.

  Read the whole post before making silly replies.

I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.

** There is no possible math based only on the figures you gave.

False. Given all the parameters it can be solved. Some of those
paramaters I do not have like transformer core saturation values,
others can be resolved experimentally but that can still be expressed
in an equation.

The whole
topic is about ACTUAL temp rise of the windings of the particular
transformer under particular conditions.

This is getting amusing. The whole topic is not about building one
thing with one transformer and what that one transformer's rating
should be. The whole topic which I started (least you forget?) was
asking for equations. If you don't know those, join the club- neither
do I. I could guess, have some theories but it would be better to
hear from others who can stay focused on the topic.

Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter..
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core ) but is in the range of 0.5 to 0.7.

So, if you really do need 1.67 amps DC at 16 volts - then pick a 40 VA
transformer.

As mentioned I need to know the transformer's capability, not 1.67A at
16V.  Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.

** I just supplied the derating figures and the reason why it cannot be
precise.

You used the term "derating", are you saying it is not a derating
figure but rather a conversion figure? I ask because you have gone on
and on about derating for temp, when I am very specifically looking
for only a mathematical equation for maximum transformer capability, a
theoretical one in a model that ignores temperature because the answer
is not going to be directly used to build something!

So are you claiming that if I short circuited the rectified output of
a transformer rated for 1.67A @ 12VAC, that the output would be 20VA *
0.5 to 0.7? I do not think that is the correct answer, I vaguely
recall it should be higher than that.

If the shorted output power is not 20VA * 0.5 to 0.7V, then you have
not answered the question, because I am not asking about
implementation details for any transformer, only how to convert
mathematically to actual peak, not prudent design values. Perhaps I
did not initially describe the question well enough. If so that is my
fault, but I hope by now it is clear I am not looking to directly
apply any info directly towards choosing the mentioned transformer for
anything in particular based only on shorted power.

I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

** Neither is possible with simple math and only the VA rating to go on.

If it's complex math so be it. If it were simple I expect I'd have
known already, which was why it was asked regardless of how complex
the answer would be, but perhaps to the wrong group, perhaps it is not
a sci.electronics.basics level question. Ultimately if I find a spare
rectifier board lying around I'll hook that up and short it to get an
answer experimentally, at least for one transformer but it still
doesn't provide the sought after equation, I'd have to speculate about
whether similar E-Cores (which the example transformer was) behave the
same and collect more data about other E-cores in the same scenario,
hoping all the while that their manufacturers rated them equivalently,
accurately, then try to find some relationship.

I have written a ( non simple) program that gets fairly close to predicting
the results ( ie the DC output voltage at any specified current )  -
providing you have much more data on the transformer like the primary and
secondary resistances, the size of the filter cap and a figure for
transformer leakage inductance.

Then the equation(s) used in that would be much closer to what I was
after. Variables unresolvable could be assumed as constants then
experimentally compared to determine the level of deviation.

You are WAY over-analysing the problem.

You only need to be sure the voltage is high enough, but not too high,  for
you charger to work and the transformer is not overloaded under any
operating condition.

Ok, back to the charger. I have too many transformers to count that
could be used there. The transformer heat is not a problem. Using a
fan would be fine, having it shut down during regular use would even
be fine through thermal cutoff circuit(s). I would rather a warmish
running transformer with a fan cooling it than a burnt out regulator,
a bigger case for a giant heatsink plus added weight, or hot project
case.

With all due respect Phil, you have no idea of the project
requirements beyond what I had told you. Three manditory ones are low
temp density, small size, light weight. I'd MUCH rather put a thermal
cutoff or two in than dump a lot of heat on a regulator. A 40VA
transformer is not going to work, unless it's one rated slightly below
12V which is an odd figure. I guess I could strip some windings off
of one but that is such an ugly solution.

Ultimately for that particular project I will most likely use a SMPS
instead, but that project was only an example not the question asked
which is what the equation is to determine AC to DC spec conversion.
I do not need an answer based on the limited specs of the transformer
info I had given, I only need the equation as complete as it can be
for future uses.

Perhaps we got off on the wrong foot Phil, but please believe someone
when they state plainly what they don't care about and focus on what
they asked. This is NOT a topic about building one project, this is
about actual vis-a-vis rating conversion. I wouldn't have ran a 12VAC
1.67A transformer at 1.67A, AC output either, I know about derating
and temperature issues already which is why I stated several times
that this was not what I asked about, not what the question was about,
not what I care about. I care only about the equation. If you have
the answer because of your computer application that was relevant.

I hope I've made things clearer. I do not care about anything except
an equation to determine exactly what I asked about. It was why I
asked that question instead of some other question.

 

[Baron]

emailaddress@insightbb.com Inscribed thus:

Phil is giving you good advice !

Watts in equals watts out less losses.

Its not his fault you don't understand what Phil is telling you !

--
Best Reagrds:
Baron.

 

[whit3rd]

On Jul 9, 10:25 pm, emailaddr...@insightbb.com wrote:

On Jul 9, 10:57 pm, "Phil Allison" <philalli...@tpg.com.au> wrote:

emailaddr...@insightbb.com
"Phil Allison"
** This is your error.
The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.

Ok, but I'm not asking what it's not, I'm asking what it is.

None of the information given is useful in making a determination on
this issue. Phil is telling you that the AC
rating is for 1.67A RMS, which amounts to 2.3A peak, but
that your rectifier will only conduct for a fraction of the duty
cycle, so
the 1.67A average output of the supply is all delivered in very
short time intervals (when the transformer output exceeds
the capacitor charge plus the diode forward drop). The
rectifier current could be 18A repeating peaks (even though
the average is lower), because the rectifier is almost always
turned off.

18A can melt the copper windings.

The ratings for rectifier transformers are different than those
for AC transformers. If you used a choke filter instead of a
capacitor, the rectifier would have a larger conduction angle (and you
can get closer to the AC current rating, safely).

 

[Peter Bennett]

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress@insightbb.com
wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

Go to http://www.hammondmfg.com/5cpwr.htm and download their Power
Transformer Selection Guide - it is a one page .pdf that shows various
circuit configurations and resulting voltage/current relations.



--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

On Jul 10, 1:50 am, ehsjr
<e.h.s.j.r.removethespampunctuat...@bellatlantic.net> wrote:

There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.

Ok, but this should be resolvable to a reasonable level given the
example context of it being a typical E-core transformer and basic
silicon bridge rectified and capacitive filtered circuit. The output
at that point is the question. Even if some formula has an unknown
variable, or two or one hundred, the start would be to resolve those
which requires a working equation in which to place those variables.

I'll mention some general things.

First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.

This seems to counter the majority of small transformer examples out
there, does it not? Consider products powering just about anything
that uses a wall wart.
Consider a sub-20VA transformer rated in a wall wart for 12VDC, 1A
output. Granted, we could call that a little over 1/2 the rating but
just how much or little is the crucial issue, what variables determine
how much or little and how to express those mathematically.

Next, figure the ripple voltage, Vr. To figure Vr, seehttp://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-S...

Thank you for the link, though since I am not mass producing equipment
and so not overly concerned about small differences in component cost
or size, I'm essentially going to consider ripple current a constant
resolved later as would apply to any one design, while I am asking
about a general formula for conversion without regard to variables
that would change in different designs beyond the basic assumption of
a capacitor large enough to achieve acceptibly low ripple for example.

Next, find out how much headroom your regulator needs, usually
available from the datasheet. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the datasheet. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.

By the way, the 12.3 volt figure is odd - how did you arrive
at it?

This is what the example circuit would need at a bare minimum by
calculating similarly to what you have above plus other drops in the
circuit @ expected current levels, to still retain the necessary
minimum input voltage, plus or minus a margin of error as median
datasheet values where used.

However, I am not concerned about the entire circuit, not about drop
over a regulator, I am only concerned about resolving how the AC
transformer rating relates to DC output before anything further in a
circuit beyond a typical bridge rectifier and capacitor sufficient to
smooth to a hypothetical 0V ripple, and I am also accounting for the
(Often negligable) different in forward voltage over the rectifier(s)
at different current levels, but this too can be expressed
mathematically.

Essentially, I don't want information more applicable to one project
than to another. Only what remains true mathetically for all projects
which employ AC spec'd transformers of typical design and through
bridge rectification by the most common silicon diodes. In other
words, as you'd see in the vast majority of electronics already if
they're not using switching PSU.

 

[Hammy]

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress@insightbb.com
wrote:

[snip]

Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.

Thanks,
JC

The info you need can be found in any electronic text, or in numerous
sources on the web. If I understand what you want transformer sizing
and bulk filter cap "unregulated supply".

Here is just a couple sources.

Onsemis "Linear & Switching Voltage Regulator Handbook" see SECTION 8
" DESIGNING THE INPUT SUPPLY".

http://www.onsemi.com/pub_link/Collateral/HB206-D.PDF

Other free HB from onsemi

http://www.onsemi.com/PowerSolutions/supportDoc.do?type=manuals

Midcom also has an extensive collection of tech notes.

http://www.midcom-inc.com/Tech/TechNotes.asp

More transformer info

http://www.atc-frost.com/products/design/

Similar to onsemis

http://elearning.algonquincollege.com/coursemat/saurioc/ELE-3/ELE-3-NOTES/VOLT-REG.pdf

 

On Jul 10, 2:12 am, "Phil Allison" <philalli...@tpg.com.au> wrote:

emailaddr...@insightbb.com

"Phil Allison"
** This is your error.

Phil, isn't this a bit of irony since I clearly asked for something
specific

** And impossible.

Apparently not, because plenty of products out there use wall warts,
are you suggesting they all have to build prototypes for even a simple
AC-DC wall wart? You gave a figure of 0.5 to 0.7 times the rating but
clearly darn near every wart here (quite a big box full) don't adhere
to this, they're all using transformers smaller than your suggestion
would imply and they aren't all bursting into flames or melting
things.

Cos you know NOTHING about the topic - fuckhead.

** Don't complain when you get free, good advice - fuckhead.

Phil, how do I put it so you understand?

** Do not ever try that shit on - fuckhead.

You need a pet Phil, or something. If we were in church then the
cursing might seem dramatic but on usenet does it really have any
useful purpose? I suppose you're trying to incite me to call you a
fuckhead too. Will it make you happy, can we move on then or will you
still be stewing for no good reason? I suspect the latter.

I am not interested in what overloads it,

** You need to be - for safety and functional reasons.

This is not a fixed plan for a one-off design Phil, this is a generic
question about transformers.

** You still have not read the whole post - have you ?

Yes

Wasn't it obvious when I replied in a way related to what you had
written throughout the whole post?

What an utter ass you are.

Thank you Phil, but that's not quite enough. You hassle so many
people on usenet that calling me an ass just isn't special or rare
enough to be satisfying.

** You are one, colossally stubborn fool.

Yes, I generally keep at it until I achieve the desired result. I'll
have an equation even if I have to invent one of my own and
experimentally test it. I was just hoping someone out there had more
related info than I did besides generalizations of jumping up in
transformer size to create some margin, which is what I'd been doing
all along and is not so desirable as knowing exactly what that margin
is instead of just a ballpark figure, or at least to get much closer
in accuracy. Even if "use a bigger hammer" is a popular phrase there
has to be a method others are using when they start out with AC
transformer specs, as they manage to produce gear that works fine but
is not rated as high as your conservative estimate suggested would be
necessary. If real world examples and my wanting to find the
discrepancy between them and what you claim, is upsetting, all i can
say is whatever you want to insist won't change the box of
transformers I'm staring at. I suppose we could call the entire wall
wart industry a massive fraud for being overrated but I doubt it's
that simple.

** I just supplied the derating figures and the reason why it cannot be
precise.

You used the term "derating", are you saying it is not a derating
figure but rather a conversion figure?

** The figure was fully explained.

It derates the AC current figure to give the available DC current while not
exceeding the VA rating.

Thank you for clarifying, but I'm still in disbelief. Take this
example-

I have two typical looking E-core transformers rated for 32VA.
One of which was in a cordless drill quickcharger, it charged a pack
in an hour and also has an appropriate spec on it to accomplish that
per the battery pack cell capacity, 1.8A @ 12V Output (after going
through the control circuit).

This charger was used frequently for over 10 years. No failure on it,
no discolorations or smell like you'd have from excessive heat, looks
new inside except for a little bit of sawdust from the environment it
was occasionally used in. 'Twas only scrapped because it became less
viable to rebuild the battery packs another time on a drill after many
years of wear.

Based on your suggestion, even the upper end of getting 0.7 * AC
capacity, 0.7 * 32VA = 22.4VA. Output from the entire charger after
charge control board losses was 12V * 1.8A = 21.6. I think we both
know an old charger board is not 21.6/22.4 = 96.4% efficient!

I ask because you have gone on
and on about derating for temp, when I am very specifically looking
for only a mathematical equation for maximum transformer capability, a
theoretical one in a model that ignores temperature because the answer
is not going to be directly used to build something!

** You Q is just plain absurd.

Not in the context of using it as a starting point for an equation
upon which other variables are plugged in per the transformer and
application.

Transformer current and VA ratings are ALL about temp rise !!!!!!

ALL? Phil you seem to be stuck in infinite loop mode or something.
Regardless of whether you feel that way, I am talking about like
specs, there had to be a constant involved in the inital AC spec so
that remains the constant in the DC spec with regards to temperature.
What DC voltage and current is there, resolved mathematically, if we
allow the transformer to rise to the temp it would attain operating at
the AC voltage and current but not hotter than that? Wouldn't you
concede that in this case, a fair comparison would be if they are the
same temp so it isn't about temp rise at all? Some of the example
wall warts I have would tend to run quite a bit hotter than their free
air rating once placed in a sealed (or barely, passively ventilated)
plastic shell, and yet they don't have the thermal issues you imply
they should.

The regulation curve ( whether AC or DC plus rectifier /filter ) is a
function of VA and a number of other parameters plus the type of
construction used.

Agreed. Since the transformer itself is a constant, including it's
parameters and construction, I am looking for the remaining equation
of the AC voltage & current to DC voltage and current. I am looking
for the basic equation in which things that don't have to be
different, that can be held as constants like temperature and the
transformer being the same, are held as constants.

So are you claiming that if I short circuited the rectified output of
a transformer

** Never do that and I have never mentioned any such idiocy.

You are hee hawing like some barnyard ass now - pal.

Phil, I have a box of transformers and it's not a big deal to short
one of the smaller ones. Yes it'll smoke, maybe blow an internal fuse
or something but until it gets up to the high temps you fear it might
just last long enough to get a current reading. I was hoping to avoid
doing such things when I came here asking for an equation, but it
looks like testing is the only way to resolve this. Dead shorts
aren't necessary of course, but plotting data on several transformers
would seem useful if you were the only source of info on this topic.

If the shorted output power is not 20VA

** The power delivered into a short is always zero - fuckhead.

Are you saying it won't create heat, at least until something blows?
It sure seems like all that energy is going somewhere, and that it's
measurable.

But massive heat build up in the transformer will destroy it - pronto.

I'm sure it won't run long like that, but nevertheless you got the
jist of what I meant so who is being stubborn now?

because I am not asking about
implementation details for any transformer, only how to convert
mathematically to actual peak, not prudent design values.

** Utterly stupid question, only asked by fools.

If you had emailed me ahead of time before I ever started the thread
then I would have known what you wanted me to ask.
Unfortunately you failed to do so.

I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.

** Neither is possible with simple math and only the VA rating to go on.

If it's complex math so be it.

** You do not need any complex math to size a tranny to do a given job.

Ah, but I do if I don't want to be half ignorant about it, if I have a
need to use one only large enough for the job, or if I need to know
whether one I had available is suitable, at least long enough to get a
prototype running or whatever the purpose.

There are lots of devices out there in the world that don't appear to
just size a transformer by throwing in one that's definitely larger
than it needs to be to supply the current and (your biggest concern)
stay cool enough not to fail or cause thermal stress to anything in
the vicinity), to have an acceptibly long service life without any
issues, unless the load itself causes excessive current and it blows
the thermal fuse.

Problem here is, you cannot even specify what the job is.

I especially do not want an equation that is only applicable to one
job. That would defeat the whole purpose.

I have written a ( non simple) program that gets fairly close to
predicting
the results ( ie the DC output voltage at any specified current ) -
providing you have much more data on the transformer like the primary and
secondary resistances, the size of the filter cap and a figure for
transformer leakage inductance.

Then the equation(s) used in that would be much closer to what I was
after.

** I don't give a shit what garbage you are after - fuckhead.

What you NEED is a damn good kick up the arse.

Pot and Kettle, Phil?

You are WAY over-analysing the problem.

You only need to be sure the voltage is high enough, but not too high, for
you charger to work and the transformer is not overloaded under any
operating condition.

With all due respect Phil, you have no idea of the project
requirements beyond what I had told you.

** So you are now making then up as you go along - right ?

I am declaring them irrelevant constants because I am looking for an
equation that is independent of any one project. That's why I have
mentioned VARIABLES quite a few times, that it is not a problem to
have them to plug in later per any specific project.

A 40VA transformer is not going to work, unless it's one rated slightly
below 12V which is an odd figure.

** Totally insane crap.

Not work as-in, not meet the other criteria, I'm sure it'll supply the
minimal current in the example and then some but you still aren't
seeing that I am not looking for a recommendation for one project, it
was only an example posed towards coming up with a generic equation.
Again I will state that I should not have mentioned any specifics
about any project because it has distracted you too much.

Ultimately for that particular project I will most likely use a SMPS
instead, but that project was only an example not the question asked
which is what the equation is to determine AC to DC spec conversion.
I do not need an answer based on the limited specs of the transformer
info I had given, I only need the equation as complete as it can be
for future uses.

** The detailed specs you need are NOT published by makers.

You HAVE to measure individually them after you BUY a transformer !!!

I think you are right, to resolve some variables that will be
necessary. Even then, these seem to be less relevant variables as
there is a typical ratio of size to current rating that is above what
your 0.5 to 0.7 derating suggesting would dictate.

Just as quick to run a couple of bench tests if you have the damn tranny
right there in front of you.

Yes, I could run a test on every transformer but at some point one has
to pick the transformer to test, make a primary selection of
candidates and based on what I've seen there has to be a more accurate
way. Consider the example I posed above about a drill battery
recharger (which is not the same example as the other battery charger
project I initially posed), where the 32VA transformer is producing
enough power that if you were right, it should be too hot but it
isn't.

But I do not answer ridiculous Qs as specified by arrogant, bullshitting
novices

I'm starting to wonder if you've built nearly as many power supplies
as I have Phil. I have done similar to what you suggest in the past,
using vague ranges to derate as rules of thumb but I know there has to
be a better way and I observe far too many real world products that
contradict your claims, even if they are not so different than my
prior beliefs, and that they contradict my prior beliefs as well as
yours is part of what brought me here in the newsgroup to ask in the
first place.

No-one on usenet EVER has to do that.

I hope I've made things clearer.

** What you have made very clear is that you are an utterly autistic, rabid
nut case.

Are you sure it's transformers that are overheating?

Have a nice day Phil. It's not worth getting upset over.

 

[Baron]

emailaddress@insightbb.com Inscribed thus:

On Jul 10, 5:49 am, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:
emailaddr...@insightbb.com Inscribed thus:

Phil is giving you good advice !

Watts in equals watts out less losses.

Its not his fault you don't understand what Phil is telling you !

--
Best Reagrds:
Baron.

I agree, and understand what Phil was saying, but that's a generic
guesstimation that I'm looking to move beyond. Given enough time most
generic guesstimations can be resolved mathematically and there are
plenty of products out there that don't go to overkill Phil suggests
is necessary. Yes there are unresolved variables but this is a
science after all, and equations can have variables in them and be
valid expressions. Remember I asked for an equation, if Phil can't
provide one that's not something to be ashamed of, neither can I, so I
asked hoping for someone who knows rather than someone who thinks I
don't need to know.

I would openly admit that I couldn't give an equation either ! I would
fall back to first principles and go from there... I hate math but
thats my curse.

If Phil insists he is right and the rest of the world is wrong -
having produced working electronics that seem an aweful lot like
evidence, it would be nice to have some of the pixie dust they must be
sprinking on these things to keep them working.

There HAS to be a better way!

I think that the Chinese engineers have learnt how to work up against
the tightest of margins to produce some of the products we see today !
Probably born of the culture / nature of their country.

--
Best Reagrds:
Baron.

 

On Jul 10, 5:49 am, Baron <baron.nos...@linuxmaniac.nospam.net> wrote:

emailaddr...@insightbb.com Inscribed thus:

Phil is giving you good advice !

Watts in equals watts out less losses.

Its not his fault you don't understand what Phil is telling you !

--
Best Reagrds:
                        Baron.

I agree, and understand what Phil was saying, but that's a generic
guesstimation that I'm looking to move beyond. Given enough time most
generic guesstimations can be resolved mathematically and there are
plenty of products out there that don't go to overkill Phil suggests
is necessary. Yes there are unresolved variables but this is a
science after all, and equations can have variables in them and be
valid expressions. Remember I asked for an equation, if Phil can't
provide one that's not something to be ashamed of, neither can I, so I
asked hoping for someone who knows rather than someone who thinks I
don't need to know.

If Phil insists he is right and the rest of the world is wrong -
having produced working electronics that seem an aweful lot like
evidence, it would be nice to have some of the pixie dust they must be
sprinking on these things to keep them working.

There HAS to be a better way!

 

On Jul 10, 1:08 pm, Peter Bennett <pete...@somewhere.invalid> wrote:

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddr...@insightbb.com
wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC.  I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

Go tohttp://www.hammondmfg.com/5cpwr.htmand download their Power
Transformer Selection Guide - it is a one page .pdf that shows various
circuit configurations and resulting voltage/current relations.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info:http://vancouver-webpages.com/peter
Vancouver Power Squadron:http://vancouver.powersquadron.ca

Thank you. This is a lot closer to what I was looking for. While it
may not resolve all variables it begins to explain the discrepancies
between what a certain someone insisted and what has been observed
about existing designs field proven to work acceptibly.

 

[ehsjr]

emailaddress@insightbb.com wrote:

On Jul 10, 1:50 am, ehsjr
e.h.s.j.r.removethespampunctuat...@bellatlantic.net> wrote:

There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.


Ok, but this should be resolvable to a reasonable level given the
example context of it being a typical E-core transformer and basic
silicon bridge rectified and capacitive filtered circuit. The output
at that point is the question. Even if some formula has an unknown
variable, or two or one hundred, the start would be to resolve those
which requires a working equation in which to place those variables.



I'll mention some general things.

First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.


This seems to counter the majority of small transformer examples out
there, does it not? Consider products powering just about anything
that uses a wall wart.
Consider a sub-20VA transformer rated in a wall wart for 12VDC, 1A
output. Granted, we could call that a little over 1/2 the rating but
just how much or little is the crucial issue, what variables determine
how much or little and how to express those mathematically.



Next, figure the ripple voltage, Vr. To figure Vr, seehttp://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-S...



Thank you for the link, though since I am not mass producing equipment
and so not overly concerned about small differences in component cost
or size, I'm essentially going to consider ripple current a constant
resolved later as would apply to any one design, while I am asking
about a general formula for conversion without regard to variables
that would change in different designs beyond the basic assumption of
a capacitor large enough to achieve acceptibly low ripple for example.


Next, find out how much headroom your regulator needs, usually
available from the datasheet. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the datasheet. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.

By the way, the 12.3 volt figure is odd - how did you arrive
at it?


This is what the example circuit would need at a bare minimum by
calculating similarly to what you have above plus other drops in the
circuit @ expected current levels, to still retain the necessary
minimum input voltage, plus or minus a margin of error as median
datasheet values where used.

However, I am not concerned about the entire circuit, not about drop
over a regulator, I am only concerned about resolving how the AC
transformer rating relates to DC output before anything further in a
circuit beyond a typical bridge rectifier and capacitor sufficient to
smooth to a hypothetical 0V ripple, and I am also accounting for the
(Often negligable) different in forward voltage over the rectifier(s)
at different current levels, but this too can be expressed
mathematically.

Essentially, I don't want information more applicable to one project
than to another. Only what remains true mathetically for all projects
which employ AC spec'd transformers of typical design and through
bridge rectification by the most common silicon diodes. In other
words, as you'd see in the vast majority of electronics already if
they're not using switching PSU.

I don't want to argue - that seems your intent.

I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.

We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.


Ed

 

[The Phantom]

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress@insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.

How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?

Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.

Thanks,
JC

You should go to the library and make a copy of the paper:

"Analysis of Rectifier Operation", O. H. Schade, Proceedings of the IRE,
July 1943

This will give you an idea of what's involved in a detailed analysis of the
sort you're asking for.

 

[John Fields]

On Fri, 11 Jul 2008 12:29:09 -0700, The Phantom <phantom@aol.com>
wrote:

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress@insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.

How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?

Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.

Thanks,
JC

You should go to the library and make a copy of the paper:

"Analysis of Rectifier Operation", O. H. Schade, Proceedings of the IRE,
July 1943

This will give you an idea of what's involved in a detailed analysis of the
sort you're asking for.

---
There's also Motorola's Silicon Rectifier Manual, an excellent
reference and cheap at:

http://www.amazon.com/gp/offer-listing/B000HVEWGK/ref=dp_olp_3/103-7647257-4204638

JF

 

[John Fields]

On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress@insightbb.com
wrote:

I have an application in which I need to determine if a transformer is
suitable.

Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?

I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?

Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries.

---
Why can't you just used the raw unfiltered DC and current limit the
input to the battery?


JF

 

[Hammy]

On Fri, 11 Jul 2008 15:34:47 -0500, John Fields
<jfields@austininstruments.com> wrote:

You should go to the library and make a copy of the paper:

"Analysis of Rectifier Operation", O. H. Schade, Proceedings of the IRE,
July 1943

This will give you an idea of what's involved in a detailed analysis of the
sort you're asking for.

---
There's also Motorola's Silicon Rectifier Manual, an excellent
reference and cheap at:

http://www.amazon.com/gp/offer-listing/B000HVEWGK/ref=dp_olp_3/103-7647257-4204638

JF

The hanbook from onsemi as well as the link I posted to algonquin
(volt-reg) provide work examples using Schade curves.

Buying a book for future refrence is a good idea though. Hell for five
bucks I might get that John. We used to have a shelf full of those
motorola data books and manuals in school.

 

On Jul 9, 6:39 pm, emailaddr...@insightbb.com wrote:

On Jul 9, 9:26 pm, emailaddr...@insightbb.com wrote:

I have an application in which I need to determine if a transformer is
suitable.
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply.

Another thing I am wondering is if a basic, typical small brick AC-DC
switching supply is stable powering this kind of load. Stripping down
the circuit to the basic topology, what if it had for example:

AC-DC SMPS -> LM317 -> 1.25 Ohm resistor (LM317 in current regulating
config) -> 1A into Battery Pack + LM317 feedback

Since the SMPS is trying to regulate to it's spec'd voltage, let's say
that is 13V even if an uncommon value, will it run stable doing so and
if so, is that at the same constant output wattage from the SMPS the
whole time, with the linear regulator simply dropping more voltage,
creating more heat at the beginning of a battery recharge cycle since
it's suppling constant current to batteries that are at a lower
initial, low state-of-charge voltage?

Why so many LM317s? They will generate heat too, and waste energy.

What are you charging? What supply voltage do you need?

Lots of surplus 12, 15 or 18VDC switching power supplies from old
laptops - see a thrift store, or a surplus electronics store

Michael